ABCDEFGHIJKLMNOPQRSTUVWXYZAA
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Totals ==>4525170123329900130
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NotesOk?BPBSIBSQIFIGOIWUEUXXRV
Lasso
PAnswerExplanationIDFormulaTraceExpected
3
1NoThe red light is not on in the first state.917uRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
4
1NoIn the first step, Red is not lit. Since there are no temporal operators in the formula, we're only considering that stpe.qMyfRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
5
1NoRed isn't present in the first stategbrVRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
6
1YesThere is at least one state in which the red light is on.rahcRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
7
1YesI think the formula says that the red light has to be on in at least one state, which is satisfied by the {RGB} states.N3CwRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
8
1NoRed is not in the first state's lit colors.ydkmRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
9
1NoBy default it refers to the first state, and in that state red is not on.31qwRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
10
1NoThe 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.c9jyRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
11
1YesRed is on in one states.3r04Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
12
1NoThis 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}.ptykRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
13
1NoIn the first state, the red light is not lit.KXjzRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
14
1NoRed is not in Panel.lit in the first stateiVk5Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
15
1NoThe 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).mxadRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
16
1NoRed is not on in the first staternC9Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
17
1YesThe Red light was present in lit in at least one trace.8E1hRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
18
1NoRed not in first state {GB}sainRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
19
1NoBecause the first state does not have the Red light on.niehRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
20
11NoThere is a trace where Red is not lituOG8Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
21
1NoThe red light is not on in the initial state.fsqwRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
22
1NoBecause the formula is not temporal-ly quantified and the Red light is not on in the first state ({GB}).dznhRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
23
1NoRed is not on in the first state7opyRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
24
1YesEven 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.55oaRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
25
1NoRed is not on in the current (first) state.vz2gRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
26
1NoThe formula requires that the red light be turned on in the first state, and this is not the case in the given trace.cirfRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
27
1NoThe Red light is not on in the first state, and the first state is the one referenced by the formula.fojfRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
28
1NoThe first { GB } has no R in it7jm2Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
29
1NoRed is not on in first traceeolWRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
30
1NoThe 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.c1fnRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
31
1NoRed is not lit in the initial state.nr65Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
32
1NoIn the first state, red is not in the panel.lit.m0p6Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
33
1YesBecause it doesn't talk about red always litableRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
34
1NoRed is not on in the current state.1vwvRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
35
1NoState 1 doesn't have Red lit.yoxyRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
36
1NoBecause R is off in first stateDzoDRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
37
1NoRed is not on in the first state, which is the only one checked.xqZARed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
38
1NoBecause the first state has Red to be turned off.bxd4Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
39
1NoThe red light is not on in the first state, which is the default state for the predicates when there are no temporal modifiersbx1rRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
40
1NoThe formula checks if Red light is on in the first state, which it isn't.w4t3Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
41
1NoThe Red light is not lit in the first state.clU0Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
42
1NoRed light is not on in the first state.ndijRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
43
1YesIt doesn't say always so I just looked to find a single trace with `Red in Panel.lit` - e. g. {RGB}tyosRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
44
1NoThis formula only constrains the first state, where R is not present.eknyRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
45
1NoRed is not lit in the first state.hDZGRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
46
1NoBecause the Red light is not lit in the initial state.XU9xRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
47
1NoIn the first state, the red light is not lit.zf66Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
48
1NoRed light is not on in the initial state.kkzxRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
49
1NoRed in Panel.lit only applies to the first state in the trace, where R is not present.yypqRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
50
1NoRed is not lit in the first state.pfddRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
51
1NoThis formula only talks about the first trace, and we see that red is NOT lit in the first state {GB}.s0nvRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
52
1NoI'm not sure how statements that don't use any temporal operators should be interpreted.j9mqRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
53
1NoBecause the formula is about the first state, where Red light is not onj7t9Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
54
1NoI assume if no electrum keywords are present then the predicate is applied to the first instance.n4vdRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
55
1NoBecause Red light is not lit in the first stateqjpxRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
56
1NoBecause Red is not in Panel.lit in the first statekuaaRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
57
1YesRed is in the last four traces.XLuyRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
58
1NoRed is not on in the first statexeecRed in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
59
1NoRed in Panel.lit is an event that happens in the first state, but the first state only has green and blue.duq8Red in Panel.lit{GB} {RGB} {RGB} {RGB} {RGB}No
60
11YesThe red light is on three states from the initial state.917u
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
61
11YesWe'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
1NoFor the second step in the trace this doesn't hold as after after after is nonegbrV
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
63
11YesThere 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
1YesIn 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
1YesIn the 4th state, 3 states after state 0, the Red is lit.ydkm
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
66
1YesRed is on in the fourth state31qw
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
67
1YesThe red light is on in the fourth state (indicated by three "afters").c9jy
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
68
1YesAfter the first three states, red light is on.3r04
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
69
1YesGiven 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
1NoThe red light is not lit in the fourth state.KXjz
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
71
1YesThis 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
1YesRegardless 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
1YesThe red light is on three states after the initial state.rnC9
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
74
11YesRed was lit in one state that is 3 states after some state.8E1h
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
75
1YesRed in 4th statesain
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
76
1YesBecause the Red light is on on the third state.nieh
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
77
1Yesthe fourth trace has red presentuOG8
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
78
1YesThe red light is on in the fourth state.fsqw
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
79
1YesBecause 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
1YesThe 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
11YesAgain, since this formula never specifies that it should hold in all states, the fact that it holds in states 2-4 is sufficient55oa
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
82
1Yesafter 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
1YesThis 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
1YesThe Red light is on the third state.fojf
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
85
1YesThe fourth part of the trace is { R }, which contains a R.7jm2
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
86
1YesCurrent trace + 3 after = 4th Trace, Red is oneolW
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
87
1YesThe 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
1YesRed is lit in 3 states (so 4th state beginning on 1st state).nr65
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
89
1YesRed is in the 4th statem0p6
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
90
1YesThe 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
1YesRed is on in the 3rd state in the future (the 4th state).1vwv
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
92
1YesRed is lit in state 4.yoxy
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
93
1YesBecause R is lit in 4th state and there are 3 "after" statementsDzoD
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
94
1YesFirst 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
1YesBecause 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
1YesRed light is lit in the fourth state, which is 3 states after the initial statebx1r
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
97
1YesIt 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
1YesThe 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
1YesRed light is on in the 4th state.ndij
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
100
11Yes{R} plus 3 jumps gets you back to {R}tyos
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
101
1Yes3 afters is the 4th state, in which R is lit.ekny
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
102
1YesRed 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
1Yesafter {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
1YesThe red light is lit in the fourth state.zf66
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
105
1NoI 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
1YesThe 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
1YesRed it lit in the 4th state.pfdd
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
108
1YesThis 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
11YesThe state 2 states after the initial state has red lit.j9mq
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
110
1YesBecause Red light is on in the 4th statej7t9
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
111
1YesThe red light is lit in instance 4n4vd
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
112
1YesRed light is lit in the fourth stateqjpx
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
113
11YesIt 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
1YesRed is in the third state after the first.XLuy
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
115
1YesRed is on in the fourth statexeec
after { after { after { Red in Panel.lit } } }
{R} {} {} {R} {}Yes
116
1YesThe 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
1YesWhenever 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
1YesWe 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
1YesIn 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 ongbrV
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
120
1YesIt 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
11YesIn 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
11YesSince 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
1YesAll of the states that have R on have R on also in the next of next of next state31qw
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
124
1YesThe 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
11YesFor 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
1YesBecause 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
11YesFor 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
1Nothe 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.litiVk5
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
129
1YesThe 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
1YesFor 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
1YesIt 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
1Yesif a state has red in it, the state 3 changes after also has red because the final state repeats foreversain
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
133
1YesBecause 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
1Yeswhen red was present it was present three traces lateruOG8
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
135
1YesThe 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
1Yesbecause 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
1YesWhenever 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
1YesThe 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
1YesAssuming 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
1YesWhenever 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
1YesFor 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
1NoRed 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
1YesWhenever 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
1YesThe 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
1YesWhenever 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
1YesWhen there is red, it is on 3 states afterm0p6
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
147
1NoBecause although there is a red for the first RGB but not for the second RGBable
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
148
1YesRed 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
1YesRed 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
11YesState 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
1YesEach 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
1YesThe 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
1YesOnce the red light turns on, then it must be on again three states after thatbx1r
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
154
1YesSince 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
1YesWhenever 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
1YesIf 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
11YesI'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
1YesThis 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
1YesIn 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
1YesBecause 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
1YesIn 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
1YesThe 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
1NoThe 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
1NoThe 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
1YesNow, 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
1NoThis 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
1NoIn the 3rd state, Red light is on, but 3 states after, it is not onj7t9
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
168
1YesSince 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
1YesAny Red state is followed by a Red state three states after, including the final lasso stateqjpx
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
170
11YesRed is not always in Panel.lit; therefore, any set of conditions can be truekuaa
always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } }
{} {RGB} {RGB} {} {RGB}Yes
171
1YesRed 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
1YesRed 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
1YesIn 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
1NoThe 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
1YesThe 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
1YesRed is always on until Green is turned ongbrV
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
177
1NoI'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
1YesFor 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
1YesYes, 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
1YesRed is always on in the next state until green is on in the next state31qw
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
181
1YesThe 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
1YesIn 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
1Noafter {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
1Yesyes, 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
1Yes(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
1YesThe 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
1YesState 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
1NoI 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
1Yesthe two first states satisfy the condition that the next state has red in itsain
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
190
1NoI'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
1Yesuntil does not guarantee that statement A does not hold in traces afteruOG8
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
192
1YesThe 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
1YesThe 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
1YesThe 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
1NoI 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
1NoState 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
1YesThe 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
1YesAt 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
1NoRed 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
1NoRed 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
1YesThe 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
1NoAt 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
1NoThe 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 redm0p6
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
204
1Yesbecause 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
1YesRed 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
1NoRed 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
1Yesbecause of the after statements, Red and Green can both be on one timeDzoD
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
208
1YesThe 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
1YesBecause 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
1YesThere 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
1YesFor 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
1YesFor 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
1YesThe 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
1YesThis might be a tautology.tyos
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
215
1YesThe 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
1YesYes. 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
1YesEven 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
1YesThe 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
1YesIf 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
1YesThe "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
1YesIt 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
1YesThe 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
1YesWhen 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
1NoIn the 4th state, Green light is already on but Red light is not offj7t9
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
225
1YesThe 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
1YesRed stay lit until the green light turns onqjpx
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
227
1YesBecause 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 tracekuaa
after { Red in Panel.lit } until after { Green in Panel.lit }
{RB} {RB} {RB} {RGB} {B}Yes
228
1YesRed 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
1YesBecause 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
1YesThe 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
1YesThere 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
1YesRed 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
1YesGreen 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
1YesThe 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
1YesThe 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
1YesBoth red and green are eventually litydkm
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
237
1Yesat 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
1YesBoth 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
11YesRed 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
1YesEventually {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
1Yesboth red and green are eventually litKXjz
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
242
1YesThere are traces where Red and Green are in Panel.litiVk5
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
243
1YesYes, 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
1YesThe green and red lights both turn onrnC9
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
245
1YesThere 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
1Yesboth the red and green lights are lit at some pointsain
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
247
1YesEventually 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
1Yeseventually both were lituOG8
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
249
1YesThe 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
1YesBoth 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
1YesRed is on in some states. So is Green.7opy
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
252
1YesEventually 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
1YesBoth Green and Red are lit at some point.vz2g
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
254
1YesThis 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
1YesThe 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
1YesR and G are in the trace.7jm2
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
257
1YesGreen and Red both turn on eventually.eolW
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
258
1YesThe 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
1YesEventually, 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
1YesEventually both red and green appearedm0p6
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
261
1Yesbecause 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
1YesFrom 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
1YesGreen and Red are both lit at some point.yoxy
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
264
1YesBoth lights turn on eventuallyDzoD
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
265
1YesRed and Green are both on at some point.xqZA
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
266
1YesGreen and Red should have happened eventually during the states. They both turned on eventually! :) 2nd and fifthbxd4
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
267
1YesBoth 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
1YesFrom 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
1YesThe 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
1YesAt 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
1Yes{G} and {R}tyos
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
272
1YesThe 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
1YesRed 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
1YesThey 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
1YesThe 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
1YesBoth 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
1YesThe 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
1YesEventually 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
1YesThe 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
1YesBoth "eventually" predicates hold at some pointj9mq
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
281
1YesRed light is on at some state, so is the Green lightj7t9
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
282
1YesBoth red and green are eventually litn4vd
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
283
1YesRed and Green eventually lit in some statesqjpx
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
284
1YesBoth Green and Red eventually appear in the traceskuaa
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
285
1YesBoth 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
1YesIn the first state, both green and red lights are eventually on in subsequent statesxeec
eventually { Red in Panel.lit} and eventually { Green in Panel.lit }
{} {G} {} {} {R}Yes
287
1YesThe 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
1Yesthere is some state future to two states from now where the red light is on917u
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
289
1YesRed 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
1YesRed is always on so basically any surrounding condition is irrelevant heregbrV
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
291
1YesFor 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
1YesIn 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
1YesRed 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
1YesLooking 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
1YesThe 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
1Yesred 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
1YesThis 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
1Yesthe 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
1YesRed 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
1YesYes, 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
1YesRed is on in the lasso thus the eventually is always satisfiedrnC9
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
302
1YesThere is a trace for which two traces later, eventually Red is lit8E1h
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
303
1Yesred is always litsain
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
304
1YesRed light turns on eventually after 2 turns.nieh
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
305
1Yesred was eventually lituOG8
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
306
1YesThe 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
1YesBecause, 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
1YesRed is always on.7opy
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
309
1YesSince Red is in every trace, it will always satisfy this55oa
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
310
1YesRed is lit after the second state.vz2g
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
311
1YesSince 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
1YesStarting 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
1YesR 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
1YesRed is on in 4th trace.eolW
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
315
1YesThe 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
1YesIn 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
1YesRed is in the 4th and 5th state, which makes this truem0p6
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
318
1YesBecause 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
1YesAfter 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
1YesStarting 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
1YesRed is on in state 5DzoD
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
322
1YesRed is on at some point after/including state 3.xqZA
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
323
1YesFor 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
1YesTwo states after the first, there is some state after that at some point where the Red panel is litbx1r
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
325
1YesIn 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
1YesIn 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
1YesTwo 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
1YesTheres 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
1YesRed is lit after the 2nd state.ekny
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
330
1YesRed 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
1YesWell 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
1Yesthe red light is always litzf66
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
333
1YesThe 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
1YesAll of the states have the red light lit up.yypq
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
335
1YesRed 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
1YesSince 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
1YesRed 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
1Yesstarting from the third state, Red light is on at some statej7t9
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
339
1YesYes, since red is lit in all isntancesn4vd
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
340
1YesRed is always lit, so it will eventually lit from the third stateqjpx
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
341
1YesRed eventually appears in the lit panelkuaa
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
342
1NoRed is turned on in the first two states.XLuy
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
343
1YesRed is on in states after the third statexeec
after { after { eventually { Red in Panel.lit } } }
{RGB} {RGB} {RGB} {RGB} {RGB}Yes
344
1YesThe 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
1YesThe 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
1YesSince 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
1YesBlue is never turned on so red always stays ongbrV
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
348
1NoThe 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
1NoI 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
1NoFor 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
1NoBlue never turns on.31qw
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
352
1YesRed 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
1YesBlue 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
1NoThis 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
1Yesthe 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
1NoUntil specifies that the RHS must hold at some point, and Blue is never in Panel.lit in this traceiVk5
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
357
1YesThis 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
1NoAs 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
1YesI'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
1Yesred is always litsain
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
361
1YesThe 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
1Yesjust cause we dont see the blue doesnt mean it wont be there in the futureuOG8
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
363
1NoThe 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
1NoBecause the Blue light never turns on.dznh
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
365
1NoBlue 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
1YesThe 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
1YesBlue 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
1NoThe 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
1NoUntil 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
1YesRed 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
1YesBlue is never oneolW
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
372
1YesThe '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
1Nofmla-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
1YesThere is no blue in panel.lit, so it's fine that red is always in panel.litm0p6
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
375
1NoBecause blue is not there in any stateable
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
376
1NoBlue 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
1YesThe 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
1YesBlue never turns on, so red never needs to turn offDzoD
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
379
1NoBlue is never on.xqZA
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
380
1NoThere 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
1NoThe blue light being on is required for the until operator to be satisfiedbx1r
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
382
1NoFormula 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
1NoA 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
1YesBecause 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
1YesRed in Panel.lit was enough to satisfy the formulatyos
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
386
1YesThe first condition is always true.ekny
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
387
1NoBlue 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
1YesBlue 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
1NoThere 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
1YesThe 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
1NoIn 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
1NoBlue is not lit in any statepfdd
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
393
1NoAs 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
1NoBlue 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
1YesBlue light is never on and Red light is always onj7t9
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
396
1YesYes, 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
1NoBlue never litsqjpx
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
398
1YesBlue never appears in the panel; therefore, red must always appearkuaa
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
399
1YesBlue 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
1YesBlue is never lit, so red should be always litxeec
Red in Panel.lit until Blue in Panel.lit
{R} {R} {R} {R} {R}No
401
1NoUntil 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
1NoThere is not state from where on the Red light is always on.917u
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
403
1NoThe 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
1NoThere is no state after which red is always ongbrV
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
405
1NoThere 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
1NoThe 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
1NoThere 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
1NoAll 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
1NoNo state occurs in which the Red light is permanently turned on.c9jy
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
410
1NoIn the fifth state, red light should turn on.3r04
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
411
1NoThis 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
1Noit 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
1NoThere is no state after which all of the following states have Red in Panel.litiVk5
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
414
1NoAfter 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
1NoAs red is not on in the lasso, it is never true that the red light will always be on after a given staternC9
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
416
1NoThere is no state for which every state after that state has Red lit.8E1h
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
417
1Nothere is no state after which all states have the red lightsain
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
418
1NoThe 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
1Noif 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 lituOG8
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
420
1NoThe red light is never always on after any of the states.fsqw
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
421
1Nobecause there is no state after which the Red light is always on.dznh
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
422
1NoThe 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
1NoThis 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
11NoOnce Red is lit, it should stay lit. (eventually always)vz2g
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
425
1NoThis 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
1NoSince 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
1NoThere 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
1NoLasso around all lights off - red is not on.eolW
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
429
1NoThe 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
1NoThere 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
11No3 and 5th state need to have red in it to make it truem0p6
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
432
1NoRed is not always on.able
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
433
1NoThere 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
1NoRed becomes lit but doesn't permanently stay on once it is.yoxy
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
435
1NoRed eventually turns off, which means the "always" statement is not satisfied.DzoD
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
436
1NoThere is no point at which Red turns on and stays on.xqZA
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
437
1NoThere 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
1Noeventually always forces at least the last state to satisfy the conditionbx1r
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
439
1NoAt no point in any of the states is the Red light always on.w4t3
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
440
1NoThere 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
1NoBecause 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
1NoThe last trace does not have Red in Panel.lit.tyos
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
443
1NoThere 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
1NoThe 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
1NoThe lasso state doesn't contain RXU9x
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
446
1NoThere is no state after which the red light is always lit.zf66
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
447
1NoThe 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
11NoThere 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
1NoWhenever 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
1NoThis 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
1NoThere is no time step such that red is always lit from that point forwardj9mq
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
452
1NoThere is no state after which Red light is always onj7t9
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
453
1NoNo, 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
1NoThe 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
11NoAccording to these set of conditions, once red appears, it must always appear.kuaa
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
456
1NoRed is never on continuously.XLuy
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
457
1NoThere is no state where red is always lit in all subsequent statesxeec
eventually { always { Red in Panel.lit } }
{} {RGB} {} {RGB} {}No
458
1NoThe 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
1YesThe 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
1YesSince 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
1YesThe LHS is always false so the expression is always truegbrV
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
462
1YesRed light is never on, so the part inside { } is always satisfied.rahc
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
463
1YesImplies 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
1YesBecause red is never lit, False implies False results in True.ydkm
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
465
1YesConditional is true when antecedent is not satisfied - red never on.31qw
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
466
1YesThe 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
1YesRed 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
1YesIn 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
1Yesno 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
1YesThe 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
1YesThis 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
1Yesas Red is never on, this is trivially satisfied.rnC9
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
473
1YesSince 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
1Yesfirst condition never holds so its always truesain
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
475
1YesRed 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
1Yesred was never there so green was never thereuOG8
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
477
1YesThe 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
1YesBecause 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
1YesRed is never on. So the implication is trivially true.7opy
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
480
1YesBecause 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
1YesIt'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
1YesSince 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
1YesThe 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
1YesRed is never lit, so this is vacuously true.7jm2
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
485
1YesRed never lit so implies never invoked,eolW
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
486
1YesThe 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
1YesRed is never lit, so this is vacuously satisfied.nr65
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
488
1YesRed is never in panel.lit, so therefore there is nothing to worry aboutm0p6
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
489
1YesBecause red is never lit itselfable
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
490
1YesRed is never on, so the implication is always satisfied.1vwv
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
491
1YesRed doesn't necessarily have to be lit.yoxy
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
492
1YesBecause 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
1YesThe implication is vacuously true since Red is never on.xqZA
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
494
1YesThere 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
1YesThe red light is never on, so the green light never has to be on either. The implication is always vacuously truebx1r
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
496
1YesThe 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
1YesIn 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
1YesBecause the red light is never on, it's vacuously true.ndij
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
499
1YesLHS of implies is false. Therefore this is true.tyos
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
500
1YesThe 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
1YesRed is never lit, so the implication is vacuously true.hDZG
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
502
1YesGreen 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
1YesThe 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
1YesThe 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
1YesThe 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
1YesRed is never lit, so the implication is always true.pfdd
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
507
1YesRed 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
1YesThe 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
1YesThe implication is satisfied vacuouslyj7t9
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
510
1YesYes, the implication is satisfied vacuously.n4vd
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
511
1YesRed is never lit, so the implies is always trueqjpx
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
512
1YesThe condition above only holds true if red appears in the panel. If there is no red, then any set of traces can appear without redkuaa
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
513
1YesRed is never lit so Green also never has to be lit.XLuy
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
514
1YesRed is not lit, so the implies statement is always truexeec
always { Red in Panel.lit implies Green in Panel.lit }
{} {} {} {} {}Yes
515
1YesThe 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
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