A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z | AA | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1 | Totals ==> | 452 | 5 | 1 | 7 | 0 | 12 | 3 | 3 | 29 | 9 | 0 | 0 | 13 | 0 | ||||||||||||
2 | Notes | Ok | ? | BP | BSI | BSQ | IF | IG | OI | WU | EU | XX | RV | Lasso | P | Answer | Explanation | ID | Formula | Trace | Expected | ||||||
3 | 1 | No | The red light is not on in the first state. | 917u | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
4 | 1 | No | In the first step, Red is not lit. Since there are no temporal operators in the formula, we're only considering that stpe. | qMyf | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
5 | 1 | No | Red isn't present in the first state | gbrV | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
6 | 1 | Yes | There is at least one state in which the red light is on. | rahc | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
7 | 1 | Yes | I think the formula says that the red light has to be on in at least one state, which is satisfied by the {RGB} states. | N3Cw | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
8 | 1 | No | Red is not in the first state's lit colors. | ydkm | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
9 | 1 | No | By default it refers to the first state, and in that state red is not on. | 31qw | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
10 | 1 | No | The red light is not on in the first state; there are no LTL operators applied to indicate that the formula is referring to any future states. | c9jy | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
11 | 1 | Yes | Red is on in one states. | 3r04 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
12 | 1 | No | This is not satisfied because the formula only refers to the initial state, not to all states in the trace, and Red is not on in the initial state. This would be satisfiable if the formula read: eventually {Red in Panel.lit}. | ptyk | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
13 | 1 | No | In the first state, the red light is not lit. | KXjz | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
14 | 1 | No | Red is not in Panel.lit in the first state | iVk5 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
15 | 1 | No | The formula is referring to the very first state of the trace, {GB}. Since Red is not lit in that state, it is not a satisfying trace (even though it is lit in every other state). | mxad | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
16 | 1 | No | Red is not on in the first state | rnC9 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
17 | 1 | Yes | The Red light was present in lit in at least one trace. | 8E1h | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
18 | 1 | No | Red not in first state {GB} | sain | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
19 | 1 | No | Because the first state does not have the Red light on. | nieh | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
20 | 1 | 1 | No | There is a trace where Red is not lit | uOG8 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | |||||||||||||||||||
21 | 1 | No | The red light is not on in the initial state. | fsqw | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
22 | 1 | No | Because the formula is not temporal-ly quantified and the Red light is not on in the first state ({GB}). | dznh | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
23 | 1 | No | Red is not on in the first state | 7opy | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
24 | 1 | Yes | Even though the first state does not have Red in it, the formula never specifies that Red has to be lit in every state, so the fact that it is lit in some states is enough to satisfy. | 55oa | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
25 | 1 | No | Red is not on in the current (first) state. | vz2g | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
26 | 1 | No | The formula requires that the red light be turned on in the first state, and this is not the case in the given trace. | cirf | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
27 | 1 | No | The Red light is not on in the first state, and the first state is the one referenced by the formula. | fojf | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
28 | 1 | No | The first { GB } has no R in it | 7jm2 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
29 | 1 | No | Red is not on in first trace | eolW | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
30 | 1 | No | The lack of any temporal keywords implies the formula only applies for the first state. However, the first state { GB } does not contain R, meaning that Red is not in Panel.lit in the first state, so the formula is not satisfied. | c1fn | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
31 | 1 | No | Red is not lit in the initial state. | nr65 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
32 | 1 | No | In the first state, red is not in the panel.lit. | m0p6 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
33 | 1 | Yes | Because it doesn't talk about red always lit | able | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
34 | 1 | No | Red is not on in the current state. | 1vwv | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
35 | 1 | No | State 1 doesn't have Red lit. | yoxy | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
36 | 1 | No | Because R is off in first state | DzoD | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
37 | 1 | No | Red is not on in the first state, which is the only one checked. | xqZA | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
38 | 1 | No | Because the first state has Red to be turned off. | bxd4 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
39 | 1 | No | The red light is not on in the first state, which is the default state for the predicates when there are no temporal modifiers | bx1r | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
40 | 1 | No | The formula checks if Red light is on in the first state, which it isn't. | w4t3 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
41 | 1 | No | The Red light is not lit in the first state. | clU0 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
42 | 1 | No | Red light is not on in the first state. | ndij | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
43 | 1 | Yes | It doesn't say always so I just looked to find a single trace with `Red in Panel.lit` - e. g. {RGB} | tyos | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
44 | 1 | No | This formula only constrains the first state, where R is not present. | ekny | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
45 | 1 | No | Red is not lit in the first state. | hDZG | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
46 | 1 | No | Because the Red light is not lit in the initial state. | XU9x | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
47 | 1 | No | In the first state, the red light is not lit. | zf66 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
48 | 1 | No | Red light is not on in the initial state. | kkzx | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
49 | 1 | No | Red in Panel.lit only applies to the first state in the trace, where R is not present. | yypq | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
50 | 1 | No | Red is not lit in the first state. | pfdd | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
51 | 1 | No | This formula only talks about the first trace, and we see that red is NOT lit in the first state {GB}. | s0nv | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
52 | 1 | No | I'm not sure how statements that don't use any temporal operators should be interpreted. | j9mq | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
53 | 1 | No | Because the formula is about the first state, where Red light is not on | j7t9 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
54 | 1 | No | I assume if no electrum keywords are present then the predicate is applied to the first instance. | n4vd | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
55 | 1 | No | Because Red light is not lit in the first state | qjpx | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
56 | 1 | No | Because Red is not in Panel.lit in the first state | kuaa | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
57 | 1 | Yes | Red is in the last four traces. | XLuy | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
58 | 1 | No | Red is not on in the first state | xeec | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
59 | 1 | No | Red in Panel.lit is an event that happens in the first state, but the first state only has green and blue. | duq8 | Red in Panel.lit | {GB} {RGB} {RGB} {RGB} {RGB} | No | ||||||||||||||||||||
60 | 1 | 1 | Yes | The red light is on three states from the initial state. | 917u | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
61 | 1 | 1 | Yes | We're looking at the 4th time step (3 afters); at which point, Red is lit. | qMyf | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
62 | 1 | No | For the second step in the trace this doesn't hold as after after after is none | gbrV | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
63 | 1 | 1 | Yes | There is a state whose third state afterwards has the red light on in that state. | rahc | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
64 | 1 | Yes | In state 1, the formula is satisfied because three states after, the Red light turns on. | N3Cw | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
65 | 1 | Yes | In the 4th state, 3 states after state 0, the Red is lit. | ydkm | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
66 | 1 | Yes | Red is on in the fourth state | 31qw | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
67 | 1 | Yes | The red light is on in the fourth state (indicated by three "afters"). | c9jy | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
68 | 1 | Yes | After the first three states, red light is on. | 3r04 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
69 | 1 | Yes | Given the 3 afters in this formula, the statement Red in Panel.lit is being evaluated at state 4. In state 4, the Red light is on, so this formula is satisfied. | ptyk | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
70 | 1 | No | The red light is not lit in the fourth state. | KXjz | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
71 | 1 | Yes | This formula is satisfied by the fourth state. There are three states before the first state so the fourth state is (after after after) the first state and Red is in Panel.lit in the fourth state. | iVk5 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
72 | 1 | Yes | Regardless of what happens in the first three states, as long as Red is lit in the fourth state, the trace should satisfy the formula. As such, this trace is valid. | mxad | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
73 | 1 | Yes | The red light is on three states after the initial state. | rnC9 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
74 | 1 | 1 | Yes | Red was lit in one state that is 3 states after some state. | 8E1h | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
75 | 1 | Yes | Red in 4th state | sain | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
76 | 1 | Yes | Because the Red light is on on the third state. | nieh | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
77 | 1 | Yes | the fourth trace has red present | uOG8 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
78 | 1 | Yes | The red light is on in the fourth state. | fsqw | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
79 | 1 | Yes | Because the Red light is on in the 4th state of the trace (3 states after the initial state). | dznh | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
80 | 1 | Yes | The Red light is on in the forth state (after three steps from the first state) | 7opy | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
81 | 1 | 1 | Yes | Again, since this formula never specifies that it should hold in all states, the fact that it holds in states 2-4 is sufficient | 55oa | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
82 | 1 | Yes | after is the 2nd, after after is the 3rd, after after after is the 4th. Red is lit in the 4th. | vz2g | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
83 | 1 | Yes | This formula requires that the red light be turned on in the fourth state, which is satisfied by this trace. | cirf | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
84 | 1 | Yes | The Red light is on the third state. | fojf | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
85 | 1 | Yes | The fourth part of the trace is { R }, which contains a R. | 7jm2 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
86 | 1 | Yes | Current trace + 3 after = 4th Trace, Red is on | eolW | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
87 | 1 | Yes | The use of 3 'after' keywords makes the formula imply that the red light should be on 3 states after the present one. The state 3 states after the first, according to the trace, is { R }, where the red light is on, thereby satisfying this requirement. | c1fn | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
88 | 1 | Yes | Red is lit in 3 states (so 4th state beginning on 1st state). | nr65 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
89 | 1 | Yes | Red is in the 4th state | m0p6 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
90 | 1 | Yes | The red should be there after the first three states which it is. voila! | able | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
91 | 1 | Yes | Red is on in the 3rd state in the future (the 4th state). | 1vwv | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
92 | 1 | Yes | Red is lit in state 4. | yoxy | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
93 | 1 | Yes | Because R is lit in 4th state and there are 3 "after" statements | DzoD | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
94 | 1 | Yes | First state is 1, after 3 time steps we are in state 4 with Red on. | xqZA | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
95 | 1 | Yes | Because the after after after state of the current state, the red light is also turned on. | bxd4 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
96 | 1 | Yes | Red light is lit in the fourth state, which is 3 states after the initial state | bx1r | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
97 | 1 | Yes | It is stating that in the fourth state the Red light is on, which is true of the trace. | w4t3 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
98 | 1 | Yes | The Red light is lit in the fourth state, i.e., three states after the first state. | clU0 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
99 | 1 | Yes | Red light is on in the 4th state. | ndij | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
100 | 1 | 1 | Yes | {R} plus 3 jumps gets you back to {R} | tyos | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
101 | 1 | Yes | 3 afters is the 4th state, in which R is lit. | ekny | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
102 | 1 | Yes | Red is lit in the state after the state after the state after the first state (i.e., the fourth state). | hDZG | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
103 | 1 | Yes | after {after { after { x }}} is checking for x three states after the initial, which is the 4th state. The red light is lit in the 4th state. | XU9x | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
104 | 1 | Yes | The red light is lit in the fourth state. | zf66 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
105 | 1 | No | I think the formula defines that the red light to be on in the third state, but the trace didn't have the red light on in the third state. | kkzx | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
106 | 1 | Yes | The formula states that Red will be lit after 3 time steps, which is the 4th state. | yypq | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
107 | 1 | Yes | Red it lit in the 4th state. | pfdd | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
108 | 1 | Yes | This formula says that three states after the first state, the red light will be on, which is represented in the trace, because the fourth state (3 states after the first) has the red light on. | s0nv | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
109 | 1 | 1 | Yes | The state 2 states after the initial state has red lit. | j9mq | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
110 | 1 | Yes | Because Red light is on in the 4th state | j7t9 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
111 | 1 | Yes | The red light is lit in instance 4 | n4vd | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
112 | 1 | Yes | Red light is lit in the fourth state | qjpx | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
113 | 1 | 1 | Yes | It can be! If we are starting on the first trace, then yes, because it appears three traces later (which is the number of afters) | kuaa | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | |||||||||||||||||||
114 | 1 | Yes | Red is in the third state after the first. | XLuy | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
115 | 1 | Yes | Red is on in the fourth state | xeec | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
116 | 1 | Yes | The statement never said anything about the first state, so whatever value in it would suffice, the same applies to 2nd, 3rd, and 5th. It only specified that the 4th state must have red. | duq8 | after { after { after { Red in Panel.lit } } } | {R} {} {} {R} {} | Yes | ||||||||||||||||||||
117 | 1 | Yes | Whenever the red light is on the red light is also on three states from now. | 917u | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
118 | 1 | Yes | We note that the last state repeats forever. In the 2nd step, we hope that In the 5th step, R will be lit, and it is. In the 3rd, we hope R is lit in the 6th, and it is. For all steps i, i >= 5, we hope that it is lit in step i+3; since it is always lit after step 5, this holds. | qMyf | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
119 | 1 | Yes | In the first state the LHS of the implies is false so the expression is true. In the second state after after after is RGB so true implies true = true. The last state keeps repeating so the RHS will always be true from this point on | gbrV | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
120 | 1 | Yes | It is always the case that if red light is on in one state, the red light is also on in the third following state. | rahc | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
121 | 1 | 1 | Yes | In any state where the Red light is on, the state three states after also has the red light on (if the final state loops). | N3Cw | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
122 | 1 | 1 | Yes | Since the traces repeat the lasso infinitely, the last two states that contain Red will meet this definition, and 3 states after the second state is the fifth state, which contains Red. | ydkm | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
123 | 1 | Yes | All of the states that have R on have R on also in the next of next of next state | 31qw | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
124 | 1 | Yes | The always indicates that the formula holds at any state. In this case, the Red in Panel.lit implies... part of the formula applies at the second state, at which R is lit three states later (in the fifth state). | c9jy | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
125 | 1 | 1 | Yes | For the given states, if in the second state red light is on, in the fifth state it must on. | 3r04 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
126 | 1 | Yes | Because this formula begins with always, we must evaluate it for every state. Red in Panel.lit is true in states 2, 3, 5, and beyond. At state 2, after after after refers to state 5. In state 5, the Red light is on, so this is satisfied. In state 3, after after after refers to state 6, which is equivalent to state 5, so again this is satisfied. This is also true in state 5. In state 5, after after after refers to state 8, which is a lasso of state 5, so it is still satisfied. Red in Panel.lit will also be true in all states beyond state 5. Three states after all of these states will be replicas of state 5, so will contain Red in Panel.lit, meaning that this will always be satisfied. | ptyk | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
127 | 1 | 1 | Yes | For all traces where red is lit, red is lit three states later unless a jump of three states exceeds the number of states checked. | KXjz | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
128 | 1 | No | the state that is three states after the third state (which has Red in Panel.lit) is the first state, which does not have Red in Panel.lit | iVk5 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
129 | 1 | Yes | The formula shows that the first time Red is lit (the second state) means that in the sixth state, Red must also be lit. This also means that every four states, starting from the second, Red will be lit. Since the fifth state shown, {RGB}, continues to infinity, Red is lit in every state to infinity, and as such it will necessarily be lit in every fourth state, starting with the second. | mxad | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
130 | 1 | Yes | For all states with the red light on, three states later it is also on. For all of the states this is covered by the lasso. | rnC9 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
131 | 1 | Yes | It is always true that Red is lit in a state 3 states after a state with Red lit. This is because for state 2, Red is lit in state 5, and for state 3 and 5, since state 5 repeats forever, Red is lit in all those states. | 8E1h | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
132 | 1 | Yes | if a state has red in it, the state 3 changes after also has red because the final state repeats forever | sain | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
133 | 1 | Yes | Because when the Red light is on, the Red light is on again in every 3 other state. | nieh | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
134 | 1 | Yes | when red was present it was present three traces later | uOG8 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
135 | 1 | Yes | The red light is on in the second state, and correspondingly, it is on in the fifth state, three states later. The red light is on in the third state, and since the final state repeats, it is also on three states later. Since the final state repeats, whenever the red light is on in a state after and including the fifth state, it will be on three states later. | fsqw | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
136 | 1 | Yes | because for each element of the trace (including the final lasso state) : 1. the Red light is off 2. Or the red light is on and the red light 3 states 'later' is on. | dznh | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
137 | 1 | Yes | Whenever Red is on in state i, the Red light is on again in state i+3. | 7opy | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
138 | 1 | Yes | The formula is saying that whenever there is a Red light, there will be another one 3 states away. The first Red light is trace 2 and 3 states away there is also one. There is also a Red light 3 traces away from state 3. Finally, we lasso on a state with the Red light lit so this will always hold. | 55oa | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
139 | 1 | Yes | Assuming the last state is the one that repeats forever as stated above in the constraints, then any state with Red lit will always have a 3rd following state that also has Red lit. | vz2g | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
140 | 1 | Yes | Whenever the red light is turned on, it is on again after three transitions, and the infinite repetition at the end ensures that the red light will always be on for all states after the fourth state of this trace, which is all that is needed since the red light is not on in the first state. | cirf | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
141 | 1 | Yes | For each state in which the Red light is on, the Red light is on in the state that is three states in the future. Additionally, the last state repeats forever with the Red light on. | fojf | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
142 | 1 | No | Red is in the 3rd stage, but not in the 6th stage (which is equivalent to the 1st stage). Alternatively, how I first thought of it is gcd(3, 5) = 1, so the only trace this is true is the one where Red is in every trace. | 7jm2 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
143 | 1 | Yes | Whenever Red is on, the 3rd state after that should have Red on. Since the last state that repeats forever has Red on, the 3rd state for every trace with red on is RGB. | eolW | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
144 | 1 | Yes | The formula implies that whenever the red light is on in a given state, then it will also be on in the state that occurs 3 states away from this original state. Consider the cases in which red is on in the given trace. 1. Second state: R is present in the second state as well as the fifth state, which occurs 3 states afterwards, thereby satisfying the formula. 2. Third state: The last state { RGB } is repeats indefinitely, which implies the sixth state (occurring 3 states after the third) also contains R. 3. Fifth state: Again, since { RGB } repeats indefinitely at the end, the eighth state, occurring 3 states after the fifth, also contains R. Therefore, this formula is always satisfied. | c1fn | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
145 | 1 | Yes | Whenever Red is lit, it will be lit in 3 states. Red lit in state 2 requires Red to be lit in state 5 (yes). Red lit in state 3 requires Red to be lit in state 6 (yes - a lasso state). After state 5, Red is always lit anyways. | nr65 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
146 | 1 | Yes | When there is red, it is on 3 states after | m0p6 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
147 | 1 | No | Because although there is a red for the first RGB but not for the second RGB | able | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
148 | 1 | Yes | Red is on in the 2nd state, so it must be on 3 states in the future (the 5th state), and it is. Red is on in the 3rd and 5th states, so it must be on in the 6th and 8th states, respectively. It is on in these states because the last state, which has Red on, repeats forever. Red is on in all states after the 5th state so the formula is always satisfied. | 1vwv | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
149 | 1 | Yes | Red is lit in the second state, and also lit in the third state from the second (state 5). | yoxy | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
150 | 1 | 1 | Yes | State 2 has it's "afters" satisfied, the other 2 where red is lit don't need to since there are only 5 total states. | DzoD | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
151 | 1 | Yes | Each Red being on state has Red on in the state 3 time steps later. | xqZA | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
152 | 1 | Yes | The second state has Red light on, and the fifth state has a red light on. The third state has the REd light on, and since the fifth state will repeat, it will also get a state with a red light on. Same for the fifth state and so on. | bxd4 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
153 | 1 | Yes | Once the red light turns on, then it must be on again three states after that | bx1r | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
154 | 1 | Yes | Since the formula uses implies, it is true when Red is not lit, so States 1 and 4 of the trace matches the formula. For State 2 and 3 (where the Red light is on), the formula checks if the Red light is on 3 states later. Since the last state is a lasso state, it will always be Red afterwards, so both State 2 and 3 will have a Red light on 3 states later. | w4t3 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
155 | 1 | Yes | Whenever the red light is on, it is also on three states in the future. The red light is on in the second state and also in the fifth state, in the third state and also in the sixth state (as part of the lasso), and whenever it is on in the lasso it will also be on three steps later in the lasso. | clU0 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
156 | 1 | Yes | If the red light is on, then the red light should be on again after three states. The red light is on in the 2nd state, and is on in the 5th state. Because the final state repeats forever, the red light is on from the 5th state forward. So the red light is on in the 3rd state and again in the 6th state; on in the 5th state and again in the 8th state. | ndij | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
157 | 1 | 1 | Yes | I'm not sure cause I can't tell where this trace is lasso'ing. I'm going to guess it lassos at the last {RGB}, hence this is true? | tyos | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
158 | 1 | Yes | This formula states that for every state where Red is lit, 3 states after, Red will still be lit. In the 2nd state, Red is lit, and is satisfied by the 5th state where it is still lit. As the last state repeats forever, all states after are satisfied too. | ekny | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
159 | 1 | Yes | In every state where Red is on, Red is also on three states in the future (repeating the end state as many times as necessary to get there). | hDZG | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
160 | 1 | Yes | Because the last state lassos, even state 2 and 5 will also be satisfy this predicte even though their "3 states away" state will be {RGB} | XU9x | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
161 | 1 | Yes | In the three states (second, third, and last) that have the red light lit, the state that comes three states later also has the red light lit. | zf66 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
162 | 1 | Yes | The formula defines that when the red light is on in (i)th state, it is on again in (i+3)th state. State i+1 and i+2 doesn't have any restrictions. From the trace we can see that the red light was on in the second state and then on again in the fifth state. | kkzx | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
163 | 1 | No | The third state has the Red light lit up, but if we travel 3 time steps ahead of that state, we end up at the first state, which does not have the Red light lit up. | yypq | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
164 | 1 | No | The 3rd state does not satisfy this, as going three states after the 3rd state gets us back to the 1st, where Red is off. | pfdd | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
165 | 1 | Yes | Now, we generalize to saying that at any point in time, if the red light is on, then three states afterward, the red light will be on. We see that in the second state, the red light is on, and we get that it is also on in the fifth state. Since the fifth state repeats forever, we see that any state i with a red light after the second state will have state i + 3 also having the red light on. | s0nv | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
166 | 1 | No | This doesn't hold when you consider the trace as a loop. | j9mq | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
167 | 1 | No | In the 3rd state, Red light is on, but 3 states after, it is not on | j7t9 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
168 | 1 | Yes | Since the last instance repeats forever, all instances with a red light also have a red light lit three instances later. | n4vd | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
169 | 1 | Yes | Any Red state is followed by a Red state three states after, including the final lasso state | qjpx | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
170 | 1 | 1 | Yes | Red is not always in Panel.lit; therefore, any set of conditions can be true | kuaa | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | |||||||||||||||||||
171 | 1 | Yes | Red is in the second trace and in the fifth trace it is also turned on. | XLuy | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
172 | 1 | Yes | Red is on in the second state, so it must be on in the state three later which it is. It is also on in the third state, and since the final state repeats forever, then this is satisfied. | xeec | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
173 | 1 | Yes | In the 2nd state, the color is red, so the 5th state must be red, which is true. In the 3rdstate, the color is red, so the 6th state must be red, which is true because the final state repeats forever. Since the repeated state has red, this applies to the 5th state too and so on. | duq8 | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } | {} {RGB} {RGB} {} {RGB} | Yes | ||||||||||||||||||||
174 | 1 | No | The red light is on in the next state without the green light being on. I read the formula as after{Red in Panel.lit and Green in Panel.lit} | 917u | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
175 | 1 | Yes | The formula basically says "Red will be lit in the next step until Green is lit in the next step". Green is lit in the 4th step; every step up to and including it, Red is lit, so this holds. | qMyf | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
176 | 1 | Yes | Red is always on until Green is turned on | gbrV | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
177 | 1 | No | I'm not sure if I understand the formula... But I think it's saying that, a state whose following state has the red light on happens until a state whose following state has the green light on. The fourth state has the red light on, but there is no state after it (including itself) that has the green light on. | rahc | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
178 | 1 | Yes | For all the states before state 3 (where the green light is on in the next state), each next state has red in it. Or, for 1 <= i < 3, state_{i + 1} has the red light on. | N3Cw | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
179 | 1 | Yes | Yes, because for states 1 and 2, the next state has red as lit, and then in state 3 the after state (state 4 has green) so the obiligation ceases to hold. Red is still lit in state 4, but that is allowed as until does not care about the value of the first expression after the state where the second expression is met. | ydkm | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
180 | 1 | Yes | Red is always on in the next state until green is on in the next state | 31qw | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
181 | 1 | Yes | The red light is on and holds for all states up until the one which "after {Green in Panel.lit}", which does not specify that the Red light has to turn off. | c9jy | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
182 | 1 | Yes | In the thrid state, the red is on in fourth state and green is on in the fourth state, so that in the following state, the next state of current state red light will never on. | 3r04 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
183 | 1 | No | after {Green in Panel.lit} is true in state 3 because in state 4, the Green light is on. This means, in state 3, after {Red in Panel.lit} cannot be true if this statement were to be satisfiable. Since, however, we see that in state 4, the Red light is on, after {Red in Panel.lit} is true in state 3. Thus this formula is not satisfied by the trace. | ptyk | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
184 | 1 | Yes | yes, state 3 is when the "after { Green in Panel.lit }" predicate holds, and in all previous states the "after { Red in Panel.lit }" predicate holds. | KXjz | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
185 | 1 | Yes | (after Red in Panel.lit) holds in all states where the following state has Red in Panel.lit. This is true for the first 3 states. (after Green in Panel.lit) holds in the third state because Green is in Panel.lit in the 4th state. So, the LHS of the until is true until the RHS holds. | iVk5 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
186 | 1 | Yes | The formula writes that Red will be lit until the state after Green is lit. Since Red is lit in states 1-4, and Green is lit in state 4, Red must not be lit in any state after the 4th. Indeed, the 5th and lasso state continues as {B} forever, meaning Red is never lit after the 4th state, so the trace is satisfied. | mxad | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
187 | 1 | Yes | State 3 satisfies the right side as in state 4 the green light is on, and for states 1 and 2 the red light is on in the next state. Thus the red light is on in the next state until the green light is on in the next state. | rnC9 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
188 | 1 | No | I don't really understand this one, but I think that since Red is lit in the after the state of a state for which Green is lit in the after state, then the "until" part doesn't hold. | 8E1h | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
189 | 1 | Yes | the two first states satisfy the condition that the next state has red in it | sain | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
190 | 1 | No | I'm not sure of either answer! | nieh | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
191 | 1 | Yes | until does not guarantee that statement A does not hold in traces after | uOG8 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
192 | 1 | Yes | The red light is on in the second state and third state, which means "after { Red in Panel.lit }" is true in the first and second state. In the fourth state, the green light is lit, which means "after { Green in Panel.lit }" is true in the third state. | fsqw | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
193 | 1 | Yes | The Red light stays on in the trace from the 2nd state until the Green light comes on. | dznh | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
194 | 1 | Yes | The third state triggers the until because Green is on in the forth state {RGB}. And Red is on in all of the first four states. | 7opy | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
195 | 1 | No | I believe this formula is saying that red will continually be in lit until the green one is. As I interpret it, the red and the green should never be lit at the same time, which is exactly what's happening in trace 4. | 55oa | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
196 | 1 | No | State 3 has Red lit, but State 4 has Green lit. Red is supposed to turn off once the state after has Green lit. | vz2g | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
197 | 1 | Yes | The red light is required to be on in the next state until the green light is turned on in the next state. Given the technical way "until" works, I don't think red even needs to be on in the fourth state, but since the green light is on in the next state when we are at the third state, the red light does not have to be on for the rest of the trace, so it is fine that only the blue light is on after the fourth state. "Until" doesn't require fmla-a to be false once fmla-b is true, just that it holds as long as fmla-b has not yet become true. | cirf | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
198 | 1 | Yes | At every state before the Green light is lit, Red is lit in the following state. | fojf | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
199 | 1 | No | Red is in the 4th state, but Green is too, but the statement says Red should dissappear when Green shows up. | 7jm2 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
200 | 1 | No | Red and Green shouldn't be on at the same time, the until will trigger at index 2 which means in index 3, Red should not be on. | eolW | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
201 | 1 | Yes | The red light is on until the fourth state, when the green light also turns on. This implies the third state { RB } was the last time the condition before the 'until' keyword had to hold. As a consequence, the indefinite repetition of { B } afterwards satisfies the formula at hand. | c1fn | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
202 | 1 | No | At state 4, there is a state where neither are true. until requires that one remains true until the other becomes true. The next state is solely a Blue state. | nr65 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
203 | 1 | No | The 4th state has a red in it, but it should not because after {green in panel.lit} is true in the 3rd state, so the 4th state should not have a red | m0p6 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
204 | 1 | Yes | because after the second R there is always are until G in the fourth state. | able | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
205 | 1 | Yes | Red is on in the 2nd state (`after` of the current state) and the 3rd state (the state whose `after` has Green on) so the `until` is satisfied. (I would change my answer to No if these statements are actually meant to be on separate lines). | 1vwv | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
206 | 1 | No | Red should not be lit in states 3 or 4. | yoxy | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
207 | 1 | Yes | because of the after statements, Red and Green can both be on one time | DzoD | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
208 | 1 | Yes | The requirement for Red to be on in the next time step is lifted in state 3, when Green is on in the next time step (state 4). | xqZA | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
209 | 1 | Yes | Because of the Green in the fourth state, the fifth state(repeated state) has no Red in it anymore. | bxd4 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
210 | 1 | Yes | There is always a red light in the next state until there is also a green light in the next state, at which point there is neither in the next state. | bx1r | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
211 | 1 | Yes | For State 1 and 2, State 2 and 3 both have the Red light on, so it satisfies the requirement for formula-a. This requirement ceases to hold for State 3, where the state afterwards (State 4) has the Green light turned on. However, the trace can still meet this condition (just not required to), so the formula is still true for State 3. For State 4 and onwards, it no longer needs to meet any conditions, so anything afterwards is valid. | w4t3 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
212 | 1 | Yes | For states 1 and 2, it is true that Red is lit in the next state, and in state 3, it is true that Green is lit in the next state. Together, these fulfill the obligations of the until statement. | clU0 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
213 | 1 | Yes | The red light is on in the 2nd state, so the part before until (after { Red in Panel.lit }) is satisfied. The green light is on in the 4th state, and the red light is on in the 2nd, 3rd and 4th states. | ndij | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
214 | 1 | Yes | This might be a tautology. | tyos | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
215 | 1 | Yes | The formula is saying that Red will be lit until the state after Green is lit. Red is lit in all the states except the fifth one, which is okay because in the 4th state Green is lit. | ekny | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
216 | 1 | Yes | Yes. The second part, “after { Green in Panel.lit }”, is satisfied by the third state (since Green is lit in the fourth state). In both of the states before this, “after { Red in Panel.lit }” is satisfied, because Red is lit in the second and third states (the “after”s of the first and second states). These two facts make the “until” true. | hDZG | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
217 | 1 | Yes | Even though {RGB} breaks the pattern (because {B} doesn't have Red), {RGB} contains G so the requirement to see R in subsequent states is over. | XU9x | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
218 | 1 | Yes | The first three states satisfy the left side of the until. The fourth state satisfies the right side of the until, freeing the fifth and later states from any obligations. | zf66 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
219 | 1 | Yes | If initial state is i, the red light should be on from (i+1)th state to the (j)th state where the green light appears, and it should be off from the (j+1)th state. Other states are not specified. From the trace we see the red light was on until the last state - which is the (j+1)th state. So the formula is satisfied by this trace. | kkzx | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
220 | 1 | Yes | The "after { Green in Panel.lit }" condition is fulfilled in the 3rd state and all states from the 1st state to the 3rd state fulfill the condition "after { Red in Panel.lit }" | yypq | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
221 | 1 | Yes | It is true that from the first state until the third state (after which Green is lit) Red will be lit in the next state. | pfdd | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
222 | 1 | Yes | The red light is continually on until the state after the green light turns on, and once this event occurs, the formula expires and we are free to repeat the last state infinitely. | s0nv | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
223 | 1 | Yes | When green is in the next state, the predicate on red is released. That predicate also holds up to this point. | j9mq | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
224 | 1 | No | In the 4th state, Green light is already on but Red light is not off | j7t9 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
225 | 1 | Yes | The red light is always lit in the next trace until the green light is lit in the next trace. | n4vd | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
226 | 1 | Yes | Red stay lit until the green light turns on | qjpx | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
227 | 1 | Yes | Because red must be in Panel.lit in the following state until Green is in Panel.lit in the next state. Since both conditions are met, it satisfies the trace | kuaa | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
228 | 1 | Yes | Red is not turned on in the fifth state once green turns on in the fourth. | XLuy | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
229 | 1 | Yes | Because red is on in every state until the state after the first time green is on. | xeec | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
230 | 1 | Yes | The first part specifies that red should be in the state after the first state, which happens. And each state should have red until green shows up in the state after the until state. Since green showed up in the 4th state, this means that the 2nd and 3rd must have red, which they do. | duq8 | after { Red in Panel.lit } until after { Green in Panel.lit } | {RB} {RB} {RB} {RGB} {B} | Yes | ||||||||||||||||||||
231 | 1 | Yes | There is some future state where the green light is on and some future state where the red light is on. | 917u | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
232 | 1 | Yes | Red and Green are both lit at some point during the trace, so they are both eventually lit. | qMyf | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
233 | 1 | Yes | Green is turned on in the seconds state and red in the last, satisfying the eventually construct. | gbrV | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
234 | 1 | Yes | The formula requires that eventually red light is lit and eventually green light is lit. | rahc | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
235 | 1 | Yes | The red light and the green light are on at some point. | N3Cw | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
236 | 1 | Yes | Both red and green are eventually lit | ydkm | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
237 | 1 | Yes | at some point green goes on, and at some point red goes on. | 31qw | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
238 | 1 | Yes | Both the red and green lights are eventually on; the order doesn't matter due to the bracket ordering and usage of the keyword "and". | c9jy | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
239 | 1 | 1 | Yes | Red and green in on in the last four states. | 3r04 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | |||||||||||||||||||
240 | 1 | Yes | Eventually {Red in Panel.lit} means that at some point in the trace, Red will be lit. We see in state 5 (and beyond) that this is true. Similarly, eventually {Green in Panel.lit} means that at some point in the trace Green will be lit. This is true in state 2. Thus these two formulas are both true and the entire formula is also true. | ptyk | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
241 | 1 | Yes | both red and green are eventually lit | KXjz | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
242 | 1 | Yes | There are traces where Red and Green are in Panel.lit | iVk5 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
243 | 1 | Yes | Yes, we must have Green and Red appear at some point in the trace. Since they both appear, the formula is valid. | mxad | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
244 | 1 | Yes | The green and red lights both turn on | rnC9 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
245 | 1 | Yes | There is a state for which eventually in the future Green is lit and Red is also lit at some point (not necessarily both lit in the same state) | 8E1h | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
246 | 1 | Yes | both the red and green lights are lit at some point | sain | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
247 | 1 | Yes | Eventually both conditions are handled, doesn't have to be at the same time. | nieh | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
248 | 1 | Yes | eventually both were lit | uOG8 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
249 | 1 | Yes | The red light is on in some state, and the green light is on in some state. | fsqw | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
250 | 1 | Yes | Both the Green and Red lights turn on at some point. | dznh | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
251 | 1 | Yes | Red is on in some states. So is Green. | 7opy | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
252 | 1 | Yes | Eventually means that it needs to happen in at least one state in the future. Since Green is lit in state 2 and Red is lit in the lasso, this holds. | 55oa | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
253 | 1 | Yes | Both Green and Red are lit at some point. | vz2g | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
254 | 1 | Yes | This only requires that the eventually statements hold in the first state, so as long as there is at least one state in which the green light is on and at least one state in which the red light is on, the formula should be satisfied. Since this is the case in the given trace, the formula should be satisfied. | cirf | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
255 | 1 | Yes | The Red light is lit at some point on or after the first state, and so is the Green light. | fojf | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
256 | 1 | Yes | R and G are in the trace. | 7jm2 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
257 | 1 | Yes | Green and Red both turn on eventually. | eolW | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
258 | 1 | Yes | The absence of the 'always' and 'until' keywords implies that the conditions within either 'eventually' block need only hold once. Since both the green and red lights turn on at different points within this trace, this formula is satisfied. | c1fn | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
259 | 1 | Yes | Eventually, there are states for which Red is lit and independently for which Green is lit (state 5 and 2 respectively). It doesn't require that they are lit in perpetuity. | nr65 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
260 | 1 | Yes | Eventually both red and green appeared | m0p6 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
261 | 1 | Yes | because after the start one red and one green is present. | able | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
262 | 1 | Yes | From the current state, Red and Green are both eventually on. | 1vwv | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
263 | 1 | Yes | Green and Red are both lit at some point. | yoxy | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
264 | 1 | Yes | Both lights turn on eventually | DzoD | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
265 | 1 | Yes | Red and Green are both on at some point. | xqZA | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
266 | 1 | Yes | Green and Red should have happened eventually during the states. They both turned on eventually! :) 2nd and fifth | bxd4 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
267 | 1 | Yes | Both eventually operators don't have to happen at once, they just both have to eventually happen at some point. | bx1r | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
268 | 1 | Yes | From the first state, the green light is eventually on and the red light is eventually on. | w4t3 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
269 | 1 | Yes | The Red light is eventually lit (e.g. in the fifth state), and the Green light is also eventually lit (in the second state). | clU0 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
270 | 1 | Yes | At some point the red light is on, and at some point the green light is on. Both conditions are satisfied. | ndij | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
271 | 1 | Yes | {G} and {R} | tyos | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
272 | 1 | Yes | The last state satisfies the first condition and the second state satisfies the second condition. | ekny | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
273 | 1 | Yes | Red is eventually lit (in the fifth state), and, separately, Green is eventually lit (in the second state). | hDZG | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
274 | 1 | Yes | They are both eventually lit, the formula never said it had to be in the same places. | XU9x | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
275 | 1 | Yes | The red light is lit in at least one state, as is the green light. | zf66 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
276 | 1 | Yes | Both the red light and the green light are on at some point in the trace. | kkzx | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
277 | 1 | Yes | The formula tells us that the Red light must eventually be lit and the Green light must also eventually be lit. The Red light is lit in the 5th state and the Green light is lit in the 2nd state. | yypq | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
278 | 1 | Yes | Eventually red is lit (5th state) and eventually green is lit (2nd state) | pfdd | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
279 | 1 | Yes | The formula states that at some point, the green light must be on and the right light must be on. It does not enforce that this must ALWAYS be the case. So, we see that the second state has the green light, and all states after and including the fifth state have the red light, so our formula is satisfied. | s0nv | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
280 | 1 | Yes | Both "eventually" predicates hold at some point | j9mq | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
281 | 1 | Yes | Red light is on at some state, so is the Green light | j7t9 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
282 | 1 | Yes | Both red and green are eventually lit | n4vd | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
283 | 1 | Yes | Red and Green eventually lit in some states | qjpx | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
284 | 1 | Yes | Both Green and Red eventually appear in the traces | kuaa | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
285 | 1 | Yes | Both green and red eventually become turned on in the second and fifth states respectively. | XLuy | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
286 | 1 | Yes | In the first state, both green and red lights are eventually on in subsequent states | xeec | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
287 | 1 | Yes | The problem wants it so that eventually the panel has red and eventually the panel has green, but not necessarily that they should happen at the same time. The trace has green and red, so this is satisfied. | duq8 | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } | {} {G} {} {} {R} | Yes | ||||||||||||||||||||
288 | 1 | Yes | there is some state future to two states from now where the red light is on | 917u | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
289 | 1 | Yes | Red is lit in every step in the trace, so it is lit after the 3rd step. | qMyf | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
290 | 1 | Yes | Red is always on so basically any surrounding condition is irrelevant here | gbrV | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
291 | 1 | Yes | For all the states, its second following state is such that eventually the red light is on. | rahc | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
292 | 1 | Yes | In state 1 the formula holds, because in state 3, the red light is eventually on (in fact, the red light is on in state 3). | N3Cw | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
293 | 1 | Yes | Red is lit in at least one state after state 2, which satisfies eventually. | ydkm | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
294 | 1 | Yes | Looking at the third state onwards, it's true that red comes on at some point. | 31qw | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
295 | 1 | Yes | The red light is on at some point at/after the third state (indicated by the two "after"). | c9jy | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
296 | 1 | Yes | red light is on in all states. The formula doesn't constran the first two states. | 3r04 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
297 | 1 | Yes | This statement is saying that beginning in state 3, there will be some state greater than or equal to 3 in which the Red light is on. We see that in states 3, 4, 5, and beyond the Red light is on, so this formula is satisfiable. | ptyk | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
298 | 1 | Yes | the red light is "eventually" lit after the 3rd state, since it's lit in the 4th state. | KXjz | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
299 | 1 | Yes | Red is always in Panel.lit, so regardless of the number of "after"s, Red will always be in Panel.lit in the next state. So, the eventually always holds. | iVk5 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
300 | 1 | Yes | Yes, the formula requires that Red is lit at some point from the 3rd state onwards, but every state in this trace (including the lasso) lights all three colors, so Red will necessarily always be lit. | mxad | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
301 | 1 | Yes | Red is on in the lasso thus the eventually is always satisfied | rnC9 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
302 | 1 | Yes | There is a trace for which two traces later, eventually Red is lit | 8E1h | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
303 | 1 | Yes | red is always lit | sain | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
304 | 1 | Yes | Red light turns on eventually after 2 turns. | nieh | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
305 | 1 | Yes | red was eventually lit | uOG8 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
306 | 1 | Yes | The red light is on in some state after and including the third state. | fsqw | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
307 | 1 | Yes | Because, at some point after the 2nd state in the trace, the Red light is on. | dznh | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
308 | 1 | Yes | Red is always on. | 7opy | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
309 | 1 | Yes | Since Red is in every trace, it will always satisfy this | 55oa | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
310 | 1 | Yes | Red is lit after the second state. | vz2g | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
311 | 1 | Yes | Since the red light is always on in this trace, any requirement that the red light be on at some point in time (without untils or requirements that the red light be off at some point) is satisfied by it. | cirf | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
312 | 1 | Yes | Starting in the third state, there is some state there or in the future in which the Red light is on. | fojf | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
313 | 1 | Yes | R is in the 3rd time unit (and the 4th and 5th) | 7jm2 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
314 | 1 | Yes | Red is on in 4th trace. | eolW | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
315 | 1 | Yes | The formula necessitates that there is a state at or beyond the third state (i.e. second state after the first) where the red light turns on. Since, in this trace, the red light is always on, this is automatically satisfied. | c1fn | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
316 | 1 | Yes | In 2 state, Red is eventually lit. It is lit in state 3. | nr65 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
317 | 1 | Yes | Red is in the 4th and 5th state, which makes this true | m0p6 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
318 | 1 | Yes | Because after two states red is present always although its required to appear once. | able | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
319 | 1 | Yes | After the 2nd state, Red is eventually on (it is on in the 3rd state, in fact). | 1vwv | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
320 | 1 | Yes | Starting from the third state, red is lit at least once. | yoxy | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
321 | 1 | Yes | Red is on in state 5 | DzoD | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
322 | 1 | Yes | Red is on at some point after/including state 3. | xqZA | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
323 | 1 | Yes | For every state the Red light is turned on at some point after after that state (since Red is already turned on in every state) | bxd4 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
324 | 1 | Yes | Two states after the first, there is some state after that at some point where the Red panel is lit | bx1r | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
325 | 1 | Yes | In the third state, the Red light is already on, so it satisfies that it is eventually turned on. | w4t3 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
326 | 1 | Yes | In the third state (i.e., 2 states after the first state), it is the case that the Red light is eventually lit (e.g. in the third state itself). | clU0 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
327 | 1 | Yes | Two afters mean that we only need to look at traces starting from the 3rd state, and we see that the red light is on at some point (in the 3rd state). | ndij | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
328 | 1 | Yes | Theres several states where Red panel is lit following the `after after`. | tyos | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
329 | 1 | Yes | Red is lit after the 2nd state. | ekny | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
330 | 1 | Yes | Red is lit in the third state, so “eventually { Red in Panel.lit }” is true in the third state. Therefore, the original formula is true in the first state. | hDZG | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
331 | 1 | Yes | Well R is lit in all the traces, so this is pretty trivial. | XU9x | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
332 | 1 | Yes | the red light is always lit | zf66 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
333 | 1 | Yes | The formula defines that the red light should be on at some point. The trace given has the red light on in every state and therefore the formula is sat. | kkzx | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
334 | 1 | Yes | All of the states have the red light lit up. | yypq | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
335 | 1 | Yes | Red is lit in the 5th state, which occurs after the 3rd state. | pfdd | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
336 | 1 | Yes | Since the red light is on in every state, we see that every state satisfies the `eventually` clause. `After after` is referring to the third state in our trace, and we see that the third state itself satisfies the `eventually` clause. | s0nv | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
337 | 1 | Yes | Red is always lit, so it's necessarily lit eventually regardless of where you start in the trace. | j9mq | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
338 | 1 | Yes | starting from the third state, Red light is on at some state | j7t9 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
339 | 1 | Yes | Yes, since red is lit in all isntances | n4vd | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
340 | 1 | Yes | Red is always lit, so it will eventually lit from the third state | qjpx | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
341 | 1 | Yes | Red eventually appears in the lit panel | kuaa | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
342 | 1 | No | Red is turned on in the first two states. | XLuy | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
343 | 1 | Yes | Red is on in states after the third state | xeec | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
344 | 1 | Yes | The problem wants it so that after the first two states, there exists a state where red is in the panel, which happens in the third state. | duq8 | after { after { eventually { Red in Panel.lit } } } | {RGB} {RGB} {RGB} {RGB} {RGB} | Yes | ||||||||||||||||||||
345 | 1 | Yes | The red light is always on because the blue light is never turned on. So the formula is not violated. | 917u | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
346 | 1 | Yes | Since Blue is never lit, Red can continue to be lit forever. | qMyf | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
347 | 1 | Yes | Blue is never turned on so red always stays on | gbrV | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
348 | 1 | No | The formula requires that the blue light is on at some point, and at this point onwards the red light is never on. This trace doesn't have any state where blue light is on. | rahc | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
349 | 1 | No | I think using "until" means that the "Blue in Panel.lit" has to hold at some point, but the blue light never turns on. | N3Cw | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
350 | 1 | No | For until to work there must be some state where Blue is lit. | ydkm | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
351 | 1 | No | Blue never turns on. | 31qw | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
352 | 1 | Yes | Red in Panel.lit always holds, since the "Blue in Panel.lit" (blue light on) state never occurs. | c9jy | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
353 | 1 | Yes | Blue light is never on, so red can always on. | 3r04 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
354 | 1 | No | This is not satisfiable because the until operator guarantees that Blue in Panel.lit will hold at some point in the trace, but we see here that it never does. | ptyk | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
355 | 1 | Yes | the first predicate always can hold since the second never does. | KXjz | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
356 | 1 | No | Until specifies that the RHS must hold at some point, and Blue is never in Panel.lit in this trace | iVk5 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
357 | 1 | Yes | This is true because Blue never appears lit in the trace. As shown, Red will stay lit to infinity, and so Blue will never light up and thus Red should never shut off. | mxad | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
358 | 1 | No | As blue is never on, the right side of the until can never be true. | rnC9 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
359 | 1 | Yes | I'm not as sure about this one, but I think that the idea that Red is always lit until Blue is lit, and since Blue isn't lit, it's true. | 8E1h | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
360 | 1 | Yes | red is always lit | sain | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
361 | 1 | Yes | The second clause of the until statement is never satisfied so it can keep being red. | nieh | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
362 | 1 | Yes | just cause we dont see the blue doesnt mean it wont be there in the future | uOG8 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
363 | 1 | No | The blue light is never on in any state. | fsqw | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
364 | 1 | No | Because the Blue light never turns on. | dznh | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
365 | 1 | No | Blue is never on. But the until formula claims that it should be on in some state. | 7opy | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
366 | 1 | Yes | The only time Red should stop being lit is if Blue is. Seeing as Blue is never lit, that means that Red should always be lit (which it is) | 55oa | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
367 | 1 | Yes | Blue is never lit, so Red is always lit. | vz2g | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
368 | 1 | No | The documentation page states that until requires that the second formula be true at some point (in other words, this formula implies that eventually Blue in Panel.lit). Since the blue light never turns on this trace does not satisfy the formula (even though I think the intuitive reading of until would indicate that it should). | cirf | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
369 | 1 | No | Until requires here that the Blue light is eventually lit, but it is never lit. | fojf | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
370 | 1 | Yes | Red is always lit, and Blue is never, so this is always true. | 7jm2 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
371 | 1 | Yes | Blue is never on | eolW | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
372 | 1 | Yes | The 'until' keyword implies the red light should always stay on until the blue light turns on for the first time. Since the blue light never turns on and the red light always stays on, this formula is satisfied. | c1fn | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
373 | 1 | No | fmla-b (Blue in Panel.lit) must hold in some finite state after the current state. | nr65 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
374 | 1 | Yes | There is no blue in panel.lit, so it's fine that red is always in panel.lit | m0p6 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
375 | 1 | No | Because blue is not there in any state | able | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
376 | 1 | No | Blue is not eventually in Panel.lit, so `until` is not satisfied. | 1vwv | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
377 | 1 | Yes | The until clause doesn't ever have to become true. | yoxy | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
378 | 1 | Yes | Blue never turns on, so red never needs to turn off | DzoD | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
379 | 1 | No | Blue is never on. | xqZA | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
380 | 1 | No | There is no Blue in the traces. Blue should happen in the state but since fmla-b doesn't hold, the formula cannot be satisfied by the tracee. | bxd4 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
381 | 1 | No | The blue light being on is required for the until operator to be satisfied | bx1r | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
382 | 1 | No | Formula b must hold in some state afterwards in the trace. | w4t3 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
383 | 1 | No | A statement of the form <x> until <y> requires <y> to eventually hold, but it is never the case that the Blue light is on. | clU0 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
384 | 1 | Yes | Because the blue light is never on, it's vacuously true. | ndij | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
385 | 1 | Yes | Red in Panel.lit was enough to satisfy the formula | tyos | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
386 | 1 | Yes | The first condition is always true. | ekny | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
387 | 1 | No | Blue is never lit, so the second part of the “until” is not satisfied. | hDZG | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
388 | 1 | Yes | Blue is never lit, so Red must stay lit. | XU9x | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
389 | 1 | No | There is no state where the blue light is lit. | zf66 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
390 | 1 | Yes | The formula says the red light should be on until the blue light is on. In the trace the blue light is never on and the red light is always on - it satisfies the formula since the formula doesn't say blue light must be on at some point. Since the until condition is never hit it makes sense that the red light is on throughout the trace. | kkzx | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
391 | 1 | No | In order to use "until" and return true, the second condition must be satisfied somewhere. The Blue light is never true, so we can't be certain that the Red light is on until the Blue light is on. | yypq | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
392 | 1 | No | Blue is not lit in any state | pfdd | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
393 | 1 | No | As per the Forge documentation: "<fmla-a> until <fmla-b> is true in a state i if and only if: fmla-b holds in some state j>=i and fmla-a holds in all states k where i <= k < j." In this case, <fmla-b> is {Blue in Panel.lit}, but we see that there is no state in this trace (including the lasso step) in which fmla-b holds, since the light is never blue. Therefore, fmla-b does not hold in some state j>=i, which means that this trace does not satisfy the formula. | s0nv | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
394 | 1 | No | Blue is never lit in the trace, which falsifies the 'until'. | j9mq | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
395 | 1 | Yes | Blue light is never on and Red light is always on | j7t9 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
396 | 1 | Yes | Yes, since blue is never lit the until statement is equivalent to just the antecedent. | n4vd | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
397 | 1 | No | Blue never lits | qjpx | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
398 | 1 | Yes | Blue never appears in the panel; therefore, red must always appear | kuaa | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
399 | 1 | Yes | Blue is never turned on so Red can stay on for all 5 traces. | XLuy | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
400 | 1 | Yes | Blue is never lit, so red should be always lit | xeec | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
401 | 1 | No | Until assumes the 'Red in Panel.lit' is only held for a finite number of steps, but since trace will make Red with infinite steps. | duq8 | Red in Panel.lit until Blue in Panel.lit | {R} {R} {R} {R} {R} | No | ||||||||||||||||||||
402 | 1 | No | There is not state from where on the Red light is always on. | 917u | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
403 | 1 | No | The empty state repeats forever, so Red is only lit twice. Thus, Red is never always lit. | qMyf | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
404 | 1 | No | There is no state after which red is always on | gbrV | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
405 | 1 | No | There is no state such that eventually there is a state following it where all the subsequent states have the red light on. | rahc | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
406 | 1 | No | The formula says that eventually the Red light will turn on and stay on forever, which doesn't happen. | N3Cw | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
407 | 1 | No | There is no state such that every state after has red lit, as the last state does not have red lit. | ydkm | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
408 | 1 | No | All states not shown are {}, so there's no point after which red is always on. | 31qw | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
409 | 1 | No | No state occurs in which the Red light is permanently turned on. | c9jy | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
410 | 1 | No | In the fifth state, red light should turn on. | 3r04 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
411 | 1 | No | This trace is looking for a state after which Red is always lit. If Red were lit in state 5 (and therefore in all future states), this would be true. Since it is not, we see that this formula is not satisfiable. | ptyk | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
412 | 1 | No | it is never the case that red is always lit since in the two instances it is, it isn't in the following state. | KXjz | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
413 | 1 | No | There is no state after which all of the following states have Red in Panel.lit | iVk5 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
414 | 1 | No | After Red appears at some point, it must be the case the Red will always appear after that. There are two opportunities here to satisfy this formula, but Red doesn't stay lit for even two states in a row. Since the final and lasso state has nothing lit, there are no more opportunities to satisfy the formula, so this trace does not satisfy it. | mxad | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
415 | 1 | No | As red is not on in the lasso, it is never true that the red light will always be on after a given state | rnC9 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
416 | 1 | No | There is no state for which every state after that state has Red lit. | 8E1h | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
417 | 1 | No | there is no state after which all states have the red light | sain | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
418 | 1 | No | The last state does not contain the red light therefore the always statement is never eventually satisfied. | nieh | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
419 | 1 | No | if there were more traces in the future, this could hold, but in this set of traces there is no point in which red is always lit | uOG8 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
420 | 1 | No | The red light is never always on after any of the states. | fsqw | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
421 | 1 | No | because there is no state after which the Red light is always on. | dznh | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
422 | 1 | No | The trace ends with a repetition of {}, which makes it impossible to have Red always on. | 7opy | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
423 | 1 | No | This is saying that at some point in time, Red will become lit and stay that way forever. While Red does become lit twice, in the next states it immediately turns off. If the lasso state had the Red light lit this would be satisfied, but seeing as it doesn't, this is definitely not satisfied. | 55oa | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
424 | 1 | 1 | No | Once Red is lit, it should stay lit. (eventually always) | vz2g | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | |||||||||||||||||||
425 | 1 | No | This formula requires that at some point the red light turns on and stays on forever, but since this trace ends with an infinite sequence of states where all the lights are off, this trace does not satisfy the formula. | cirf | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
426 | 1 | No | Since the empty state (in which no light, including Red, is lit) cycles forever at the end, there is no state after or on the first state such that the Red light will be on at every state in the future. | fojf | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
427 | 1 | No | There is never a time where Red is always lit, since at least every other time has Red not being lit. | 7jm2 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
428 | 1 | No | Lasso around all lights off - red is not on. | eolW | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
429 | 1 | No | The final lasso state, { }, repeats indefinitely, and the red light is not on in that state. Therefore, we never reach a state, at and after which the red light always stays on, so the formula is not satisfied. | c1fn | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
430 | 1 | No | There is no point at which every state after that state has Red lit. The trace lassos on nothing being lit. | nr65 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
431 | 1 | 1 | No | 3 and 5th state need to have red in it to make it true | m0p6 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | |||||||||||||||||||
432 | 1 | No | Red is not always on. | able | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
433 | 1 | No | There is no point beyond which Red is always on because the final empty state repeats forever. | 1vwv | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
434 | 1 | No | Red becomes lit but doesn't permanently stay on once it is. | yoxy | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
435 | 1 | No | Red eventually turns off, which means the "always" statement is not satisfied. | DzoD | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
436 | 1 | No | There is no point at which Red turns on and stays on. | xqZA | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
437 | 1 | No | There isn't a trace of states where all the traces have Red in the Panel.lit. It would've been true if the fifth state had Red in it because it repeats. | bxd4 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
438 | 1 | No | eventually always forces at least the last state to satisfy the condition | bx1r | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
439 | 1 | No | At no point in any of the states is the Red light always on. | w4t3 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
440 | 1 | No | There is no point in the trace at which the Red light stays on forever. | clU0 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
441 | 1 | No | Because the last state (all lights off) repeats forever, we are never getting to a point where the red light stays on. | ndij | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
442 | 1 | No | The last trace does not have Red in Panel.lit. | tyos | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
443 | 1 | No | There is never a point after which Red is always lit (it loops on a state where no lights are lit). | ekny | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
444 | 1 | No | The last state does not have the red light lit, and since that repeats forever, that means that Red is off forever. Therefore, there is no state after which Red is always lit. | hDZG | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
445 | 1 | No | The lasso state doesn't contain R | XU9x | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
446 | 1 | No | There is no state after which the red light is always lit. | zf66 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
447 | 1 | No | The red light is not on in the last state, but the formula states that at some point the red light will always be on. | kkzx | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
448 | 1 | 1 | No | There is never a point where Red is always lit afterwards. The Red light will always flicker on and off based on the given states. | yypq | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | |||||||||||||||||||
449 | 1 | No | Whenever red is lit, it turns off immediately, so it is never the case that red is always lit. | pfdd | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
450 | 1 | No | This states that there exists a state such that all states afterward will have the Red light on. However, we see that this does not occur in the first four states, and in the last repeating state, no lights are on, so this formula will never hold in the future. | s0nv | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
451 | 1 | No | There is no time step such that red is always lit from that point forward | j9mq | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
452 | 1 | No | There is no state after which Red light is always on | j7t9 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
453 | 1 | No | No, because there is no instance where the red panel is always lit from then on out. | n4vd | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
454 | 1 | No | The final state does not have Red in it. Thus it cannot be forever lit. | qjpx | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
455 | 1 | 1 | No | According to these set of conditions, once red appears, it must always appear. | kuaa | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | |||||||||||||||||||
456 | 1 | No | Red is never on continuously. | XLuy | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
457 | 1 | No | There is no state where red is always lit in all subsequent states | xeec | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
458 | 1 | No | The formula statements that eventually there exists a state where all states including that one has red in it. This clearly does not happen since the last state getting repeated has no colors. | duq8 | eventually { always { Red in Panel.lit } } | {} {RGB} {} {RGB} {} | No | ||||||||||||||||||||
459 | 1 | Yes | The red light is never turned on so the green light also does not need to be turned on. The formula is never violated. | 917u | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
460 | 1 | Yes | Since Red is never lit, Green never has an obligation to be lit, so this formula isn't violated. | qMyf | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
461 | 1 | Yes | The LHS is always false so the expression is always true | gbrV | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
462 | 1 | Yes | Red light is never on, so the part inside { } is always satisfied. | rahc | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
463 | 1 | Yes | Implies statements are true if the left hand is always false, which is the case here (red light never turns on). | N3Cw | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
464 | 1 | Yes | Because red is never lit, False implies False results in True. | ydkm | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
465 | 1 | Yes | Conditional is true when antecedent is not satisfied - red never on. | 31qw | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
466 | 1 | Yes | The red light is never turned on, so it doesn't matter whether the green light is turned on or not. | c9jy | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
467 | 1 | Yes | Red light is never turned on, and green lit can be turn on or not. | 3r04 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
468 | 1 | Yes | In all states, Red is not lit. Therefore the first portion of this implies statement is not true. That means that the entire statement is true. An implies is only false if the first portion holds but the second does not. | ptyk | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
469 | 1 | Yes | no lights are ever lit, so there's no obligation to satisfy that implication, and thus it always holds. | KXjz | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
470 | 1 | Yes | The LHS of the implies never holds since Red is never in Panel.lit. So, the RHS never has to hold for the formula to be true. | iVk5 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
471 | 1 | Yes | This is an infinitely empty trace. Since Red is never in Panel.lit, it doesn't constrain Green at all. The implies statement is still true because the left-side of the expression is false. | mxad | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
472 | 1 | Yes | as Red is never on, this is trivially satisfied. | rnC9 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
473 | 1 | Yes | Since Red is never lit, the statement within the always {} is always true. | 8E1h | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
474 | 1 | Yes | first condition never holds so its always true | sain | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
475 | 1 | Yes | Red light is never turned on therefore we don't ever have to satisfy the latter part of the implication. | nieh | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
476 | 1 | Yes | red was never there so green was never there | uOG8 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
477 | 1 | Yes | The red light is never on, so there's no condition that the trace needs to satisfy. | fsqw | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
478 | 1 | Yes | Because the Right light never turns on. As a result, the implication is always true. | dznh | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
479 | 1 | Yes | Red is never on. So the implication is trivially true. | 7opy | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
480 | 1 | Yes | Because of how "implies" works, this formula is saying that the Green light will be lit only if the Red light is. It also says that if the Red light isn't lit, nothing happens. Therefore, this trace holds because False implies <anything> is always True. | 55oa | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
481 | 1 | Yes | It's an implies statement. Nothing requires Red to be lit. Red and Green both never being lit is therefore satisfactory. | vz2g | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
482 | 1 | Yes | Since the red light is never on, the implication always holds true, so the trace satisfies the formula. | cirf | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
483 | 1 | Yes | The Red light is never lit. When the first part of an implies statement is false, the implies statement is true. Sine the first part of this implies statement is always false, the implies statement is always true. | fojf | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
484 | 1 | Yes | Red is never lit, so this is vacuously true. | 7jm2 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
485 | 1 | Yes | Red never lit so implies never invoked, | eolW | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
486 | 1 | Yes | The given trace shows that in all states, all lights are off. Therefore, the precondition for the implies statement (i.e. the red light being on) is never satisfied, so the implies statement evaluates to true. Since this is always the case, the formula is satisfied. | c1fn | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
487 | 1 | Yes | Red is never lit, so this is vacuously satisfied. | nr65 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
488 | 1 | Yes | Red is never in panel.lit, so therefore there is nothing to worry about | m0p6 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
489 | 1 | Yes | Because red is never lit itself | able | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
490 | 1 | Yes | Red is never on, so the implication is always satisfied. | 1vwv | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
491 | 1 | Yes | Red doesn't necessarily have to be lit. | yoxy | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
492 | 1 | Yes | Because red never turns on, there is never an opportunity to violate the statement. | DzoD | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
493 | 1 | Yes | The implication is vacuously true since Red is never on. | xqZA | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
494 | 1 | Yes | There was no Red in the trace, so whether Green is in the trace or not doesn't matter. The trace doesn't 'unsatisfy' the formula. | bxd4 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
495 | 1 | Yes | The red light is never on, so the green light never has to be on either. The implication is always vacuously true | bx1r | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
496 | 1 | Yes | The implies statement returns true when formula-a is not met. Since the Red light is never on, the implies statement always returns true. | w4t3 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
497 | 1 | Yes | In every state, the Red light is not lit, so no obligation is incurred on the Green light, meaning the implication always (vacuously) holds. | clU0 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
498 | 1 | Yes | Because the red light is never on, it's vacuously true. | ndij | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
499 | 1 | Yes | LHS of implies is false. Therefore this is true. | tyos | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
500 | 1 | Yes | The formula is vacuously true because the first part of the implies is never satisfied. | ekny | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
501 | 1 | Yes | Red is never lit, so the implication is vacuously true. | hDZG | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
502 | 1 | Yes | Green has no obligation to be lit if Red isn't (and it never gets lit) | XU9x | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
503 | 1 | Yes | The red light is never lit, so the `implies` statement is always true by default. | zf66 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
504 | 1 | Yes | The formula says whenever the red light is on, the green light is also on. Since in the trace the red light is never on, the behavior is unspecified and therefore all lights off is valid. | kkzx | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
505 | 1 | Yes | The Red light is never lit, so the implies statement is always satisfied. Since, the implies is always satisfied, the formula is satisfied. | yypq | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
506 | 1 | Yes | Red is never lit, so the implication is always true. | pfdd | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
507 | 1 | Yes | Red is never in Panel.lit, so the implication is vacuously true for all states, thus satisfying our formula for all states. | s0nv | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
508 | 1 | Yes | The first half of the implication is always false, so the expression as a whole is always true. | j9mq | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
509 | 1 | Yes | The implication is satisfied vacuously | j7t9 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
510 | 1 | Yes | Yes, the implication is satisfied vacuously. | n4vd | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
511 | 1 | Yes | Red is never lit, so the implies is always true | qjpx | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
512 | 1 | Yes | The condition above only holds true if red appears in the panel. If there is no red, then any set of traces can appear without red | kuaa | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
513 | 1 | Yes | Red is never lit so Green also never has to be lit. | XLuy | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
514 | 1 | Yes | Red is not lit, so the implies statement is always true | xeec | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
515 | 1 | Yes | The formula only needs to ensure that whenever red is in the traces, green should be in it, but red doesn't have to be in the traces, so all empty colors would suffice. | duq8 | always { Red in Panel.lit implies Green in Panel.lit } | {} {} {} {} {} | Yes | ||||||||||||||||||||
516 | |||||||||||||||||||||||||||
517 | |||||||||||||||||||||||||||
518 | |||||||||||||||||||||||||||
519 | |||||||||||||||||||||||||||
520 | |||||||||||||||||||||||||||
521 | |||||||||||||||||||||||||||
522 | |||||||||||||||||||||||||||
523 | |||||||||||||||||||||||||||
524 | |||||||||||||||||||||||||||
525 | |||||||||||||||||||||||||||
526 | |||||||||||||||||||||||||||
527 | |||||||||||||||||||||||||||
528 | |||||||||||||||||||||||||||
529 | |||||||||||||||||||||||||||
530 | |||||||||||||||||||||||||||
531 | |||||||||||||||||||||||||||
532 | |||||||||||||||||||||||||||
533 | |||||||||||||||||||||||||||
534 | |||||||||||||||||||||||||||
535 | |||||||||||||||||||||||||||
536 | |||||||||||||||||||||||||||
537 | |||||||||||||||||||||||||||
538 | |||||||||||||||||||||||||||
539 | |||||||||||||||||||||||||||
540 | |||||||||||||||||||||||||||
541 | |||||||||||||||||||||||||||
542 | |||||||||||||||||||||||||||
543 | |||||||||||||||||||||||||||
544 | |||||||||||||||||||||||||||
545 | |||||||||||||||||||||||||||
546 | |||||||||||||||||||||||||||
547 | |||||||||||||||||||||||||||
548 | |||||||||||||||||||||||||||
549 | |||||||||||||||||||||||||||
550 | |||||||||||||||||||||||||||
551 | |||||||||||||||||||||||||||
552 | |||||||||||||||||||||||||||
553 | |||||||||||||||||||||||||||
554 | |||||||||||||||||||||||||||
555 | |||||||||||||||||||||||||||
556 | |||||||||||||||||||||||||||
557 | |||||||||||||||||||||||||||
558 | |||||||||||||||||||||||||||
559 | |||||||||||||||||||||||||||
560 | |||||||||||||||||||||||||||
561 | |||||||||||||||||||||||||||
562 | |||||||||||||||||||||||||||
563 | |||||||||||||||||||||||||||
564 | |||||||||||||||||||||||||||
565 | |||||||||||||||||||||||||||
566 | |||||||||||||||||||||||||||
567 | |||||||||||||||||||||||||||
568 | |||||||||||||||||||||||||||
569 | |||||||||||||||||||||||||||
570 | |||||||||||||||||||||||||||
571 | |||||||||||||||||||||||||||
572 | |||||||||||||||||||||||||||
573 | |||||||||||||||||||||||||||
574 | |||||||||||||||||||||||||||
575 | |||||||||||||||||||||||||||
576 | |||||||||||||||||||||||||||
577 | |||||||||||||||||||||||||||
578 | |||||||||||||||||||||||||||
579 | |||||||||||||||||||||||||||
580 | |||||||||||||||||||||||||||
581 | |||||||||||||||||||||||||||
582 | |||||||||||||||||||||||||||
583 | |||||||||||||||||||||||||||
584 | |||||||||||||||||||||||||||
585 | |||||||||||||||||||||||||||
586 | |||||||||||||||||||||||||||
587 | |||||||||||||||||||||||||||
588 | |||||||||||||||||||||||||||
589 | |||||||||||||||||||||||||||
590 | |||||||||||||||||||||||||||
591 | |||||||||||||||||||||||||||
592 | |||||||||||||||||||||||||||
593 | |||||||||||||||||||||||||||
594 | |||||||||||||||||||||||||||
595 | |||||||||||||||||||||||||||
596 | |||||||||||||||||||||||||||
597 | |||||||||||||||||||||||||||
598 | |||||||||||||||||||||||||||
599 | |||||||||||||||||||||||||||
600 | |||||||||||||||||||||||||||
601 | |||||||||||||||||||||||||||
602 | |||||||||||||||||||||||||||
603 | |||||||||||||||||||||||||||
604 | |||||||||||||||||||||||||||
605 | |||||||||||||||||||||||||||
606 | |||||||||||||||||||||||||||
607 | |||||||||||||||||||||||||||
608 | |||||||||||||||||||||||||||
609 | |||||||||||||||||||||||||||
610 | |||||||||||||||||||||||||||
611 | |||||||||||||||||||||||||||
612 | |||||||||||||||||||||||||||
613 | |||||||||||||||||||||||||||
614 | |||||||||||||||||||||||||||
615 | |||||||||||||||||||||||||||
616 | |||||||||||||||||||||||||||
617 | |||||||||||||||||||||||||||
618 | |||||||||||||||||||||||||||
619 | |||||||||||||||||||||||||||
620 | |||||||||||||||||||||||||||
621 | |||||||||||||||||||||||||||
622 | |||||||||||||||||||||||||||
623 | |||||||||||||||||||||||||||
624 | |||||||||||||||||||||||||||
625 | |||||||||||||||||||||||||||
626 | |||||||||||||||||||||||||||
627 | |||||||||||||||||||||||||||
628 | |||||||||||||||||||||||||||
629 | |||||||||||||||||||||||||||
630 | |||||||||||||||||||||||||||
631 | |||||||||||||||||||||||||||
632 | |||||||||||||||||||||||||||
633 | |||||||||||||||||||||||||||
634 | |||||||||||||||||||||||||||
635 | |||||||||||||||||||||||||||
636 | |||||||||||||||||||||||||||
637 | |||||||||||||||||||||||||||
638 | |||||||||||||||||||||||||||
639 | |||||||||||||||||||||||||||
640 | |||||||||||||||||||||||||||
641 | |||||||||||||||||||||||||||
642 | |||||||||||||||||||||||||||
643 | |||||||||||||||||||||||||||
644 | |||||||||||||||||||||||||||
645 | |||||||||||||||||||||||||||
646 | |||||||||||||||||||||||||||
647 | |||||||||||||||||||||||||||
648 | |||||||||||||||||||||||||||
649 | |||||||||||||||||||||||||||
650 | |||||||||||||||||||||||||||
651 | |||||||||||||||||||||||||||
652 | |||||||||||||||||||||||||||
653 | |||||||||||||||||||||||||||
654 | |||||||||||||||||||||||||||
655 | |||||||||||||||||||||||||||
656 | |||||||||||||||||||||||||||
657 | |||||||||||||||||||||||||||
658 | |||||||||||||||||||||||||||
659 | |||||||||||||||||||||||||||
660 | |||||||||||||||||||||||||||
661 | |||||||||||||||||||||||||||
662 | |||||||||||||||||||||||||||
663 | |||||||||||||||||||||||||||
664 | |||||||||||||||||||||||||||
665 | |||||||||||||||||||||||||||
666 | |||||||||||||||||||||||||||
667 | |||||||||||||||||||||||||||
668 | |||||||||||||||||||||||||||
669 | |||||||||||||||||||||||||||
670 | |||||||||||||||||||||||||||
671 | |||||||||||||||||||||||||||
672 | |||||||||||||||||||||||||||
673 | |||||||||||||||||||||||||||
674 | |||||||||||||||||||||||||||
675 | |||||||||||||||||||||||||||
676 | |||||||||||||||||||||||||||
677 | |||||||||||||||||||||||||||
678 | |||||||||||||||||||||||||||
679 | |||||||||||||||||||||||||||
680 | |||||||||||||||||||||||||||
681 | |||||||||||||||||||||||||||
682 | |||||||||||||||||||||||||||
683 | |||||||||||||||||||||||||||
684 | |||||||||||||||||||||||||||
685 | |||||||||||||||||||||||||||
686 | |||||||||||||||||||||||||||
687 | |||||||||||||||||||||||||||
688 | |||||||||||||||||||||||||||
689 | |||||||||||||||||||||||||||
690 | |||||||||||||||||||||||||||
691 | |||||||||||||||||||||||||||
692 | |||||||||||||||||||||||||||
693 | |||||||||||||||||||||||||||
694 | |||||||||||||||||||||||||||
695 | |||||||||||||||||||||||||||
696 | |||||||||||||||||||||||||||
697 | |||||||||||||||||||||||||||
698 | |||||||||||||||||||||||||||
699 | |||||||||||||||||||||||||||
700 | |||||||||||||||||||||||||||
701 | |||||||||||||||||||||||||||
702 | |||||||||||||||||||||||||||
703 | |||||||||||||||||||||||||||
704 | |||||||||||||||||||||||||||
705 | |||||||||||||||||||||||||||
706 | |||||||||||||||||||||||||||
707 | |||||||||||||||||||||||||||
708 | |||||||||||||||||||||||||||
709 | |||||||||||||||||||||||||||
710 | |||||||||||||||||||||||||||
711 | |||||||||||||||||||||||||||
712 | |||||||||||||||||||||||||||
713 | |||||||||||||||||||||||||||
714 | |||||||||||||||||||||||||||
715 | |||||||||||||||||||||||||||
716 | |||||||||||||||||||||||||||
717 | |||||||||||||||||||||||||||
718 | |||||||||||||||||||||||||||
719 | |||||||||||||||||||||||||||
720 | |||||||||||||||||||||||||||
721 | |||||||||||||||||||||||||||
722 | |||||||||||||||||||||||||||
723 | |||||||||||||||||||||||||||
724 | |||||||||||||||||||||||||||
725 | |||||||||||||||||||||||||||
726 | |||||||||||||||||||||||||||
727 | |||||||||||||||||||||||||||
728 | |||||||||||||||||||||||||||
729 | |||||||||||||||||||||||||||
730 | |||||||||||||||||||||||||||
731 | |||||||||||||||||||||||||||
732 | |||||||||||||||||||||||||||
733 | |||||||||||||||||||||||||||
734 | |||||||||||||||||||||||||||
735 | |||||||||||||||||||||||||||
736 | |||||||||||||||||||||||||||
737 | |||||||||||||||||||||||||||
738 | |||||||||||||||||||||||||||
739 | |||||||||||||||||||||||||||
740 | |||||||||||||||||||||||||||
741 | |||||||||||||||||||||||||||
742 | |||||||||||||||||||||||||||
743 | |||||||||||||||||||||||||||
744 | |||||||||||||||||||||||||||
745 | |||||||||||||||||||||||||||
746 | |||||||||||||||||||||||||||
747 | |||||||||||||||||||||||||||
748 | |||||||||||||||||||||||||||
749 | |||||||||||||||||||||||||||
750 | |||||||||||||||||||||||||||
751 | |||||||||||||||||||||||||||
752 | |||||||||||||||||||||||||||
753 | |||||||||||||||||||||||||||
754 | |||||||||||||||||||||||||||
755 | |||||||||||||||||||||||||||
756 | |||||||||||||||||||||||||||
757 | |||||||||||||||||||||||||||
758 | |||||||||||||||||||||||||||
759 | |||||||||||||||||||||||||||
760 | |||||||||||||||||||||||||||
761 | |||||||||||||||||||||||||||
762 | |||||||||||||||||||||||||||
763 | |||||||||||||||||||||||||||
764 | |||||||||||||||||||||||||||
765 | |||||||||||||||||||||||||||
766 | |||||||||||||||||||||||||||
767 | |||||||||||||||||||||||||||
768 | |||||||||||||||||||||||||||
769 | |||||||||||||||||||||||||||
770 | |||||||||||||||||||||||||||
771 | |||||||||||||||||||||||||||
772 | |||||||||||||||||||||||||||
773 | |||||||||||||||||||||||||||
774 | |||||||||||||||||||||||||||
775 | |||||||||||||||||||||||||||
776 | |||||||||||||||||||||||||||
777 | |||||||||||||||||||||||||||
778 | |||||||||||||||||||||||||||
779 | |||||||||||||||||||||||||||
780 | |||||||||||||||||||||||||||
781 | |||||||||||||||||||||||||||
782 | |||||||||||||||||||||||||||
783 | |||||||||||||||||||||||||||
784 | |||||||||||||||||||||||||||
785 | |||||||||||||||||||||||||||
786 | |||||||||||||||||||||||||||
787 | |||||||||||||||||||||||||||
788 | |||||||||||||||||||||||||||
789 | |||||||||||||||||||||||||||
790 | |||||||||||||||||||||||||||
791 | |||||||||||||||||||||||||||
792 | |||||||||||||||||||||||||||
793 | |||||||||||||||||||||||||||
794 | |||||||||||||||||||||||||||
795 | |||||||||||||||||||||||||||
796 | |||||||||||||||||||||||||||
797 | |||||||||||||||||||||||||||
798 | |||||||||||||||||||||||||||
799 | |||||||||||||||||||||||||||
800 | |||||||||||||||||||||||||||
801 | |||||||||||||||||||||||||||
802 | |||||||||||||||||||||||||||
803 | |||||||||||||||||||||||||||
804 | |||||||||||||||||||||||||||
805 | |||||||||||||||||||||||||||
806 | |||||||||||||||||||||||||||
807 | |||||||||||||||||||||||||||
808 | |||||||||||||||||||||||||||
809 | |||||||||||||||||||||||||||
810 | |||||||||||||||||||||||||||
811 | |||||||||||||||||||||||||||
812 | |||||||||||||||||||||||||||
813 | |||||||||||||||||||||||||||
814 | |||||||||||||||||||||||||||
815 | |||||||||||||||||||||||||||
816 | |||||||||||||||||||||||||||
817 | |||||||||||||||||||||||||||
818 | |||||||||||||||||||||||||||
819 | |||||||||||||||||||||||||||
820 | |||||||||||||||||||||||||||
821 | |||||||||||||||||||||||||||
822 | |||||||||||||||||||||||||||
823 | |||||||||||||||||||||||||||
824 | |||||||||||||||||||||||||||
825 | |||||||||||||||||||||||||||
826 | |||||||||||||||||||||||||||
827 | |||||||||||||||||||||||||||
828 | |||||||||||||||||||||||||||
829 | |||||||||||||||||||||||||||
830 | |||||||||||||||||||||||||||
831 | |||||||||||||||||||||||||||
832 | |||||||||||||||||||||||||||
833 | |||||||||||||||||||||||||||
834 | |||||||||||||||||||||||||||
835 | |||||||||||||||||||||||||||
836 | |||||||||||||||||||||||||||
837 | |||||||||||||||||||||||||||
838 | |||||||||||||||||||||||||||
839 | |||||||||||||||||||||||||||
840 | |||||||||||||||||||||||||||
841 | |||||||||||||||||||||||||||
842 | |||||||||||||||||||||||||||
843 | |||||||||||||||||||||||||||
844 | |||||||||||||||||||||||||||
845 | |||||||||||||||||||||||||||
846 | |||||||||||||||||||||||||||
847 | |||||||||||||||||||||||||||
848 | |||||||||||||||||||||||||||
849 | |||||||||||||||||||||||||||
850 | |||||||||||||||||||||||||||
851 | |||||||||||||||||||||||||||
852 | |||||||||||||||||||||||||||
853 | |||||||||||||||||||||||||||
854 | |||||||||||||||||||||||||||
855 | |||||||||||||||||||||||||||
856 | |||||||||||||||||||||||||||
857 | |||||||||||||||||||||||||||
858 | |||||||||||||||||||||||||||
859 | |||||||||||||||||||||||||||
860 | |||||||||||||||||||||||||||
861 | |||||||||||||||||||||||||||
862 | |||||||||||||||||||||||||||
863 | |||||||||||||||||||||||||||
864 | |||||||||||||||||||||||||||
865 | |||||||||||||||||||||||||||
866 | |||||||||||||||||||||||||||
867 | |||||||||||||||||||||||||||
868 | |||||||||||||||||||||||||||
869 | |||||||||||||||||||||||||||
870 | |||||||||||||||||||||||||||
871 | |||||||||||||||||||||||||||
872 | |||||||||||||||||||||||||||
873 | |||||||||||||||||||||||||||
874 | |||||||||||||||||||||||||||
875 | |||||||||||||||||||||||||||
876 | |||||||||||||||||||||||||||
877 | |||||||||||||||||||||||||||
878 | |||||||||||||||||||||||||||
879 | |||||||||||||||||||||||||||
880 | |||||||||||||||||||||||||||
881 | |||||||||||||||||||||||||||
882 | |||||||||||||||||||||||||||
883 | |||||||||||||||||||||||||||
884 | |||||||||||||||||||||||||||
885 | |||||||||||||||||||||||||||
886 | |||||||||||||||||||||||||||
887 | |||||||||||||||||||||||||||
888 | |||||||||||||||||||||||||||
889 | |||||||||||||||||||||||||||
890 | |||||||||||||||||||||||||||
891 | |||||||||||||||||||||||||||
892 | |||||||||||||||||||||||||||
893 | |||||||||||||||||||||||||||
894 | |||||||||||||||||||||||||||
895 | |||||||||||||||||||||||||||
896 | |||||||||||||||||||||||||||
897 | |||||||||||||||||||||||||||
898 | |||||||||||||||||||||||||||
899 | |||||||||||||||||||||||||||
900 | |||||||||||||||||||||||||||
901 | |||||||||||||||||||||||||||
902 | |||||||||||||||||||||||||||
903 | |||||||||||||||||||||||||||
904 | |||||||||||||||||||||||||||
905 | |||||||||||||||||||||||||||
906 | |||||||||||||||||||||||||||
907 | |||||||||||||||||||||||||||
908 | |||||||||||||||||||||||||||
909 | |||||||||||||||||||||||||||
910 | |||||||||||||||||||||||||||
911 | |||||||||||||||||||||||||||
912 | |||||||||||||||||||||||||||
913 | |||||||||||||||||||||||||||
914 | |||||||||||||||||||||||||||
915 | |||||||||||||||||||||||||||
916 | |||||||||||||||||||||||||||
917 | |||||||||||||||||||||||||||
918 | |||||||||||||||||||||||||||
919 | |||||||||||||||||||||||||||
920 | |||||||||||||||||||||||||||
921 | |||||||||||||||||||||||||||
922 | |||||||||||||||||||||||||||
923 | |||||||||||||||||||||||||||
924 | |||||||||||||||||||||||||||
925 | |||||||||||||||||||||||||||
926 | |||||||||||||||||||||||||||
927 | |||||||||||||||||||||||||||
928 | |||||||||||||||||||||||||||
929 | |||||||||||||||||||||||||||
930 | |||||||||||||||||||||||||||
931 | |||||||||||||||||||||||||||
932 | |||||||||||||||||||||||||||
933 | |||||||||||||||||||||||||||
934 | |||||||||||||||||||||||||||
935 | |||||||||||||||||||||||||||
936 | |||||||||||||||||||||||||||
937 | |||||||||||||||||||||||||||
938 | |||||||||||||||||||||||||||
939 | |||||||||||||||||||||||||||
940 | |||||||||||||||||||||||||||
941 | |||||||||||||||||||||||||||
942 | |||||||||||||||||||||||||||
943 | |||||||||||||||||||||||||||
944 | |||||||||||||||||||||||||||
945 | |||||||||||||||||||||||||||
946 | |||||||||||||||||||||||||||
947 | |||||||||||||||||||||||||||
948 | |||||||||||||||||||||||||||
949 | |||||||||||||||||||||||||||
950 | |||||||||||||||||||||||||||
951 | |||||||||||||||||||||||||||
952 | |||||||||||||||||||||||||||
953 | |||||||||||||||||||||||||||
954 | |||||||||||||||||||||||||||
955 | |||||||||||||||||||||||||||
956 | |||||||||||||||||||||||||||
957 | |||||||||||||||||||||||||||
958 | |||||||||||||||||||||||||||
959 | |||||||||||||||||||||||||||
960 | |||||||||||||||||||||||||||
961 | |||||||||||||||||||||||||||
962 | |||||||||||||||||||||||||||
963 | |||||||||||||||||||||||||||
964 | |||||||||||||||||||||||||||
965 | |||||||||||||||||||||||||||
966 | |||||||||||||||||||||||||||
967 | |||||||||||||||||||||||||||
968 | |||||||||||||||||||||||||||
969 | |||||||||||||||||||||||||||
970 | |||||||||||||||||||||||||||
971 | |||||||||||||||||||||||||||
972 | |||||||||||||||||||||||||||
973 | |||||||||||||||||||||||||||
974 | |||||||||||||||||||||||||||
975 | |||||||||||||||||||||||||||
976 | |||||||||||||||||||||||||||
977 | |||||||||||||||||||||||||||
978 | |||||||||||||||||||||||||||
979 | |||||||||||||||||||||||||||
980 | |||||||||||||||||||||||||||
981 | |||||||||||||||||||||||||||
982 | |||||||||||||||||||||||||||
983 | |||||||||||||||||||||||||||
984 | |||||||||||||||||||||||||||
985 | |||||||||||||||||||||||||||
986 | |||||||||||||||||||||||||||
987 | |||||||||||||||||||||||||||
988 | |||||||||||||||||||||||||||
989 | |||||||||||||||||||||||||||
990 | |||||||||||||||||||||||||||
991 | |||||||||||||||||||||||||||
992 | |||||||||||||||||||||||||||
993 | |||||||||||||||||||||||||||
994 | |||||||||||||||||||||||||||
995 | |||||||||||||||||||||||||||
996 | |||||||||||||||||||||||||||
997 | |||||||||||||||||||||||||||
998 | |||||||||||||||||||||||||||
999 | |||||||||||||||||||||||||||
1000 |