ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>774400020050032001000096
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRV
Scope
Answer
Explanation
IDFormulaExpected
3
1if there is a red state, some time later there is a not red state
R_3MS5InthwHWtaZT
F(r => F(!r))
if red, must have !red in future
4
R_3G88D8gpZ7v63nz
F(r => F(!r))
if red, must have !red in future
5
R_3Dw4BW8x0ynmmQm
F(r => F(!r))
if red, must have !red in future
6
1Eventually, if a state makes Red true, then some future state must make Red false
R_DSlVV5XN50E2Mzn
F(r => F(!r))
if red, must have !red in future
7
R_WByy6AuVYHXV65H
F(r => F(!r))
if red, must have !red in future
8
1
There exists an instant where, if the Red light is on, then there exists a future instant with the Red light off. It is satisfied by an infinite trace with the Red light off at the first instant (under the reflexive interpretation of "F", which I am assuming).
R_vvG6iiXnXM4vZJv
F(r => F(!r))
if red, must have !red in future
9
R_9Td5mdNtgffkvJL
F(r => F(!r))
if red, must have !red in future
10
1The red light is not always on.
R_9ukqO8kCEIR4pLb
F(r => F(!r))
if red, must have !red in future
11
1not sure
R_2saTh9ICAgM0BHV
F(r => F(!r))
if red, must have !red in future
12
1
There is an instant in which the light is red. Following, there will be an instant in which the light is not red.
R_3J3NZWyA2hcl5by
F(r => F(!r))
if red, must have !red in future
13
1There is a non-red state following a red state, or all states are non-red after some point.
R_1mfLnqhYsR97vhv
F(r => F(!r))
if red, must have !red in future
14
ok1
At some state, red has to hold. After that state, at some point, red has to not hold. when this has happened, the rest of trace is irrelevant Otherwise, red never holds
R_VHZdlnCi0wPqPGF
F(r => F(!r))
if red, must have !red in future
15
R_0ize6N4UFi1irrb
F(r => F(!r))
if red, must have !red in future
16
R_3kpTXi28NBbr8mO
F(r => F(!r))
if red, must have !red in future
17
R_3Rt21EZCScgM3oC
F(r => F(!r))
if red, must have !red in future
18
R_enENzp0guuxVGwh
F(r => F(!r))
if red, must have !red in future
19
1Red is false at a state
R_1g8tSbzTTgxijrd
F(r => F(!r))
if red, must have !red in future
20
R_3M5V4sL4b6COyD2
F(r => F(!r))
if red, must have !red in future
21
R_23fQaq1kM65aSP7
F(r => F(!r))
if red, must have !red in future
22
1There is a state in which red light is on and from there on there is a state in which red light is off
R_1QFcce33iJrOdLO
F(r => F(!r))
if red, must have !red in future
23
1Eventually there is a state where red is false after some states where it was true
R_2EnQzWu9HQsIjn8
F(r => F(!r))
if red, must have !red in future
24
1
At a certain point, either Red is always off or will eventually be off after it is on. It should be similar to asking someone to turn off the light after (and if) he turns it on, from a certain moment.
R_R9tc8AnNRKwPdkd
F(r => F(!r))
if red, must have !red in future
25
R_2QPzqLc2iRGU0Jd
F(r => F(!r))
if red, must have !red in future
26
R_3EO18al85qREqgF
F(r => F(!r))
if red, must have !red in future
27
R_3MS5InthwHWtaZT
(r U b) & G(r)
always r, eventually b
28
R_3G88D8gpZ7v63nz
(r U b) & G(r)
always r, eventually b
29
1Red must always stay on. At some point in the future, blue will come on
R_3Dw4BW8x0ynmmQm
(r U b) & G(r)
always r, eventually b
30
1Holds if Red is true in every state and Blue is true in same state
R_DSlVV5XN50E2Mzn
(r U b) & G(r)
always r, eventually b
31
1The red light is always on, and at some point blue is on.
R_WByy6AuVYHXV65H
(r U b) & G(r)
always r, eventually b
32
R_vvG6iiXnXM4vZJv
(r U b) & G(r)
always r, eventually b
33
R_9Td5mdNtgffkvJL
(r U b) & G(r)
always r, eventually b
34
R_9ukqO8kCEIR4pLb
(r U b) & G(r)
always r, eventually b
35
11
Red is true until blue is true (which might never happen) also, forever red must be true. Is the same as just G(Red)
R_2saTh9ICAgM0BHV
(r U b) & G(r)
always r, eventually b
36
1The light is red on every state, and there exists a state in which the light is blue
R_3J3NZWyA2hcl5by
(r U b) & G(r)
always r, eventually b
37
R_1mfLnqhYsR97vhv
(r U b) & G(r)
always r, eventually b
38
R_VHZdlnCi0wPqPGF
(r U b) & G(r)
always r, eventually b
39
1all red at least one blue
R_0ize6N4UFi1irrb
(r U b) & G(r)
always r, eventually b
40
R_3kpTXi28NBbr8mO
(r U b) & G(r)
always r, eventually b
41
R_3Rt21EZCScgM3oC
(r U b) & G(r)
always r, eventually b
42
1Red always holds and blue will be true eventually.
R_enENzp0guuxVGwh
(r U b) & G(r)
always r, eventually b
43
1Red occurs at every state, and blue occurs at a state.
R_1g8tSbzTTgxijrd
(r U b) & G(r)
always r, eventually b
44
1Red is Always true and Blue is true at some point
R_3M5V4sL4b6COyD2
(r U b) & G(r)
always r, eventually b
45
R_23fQaq1kM65aSP7
(r U b) & G(r)
always r, eventually b
46
R_1QFcce33iJrOdLO
(r U b) & G(r)
always r, eventually b
47
R_2EnQzWu9HQsIjn8
(r U b) & G(r)
always r, eventually b
48
R_R9tc8AnNRKwPdkd
(r U b) & G(r)
always r, eventually b
49
R_2QPzqLc2iRGU0Jd
(r U b) & G(r)
always r, eventually b
50
1
Since we have RED for each position, the formula (RED U BLUE) cannot be true because BLUE cannot never be satisfied. The formula is always false.
R_3EO18al85qREqgF
(r U b) & G(r)
always r, eventually b
51
R_3MS5InthwHWtaZT
r & !X(b)r now, !b next
52
1Red is on now and in the next step blue is off.
R_3G88D8gpZ7v63nz
r & !X(b)r now, !b next
53
1Red is on at first instant, blue must be off at the next time instant.
R_3Dw4BW8x0ynmmQm
r & !X(b)r now, !b next
54
R_DSlVV5XN50E2Mzn
r & !X(b)r now, !b next
55
R_WByy6AuVYHXV65H
r & !X(b)r now, !b next
56
R_vvG6iiXnXM4vZJv
r & !X(b)r now, !b next
57
1Red initially and second state in not blue
R_9Td5mdNtgffkvJL
r & !X(b)r now, !b next
58
1The red light is on in the first state and the blue light is not on in the second state.
R_9ukqO8kCEIR4pLb
r & !X(b)r now, !b next
59
R_2saTh9ICAgM0BHV
r & !X(b)r now, !b next
60
R_3J3NZWyA2hcl5by
r & !X(b)r now, !b next
61
1The first state is red and the second state is not blue.
R_1mfLnqhYsR97vhv
r & !X(b)r now, !b next
62
1
red must always hold in the first state, second state must not be blue, states after that are not important to the formula
R_VHZdlnCi0wPqPGF
r & !X(b)r now, !b next
63
R_0ize6N4UFi1irrb
r & !X(b)r now, !b next
64
1The first state has Red, the second does not have blue
R_3kpTXi28NBbr8mO
r & !X(b)r now, !b next
65
1The light starts as red and is not blue in the next state
R_3Rt21EZCScgM3oC
r & !X(b)r now, !b next
66
R_enENzp0guuxVGwh
r & !X(b)r now, !b next
67
R_1g8tSbzTTgxijrd
r & !X(b)r now, !b next
68
R_3M5V4sL4b6COyD2
r & !X(b)r now, !b next
69
1The red light is in a state and then there must be a state in which the light is not blue
R_23fQaq1kM65aSP7
r & !X(b)r now, !b next
70
1Red light is on in the first state and does not exist a next state which has Blue light on
R_1QFcce33iJrOdLO
r & !X(b)r now, !b next
71
1
Red is true in current state and it does not exist a next state where blue hold or if it exists blue may not hold
R_2EnQzWu9HQsIjn8
r & !X(b)r now, !b next
72
R_R9tc8AnNRKwPdkd
r & !X(b)r now, !b next
73
1the first state must be red and the second must be not blue
R_2QPzqLc2iRGU0Jd
r & !X(b)r now, !b next
74
R_3EO18al85qREqgF
r & !X(b)r now, !b next
75
1every red state is followed by non- red followed by red
R_3MS5InthwHWtaZT
G(r => X(!r & X(r)))
whenever r, off/on evermore
76
1whenever red holds, it also holds two steps later.
R_3G88D8gpZ7v63nz
G(r => X(!r & X(r)))
whenever r, off/on evermore
77
R_3Dw4BW8x0ynmmQm
G(r => X(!r & X(r)))
whenever r, off/on evermore
78
R_DSlVV5XN50E2Mzn
G(r => X(!r & X(r)))
whenever r, off/on evermore
79
1
Either the red light is never on, or it becomes on in state k, and from that point onwards it is on in state k+m iff m is even.
R_WByy6AuVYHXV65H
G(r => X(!r & X(r)))
whenever r, off/on evermore
80
1
In every instant, if the Red light is on, then the next instant has the Red light off and its subsequent instant has the Red light on. It is satisfied also by an infinite trace where the Red light is always off.
R_vvG6iiXnXM4vZJv
G(r => X(!r & X(r)))
whenever r, off/on evermore
81
1Red and not red states alternate ad infinitum
R_9Td5mdNtgffkvJL
G(r => X(!r & X(r)))
whenever r, off/on evermore
82
R_9ukqO8kCEIR4pLb
G(r => X(!r & X(r)))
whenever r, off/on evermore
83
R_2saTh9ICAgM0BHV
G(r => X(!r & X(r)))
whenever r, off/on evermore
84
R_3J3NZWyA2hcl5by
G(r => X(!r & X(r)))
whenever r, off/on evermore
85
R_1mfLnqhYsR97vhv
G(r => X(!r & X(r)))
whenever r, off/on evermore
86
R_VHZdlnCi0wPqPGF
G(r => X(!r & X(r)))
whenever r, off/on evermore
87
1alternating red
R_0ize6N4UFi1irrb
G(r => X(!r & X(r)))
whenever r, off/on evermore
88
1
All states that have red are followed by a state that does not have red and is itself followed by a state that does have red. I.e. after the first red in the trace, the trace alternates between !Red and Red
R_3kpTXi28NBbr8mO
G(r => X(!r & X(r)))
whenever r, off/on evermore
89
could be IF1After the first time the red light is on, it will be alternating between on and off
R_3Rt21EZCScgM3oC
G(r => X(!r & X(r)))
whenever r, off/on evermore
90
1If red holds, red should be false at the next state and then turns back to true again
R_enENzp0guuxVGwh
G(r => X(!r & X(r)))
whenever r, off/on evermore
91
R_1g8tSbzTTgxijrd
G(r => X(!r & X(r)))
whenever r, off/on evermore
92
1Alternate forever Red and !Red from the first occurrence of Red
R_3M5V4sL4b6COyD2
G(r => X(!r & X(r)))
whenever r, off/on evermore
93
1This can be described as a sequence of red, followed by a not red and a next red state
R_23fQaq1kM65aSP7
G(r => X(!r & X(r)))
whenever r, off/on evermore
94
R_1QFcce33iJrOdLO
G(r => X(!r & X(r)))
whenever r, off/on evermore
95
R_2EnQzWu9HQsIjn8
G(r => X(!r & X(r)))
whenever r, off/on evermore
96
1The Red light is alternated. When it is on, at the next state is off, and vice versa.
R_R9tc8AnNRKwPdkd
G(r => X(!r & X(r)))
whenever r, off/on evermore
97
1A red state is always followed by a not-red and a red states, in sequence.
R_2QPzqLc2iRGU0Jd
G(r => X(!r & X(r)))
whenever r, off/on evermore
98
1
The trace which satisfies the formula is of this type: RED->!RED->RED->!RED-> .... ->RED->!RED-> .... or (if it starts with !RED) !RED->RED->!RED->RED-> .... ->!RED->RED-> ....
R_3EO18al85qREqgF
G(r => X(!r & X(r)))
whenever r, off/on evermore
99
1same
R_3MS5InthwHWtaZT
F(r => F(!r))same
100
R_3G88D8gpZ7v63nz
F(r => F(!r))same
101
R_3Dw4BW8x0ynmmQm
F(r => F(!r))same
102
1Same as LTL
R_DSlVV5XN50E2Mzn
F(r => F(!r))same
103
R_WByy6AuVYHXV65H
F(r => F(!r))same
104
1Same.
R_vvG6iiXnXM4vZJv
F(r => F(!r))same
105
R_9Td5mdNtgffkvJL
F(r => F(!r))same
106
1same
R_9ukqO8kCEIR4pLb
F(r => F(!r))same
107
1not sure
R_2saTh9ICAgM0BHV
F(r => F(!r))same
108
1same
R_3J3NZWyA2hcl5by
F(r => F(!r))same
109
1Same as above
R_1mfLnqhYsR97vhv
F(r => F(!r))same
110
1same
R_VHZdlnCi0wPqPGF
F(r => F(!r))same
111
R_0ize6N4UFi1irrb
F(r => F(!r))same
112
R_3kpTXi28NBbr8mO
F(r => F(!r))same
113
R_3Rt21EZCScgM3oC
F(r => F(!r))same
114
R_enENzp0guuxVGwh
F(r => F(!r))same
115
1same
R_1g8tSbzTTgxijrd
F(r => F(!r))same
116
R_3M5V4sL4b6COyD2
F(r => F(!r))same
117
R_23fQaq1kM65aSP7
F(r => F(!r))same
118
1same
R_1QFcce33iJrOdLO
F(r => F(!r))same
119
1same
R_2EnQzWu9HQsIjn8
F(r => F(!r))same
120
1It should be the same interpretation.
R_R9tc8AnNRKwPdkd
F(r => F(!r))same
121
R_2QPzqLc2iRGU0Jd
F(r => F(!r))same
122
R_3EO18al85qREqgF
F(r => F(!r))same
123
R_3MS5InthwHWtaZT
(r U b) & G(r)
r for all states, b for some
124
R_3G88D8gpZ7v63nz
(r U b) & G(r)
r for all states, b for some
125
1Red must always stay on and before the trace ends, blue must come on
R_3Dw4BW8x0ynmmQm
(r U b) & G(r)
r for all states, b for some
126
1Same as LTL
R_DSlVV5XN50E2Mzn
(r U b) & G(r)
r for all states, b for some
127
1same
R_WByy6AuVYHXV65H
(r U b) & G(r)
r for all states, b for some
128
R_vvG6iiXnXM4vZJv
(r U b) & G(r)
r for all states, b for some
129
R_9Td5mdNtgffkvJL
(r U b) & G(r)
r for all states, b for some
130
R_9ukqO8kCEIR4pLb
(r U b) & G(r)
r for all states, b for some
131
1
Red is true until blue is true (which must eventually happen before the trace finishes), also in every point until the end red is true. Is different that G(Read) since blue must eventually hold.
R_2saTh9ICAgM0BHV
(r U b) & G(r)
r for all states, b for some
132
1
The lights is red on every state until the end of the trace, and there exists a state, before the end of the trace, in which the light is blue
R_3J3NZWyA2hcl5by
(r U b) & G(r)
r for all states, b for some
133
R_1mfLnqhYsR97vhv
(r U b) & G(r)
r for all states, b for some
134
R_VHZdlnCi0wPqPGF
(r U b) & G(r)
r for all states, b for some
135
1same
R_0ize6N4UFi1irrb
(r U b) & G(r)
r for all states, b for some
136
R_3kpTXi28NBbr8mO
(r U b) & G(r)
r for all states, b for some
137
R_3Rt21EZCScgM3oC
(r U b) & G(r)
r for all states, b for some
138
1same
R_enENzp0guuxVGwh
(r U b) & G(r)
r for all states, b for some
139
1same
R_1g8tSbzTTgxijrd
(r U b) & G(r)
r for all states, b for some
140
1same
R_3M5V4sL4b6COyD2
(r U b) & G(r)
r for all states, b for some
141
R_23fQaq1kM65aSP7
(r U b) & G(r)
r for all states, b for some
142
R_1QFcce33iJrOdLO
(r U b) & G(r)
r for all states, b for some
143
R_2EnQzWu9HQsIjn8
(r U b) & G(r)
r for all states, b for some
144
R_R9tc8AnNRKwPdkd
(r U b) & G(r)
r for all states, b for some
145
R_2QPzqLc2iRGU0Jd
(r U b) & G(r)
r for all states, b for some
146
1Same reasoning of LTL.
R_3EO18al85qREqgF
(r U b) & G(r)
r for all states, b for some
147
R_3MS5InthwHWtaZT
r & !X(b)
r now, if next !b
148
1red is on now and either the trace ends here, or it has a next step and in this case blue is off.
R_3G88D8gpZ7v63nz
r & !X(b)
r now, if next !b
149
1Red is on at first time instant and the trace can end. If time step 2 exists, then blue must be off
R_3Dw4BW8x0ynmmQm
r & !X(b)
r now, if next !b
150
R_DSlVV5XN50E2Mzn
r & !X(b)
r now, if next !b
151
R_WByy6AuVYHXV65H
r & !X(b)
r now, if next !b
152
R_vvG6iiXnXM4vZJv
r & !X(b)
r now, if next !b
153
1Red initially and second star is not blue or there is no second state
R_9Td5mdNtgffkvJL
r & !X(b)
r now, if next !b
154
1
The red light is on in the first state, and it's not the case that there is a second state in which the blue light is on (either there is no second state, or the blue light is off then).
R_9ukqO8kCEIR4pLb
r & !X(b)
r now, if next !b
155
R_2saTh9ICAgM0BHV
r & !X(b)
r now, if next !b
156
R_3J3NZWyA2hcl5by
r & !X(b)
r now, if next !b
157
1The first state is red and the second state (if it exists) is blue
R_1mfLnqhYsR97vhv
r & !X(b)
r now, if next !b
158
1same but trace of length 1 with red on in the first state would also satisfy the formula
the formula on describes the first and second state, for LTLf, since there is a negative next, if no next state exist that part of the formula is satisfied. Hence, trace of length 1 satisfying red is also a valid trace.
R_VHZdlnCi0wPqPGF
r & !X(b)
r now, if next !b
159
R_0ize6N4UFi1irrb
r & !X(b)
r now, if next !b
160
1The first state has red, and if there is a second state it does not have blue.
R_3kpTXi28NBbr8mO
r & !X(b)
r now, if next !b
161
1The light is red and if there is a next state the light is not blue in that state
R_3Rt21EZCScgM3oC
r & !X(b)
r now, if next !b
162
R_enENzp0guuxVGwh
r & !X(b)
r now, if next !b
163
R_1g8tSbzTTgxijrd
r & !X(b)
r now, if next !b
164
R_3M5V4sL4b6COyD2
r & !X(b)
r now, if next !b
165
wow1In LTLf we cannot use a strong next, since there is no guarantees that there will be a next state
R_23fQaq1kM65aSP7
r & !X(b)
r now, if next !b
166
1same
R_1QFcce33iJrOdLO
r & !X(b)
r now, if next !b
167
1same
R_2EnQzWu9HQsIjn8
r & !X(b)
r now, if next !b
168
R_R9tc8AnNRKwPdkd
r & !X(b)
r now, if next !b
169
see above, "second must not be blue"
1same [[ the first state must be red and the second must be not blue ]]
R_2QPzqLc2iRGU0Jd
r & !X(b)
r now, if next !b
170
R_3EO18al85qREqgF
r & !X(b)
r now, if next !b
171
1same but only satisfiable if no red states
cannot be satisfied in ltlf if there is one red state there should be infinitely many (if strong next)
R_3MS5InthwHWtaZT
G(r => X(!r & X(r)))
never r
172
1red never holds
R_3G88D8gpZ7v63nz
G(r => X(!r & X(r)))
never r
173
R_3Dw4BW8x0ynmmQm
G(r => X(!r & X(r)))
never r
174
R_DSlVV5XN50E2Mzn
G(r => X(!r & X(r)))
never r
175
1The property can only be satisfied by traces where red is never on.
R_WByy6AuVYHXV65H
G(r => X(!r & X(r)))
never r
176
11
In every instant, if the Red light is on, then there exists the next instant with Red light off and it has a subsequent instant with the Red light on. A finite trace with just one instant that has the Red light off is enough to satisfy the formula.
R_vvG6iiXnXM4vZJv
G(r => X(!r & X(r)))
never r
177
1Not satisfiable, must always be a next and next next state
R_9Td5mdNtgffkvJL
G(r => X(!r & X(r)))
never r
178
R_9ukqO8kCEIR4pLb
G(r => X(!r & X(r)))
never r
179
R_2saTh9ICAgM0BHV
G(r => X(!r & X(r)))
never r
180
R_3J3NZWyA2hcl5by
G(r => X(!r & X(r)))
never r
181
R_1mfLnqhYsR97vhv
G(r => X(!r & X(r)))
never r
182
R_VHZdlnCi0wPqPGF
G(r => X(!r & X(r)))
never r
183
1same
R_0ize6N4UFi1irrb
G(r => X(!r & X(r)))
never r
184
1same
R_3kpTXi28NBbr8mO
G(r => X(!r & X(r)))
never r
185
1The light can never be red
R_3Rt21EZCScgM3oC
G(r => X(!r & X(r)))
never r
186
1Red never holds
R_enENzp0guuxVGwh
G(r => X(!r & X(r)))
never r
187
R_1g8tSbzTTgxijrd
G(r => X(!r & X(r)))
never r
188
1Red cannot be true
R_3M5V4sL4b6COyD2
G(r => X(!r & X(r)))
never r
189
WOW1In LTLf we cannot use a strong next, since there is no guarantee of a next state
R_23fQaq1kM65aSP7
G(r => X(!r & X(r)))
never r
190
R_1QFcce33iJrOdLO
G(r => X(!r & X(r)))
never r
191
R_2EnQzWu9HQsIjn8
G(r => X(!r & X(r)))
never r
192
1Impossible to guarantee the alternation because there's no next in the last step.
R_R9tc8AnNRKwPdkd
G(r => X(!r & X(r)))
never r
193
1
same meaning as LTL, but finite traces that satisfies the formula are only those ones that doesn't contain any red state (due to the implication)
R_2QPzqLc2iRGU0Jd
G(r => X(!r & X(r)))
never r
194
1same
R_3EO18al85qREqgF
G(r => X(!r & X(r)))
never r