ABCDEFGHIJKLMNOPQRSTUVWXYZAA
1
Totals ==>239001000020000242
2
Ok?BPBSI
BSQ
IFIGOIWUEUXXRV
Scope
AnswerExplanationIDFormulaTraceExpected
3
1Yes
R_10pfB7n4SRWFEiA
G(X(r)){gb} {rgb}*Yes
4
1Yes
R_1d6RMi3t0gkcM5M
G(X(r)){gb} {rgb}*Yes
5
1Yes
R_e8QLXoeqrle4xS9
G(X(r)){gb} {rgb}*Yes
6
1Yes
R_z5PVfNBIPKX5S4F
G(X(r)){gb} {rgb}*Yes
7
1Yes
R_1E0ztwRbCQeEjAn
G(X(r)){gb} {rgb}*Yes
8
1Yes
R_d6w1dXzacJ9KN0Z
G(X(r)){gb} {rgb}*Yes
9
1Yes
R_3MuS1ttcj8sdoRv
G(X(r)){gb} {rgb}*Yes
10
1Yes
R_1hEGqpvvTZegzQ7
G(X(r)){gb} {rgb}*Yes
11
1Yes
All paths from state 2 satisfy G(Red), so X(G(Red)) is satisfied, so G(X(Red)) is satisfied
R_3gO868rLSy0R2Ep
G(X(r)){gb} {rgb}*Yes
12
1Yes
Yes, because after the first (initial) state, Red is always on forever, and hence no matter what state you are in, the next state will have a Red on.
R_332amsBlozpU5n8
G(X(r)){gb} {rgb}*Yes
13
1Yes
It is always the case that in the next state we have Red
R_QcBt5TnllV1Ynq9
G(X(r)){gb} {rgb}*Yes
14
1Yes
R_2t50ks4XN0wxwLl
G(X(r)){gb} {rgb}*Yes
15
1Yes
Red holds in all states but the first, in which next red holds.
R_0Vg19zBXa3hIGzf
G(X(r)){gb} {rgb}*Yes
16
1Yes
R_33wl4yU4sEzJKu4
G(X(r)){gb} {rgb}*Yes
17
1Yes
R_3iWIMn2gWx1V4Lp
G(X(r)){gb} {rgb}*Yes
18
1Yes
In every state, the next state has Red on
R_W7oqYTh5ZL9V9vP
G(X(r)){gb} {rgb}*Yes
19
1Yes
R_3e8mdHJtEsNstwp
G(X(r)){gb} {rgb}*Yes
20
1Yes
R_1i3xVhhSYFfFwDR
G(X(r)){gb} {rgb}*Yes
21
1Yes
Red is always on in the next state
R_6F2ykLOE4e1S4IF
G(X(r)){gb} {rgb}*Yes
22
1Yes
R_sb9MnVHZ3rBbcuR
G(X(r)){gb} {rgb}*Yes
23
1Yes
R_3GB8QvZ8WJ7imO7
G(X(r)){gb} {rgb}*Yes
24
1Yes
R_3KSFgYqXtEshyWa
G(X(r)){gb} {rgb}*Yes
25
1No
R_10pfB7n4SRWFEiA
r{gb} {rgb}*No
26
1No
R_1d6RMi3t0gkcM5M
r{gb} {rgb}*No
27
1No
R_e8QLXoeqrle4xS9
r{gb} {rgb}*No
28
1No
R_z5PVfNBIPKX5S4F
r{gb} {rgb}*No
29
1No
R_1E0ztwRbCQeEjAn
r{gb} {rgb}*No
30
1No
R_d6w1dXzacJ9KN0Z
r{gb} {rgb}*No
31
1No
R_3MuS1ttcj8sdoRv
r{gb} {rgb}*No
32
1No
R_1hEGqpvvTZegzQ7
r{gb} {rgb}*No
33
1No
R_3gO868rLSy0R2Ep
r{gb} {rgb}*No
34
1No
The formula "Red" in LTL indicates that it should be satisfied in the current state, i.e., in state 1, which it is not.
R_332amsBlozpU5n8
r{gb} {rgb}*No
35
1No
In the initial state we do not have Red
R_QcBt5TnllV1Ynq9
r{gb} {rgb}*No
36
1No
R_2t50ks4XN0wxwLl
r{gb} {rgb}*No
37
1No
Red doesn't hold in the first state
R_0Vg19zBXa3hIGzf
r{gb} {rgb}*No
38
1No
R_33wl4yU4sEzJKu4
r{gb} {rgb}*No
39
1No
R_3iWIMn2gWx1V4Lp
r{gb} {rgb}*No
40
1No
In the first state Red is not on
R_W7oqYTh5ZL9V9vP
r{gb} {rgb}*No
41
1No
R_3e8mdHJtEsNstwp
r{gb} {rgb}*No
42
1No
R_1i3xVhhSYFfFwDR
r{gb} {rgb}*No
43
1No
Red is not on in the first state
R_6F2ykLOE4e1S4IF
r{gb} {rgb}*No
44
1No
R_sb9MnVHZ3rBbcuR
r{gb} {rgb}*No
45
1No
R_3GB8QvZ8WJ7imO7
r{gb} {rgb}*No
46
1No
R_3KSFgYqXtEshyWa
r{gb} {rgb}*No
47
1Yes
R_10pfB7n4SRWFEiA
F(r){r} {}*Yes
48
1Yes
R_1d6RMi3t0gkcM5M
F(r){r} {}*Yes
49
1Yes
R_e8QLXoeqrle4xS9
F(r){r} {}*Yes
50
1Yes
R_z5PVfNBIPKX5S4F
F(r){r} {}*Yes
51
1Yes
R_1E0ztwRbCQeEjAn
F(r){r} {}*Yes
52
1Yes
R_d6w1dXzacJ9KN0Z
F(r){r} {}*Yes
53
1Yes
R_3MuS1ttcj8sdoRv
F(r){r} {}*Yes
54
1Yes
R_1hEGqpvvTZegzQ7
F(r){r} {}*Yes
55
1Yes
Red is satisfied in state 1.
R_3gO868rLSy0R2Ep
F(r){r} {}*Yes
56
1Yes
The formula F(Red) indicates that there exists some time step t>0 such that Red is on, which in this case is trivially true as it is satisfied when t=1, the initial state.
R_332amsBlozpU5n8
F(r){r} {}*Yes
57
1Yes
We have Red in the initial state so eventually we get Red
R_QcBt5TnllV1Ynq9
F(r){r} {}*Yes
58
1Yes
R_2t50ks4XN0wxwLl
F(r){r} {}*Yes
59
1Yes
Red holds in the initial state
R_0Vg19zBXa3hIGzf
F(r){r} {}*Yes
60
1Yes
R_33wl4yU4sEzJKu4
F(r){r} {}*Yes
61
1Yes
R_3iWIMn2gWx1V4Lp
F(r){r} {}*Yes
62
1Yes
Red is on in the first step
R_W7oqYTh5ZL9V9vP
F(r){r} {}*Yes
63
1Yes
R_3e8mdHJtEsNstwp
F(r){r} {}*Yes
64
1Yes
R_1i3xVhhSYFfFwDR
F(r){r} {}*Yes
65
1Yes
Red is eventually on (this occurs in the 1st state)
R_6F2ykLOE4e1S4IF
F(r){r} {}*Yes
66
1Yes
R_sb9MnVHZ3rBbcuR
F(r){r} {}*Yes
67
1Yes
R_3GB8QvZ8WJ7imO7
F(r){r} {}*Yes
68
1Yes
R_3KSFgYqXtEshyWa
F(r){r} {}*Yes
69
1Yes
R_10pfB7n4SRWFEiA
X(X(X(r))){r} {} {} {r} {}*Yes
70
1Yes
R_1d6RMi3t0gkcM5M
X(X(X(r))){r} {} {} {r} {}*Yes
71
1Yes
R_e8QLXoeqrle4xS9
X(X(X(r))){r} {} {} {r} {}*Yes
72
1Yes
R_z5PVfNBIPKX5S4F
X(X(X(r))){r} {} {} {r} {}*Yes
73
1Yes
R_1E0ztwRbCQeEjAn
X(X(X(r))){r} {} {} {r} {}*Yes
74
1Yes
R_d6w1dXzacJ9KN0Z
X(X(X(r))){r} {} {} {r} {}*Yes
75
1Yes
R_3MuS1ttcj8sdoRv
X(X(X(r))){r} {} {} {r} {}*Yes
76
1Yes
R_1hEGqpvvTZegzQ7
X(X(X(r))){r} {} {} {r} {}*Yes
77
1Yes
R_3gO868rLSy0R2Ep
X(X(X(r))){r} {} {} {r} {}*Yes
78
1Yes
The formula X(X(X(Red))) means that a Red will be on in three time steps starting from the initial state t=1, which in this case is true since Red is on when t=4.
R_332amsBlozpU5n8
X(X(X(r))){r} {} {} {r} {}*Yes
79
1Yes
The third state has Red
R_QcBt5TnllV1Ynq9
X(X(X(r))){r} {} {} {r} {}*Yes
80
1Yes
R_2t50ks4XN0wxwLl
X(X(X(r))){r} {} {} {r} {}*Yes
81
1Yes
R_0Vg19zBXa3hIGzf
X(X(X(r))){r} {} {} {r} {}*Yes
82
1Yes
R_33wl4yU4sEzJKu4
X(X(X(r))){r} {} {} {r} {}*Yes
83
1Yes
R_3iWIMn2gWx1V4Lp
X(X(X(r))){r} {} {} {r} {}*Yes
84
1Yes
3 steps after the initial state red is on
R_W7oqYTh5ZL9V9vP
X(X(X(r))){r} {} {} {r} {}*Yes
85
1Yes
R_3e8mdHJtEsNstwp
X(X(X(r))){r} {} {} {r} {}*Yes
86
1Yes
R_1i3xVhhSYFfFwDR
X(X(X(r))){r} {} {} {r} {}*Yes
87
1No
The next next next state (from the 1st) is the 5th, in which Red is not on.
R_6F2ykLOE4e1S4IF
X(X(X(r))){r} {} {} {r} {}*Yes
88
1Yes
R_sb9MnVHZ3rBbcuR
X(X(X(r))){r} {} {} {r} {}*Yes
89
1Yes
R_3GB8QvZ8WJ7imO7
X(X(X(r))){r} {} {} {r} {}*Yes
90
1Yes
R_3KSFgYqXtEshyWa
X(X(X(r))){r} {} {} {r} {}*Yes
91
1Yes
R_10pfB7n4SRWFEiA
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
92
1Yes
R_1d6RMi3t0gkcM5M
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
93
1Yes
R_e8QLXoeqrle4xS9
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
94
1Yes
R_z5PVfNBIPKX5S4F
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
95
1Yes
R_1E0ztwRbCQeEjAn
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
96
1Yes
R_d6w1dXzacJ9KN0Z
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
97
1Yes
R_3MuS1ttcj8sdoRv
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
98
1Yes
R_1hEGqpvvTZegzQ7
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
99
1Yes
All paths from all red states (2,3,5) satisfy X(X(X(Red)))
R_3gO868rLSy0R2Ep
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
100
1Yes
The formula G(Red => X(X(X(Red)))) is true, since at time t=2 we see a Red is on, and 3 time steps later at t=5, Red is also on. Similarly for when time t=3, a Red is on and 3 time steps later at t=6, Red is also on. From time step 5 onwards, Red is always on and hence the formula is trivially satisfied.
R_332amsBlozpU5n8
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
101
1Yes
On each state on which we have Red three positions from there we have Red (states 5+)
R_QcBt5TnllV1Ynq9
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
102
1Yes
R_2t50ks4XN0wxwLl
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
103
1Yes
R_0Vg19zBXa3hIGzf
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
104
1Yes
R_33wl4yU4sEzJKu4
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
105
1Yes
R_3iWIMn2gWx1V4Lp
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
106
1Yes
Red is on in the 2nd, 3rd and (5+n)th times for all n>=0, and so if Red holds at a given time, it will also hold 3 time steps later
R_W7oqYTh5ZL9V9vP
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
107
1Yes
R_3e8mdHJtEsNstwp
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
108
1Yes
R_1i3xVhhSYFfFwDR
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
109
1Yes
Every time Red is on, Red will be on three states later.
R_6F2ykLOE4e1S4IF
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
110
1Yes
R_sb9MnVHZ3rBbcuR
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
111
1Yes
R_3GB8QvZ8WJ7imO7
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
112
1Yes
R_3KSFgYqXtEshyWa
G(r => X(X(X(r))))
{} {rgb} {rgb} {} {rgb}*
Yes
113
1Yes
R_10pfB7n4SRWFEiA
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
114
1Yes
R_1d6RMi3t0gkcM5M
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
115
1Yes
R_e8QLXoeqrle4xS9
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
116
1Yes
R_z5PVfNBIPKX5S4F
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
117
1Yes
R_1E0ztwRbCQeEjAn
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
118
1Yes
R_d6w1dXzacJ9KN0Z
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
119
1Yes
R_3MuS1ttcj8sdoRv
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
120
1Yes
R_1hEGqpvvTZegzQ7
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
121
1Yes
State 4 releases the until. Before state 4 is reached red is satisfied by all paths from 1
R_3gO868rLSy0R2Ep
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
122
1Yes
The formula states that Red is on until the first occurrence of Green being on, which is true in this case as Green is first on at t=4, and from t=1 to t=3, Red is always on.
R_332amsBlozpU5n8
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
123
1Yes
State 1 to 3 are Red and then we have Green
R_QcBt5TnllV1Ynq9
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
124
1Yes
R_2t50ks4XN0wxwLl
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
125
1Yes
R_0Vg19zBXa3hIGzf
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
126
1Yes
R_33wl4yU4sEzJKu4
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
127
1Yes
R_3iWIMn2gWx1V4Lp
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
128
1Yes
Green is first on in the 4th step, and before that red always holds
R_W7oqYTh5ZL9V9vP
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
129
1Yes
R_3e8mdHJtEsNstwp
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
130
1Yes
R_1i3xVhhSYFfFwDR
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
131
1Yes
Red is on until Green is on, which does occur.
R_6F2ykLOE4e1S4IF
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
132
1Yes
R_sb9MnVHZ3rBbcuR
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
133
1Yes
R_3GB8QvZ8WJ7imO7
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
134
1Yes
R_3KSFgYqXtEshyWa
r U g
{rb} {rb} {rb} {rgb} {b}*
Yes
135
1Yes
R_10pfB7n4SRWFEiA
F(r) & F(g){} {g} {} {} {r}*Yes
136
1Yes
R_1d6RMi3t0gkcM5M
F(r) & F(g){} {g} {} {} {r}*Yes
137
1Yes
R_e8QLXoeqrle4xS9
F(r) & F(g){} {g} {} {} {r}*Yes
138
1Yes
R_z5PVfNBIPKX5S4F
F(r) & F(g){} {g} {} {} {r}*Yes
139
1Yes
R_1E0ztwRbCQeEjAn
F(r) & F(g){} {g} {} {} {r}*Yes
140
1Yes
R_d6w1dXzacJ9KN0Z
F(r) & F(g){} {g} {} {} {r}*Yes
141
1Yes
R_3MuS1ttcj8sdoRv
F(r) & F(g){} {g} {} {} {r}*Yes
142
1Yes
R_1hEGqpvvTZegzQ7
F(r) & F(g){} {g} {} {} {r}*Yes
143
1Yes
Both conjuncts are true
R_3gO868rLSy0R2Ep
F(r) & F(g){} {g} {} {} {r}*Yes
144
1Yes
Yes, as both Red and Green is eventually on starting from t=1, where Red is on at t=5 onwards and Green is on at t=2.
R_332amsBlozpU5n8
F(r) & F(g){} {g} {} {} {r}*Yes
145
1Yes
We eventually get Red (state 5) and eventually get Green (state 2)
R_QcBt5TnllV1Ynq9
F(r) & F(g){} {g} {} {} {r}*Yes
146
1Yes
R_2t50ks4XN0wxwLl
F(r) & F(g){} {g} {} {} {r}*Yes
147
1Yes
R_0Vg19zBXa3hIGzf
F(r) & F(g){} {g} {} {} {r}*Yes
148
1Yes
R_33wl4yU4sEzJKu4
F(r) & F(g){} {g} {} {} {r}*Yes
149
1Yes
R_3iWIMn2gWx1V4Lp
F(r) & F(g){} {g} {} {} {r}*Yes
150
1Yes
Red is on after 4 steps and green is on after 1 step
R_W7oqYTh5ZL9V9vP
F(r) & F(g){} {g} {} {} {r}*Yes
151
1Yes
R_3e8mdHJtEsNstwp
F(r) & F(g){} {g} {} {} {r}*Yes
152
1Yes
R_1i3xVhhSYFfFwDR
F(r) & F(g){} {g} {} {} {r}*Yes
153
1Yes
Eventually Green will be on and eventually Red will be on.
R_6F2ykLOE4e1S4IF
F(r) & F(g){} {g} {} {} {r}*Yes
154
1Yes
R_sb9MnVHZ3rBbcuR
F(r) & F(g){} {g} {} {} {r}*Yes
155
1Yes
R_3GB8QvZ8WJ7imO7
F(r) & F(g){} {g} {} {} {r}*Yes
156
1Yes
R_3KSFgYqXtEshyWa
F(r) & F(g){} {g} {} {} {r}*Yes
157
1Yes
R_10pfB7n4SRWFEiA
X(X(F(r))){rgb}*Yes
158
1Yes
R_1d6RMi3t0gkcM5M
X(X(F(r))){rgb}*Yes
159
1Yes
R_e8QLXoeqrle4xS9
X(X(F(r))){rgb}*Yes
160
1Yes
R_z5PVfNBIPKX5S4F
X(X(F(r))){rgb}*Yes
161
1Yes
R_1E0ztwRbCQeEjAn
X(X(F(r))){rgb}*Yes
162
1Yes
R_d6w1dXzacJ9KN0Z
X(X(F(r))){rgb}*Yes
163
1Yes
R_3MuS1ttcj8sdoRv
X(X(F(r))){rgb}*Yes
164
1Yes
R_1hEGqpvvTZegzQ7
X(X(F(r))){rgb}*Yes
165
1Yes
All paths from initial state are in state 3 after 2 transitions, at which point Red is satisfied, so F(Red) is
R_3gO868rLSy0R2Ep
X(X(F(r))){rgb}*Yes
166
1Yes
Yes, as Red is always on forever starting from the initial state t=1, and hence the formula is trivially satisfied.
R_332amsBlozpU5n8
X(X(F(r))){rgb}*Yes
167
1Yes
From state 3 we eventually get Red (right in state 3)
R_QcBt5TnllV1Ynq9
X(X(F(r))){rgb}*Yes
168
1Yes
R_2t50ks4XN0wxwLl
X(X(F(r))){rgb}*Yes
169
1Yes
R_0Vg19zBXa3hIGzf
X(X(F(r))){rgb}*Yes
170
1Yes
R_33wl4yU4sEzJKu4
X(X(F(r))){rgb}*Yes
171
1Yes
R_3iWIMn2gWx1V4Lp
X(X(F(r))){rgb}*Yes
172
1Yes
Any formula with no negations will hold trivially when all lights are on at all times
R_W7oqYTh5ZL9V9vP
X(X(F(r))){rgb}*Yes
173
1Yes
R_3e8mdHJtEsNstwp
X(X(F(r))){rgb}*Yes
174
1Yes
R_1i3xVhhSYFfFwDR
X(X(F(r))){rgb}*Yes
175
1Yes
In the next next state (3rd), Red will be on.
R_6F2ykLOE4e1S4IF
X(X(F(r))){rgb}*Yes
176
1Yes
R_sb9MnVHZ3rBbcuR
X(X(F(r))){rgb}*Yes
177
1Yes
R_3GB8QvZ8WJ7imO7
X(X(F(r))){rgb}*Yes
178
1Yes
R_3KSFgYqXtEshyWa
X(X(F(r))){rgb}*Yes
179
1No
R_10pfB7n4SRWFEiA
r U b{r}*No
180
1No
R_1d6RMi3t0gkcM5M
r U b{r}*No
181
1No
R_e8QLXoeqrle4xS9
r U b{r}*No
182
1No
R_z5PVfNBIPKX5S4F
r U b{r}*No
183
1No
R_1E0ztwRbCQeEjAn
r U b{r}*No
184
1No
R_d6w1dXzacJ9KN0Z
r U b{r}*No
185
1No
R_3MuS1ttcj8sdoRv
r U b{r}*No
186
1No
R_1hEGqpvvTZegzQ7
r U b{r}*No
187
1No
The Until operator here requires F(Blue) to hold, which it does not
R_3gO868rLSy0R2Ep
r U b{r}*No
188
1No
No, as even though Red is on forever, but Blue is never on, and hence it does not satisfy the strong until operator.
R_332amsBlozpU5n8
r U b{r}*No
189
1No
We never get Blue because no state has Blue
R_QcBt5TnllV1Ynq9
r U b{r}*No
190
1No
R_2t50ks4XN0wxwLl
r U b{r}*No
191
1Nono blue
R_0Vg19zBXa3hIGzf
r U b{r}*No
192
1No
R_33wl4yU4sEzJKu4
r U b{r}*No
193
1No
R_3iWIMn2gWx1V4Lp
r U b{r}*No
194
1No
Strong until requires Blue to be on at some time step, but it is never on
R_W7oqYTh5ZL9V9vP
r U b{r}*No
195
1Yes
R_3e8mdHJtEsNstwp
r U b{r}*No
196
1No
R_1i3xVhhSYFfFwDR
r U b{r}*No
197
1Yes
Yes, Red is on in the first state.
R_6F2ykLOE4e1S4IF
r U b{r}*No
198
1No
R_sb9MnVHZ3rBbcuR
r U b{r}*No
199
1No
R_3GB8QvZ8WJ7imO7
r U b{r}*No
200
1No
R_3KSFgYqXtEshyWa
r U b{r}*No
201
1No
R_10pfB7n4SRWFEiA
F(G(r))
{} {rgb} {} {rgb} {}*
No
202
1No
R_1d6RMi3t0gkcM5M
F(G(r))
{} {rgb} {} {rgb} {}*
No
203
1No
R_e8QLXoeqrle4xS9
F(G(r))
{} {rgb} {} {rgb} {}*
No
204
1No
R_z5PVfNBIPKX5S4F
F(G(r))
{} {rgb} {} {rgb} {}*
No
205
1No
R_1E0ztwRbCQeEjAn
F(G(r))
{} {rgb} {} {rgb} {}*
No
206
1No
R_d6w1dXzacJ9KN0Z
F(G(r))
{} {rgb} {} {rgb} {}*
No
207
1No
R_3MuS1ttcj8sdoRv
F(G(r))
{} {rgb} {} {rgb} {}*
No
208
1No
R_1hEGqpvvTZegzQ7
F(G(r))
{} {rgb} {} {rgb} {}*
No
209
1No
All paths from state 5 satisfy G(Not(Red)), and state 5 is reachable from state 1
R_3gO868rLSy0R2Ep
F(G(r))
{} {rgb} {} {rgb} {}*
No
210
1No
No, as F(G(Red)) means that there exists t>0 such that every time step from t onwards, Red will always be on, which is not the case as from t=5 onwards, no lights are ever on.
R_332amsBlozpU5n8
F(G(r))
{} {rgb} {} {rgb} {}*
No
211
1No
We do not eventually get always red because from state 5 we do not have Red
R_QcBt5TnllV1Ynq9
F(G(r))
{} {rgb} {} {rgb} {}*
No
212
1No
R_2t50ks4XN0wxwLl
F(G(r))
{} {rgb} {} {rgb} {}*
No
213
1No
you end up in state 5 forever where red doesn't hold .
R_0Vg19zBXa3hIGzf
F(G(r))
{} {rgb} {} {rgb} {}*
No
214
1No
R_33wl4yU4sEzJKu4
F(G(r))
{} {rgb} {} {rgb} {}*
No
215
1No
R_3iWIMn2gWx1V4Lp
F(G(r))
{} {rgb} {} {rgb} {}*
No
216
1No
There is no state after which Red is always on
R_W7oqYTh5ZL9V9vP
F(G(r))
{} {rgb} {} {rgb} {}*
No
217
1No
R_3e8mdHJtEsNstwp
F(G(r))
{} {rgb} {} {rgb} {}*
No
218
1No
R_1i3xVhhSYFfFwDR
F(G(r))
{} {rgb} {} {rgb} {}*
No
219
1No
It is not eventually the case that Red is always on.
R_6F2ykLOE4e1S4IF
F(G(r))
{} {rgb} {} {rgb} {}*
No
220
1No
R_sb9MnVHZ3rBbcuR
F(G(r))
{} {rgb} {} {rgb} {}*
No
221
1No
R_3GB8QvZ8WJ7imO7
F(G(r))
{} {rgb} {} {rgb} {}*
No
222
1No
R_3KSFgYqXtEshyWa
F(G(r))
{} {rgb} {} {rgb} {}*
No
223
1Yes
R_10pfB7n4SRWFEiA
F(r => g){}*Yes
224
1Yes
R_1d6RMi3t0gkcM5M
F(r => g){}*Yes
225
1Yes
R_e8QLXoeqrle4xS9
F(r => g){}*Yes
226
1Yes
R_z5PVfNBIPKX5S4F
F(r => g){}*Yes
227
1Yes
R_1E0ztwRbCQeEjAn
F(r => g){}*Yes
228
1Yes
R_d6w1dXzacJ9KN0Z
F(r => g){}*Yes
229
1Yes
R_3MuS1ttcj8sdoRv
F(r => g){}*Yes
230
1Yes
R_1hEGqpvvTZegzQ7
F(r => g){}*Yes
231
1Yes
The implication is vacuously true in state 1, as it does not satisfy Red
R_3gO868rLSy0R2Ep
F(r => g){}*Yes
232
1Yes
Yes, as F(Red => Green) can be converted into F(!Red || Green), and we can see that !Red is satisfied for all time steps, and hence the formula is trivially satisfied.
R_332amsBlozpU5n8
F(r => g){}*Yes
233
1Yes
This is equivalent to F(Green or !Red) and this is true because state 1 does not have Red
R_QcBt5TnllV1Ynq9
F(r => g){}*Yes
234
1Yes
R_2t50ks4XN0wxwLl
F(r => g){}*Yes
235
1Yes
red never holds
R_0Vg19zBXa3hIGzf
F(r => g){}*Yes
236
1Yes
R_33wl4yU4sEzJKu4
F(r => g){}*Yes
237
1Yes
R_3iWIMn2gWx1V4Lp
F(r => g){}*Yes
238
1Yes
The implication is vacuously true as Red is never on, and so the outer F operator will hold trivially
R_W7oqYTh5ZL9V9vP
F(r => g){}*Yes
239
1Yes
R_3e8mdHJtEsNstwp
F(r => g){}*Yes
240
1Yes
R_1i3xVhhSYFfFwDR
F(r => g){}*Yes
241
1Yes
Vacuously. Eventually (in fact, always) Red being on implies Green being on.
R_6F2ykLOE4e1S4IF
F(r => g){}*Yes
242
1Yes
R_sb9MnVHZ3rBbcuR
F(r => g){}*Yes
243
1Yes
R_3GB8QvZ8WJ7imO7
F(r => g){}*Yes
244
1Yes
R_3KSFgYqXtEshyWa
F(r => g){}*Yes
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
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280