ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>121102000300600020010000144
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRV
Scope
Answer
Explanation
IDFormulaExpected
3
1It is equivalent to F(!Red): eventually Red becomes false.
R_3zr1TPsbL8Fh01g
F(r => F(!r))
if red, must have !red in future
4
1The red light is off in some state.
R_7t8OynItMvsY3qS
F(r => F(!r))
if red, must have !red in future
5
1Either there is no state with the Red light on, or Red will be off after being on.
R_5pl5ovreUYilD5z
F(r => F(!r))
if red, must have !red in future
6
1There is a state which is not red
R_6ffI4bTyX5zoEHh
F(r => F(!r))
if red, must have !red in future
7
1only satisfied if !Red is true in some state (equivalent to F !Red)
R_4mVF8KOBt6MErhn
F(r => F(!r))
if red, must have !red in future
8
1Eventually, a red light must be followed (in >=0 states) by no red light.
R_2pKyoW6gWKFoQnZ
F(r => F(!r))
if red, must have !red in future
9
1The Red light is off on some state in the infinite trace.
R_1kn76aEhDPLve1a
F(r => F(!r))
if red, must have !red in future
10
1
There is a state where Red being on implies that after zero or more states it becomes off.
R_5PdIR5FnMDzrVUu
F(r => F(!r))
if red, must have !red in future
11
1
There must be at least one state where when Red is on, then there is at least one state where Red is off.
R_3p46SBAhKNhlNvS
F(r => F(!r))
if red, must have !red in future
12
1Eventually, a Red light being on will imply that it will eventually turn off
R_47BVZHw0XbaticF
F(r => F(!r))
if red, must have !red in future
13
1If the Red light is on in the first state then it will eventually turn off.
R_51nrC3cVmGQGCbZ
F(r => F(!r))
if red, must have !red in future
14
1Eventually, if the Red light is on, then eventually, it will be switched off.
R_2rjQI459Ka4U504
F(r => F(!r))
if red, must have !red in future
15
1It will eventually be true that if Red is on, then eventually the Red will be off.
R_10WMbhmzVMMfCQ9
F(r => F(!r))
if red, must have !red in future
16
1
there is at least one state in which red is on, and after this state there is at least one state in which Red is off
R_3p2LrOwAHaKfYYG
F(r => F(!r))
if red, must have !red in future
17
1
Eventually, we have a state where Red is true and there exists a future state where Red is False.
R_2EPSIkVcvTwusZ1
F(r => F(!r))
if red, must have !red in future
18
1
Eventually, the red light will be on and, sometime eventually after that, the red light will be off.
R_6ZBMLj5oxGtGYyV
F(r => F(!r))
if red, must have !red in future
19
1
There is a state in the future, such that if in that state the Red light is on, then there is a state where it is off after that as well.
R_7gRsvfLpsHGFun8
F(r => F(!r))
if red, must have !red in future
20
1
Eventually if Red is on the there must exist a point in the futere included the current one where Red turns off. (if in the first state Red is of it is already satisfied)
R_2wQrFCdtVBTz165
F(r => F(!r))
if red, must have !red in future
21
1Red is true all along the trace and eventually Blue also becomes true.
R_3zr1TPsbL8Fh01g
(r U b) & G(r)
always r, eventually b
22
1Red light is always on and blue light turns on in some state.
R_7t8OynItMvsY3qS
(r U b) & G(r)
always r, eventually b
23
1Red is always on and Blue is on in at least one state.
R_5pl5ovreUYilD5z
(r U b) & G(r)
always r, eventually b
24
1All states are red and there exists a state that is blue as well
R_6ffI4bTyX5zoEHh
(r U b) & G(r)
always r, eventually b
25
1red always holds and blue holds at some point
R_4mVF8KOBt6MErhn
(r U b) & G(r)
always r, eventually b
26
1The red light must be on until the blue light is on, AND the red light must always be on.
R_2pKyoW6gWKFoQnZ
(r U b) & G(r)
always r, eventually b
27
1The Red light is on in every state, and the Blue light is on in some state
R_1kn76aEhDPLve1a
(r U b) & G(r)
always r, eventually b
28
1Red is always on and, at least in one state, Blue is on.
R_5PdIR5FnMDzrVUu
(r U b) & G(r)
always r, eventually b
29
1Blue is on at one state, and Red is on in every state before and after it.
R_3p46SBAhKNhlNvS
(r U b) & G(r)
always r, eventually b
30
1Red is always on and blue will eventually turn on
R_47BVZHw0XbaticF
(r U b) & G(r)
always r, eventually b
31
1
The Blue light eventually turns on and the Red light was always on before that state. Moreover, the Green light is always on.
R_51nrC3cVmGQGCbZ
(r U b) & G(r)
always r, eventually b
32
1
It is the case that the Red light is on forever and the Blue light is off until a certain state, where the Blue light will be turned on.
R_2rjQI459Ka4U504
(r U b) & G(r)
always r, eventually b
33
1Blue is on for some state and Red is on for every state in the trace.
R_10WMbhmzVMMfCQ9
(r U b) & G(r)
always r, eventually b
34
11Red is always on and Blue is never on
R_3p2LrOwAHaKfYYG
(r U b) & G(r)
always r, eventually b
35
1
We always have Red to be true, and we must have Blue to be true in some future state.
R_2EPSIkVcvTwusZ1
(r U b) & G(r)
always r, eventually b
36
1The red light is always on. At some point, the blue light will be on.
R_6ZBMLj5oxGtGYyV
(r U b) & G(r)
always r, eventually b
37
1Globally red is true and at some point blue is true as well.
R_7gRsvfLpsHGFun8
(r U b) & G(r)
always r, eventually b
38
1Red is on until Blue is on, and Red is always on.
R_2wQrFCdtVBTz165
(r U b) & G(r)
always r, eventually b
39
1Red is true at the beginning and Blue is false next.
R_3zr1TPsbL8Fh01g
r & !X(b)r now, !b next
40
1Red light is on in this state and blue light is off in the next state.
R_7t8OynItMvsY3qS
r & !X(b)r now, !b next
41
1Red is on in the first state, and in the second state Blue is not on.
R_5pl5ovreUYilD5z
r & !X(b)r now, !b next
42
1First state is red, and the next one is not blue
R_6ffI4bTyX5zoEHh
r & !X(b)r now, !b next
43
1red holds in state 1 and blue must not hold in state 2.
R_4mVF8KOBt6MErhn
r & !X(b)r now, !b next
44
1The red light is on initially, AND the following state has no blue light.
R_2pKyoW6gWKFoQnZ
r & !X(b)r now, !b next
45
1In the first state, the Red light is on and in the second state, the Blue light is off
R_1kn76aEhDPLve1a
r & !X(b)r now, !b next
46
1Red is on at the initial state and Blue is off at the next state.
R_5PdIR5FnMDzrVUu
r & !X(b)r now, !b next
47
1Red is on in the initial state, and there doesn't exist a next state which is blue.
R_3p46SBAhKNhlNvS
r & !X(b)r now, !b next
48
1Red is on in the current state and blue is off in the next state
R_47BVZHw0XbaticF
r & !X(b)r now, !b next
49
1The Red light is on in the first state and Blue is off in the second state.
R_51nrC3cVmGQGCbZ
r & !X(b)r now, !b next
50
1
It is the case that, in the first state, the Red light is on and in the next state, the Blue light will be off.
R_2rjQI459Ka4U504
r & !X(b)r now, !b next
51
1Red is on in the initial state and Blue is off in the next state.
R_10WMbhmzVMMfCQ9
r & !X(b)r now, !b next
52
1Red is on in the initial state and Blue is not on in the state after
R_3p2LrOwAHaKfYYG
r & !X(b)r now, !b next
53
1Current state has Red to be true and the next state to have Blue to be false.
R_2EPSIkVcvTwusZ1
r & !X(b)r now, !b next
54
1
At some state in the trace, the red light is on and, in the next state, the blue light is not on.
R_6ZBMLj5oxGtGYyV
r & !X(b)r now, !b next
55
1The current state is red and in the next state blue is false.
R_7gRsvfLpsHGFun8
r & !X(b)r now, !b next
56
1Red is on at the first state, and in the next state Blue is off.
R_2wQrFCdtVBTz165
r & !X(b)r now, !b next
57
not english1Always(if Red then next not Red and then Red again).
R_3zr1TPsbL8Fh01g
G(r => X(!r & X(r)))
whenever r, off/on evermore
58
1Whenever red light turns on, it is off in the next state and on again in the state after.
R_7t8OynItMvsY3qS
G(r => X(!r & X(r)))
whenever r, off/on evermore
59
1
At every state, if Red is on then in the next state Red is not on and in the state after that Red is on again.
R_5pl5ovreUYilD5z
G(r => X(!r & X(r)))
whenever r, off/on evermore
60
1Every appearance of red is followed by a state which is not red and then a red state
R_6ffI4bTyX5zoEHh
G(r => X(!r & X(r)))
whenever r, off/on evermore
61
1
whenever red holds, it must not hold in the next state and must hold in the state after the next state. this is the same as saying, either red never holds or, it holds at some point in the trace and from that point on red alternates between true and false.
R_4mVF8KOBt6MErhn
G(r => X(!r & X(r)))
whenever r, off/on evermore
62
1
Always, a red light is immediately followed by no red light, which is immediately followed by a red light.
R_2pKyoW6gWKFoQnZ
G(r => X(!r & X(r)))
whenever r, off/on evermore
63
1
Whenever the Red light is on, it is off in the next state, and on again in the state after that. This amounts to: From the first time the Red light is on (if any), the Red light is on in exactly every other state
R_1kn76aEhDPLve1a
G(r => X(!r & X(r)))
whenever r, off/on evermore
64
1
Whenever the red light is on, it is off at the next state and back on at the state after that.
R_5PdIR5FnMDzrVUu
G(r => X(!r & X(r)))
whenever r, off/on evermore
65
1
In every state, when Red is on, then there must exist a next state where Red is off and the next state of that state is Red.
R_3p46SBAhKNhlNvS
G(r => X(!r & X(r)))
whenever r, off/on evermore
66
1
Globally, Red being on implies that it will be off in the next state and back on in the one after
R_47BVZHw0XbaticF
G(r => X(!r & X(r)))
whenever r, off/on evermore
67
1
If the Red light is on in a state, then it is off in the next state and on again in the state after that.
R_51nrC3cVmGQGCbZ
G(r => X(!r & X(r)))
whenever r, off/on evermore
68
1
It is always the case that, whenever the Red light is on in a state, then in the next state, the Red light is off, and the Red light is back on in the state following that.
R_2rjQI459Ka4U504
G(r => X(!r & X(r)))
whenever r, off/on evermore
69
1
Whenever the Red light is on, the Red light is off in the next state and again on in the next of the next state.
R_10WMbhmzVMMfCQ9
G(r => X(!r & X(r)))
whenever r, off/on evermore
70
1
Once Red is on at least once, then it alternates between being on and off in subsequent states
R_3p2LrOwAHaKfYYG
G(r => X(!r & X(r)))
whenever r, off/on evermore
71
1
Always, whenever we have Red to be true, the next state has Red to be false and the next to next state has Red to be true.
R_2EPSIkVcvTwusZ1
G(r => X(!r & X(r)))
whenever r, off/on evermore
72
1
Whenever the red light is on, in the next state the red light is off and in the state after that the red light is on again. So, the red light alternates on and off.
R_6ZBMLj5oxGtGYyV
G(r => X(!r & X(r)))
whenever r, off/on evermore
73
1
Whenever there red light is on, it is off in the next state and on in the state that follows.
R_7gRsvfLpsHGFun8
G(r => X(!r & X(r)))
whenever r, off/on evermore
74
1
Whenever Red is on, then in the next state Red turns off and in the next REd is on again.
R_2wQrFCdtVBTz165
G(r => X(!r & X(r)))
whenever r, off/on evermore
75
1Same
R_3zr1TPsbL8Fh01g
F(r => F(!r))same
76
1same
R_7t8OynItMvsY3qS
F(r => F(!r))same
77
1Either there is no state with the Red light on, or Red will be off after being on.
R_5pl5ovreUYilD5z
F(r => F(!r))same
78
1Same
R_6ffI4bTyX5zoEHh
F(r => F(!r))same
79
1same
R_4mVF8KOBt6MErhn
F(r => F(!r))same
80
1same
R_2pKyoW6gWKFoQnZ
F(r => F(!r))same
81
1Pretty much the same, the Red light is off on some state in the finite trace
R_1kn76aEhDPLve1a
F(r => F(!r))same
82
1same
R_5PdIR5FnMDzrVUu
F(r => F(!r))same
83
1
There must be at least one state where when Red is on, then there is at least one state where Red is off.
R_3p46SBAhKNhlNvS
F(r => F(!r))same
84
1Finally, if Red holds, then eventually, Red does not hold.
R_47BVZHw0XbaticF
F(r => F(!r))same
85
1same
R_51nrC3cVmGQGCbZ
F(r => F(!r))same
86
1
Eventually, if the Red light is on, then eventually, it will be switched off in a finite number of steps.
R_2rjQI459Ka4U504
F(r => F(!r))same
87
1Same.
Eventually has the same meaning in both.
R_10WMbhmzVMMfCQ9
F(r => F(!r))same
88
1Same
R_3p2LrOwAHaKfYYG
F(r => F(!r))same
89
1Same
R_2EPSIkVcvTwusZ1
F(r => F(!r))same
90
1
Eventually, the red light will be on and, sometime eventually after that, the red light will be off.
R_6ZBMLj5oxGtGYyV
F(r => F(!r))same
91
1Same
R_7gRsvfLpsHGFun8
F(r => F(!r))same
92
1Same as LTL.
R_2wQrFCdtVBTz165
F(r => F(!r))same
93
1Same
R_3zr1TPsbL8Fh01g
(r U b) & G(r)
r for all states, b for some
94
1Same
R_7t8OynItMvsY3qS
(r U b) & G(r)
r for all states, b for some
95
1Red is always on and Blue is on in at least one state.
R_5pl5ovreUYilD5z
(r U b) & G(r)
r for all states, b for some
96
1Same
R_6ffI4bTyX5zoEHh
(r U b) & G(r)
r for all states, b for some
97
1same
R_4mVF8KOBt6MErhn
(r U b) & G(r)
r for all states, b for some
98
1same
R_2pKyoW6gWKFoQnZ
(r U b) & G(r)
r for all states, b for some
99
1Same
R_1kn76aEhDPLve1a
(r U b) & G(r)
r for all states, b for some
100
1same
R_5PdIR5FnMDzrVUu
(r U b) & G(r)
r for all states, b for some
101
1Blue is on at one state, and Red is on in every state before and after it.
R_3p46SBAhKNhlNvS
(r U b) & G(r)
r for all states, b for some
102
1Red is always on and blue will eventually turn on
R_47BVZHw0XbaticF
(r U b) & G(r)
r for all states, b for some
103
1same
R_51nrC3cVmGQGCbZ
(r U b) & G(r)
r for all states, b for some
104
1
It is the case that the Red light is on in every state and the Blue light is off until a certain state, where the Blue light will be turned on.
R_2rjQI459Ka4U504
(r U b) & G(r)
r for all states, b for some
105
1Same.
R_10WMbhmzVMMfCQ9
(r U b) & G(r)
r for all states, b for some
106
1Same
R_3p2LrOwAHaKfYYG
(r U b) & G(r)
r for all states, b for some
107
1Same
R_2EPSIkVcvTwusZ1
(r U b) & G(r)
r for all states, b for some
108
1The red light is on at every state. The blue light is on for some state in the trace.
R_6ZBMLj5oxGtGYyV
(r U b) & G(r)
r for all states, b for some
109
1Same.
R_7gRsvfLpsHGFun8
(r U b) & G(r)
r for all states, b for some
110
1Same as LTL.
R_2wQrFCdtVBTz165
(r U b) & G(r)
r for all states, b for some
111
1Red is true at the beginning and if there exist a next instant Blue is false in it.
R_3zr1TPsbL8Fh01g
r & !X(b)
r now, if next !b
112
1
Red light is on in this state. If the trace is of length more than 1 then blue light is off in the next state.
R_7t8OynItMvsY3qS
r & !X(b)
r now, if next !b
113
1
Red is on in the first state, and either the trace has only one state or Blue is off in the second state.
R_5pl5ovreUYilD5z
r & !X(b)
r now, if next !b
114
1
First state is red, and either this state is the last state, or the next one is not blue (if it exists).
R_6ffI4bTyX5zoEHh
r & !X(b)
r now, if next !b
115
1Red must hold in state 1 and EITHER the trace has length 1 OR blue holds in state 2.
R_4mVF8KOBt6MErhn
r & !X(b)
r now, if next !b
116
1The red light is on initially, AND there is no next state OR the next state is not blue
R_2pKyoW6gWKFoQnZ
r & !X(b)
r now, if next !b
117
1
In the first state, the Red light is on. In addition, either there is no second state (the trace has length 1) or in the second state the Blue light is off
R_1kn76aEhDPLve1a
r & !X(b)
r now, if next !b
118
1
Red is on at the initial state and either a next state exists on which Blue is off, or a next state doesn't exist and the trace is automatically satisfying. In other words, all traces of length 1 where Red is on at the initial state satisfy this formula.
R_5PdIR5FnMDzrVUu
r & !X(b)
r now, if next !b
119
1Red is on in the initial state, and there doesn't exist a next state which is blue.
R_3p46SBAhKNhlNvS
r & !X(b)
r now, if next !b
120
1Red is on in the current state and blue is off in the next state
R_47BVZHw0XbaticF
r & !X(b)
r now, if next !b
121
1
The Red light is on in the first state and either there is no second state or the blue light is off in the second state.
R_51nrC3cVmGQGCbZ
r & !X(b)
r now, if next !b
122
1
It is the case that, in the first state, the Red light is on and either there is no further state (i.e. the trace ends after the first state) or, if there is a second state, then Blue light is off in this state.
R_2rjQI459Ka4U504
r & !X(b)
r now, if next !b
123
1Red is on in the initial state and if there exists a next state, Blue is off in it.
Because the negative condition is before the strong next, the meaning of the formula is similar in both.
R_10WMbhmzVMMfCQ9
r & !X(b)
r now, if next !b
124
1
As long as the trace is 2 or more states long then it is satisfiable with the same behavior
R_3p2LrOwAHaKfYYG
r & !X(b)
r now, if next !b
125
1
Current state has Red to be true and the next state either does not exist or it exists and Blue is false in that state.
R_2EPSIkVcvTwusZ1
r & !X(b)
r now, if next !b
126
1
There is a state where the red light is on, and, if the next state exists, the blue light is not on in it.
I am not 100% sure of the implications of !X. I assume here that it means that X may not exist, and in that case, !X is satisfied.
R_6ZBMLj5oxGtGYyV
r & !X(b)
r now, if next !b
127
1The current state is red and if there is a next state, then blue needs to be false.
R_7gRsvfLpsHGFun8
r & !X(b)
r now, if next !b
128
1
Red is on at the first state. Trace of length 1 where Red is on satisfies the formula. Any other trace of length greater or equal to two, the definition is the same as LTL.
R_2wQrFCdtVBTz165
r & !X(b)
r now, if next !b
129
1Same.
R_3zr1TPsbL8Fh01g
G(r => X(!r & X(r)))
never r
130
1Red light is never on.
R_7t8OynItMvsY3qS
G(r => X(!r & X(r)))
never r
131
1This formula is only satisfied if the Red is never on.
R_5pl5ovreUYilD5z
G(r => X(!r & X(r)))
never r
132
1There is no red states
If there was a red state in the finite trace, then from that point we would must have an alternating sequence of red and not red states. For the last red state in sequence (which is the very last state, or the one before) one of the X expressions will fail
R_6ffI4bTyX5zoEHh
G(r => X(!r & X(r)))
never r
133
1
EITHER red never holds OR (same as LTL) it holds at some point in the trace and from that point on red alternates between true and false. However, the 2nd possibility can never be satisfied by a finite trace.
R_4mVF8KOBt6MErhn
G(r => X(!r & X(r)))
never r
134
1same (so cannot be satisfied by a finite trace)
R_2pKyoW6gWKFoQnZ
G(r => X(!r & X(r)))
never r
135
1
No finite traces where the Red light is on satisfy this formula, so the formula in LTLf amounts to the Red light is always off
R_1kn76aEhDPLve1a
G(r => X(!r & X(r)))
never r
136
1
Every state where the Red light is on is followed by a state where Red is off and by another state where it is on again. Therefore, no trace of length at most 2 can satisfy this formula.
R_5PdIR5FnMDzrVUu
G(r => X(!r & X(r)))
never r
137
1
In every state, when Red is on, then there must exist a next state where Red is off and the next state of that state is Red. This can not be satisfied by LTLf since there is not next state for the last state.
R_3p46SBAhKNhlNvS
G(r => X(!r & X(r)))
never r
138
1
Globally, Red being on implies that it will be off in the next state and back on in the one after. No finite traces satisfy this
R_47BVZHw0XbaticF
G(r => X(!r & X(r)))
never r
139
1The Red light never turns on.
R_51nrC3cVmGQGCbZ
G(r => X(!r & X(r)))
never r
140
1
For every state, if Red light is on, then in the next state, the Red light is off, and the Red light is back on in the state following that. No finite traces satisfy the formula.
R_2rjQI459Ka4U504
G(r => X(!r & X(r)))
never r
141
1
Every state where Red is on, must be followed by a state with Red off and then by a state with Red on. A finite trace can satisfy the formula only if there is no state where Red is on.
R_10WMbhmzVMMfCQ9
G(r => X(!r & X(r)))
never r
142
1this is not possible over finite traces
R_3p2LrOwAHaKfYYG
G(r => X(!r & X(r)))
never r
143
1
Same as above, but the only finite traces that can satisfy it do not have Red to be true. This is because once Red is true, it starts alternating forever and a finite trace will cause the implication to be false as X(State) gives false if we are at the last state of the finite trace.
R_2EPSIkVcvTwusZ1
G(r => X(!r & X(r)))
never r
144
1
This cannot be expressed without an infinite trace as it demands the existence of a next state for all states.
R_6ZBMLj5oxGtGYyV
G(r => X(!r & X(r)))
never r
145
1
Whenever the red light is on, there needs to be a next state where it is off and a third successor state, where it is on again. But then, this also holds for that state ==> thus this formula is not satisfiable over finite traces.
R_7gRsvfLpsHGFun8
G(r => X(!r & X(r)))
never r
146
1
The traces satisfying the formula are the ones where Red is always off, independently from the size (empty trace included). Then it is the same as LTL with the checking of the last poit of the trace.
R_2wQrFCdtVBTz165
G(r => X(!r & X(r)))
never r