ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>12180000000710070000000144
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRV
Scope
Answer
Explanation
IDFormulaExpected
3
1
If we get to a state with Red on we eventually reach a state with Red off. \\ \\ i.e. \\ If the Red light ever gets on, it eventually gets off.
R_2PtHv3KdsGqQLoB
F(r => F(!r))
if red, must have !red in future
4
1The Red light is on in some state, and the Red light is off in some state after it.
R_3079Udc3BOxNquc
F(r => F(!r))
if red, must have !red in future
5
1If red is on at some point, then red is eventually off.
R_2U5AqOKS1RFOYwc
F(r => F(!r))
if red, must have !red in future
6
1At least one state in the future must not be Red.
R_2yscxjylS1SXLu5
F(r => F(!r))
if red, must have !red in future
7
1At some state the red light goes from on to off
R_oYU5yKiaqPgXzMd
F(r => F(!r))
if red, must have !red in future
8
1Traces in which there is at least one state in which the red light is off.
R_5vtXYvxyZKlCUx3
F(r => F(!r))
if red, must have !red in future
9
1Whenever the red light is on, later at some point it should be off.
R_2aenS9mNn6Dk4Cz
F(r => F(!r))
if red, must have !red in future
10
1eventually (not red or eventually (not red)) i.e. eventually not red.
R_3s0nFuWCbCk5CjP
F(r => F(!r))
if red, must have !red in future
11
1
The Red light is eventually turned on, and from that point onwards, the Red light will be eventually turned off.
R_2uWWZejY6DyOYaH
F(r => F(!r))
if red, must have !red in future
12
1The red light turns on and then turns off in some finite time
R_2WwptK2YPEgUAGJ
F(r => F(!r))
if red, must have !red in future
13
1
There is a state that either has the Red light off or some state after it has the Red light off.
R_sLorPWGMjcKuGYx
F(r => F(!r))
if red, must have !red in future
14
1This is equivalent to !G(Red). eventually there is a state that is not Red
R_2zeuLOxEmJflmcu
F(r => F(!r))
if red, must have !red in future
15
1Eventually, if red light is on, it is followed by a state with red light off
R_1rjpSmO14SmyPCQ
F(r => F(!r))
if red, must have !red in future
16
1Eventually, if the red light is on then non-red light will be on after several steps
R_1FRNUwvnkvroHRF
F(r => F(!r))
if red, must have !red in future
17
1If red is on at some point, then red is eventually off.
R_3IXnVYuOI9QeI10
F(r => F(!r))
if red, must have !red in future
18
1The red light is on in a state and off later.
R_3KGxBOYYe7aGaXI
F(r => F(!r))
if red, must have !red in future
19
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
F(r => F(!r))
if red, must have !red in future
20
1
There will be a state with the red light on, and another state after this state with the red light off.
R_31j7Jp3yQbqaZqk
F(r => F(!r))
if red, must have !red in future
21
1Red light is always on and we eventually get the Blue light.
R_2PtHv3KdsGqQLoB
(r U b) & G(r)
always r, eventually b
22
1The Red light is on in every state, and the Blue light is on in some state.
R_3079Udc3BOxNquc
(r U b) & G(r)
always r, eventually b
23
1red is always on, and at some point blue is on.
R_2U5AqOKS1RFOYwc
(r U b) & G(r)
always r, eventually b
24
1Red is always on, and Blue is on in at least one state in the future.
R_2yscxjylS1SXLu5
(r U b) & G(r)
always r, eventually b
25
1The red light is on in every state, and the blue light is on in some state
R_oYU5yKiaqPgXzMd
(r U b) & G(r)
always r, eventually b
26
1
Traces in which in every state red light is on and there is at least one state in which the blue light is on.
R_5vtXYvxyZKlCUx3
(r U b) & G(r)
always r, eventually b
27
1The red light is always on, and eventually the blue light must be on.
R_2aenS9mNn6Dk4Cz
(r U b) & G(r)
always r, eventually b
28
1The red light is always on and eventually the blue light must turn on.
R_3s0nFuWCbCk5CjP
(r U b) & G(r)
always r, eventually b
29
1
The Red light must be turned on forever, and eventually the Blue light must be turned on.
R_2uWWZejY6DyOYaH
(r U b) & G(r)
always r, eventually b
30
1The red light is on and the blue light turns on after some finite time.
R_2WwptK2YPEgUAGJ
(r U b) & G(r)
always r, eventually b
31
1
There is a state that has the Blue light on and every state before this has the Red light on, and also every state has the Red light on.
R_sLorPWGMjcKuGYx
(r U b) & G(r)
always r, eventually b
32
1All states are Red, some state is Red and Blue
R_2zeuLOxEmJflmcu
(r U b) & G(r)
always r, eventually b
33
1Before the blue light is on, the Red light is always on. Red is on in every state as well.
R_1rjpSmO14SmyPCQ
(r U b) & G(r)
always r, eventually b
34
1Red light is always on, and the blue light is on at some point.
R_1FRNUwvnkvroHRF
(r U b) & G(r)
always r, eventually b
35
1Red is always on and at some point blue is on.
R_3IXnVYuOI9QeI10
(r U b) & G(r)
always r, eventually b
36
1The red light is always on and the blue light is on in a state.
R_3KGxBOYYe7aGaXI
(r U b) & G(r)
always r, eventually b
37
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
(r U b) & G(r)
always r, eventually b
38
1All the states are with the red light on and there will be a blue state.
R_31j7Jp3yQbqaZqk
(r U b) & G(r)
always r, eventually b
39
1Red is on initially and after that Blue is not on.
R_2PtHv3KdsGqQLoB
r & !X(b)r now, !b next
40
1The Red light is on in the first state, and the Blue light is off in the second state.
R_3079Udc3BOxNquc
r & !X(b)r now, !b next
41
1Red is on at the first state, and blue is off in the second state.
R_2U5AqOKS1RFOYwc
r & !X(b)r now, !b next
42
1Red is on, Blue is not on in the next state
R_2yscxjylS1SXLu5
r & !X(b)r now, !b next
43
1The Red light is on in the first state, and the Blue light is not on in the second state
R_oYU5yKiaqPgXzMd
r & !X(b)r now, !b next
44
1
Traces in which in the first state red light is on and in which in the second state the blue light is off.
R_5vtXYvxyZKlCUx3
r & !X(b)r now, !b next
45
1The red light is on at the first step, and the blue one should be off in the next state.
R_2aenS9mNn6Dk4Cz
r & !X(b)r now, !b next
46
1Red holds in the first state and the next state must not be blue.
R_3s0nFuWCbCk5CjP
r & !X(b)r now, !b next
47
1
The Red light is currently turned on at the initial time step, and the state in the next time step should not have the Blue light turned on.
R_2uWWZejY6DyOYaH
r & !X(b)r now, !b next
48
1The red light is on at the start and the blue light is not on in the second timestep.
R_2WwptK2YPEgUAGJ
r & !X(b)r now, !b next
49
1The first state has the Red light on and the next state does not have the Blue light on.
R_sLorPWGMjcKuGYx
r & !X(b)r now, !b next
50
1
in LTL !X(f) is X(!f) since there is always a next state. This formula means the first state is Red, and the second state is not Blue
R_2zeuLOxEmJflmcu
r & !X(b)r now, !b next
51
1The current state is with the red light on and the next state is with the blue light off.
R_1rjpSmO14SmyPCQ
r & !X(b)r now, !b next
52
1On the first state the red light is on, and the next state is not blue.
R_1FRNUwvnkvroHRF
r & !X(b)r now, !b next
53
1Red is on in the first state, blue is off in the second state.
R_3IXnVYuOI9QeI10
r & !X(b)r now, !b next
54
1The red light is on now and the blue light is off in next state.
R_3KGxBOYYe7aGaXI
r & !X(b)r now, !b next
55
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
r & !X(b)r now, !b next
56
1The first state is red, and the next state is not blue
R_31j7Jp3yQbqaZqk
r & !X(b)r now, !b next
57
1If we ever get Red on, after that it is off and then it is on again.
R_2PtHv3KdsGqQLoB
G(r => X(!r & X(r)))
whenever r, off/on evermore
58
1
Whenever the Red light is on in one state, it must alternate between off and on from the next state onwards.
R_3079Udc3BOxNquc
G(r => X(!r & X(r)))
whenever r, off/on evermore
59
1
Whenever red is on, then it gets off in the next state, and then alternates between on and off.
R_2U5AqOKS1RFOYwc
G(r => X(!r & X(r)))
whenever r, off/on evermore
60
1If ever turns on, Red will alternate on and off forever for all following states.
R_2yscxjylS1SXLu5
G(r => X(!r & X(r)))
whenever r, off/on evermore
61
1
From the first state in which the Red light is on (if there is such a state), the Red light will flicker off/on forever
R_oYU5yKiaqPgXzMd
G(r => X(!r & X(r)))
whenever r, off/on evermore
62
1
Traces in which either red light is off in every state or after the first state in which the red light is on the red light switches on and off in each state onward.
R_5vtXYvxyZKlCUx3
G(r => X(!r & X(r)))
whenever r, off/on evermore
63
1
At each moment when the red light is on, it should be off in the next state, and on again in the one after. So, after the first time the red light is on (if ever), it should be alternating between on and off.
R_2aenS9mNn6Dk4Cz
G(r => X(!r & X(r)))
whenever r, off/on evermore
64
1
The states can be any colour, but as soon as we get a state which is red, then we have to have a red state, not red state, red state, not red state,... etc.
R_3s0nFuWCbCk5CjP
G(r => X(!r & X(r)))
whenever r, off/on evermore
65
1
It is forever the case that if the Red light is turned on, then in the next time step, the Red light is not turned on, and the Red light is turned back on in the following time step.
R_2uWWZejY6DyOYaH
G(r => X(!r & X(r)))
whenever r, off/on evermore
66
1If the red light turns on, it will alternate being on and off forever
R_2WwptK2YPEgUAGJ
G(r => X(!r & X(r)))
whenever r, off/on evermore
67
1
Whenever a state has the Red light on, the next state has the Red light off and the state after it has the Red light on.
R_sLorPWGMjcKuGYx
G(r => X(!r & X(r)))
whenever r, off/on evermore
68
1
either no state is Red (all satisfy the implication vacuously) or some state is Red, and from that point onwards Red alternates strictly off and on from state to state.
R_2zeuLOxEmJflmcu
G(r => X(!r & X(r)))
whenever r, off/on evermore
69
1
Whenever the red light is on, in the next state the red light is off and in the next state after the next state the red light is on
R_1rjpSmO14SmyPCQ
G(r => X(!r & X(r)))
whenever r, off/on evermore
70
1
it is always the case that, if the red light is on, then red light is off in the next state, and red light is on in the next of the next state.
R_1FRNUwvnkvroHRF
G(r => X(!r & X(r)))
whenever r, off/on evermore
71
1If red ever gets on, it alternates between on and off.
R_3IXnVYuOI9QeI10
G(r => X(!r & X(r)))
whenever r, off/on evermore
72
1Once the red light is on in a state, the red light is on and off alternarely afterward.
R_3KGxBOYYe7aGaXI
G(r => X(!r & X(r)))
whenever r, off/on evermore
73
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
G(r => X(!r & X(r)))
whenever r, off/on evermore
74
1
For every state, if there is a red light on, the next state is with the red light off, and the state afterward is with the red light on.
R_31j7Jp3yQbqaZqk
G(r => X(!r & X(r)))
whenever r, off/on evermore
75
1Same.
R_2PtHv3KdsGqQLoB
F(r => F(!r))same
76
1same
R_3079Udc3BOxNquc
F(r => F(!r))same
77
1same
R_2U5AqOKS1RFOYwc
F(r => F(!r))same
78
1Same
R_2yscxjylS1SXLu5
F(r => F(!r))same
79
1Same as LTL
Equiv to F(Red & X(!Red))
R_oYU5yKiaqPgXzMd
F(r => F(!r))same
80
1Traces in which there is at least one state in which the red light is off.
R_5vtXYvxyZKlCUx3
F(r => F(!r))same
81
1At some point, if the red light is on, it will eventually be off.
R_2aenS9mNn6Dk4Cz
F(r => F(!r))same
82
1Eventually a state in the finite trace must be not red.
R_3s0nFuWCbCk5CjP
F(r => F(!r))same
83
1Same.
R_2uWWZejY6DyOYaH
F(r => F(!r))same
84
1Same
The red light must be on after some finite interval, and after that the red light must be off. The finally operator has the same semantics in both LTL and LTLf.
R_2WwptK2YPEgUAGJ
F(r => F(!r))same
85
1same
R_sLorPWGMjcKuGYx
F(r => F(!r))same
86
1Identical to LTL, any trace where some state has no Red satisfies it
R_2zeuLOxEmJflmcu
F(r => F(!r))same
87
1Same
R_1rjpSmO14SmyPCQ
F(r => F(!r))same
88
1
Before the last state, if the red light is on then non-red light will be on after several steps (before the last state).
R_1FRNUwvnkvroHRF
F(r => F(!r))same
89
1Same.
R_3IXnVYuOI9QeI10
F(r => F(!r))same
90
1same
R_3KGxBOYYe7aGaXI
F(r => F(!r))same
91
noop1Your LTLf descriptions may ignore the possibility of empty traces.
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
F(r => F(!r))same
92
1
There will be a state with the red light on, and another state after this state with the red light off.
R_31j7Jp3yQbqaZqk
F(r => F(!r))same
93
1Same.
R_2PtHv3KdsGqQLoB
(r U b) & G(r)
r for all states, b for some
94
1same
R_3079Udc3BOxNquc
(r U b) & G(r)
r for all states, b for some
95
1same
R_2U5AqOKS1RFOYwc
(r U b) & G(r)
r for all states, b for some
96
1Same
R_2yscxjylS1SXLu5
(r U b) & G(r)
r for all states, b for some
97
1Same as LTL
Equiv to G(Red) & F(Blue)
R_oYU5yKiaqPgXzMd
(r U b) & G(r)
r for all states, b for some
98
1
Traces in which in every state red light is on and there is at least one state in which the blue light is on.
R_5vtXYvxyZKlCUx3
(r U b) & G(r)
r for all states, b for some
99
1The red light is always on, and eventually the blue light must be on.
R_2aenS9mNn6Dk4Cz
(r U b) & G(r)
r for all states, b for some
100
1
All of the states in the trace must be red and we must have a blue state somewhere as well.
R_3s0nFuWCbCk5CjP
(r U b) & G(r)
r for all states, b for some
101
1
The Red light must be turned on until the last time step, and within those finite time steps, the Blue light must be eventually turned on.
R_2uWWZejY6DyOYaH
(r U b) & G(r)
r for all states, b for some
102
1Same
The second clause enforces that the red light is always on, and the first clause requires the blue light to turn on at some point. There is no difference to the LTLf case as Always and Until have the same semantics.
R_2WwptK2YPEgUAGJ
(r U b) & G(r)
r for all states, b for some
103
1same
R_sLorPWGMjcKuGYx
(r U b) & G(r)
r for all states, b for some
104
1Exactly the Same, all states are Red, and some state is Red and Blue
R_2zeuLOxEmJflmcu
(r U b) & G(r)
r for all states, b for some
105
1Same
R_1rjpSmO14SmyPCQ
(r U b) & G(r)
r for all states, b for some
106
1The red light is always on and the blue light is on before (and including) the last state.
R_1FRNUwvnkvroHRF
(r U b) & G(r)
r for all states, b for some
107
1Same.
R_3IXnVYuOI9QeI10
(r U b) & G(r)
r for all states, b for some
108
1same
R_3KGxBOYYe7aGaXI
(r U b) & G(r)
r for all states, b for some
109
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
(r U b) & G(r)
r for all states, b for some
110
1All the states are with the red light on and there will be a blue state.
R_31j7Jp3yQbqaZqk
(r U b) & G(r)
r for all states, b for some
111
1Red is on initially and after that Blue is on if we get a second state.
R_2PtHv3KdsGqQLoB
r & !X(b)
r now, if next !b
112
1
The Red light is on in the first state, and the Blue light is off in the second state; but the second state need not exist.
R_3079Udc3BOxNquc
r & !X(b)
r now, if next !b
113
1Red is on at the first state, and if there is a second state, then blue is off there.
R_2U5AqOKS1RFOYwc
r & !X(b)
r now, if next !b
114
1Red is on, the next state either does not exist or Blue is not on
R_2yscxjylS1SXLu5
r & !X(b)
r now, if next !b
115
1
The Red light is on in the first state. Either there is no second state, or there is a second state in which the Blue light is off
R_oYU5yKiaqPgXzMd
r & !X(b)
r now, if next !b
116
1
Traces in which in the first state red light is on and in which in the second state the blue light is off or a second state does not exist.
R_5vtXYvxyZKlCUx3
r & !X(b)
r now, if next !b
117
1
The red light is on at the first step, and either there is no next moment, or the blue light should be off in it.
R_2aenS9mNn6Dk4Cz
r & !X(b)
r now, if next !b
118
1
Red holds in the first state and either there isn't another state, or there is another state which isn't blue.
R_3s0nFuWCbCk5CjP
r & !X(b)
r now, if next !b
119
X mixup, ! mixup
1
The Red light is currently turned on at the initial time step, there must exist a next state, and the next state should not have the Blue light turned on.
R_2uWWZejY6DyOYaH
r & !X(b)
r now, if next !b
120
1
The red light is on at the start, and then if the next step exists the blue light is not on there.
R_2WwptK2YPEgUAGJ
r & !X(b)
r now, if next !b
121
same1
The first state has the Red light on and there must be a next state that doesn't have the Blue light on. Any valid trace must have at least two states.
R_sLorPWGMjcKuGYx
r & !X(b)
r now, if next !b
122
1
In LTLf, !X(f) is Xw(!f) so either the next state doesn't exist, or is not f. \\ This formula means the first state is Red, and either the second state is not Blue or doesn't exist
R_2zeuLOxEmJflmcu
r & !X(b)
r now, if next !b
123
1
The current state is with the red light on and there is no next state or the next is with the blue light off.
R_1rjpSmO14SmyPCQ
r & !X(b)
r now, if next !b
124
1
On the first state the red light is on, and either the next state does not exist or the blue light is on in the next state.
R_1FRNUwvnkvroHRF
r & !X(b)
r now, if next !b
125
1
Red is on in the first state, either there is no second state, or there is one, in which blue is off.
R_3IXnVYuOI9QeI10
r & !X(b)
r now, if next !b
126
1The red light is on now and if there is a next state, the blue light is off.
R_3KGxBOYYe7aGaXI
r & !X(b)
r now, if next !b
127
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
r & !X(b)
r now, if next !b
128
1
The first state is with the red light on, and the trace doesn't have a second state with the blue light on. (i.e. the trace has only one state with the red light on or the trace has more than one state that the first state is with the red light on and the second state is with the blue light off)
R_31j7Jp3yQbqaZqk
r & !X(b)
r now, if next !b
129
weak X1
Every state with Red on is followed by one with Red off and then alternates between on and off until the end of the trace.
R_2PtHv3KdsGqQLoB
G(r => X(!r & X(r)))
never r
130
1
Whenever the Red light is on in one state, it must alternate between off and on from the next state onwards. No finite traces satisfy the formula.
R_3079Udc3BOxNquc
G(r => X(!r & X(r)))
never r
131
1Red is never on.
R_2U5AqOKS1RFOYwc
G(r => X(!r & X(r)))
never r
132
1Red can never turn on. (If it does, no finite traces can satisfy the formula)
R_2yscxjylS1SXLu5
G(r => X(!r & X(r)))
never r
133
weak X1
From the first state in which the Red light is on (if there is such a state), the Red light will flicker off/on until and including the final state
R_oYU5yKiaqPgXzMd
G(r => X(!r & X(r)))
never r
134
1Traces in which the red light is off in every state.
R_5vtXYvxyZKlCUx3
G(r => X(!r & X(r)))
never r
135
1
The red light is never on. (Otherwise, it is never satisfied, as there does not always exist a next moment.)
R_2aenS9mNn6Dk4Cz
G(r => X(!r & X(r)))
never r
136
1
The only way a finite trace can satisfy this is by being a finite sequence of !red states.
R_3s0nFuWCbCk5CjP
G(r => X(!r & X(r)))
never r
137
weak X1
It is always the case that if the Red light is turned on, then there exists a next time step where the Red light is not turned on, and there exists a following time step where the Red light is turned back on.
R_2uWWZejY6DyOYaH
G(r => X(!r & X(r)))
never r
138
1
The red light either never turns on and the trace can terminate, or the red light turns on and then it will alternate being on and off forever with no finite trace being accepted if Red ever holds.
R_2WwptK2YPEgUAGJ
G(r => X(!r & X(r)))
never r
139
1
Whenever a state has the Red light on, there must be a next state with the Red light off and there must be a state after it that has the Red light on. \\ \\ The valid traces must have all the states with the Red light off. A state with the Red light on will require the trace to be infinite, which is not possible in this case.
R_sLorPWGMjcKuGYx
G(r => X(!r & X(r)))
never r
140
1
Only the vacuous case can hold, since the alternation must go on infinitely. The only finite traces satisfying this are those that are never Red
R_2zeuLOxEmJflmcu
G(r => X(!r & X(r)))
never r
141
1
Similar, but any trace with the last state with red on cannot satisfy since the trace is too short to compute the next state after the next state.
R_1rjpSmO14SmyPCQ
G(r => X(!r & X(r)))
never r
142
weak X1
it is always true that, if the red light is on then there must exist two more states in the trace, on which red-light is first off then on.
R_1FRNUwvnkvroHRF
G(r => X(!r & X(r)))
never r
143
1Red is never on.
R_3IXnVYuOI9QeI10
G(r => X(!r & X(r)))
never r
144
1No finite traces satisfy the formula.
R_3KGxBOYYe7aGaXI
G(r => X(!r & X(r)))
never r
145
noop1Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
G(r => X(!r & X(r)))
never r
146
weak X1
For every state, if there is a red light on, the next state is with the red light off, and the state afterward is with the red light on. The trace must at least have 3 states.
R_31j7Jp3yQbqaZqk
G(r => X(!r & X(r)))
never r