ABCDEFGHIJKLMNOPQRSTUVWXYZAA
1
Totals ==>8640000000000090
2
Ok?BPBSI
BSQ
IFIGOIWUEUXXRV
Scope
AnswerExplanationIDFormulaTraceExpected
3
1
Because it requires to have at least another instant.
R_3zr1TPsbL8Fh01g
X(G(r)){r}too short
4
1Trace is too short.
R_7t8OynItMvsY3qS
X(G(r)){r}too short
5
1
Because the trace has only one state, and the next operator requires at least two states to be present.
R_5pl5ovreUYilD5z
X(G(r)){r}too short
6
1
This trace is to short (X operator is used on the last state, which always yields false)
R_6ffI4bTyX5zoEHh
X(G(r)){r}too short
7
1
The formula requires a next state, but there is none
R_4mVF8KOBt6MErhn
X(G(r)){r}too short
8
1
Trace too short. Trace must have at least length 2.
R_2pKyoW6gWKFoQnZ
X(G(r)){r}too short
9
1
Trace is too short - requires trace of length at least 2
R_1kn76aEhDPLve1a
X(G(r)){r}too short
10
1
The trace is too short; no trace of length 1 can satisfy a formula with one X, which requires a trace of length at least 2.
R_5PdIR5FnMDzrVUu
X(G(r)){r}too short
11
1
The trace is too short. No traces of length 1 can satisfy this formula with X.
R_3p46SBAhKNhlNvS
X(G(r)){r}too short
12
1There is no next state
R_47BVZHw0XbaticF
X(G(r)){r}too short
13
1
Trace is too short. There is no next state.
R_51nrC3cVmGQGCbZ
X(G(r)){r}too short
14
1
The trace is too short. No traces of length 1 can satisfy this formula with one X.
R_2rjQI459Ka4U504
X(G(r)){r}too short
15
1
The trace is too short. There is no next state.
R_10WMbhmzVMMfCQ9
X(G(r)){r}too short
16
1Trace too short
R_3p2LrOwAHaKfYYG
X(G(r)){r}too short
17
1The trace is too short.
R_2EPSIkVcvTwusZ1
X(G(r)){r}too short
18
1
The trace is not long enough for a formula with an X.
R_6ZBMLj5oxGtGYyV
X(G(r)){r}too short
19
1
As there is no second state, any formula of the kind next phi (i.e. X(\phi)) is rejected
R_7gRsvfLpsHGFun8
X(G(r)){r}too short
20
1
Because the trace should be of length 2.
R_2wQrFCdtVBTz165
X(G(r)){r}too short
21
1
Because the formula requires that eventually Blue becomes true.
R_3zr1TPsbL8Fh01g
r U (!r & F(b)){r} {r} {}mismatch
22
1
Content mismatch. after red turns off blue never turn on.
R_7t8OynItMvsY3qS
r U (!r & F(b)){r} {r} {}mismatch
23
1
Because (!Red & F(Blue)) is not satisfied, since the blue light is never on.
R_5pl5ovreUYilD5z
r U (!r & F(b)){r} {r} {}mismatch
24
1
F Blue does not hold, as there are not blue states. This makes the whole until expression false
R_6ffI4bTyX5zoEHh
r U (!r & F(b)){r} {r} {}mismatch
25
1
the formula requires that, as soon as the red light turns off, the blue light must turn on at least once from that point on. but the blue light is never on in the trace.
R_4mVF8KOBt6MErhn
r U (!r & F(b)){r} {r} {}mismatch
26
1
Content mismatch: The formula requires that Blue is eventually true.
R_2pKyoW6gWKFoQnZ
r U (!r & F(b)){r} {r} {}mismatch
27
1
At some point in the trace, !Red & F(Blue) must be satisfied (strong until), but F(Blue) is never satisfied in this trace
R_1kn76aEhDPLve1a
r U (!r & F(b)){r} {r} {}mismatch
28
1
Content mismatch: the trace has no state where Blue is on, but the use of strong until implies a formula where any satisfying trace must have at least one state with Blue on (A satisfying trace of length 3 does exist, simply turn Blue on in this trace at state 3).
R_5PdIR5FnMDzrVUu
r U (!r & F(b)){r} {r} {}mismatch
29
1
Content mismatch: the state with Red off does not have Blue on
R_3p46SBAhKNhlNvS
r U (!r & F(b)){r} {r} {}mismatch
30
1
In state 3 Red no longer holds hence we require (!Red && F(Blue)) to be true. !Red is satisfied, however F(Blue) does not hold and hence the trace is invalid
R_47BVZHw0XbaticF
r U (!r & F(b)){r} {r} {}mismatch
31
1
The right side of the until never happens since the Blue light never turns on.
R_51nrC3cVmGQGCbZ
r U (!r & F(b)){r} {r} {}mismatch
32
1
Content mismatch: the trace has no Blue states, but the formula requires the last state to have a Blue light on.
R_2rjQI459Ka4U504
r U (!r & F(b)){r} {r} {}mismatch
33
1
Content mismatch: the trace has no Blue states, but the formula eventually requires at least one with Blue on (and Red off).
R_10WMbhmzVMMfCQ9
r U (!r & F(b)){r} {r} {}mismatch
34
1
content mismatch, blue would need to be on in state 3
R_3p2LrOwAHaKfYYG
r U (!r & F(b)){r} {r} {}mismatch
35
1(!Red & F(Blue)) is never true.
R_2EPSIkVcvTwusZ1
r U (!r & F(b)){r} {r} {}mismatch
36
1
This formula rejects this trace because the blue light is never on (though at some point it should be to satisfy F(Blue)).
R_6ZBMLj5oxGtGYyV
r U (!r & F(b)){r} {r} {}mismatch
37
1
For an until to be true, the second part (here !Red & F(Blue)) has to hold at some point. However, here it is not the case that blue is ever true.
R_7gRsvfLpsHGFun8
r U (!r & F(b)){r} {r} {}mismatch
38
1
Because in state 3 both red and blue are off, and it doesn't exist a state such that Blue is on (there is an exist)
R_2wQrFCdtVBTz165
r U (!r & F(b)){r} {r} {}mismatch
39
lots of questionable responses in here, Red must be on and off in state 3!
1
Because there are next two instants but in the latter Red is true.
R_3zr1TPsbL8Fh01g
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
40
1
Content mismatch. The red light never turns off.
R_7t8OynItMvsY3qS
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
41
1
The second part of the formula requires the third state to be not Red, which is not the case here.
R_5pl5ovreUYilD5z
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
42
1
Xw(Xw(!Red)) does not hold. Xw Xw refers to state 3, but over there Red holds, which makes the whole formula evaluate to false
R_6ffI4bTyX5zoEHh
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
43
1
The G(Red) part of the conjunction is satisfied. however, according to the formula if there is a third state, the red light must be off in it. In our trace, the red light is on in the third state.
R_4mVF8KOBt6MErhn
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
44
1
Content mismatch: From state 1, we must have that in 2 states !Red (if those states exist). i.e. State 3 should have !Red
R_2pKyoW6gWKFoQnZ
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
45
1
The trace doesn't satisfy Xw(Xw(!red)) as this would require either the third state to not exist (trace being 0,1, or 2 states long) or the third state not having the Red light on.
R_1kn76aEhDPLve1a
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
46
1
Content mismatch: the trace has Red at state 3, but the formula requires that Red is turned off at state 3 on any satisfying trace that has length at least 3 (A satisfying trace of length 3 does exist, simply turn Red off in this trace at state 3).
R_5PdIR5FnMDzrVUu
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
47
1
Content mismatch: Red is on in the third state.
R_3p46SBAhKNhlNvS
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
48
1
From the start there exists two next states hence in state 2 Xw(!Red) should hold and in state 3 !Red should b=hold, which it does not
R_47BVZHw0XbaticF
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
49
1
Trace is too long. This formula is never satisfied if there is a 3rd state.
R_51nrC3cVmGQGCbZ
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
50
1
No trace can satisfy this formula: the last state cannot both have the Red light on and off at the same time.
R_2rjQI459Ka4U504
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
51
1
Content mismatch: Red should be off in the third state.
R_10WMbhmzVMMfCQ9
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
52
1
content mismatch, the first part of the logical and is satisfied but the weak next operator can apply in state 1 where it is not satisfied.
R_3p2LrOwAHaKfYYG
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
53
1!Red is false in Xw(Xw()) state.
R_2EPSIkVcvTwusZ1
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
54
1
This formula rejects the trace because there is a trace content mismatch. For example, for the first state, it is red and all subsequent states are red. However, the formula also demands that if a state after next exists, it must be not red. For state 1, this isn't true, as in state 3, the red light is still on.
R_6ZBMLj5oxGtGYyV
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
55
1
Since there are 3 states, the double XW is the same as X i.e. requiring that in state 3 !Red is true; however in State 3, Red is true and thus !Red is false.
R_7gRsvfLpsHGFun8
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
56
also with X???1
G(Red) is satisfied, but after 2 steps red is still on. (it would have been correct also with X.)
R_2wQrFCdtVBTz165
G(r) & Xw(Xw(!r))
{r} {r} {r}too long
57
1
Because the formula requires Red to be false at the end of the trace.
R_3zr1TPsbL8Fh01g
F(G(!r)){g} {r}mismatch
58
1
Content mismatch. Red light is on in the last state.
R_7t8OynItMvsY3qS
F(G(!r)){g} {r}mismatch
59
1
Because there is no state where G(!Red) becomes true.
R_5pl5ovreUYilD5z
F(G(!r)){g} {r}mismatch
60
1
The formula would be satisfied iff the last state would not be red, but it is.
R_6ffI4bTyX5zoEHh
F(G(!r)){g} {r}mismatch
61
1
the formula needs the red light to be off from some point in the trace until the end. however, the red light is on in the last state so !Red is not true.
R_4mVF8KOBt6MErhn
F(G(!r)){g} {r}mismatch
62
1
Content mismatch: At least the final state must have !Red.
R_2pKyoW6gWKFoQnZ
F(G(!r)){g} {r}mismatch
63
1
The last state does not satisfy !Red, whereas the formula requires !Red to be satisfied for all states after a certain state - impossible with this trace.
R_1kn76aEhDPLve1a
F(G(!r)){g} {r}mismatch
64
1
Content mismatch: the formula implies that any satisfying trace must, at the very least, have Red turned off at the last state, but that is clearly not true on the given trace (A satisfying trace of length 2 does exist, simply turn Red off in state 2 on this trace).
R_5PdIR5FnMDzrVUu
F(G(!r)){g} {r}mismatch
65
1
Content mismatch: The Red is on in the last state, but the formula requires Red to be off.
R_3p46SBAhKNhlNvS
F(G(!r)){g} {r}mismatch
66
1
!Red is not true in state 2, hence G(!Red) is false and hence F(G(!Red)) is false
R_47BVZHw0XbaticF
F(G(!r)){g} {r}mismatch
67
1
There does not exist a state from which onwards the Red light is always off.
R_51nrC3cVmGQGCbZ
F(G(!r)){g} {r}mismatch
68
1
Content mismatch: there is no state such that, starting from there, the Red light is off and stays off along all the remaining states of the trace.
R_2rjQI459Ka4U504
F(G(!r)){g} {r}mismatch
69
1
Content mismatch: the trace should have finished with Red off.
R_10WMbhmzVMMfCQ9
F(G(!r)){g} {r}mismatch
70
1content mismatch
R_3p2LrOwAHaKfYYG
F(G(!r)){g} {r}mismatch
71
1
!Red is never always true at any point of the trace.
R_2EPSIkVcvTwusZ1
F(G(!r)){g} {r}mismatch
72
1
This formula reject this trace because in the final state the red light it on (i.e. content mismatch). It is not true that eventually the red light is always off.
R_6ZBMLj5oxGtGYyV
F(G(!r)){g} {r}mismatch
73
1
It is never the case, that all future states are not red, since state 2 is red.
R_7gRsvfLpsHGFun8
F(G(!r)){g} {r}mismatch
74
state 1?!1
Because it does not exist a state where at that point on red is off.
R_2wQrFCdtVBTz165
F(G(!r)){g} {r}mismatch
75
1
Because it requires Red to be true every 2 instants (not every 3 instants).
R_3zr1TPsbL8Fh01g
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
76
1
Only an infinite trace can satisfy this formula.
R_7t8OynItMvsY3qS
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
77
1
According to the second part of the formula, since state 1 has Red, state 3 should also have Red, but this is not the case.
R_5pl5ovreUYilD5z
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
78
1
The G(Red => X X Red) part does not hold. In particular it must hold for last state which is red, and so X X Red must hold, but it is the last state, hence X ... yields false
R_6ffI4bTyX5zoEHh
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
79
1
the formula says that whenever the red light is on, it must be on again after 2 steps. In the trace, the red light is on in state 1 so it should be on in state 3 as well, but it is not.
R_4mVF8KOBt6MErhn
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
80
more than "too short"
1
Trace is too short. Since the final state is Red, we must have X(X(Red)) from State 7. Since this is Strong Next, and there are no next states, then it rejects.
R_2pKyoW6gWKFoQnZ
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
81
1
Two reasons 1. As the formula uses the strong next operator, this formula con only be satisfied by an infinite trace 2. The formula requires that whenever the Red light is on, it is on again in two states, this is not the case for the first state, which Red is on, but Red is not on in two states (state 3)
R_1kn76aEhDPLve1a
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
82
1
There is a content mismatch, as Red should be on at state 3, but more importantly only an infinite trace can satisfy this formula (Red on at state 1, so state 3 must exist and Red is on at it, so state 5 must exist and Red is on at it etc).
R_5PdIR5FnMDzrVUu
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
83
1
Trace too short: Red is on in the last state which mean there must be two next states after it with the requirement of the second one to be Red, but there's no such state in this trace.
R_3p46SBAhKNhlNvS
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
84
1
State 1 is Red so we'd expect the next next state (i.e. state 3) to be red; this is not the case
R_47BVZHw0XbaticF
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
85
1
This formula cannot by satisfied by a finite trace.
R_51nrC3cVmGQGCbZ
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
86
1
Only an infinite trace can satisfy the formula, because of the use of the strong next operator within the eventually clause: since the last state of the trace has Red on, then it should be the case that after 2 transitions, Red is also on, which can never happen, since the trace is finite.
R_2rjQI459Ka4U504
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
87
1
Content mismatch: Red should be on in states 3 and 6. Also, Red in state seven could never satisfy the condition of the strong nexts!
R_10WMbhmzVMMfCQ9
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
88
more than "too short"
1
trace too short, the implication is not satisfied by state 7 because it uses strong next instead of weak next
R_3p2LrOwAHaKfYYG
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
89
1
3rd element doesn't satisfy Red even though the 1st element does.
R_2EPSIkVcvTwusZ1
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
90
1
I think only an infinite trace can satisfy this. We have a problem in state 7 because X demands that another state exist and there are no more states beyond 7.
R_6ZBMLj5oxGtGYyV
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
91
1
In the first state, Red is true, however state 3 has !Red, but this would be required from G(Red => X(X(Red))
R_7gRsvfLpsHGFun8
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
92
1
Because in state 7 Red is on, and it does not exist a next step.
R_2wQrFCdtVBTz165
r & G(r => X(X(r)))
{r} {} {} {r} {} {} {r}
inf only [TYPO in formula]
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266