Yes or No: Trace Satisfaction
Welcome
This file contains anonymized, hand-labeled responses to the third and final segment of our Spring 2021 LTL/Electrum Survey.
- There are 9 sections below, one for each question in the survey.
- Each section presents six parts:
- an Electrum formula,
- a 5-state description of an infinite trace,
- the correct answer,
- the number of correct and incorrect answers,
- a one-sentence takeaway about the main misconception that the question addresses, and
- labeled responses.
- Labels have up to two parts: (1) is the answer correct?; (2) what misconception, if any, does the answer show?
- Click on any label to show the responses.
- Responses have four parts:
- Answer
- Anonymized learner ID
- Explanation
Participants received the following instructions:
Recall that Electrum formulas are statements about traces (sequences of
states). A non-trivial formula is satisfied by some traces, but not all.
The following questions present a formula and ask whether it is satisfied by a
specific trace. For each, choose Yes or No and briefly explain which part(s) of
the trace motivated your answer.
Notation for states and traces:
- {} is the empty state, where all lights are off
- {RGB} is the state where all lights are on
- in {G} only the Green light is lit
- ... and so on, for other combinations of R G and B
- {} {} {} {R} {} is a trace in which the Red light is lit in the fourth state, and all lights are off at other times
- all traces below show exactly five states; you can assume the final state repeats forever
Template:
HIDDEN FOR DOUBLE BLIND REVIEW
Example Q/A:
Q. Is the formula
eventually { Blue in Panel.lit }
satisfied by trace {} {RG} {RG} {RG} {G} ?
A. No
Why? Because the Blue light never turns on --- neither in the first four
states nor in the final lasso state.
Q6.3
formula: | Red in Panel.lit |
trace: | {GB} {RGB} {RGB} {RGB} {RGB} |
answer | No |
49 / 57 correct (86%)
8 students (14%) fell for implicit F, most students understand question format
correct, N = 47
- No 7jm The first { GB } has no R in it
- No w4t The formula checks if Red light is on in the first state, which it isn't.
- No 1vwv Red is not on in the current state.
- No 31qw By default it refers to the first state, and in that state red is not on.
- No 7opy Red is not on in the first state
- No 917u The red light is not on in the first state.
- No DzoD Because R is off in first state
- No KXjz In the first state, the red light is not lit.
- No XU9x Because the Red light is not lit in the initial state.
- No bx1r The red light is not on in the first state, which is the default state for the predicates when there are no temporal modifiers
- No bxd4 Because the first state has Red to be turned off.
- No c1fn The lack of any temporal keywords implies the formula only applies for the first state. However, the first state { GB } does not contain R, meaning that Red is not in Panel.lit in the first state, so the formula is not satisfied.
- No c9jy The red light is not on in the first state; there are no LTL operators applied to indicate that the formula is referring to any future states.
- No cirf The formula requires that the red light be turned on in the first state, and this is not the case in the given trace.
- No clU0 The Red light is not lit in the first state.
- No duq8 Red in Panel.lit is an event that happens in the first state, but the first state only has green and blue.
- No dznh Because the formula is not temporal-ly quantified and the Red light is not on in the first state ({GB}).
- No ekny This formula only constrains the first state, where R is not present.
- No eolW Red is not on in first trace
- No fojf The Red light is not on in the first state, and the first state is the one referenced by the formula.
- No fsqw The red light is not on in the initial state.
- No gbrV Red isn't present in the first state
- No hDZG Red is not lit in the first state.
- No iVk5 Red is not in Panel.lit in the first state
- No j7t9 Because the formula is about the first state, where Red light is not on
- No kkzx Red light is not on in the initial state.
- No kuaa Because Red is not in Panel.lit in the first state
- No m0p6 In the first state, red is not in the panel.lit.
- No mxad The formula is referring to the very first state of the trace, {GB}. Since Red is not lit in that state, it is not a satisfying trace (even though it is lit in every other state).
- No n4vd I assume if no electrum keywords are present then the predicate is applied to the first instance.
- No ndij Red light is not on in the first state.
- No nieh Because the first state does not have the Red light on.
- No nr65 Red is not lit in the initial state.
- No pfdd Red is not lit in the first state.
- No ptyk This is not satisfied because the formula only refers to the initial state, not to all states in the trace, and Red is not on in the initial state. This would be satisfiable if the formula read: eventually {Red in Panel.lit}.
- No qMyf In the first step, Red is not lit. Since there are no temporal operators in the formula, we're only considering that stpe.
- No qjpx Because Red light is not lit in the first state
- No rnC9 Red is not on in the first state
- No s0nv This formula only talks about the first trace, and we see that red is NOT lit in the first state {GB}.
- No sain Red not in first state {GB}
- No vz2g Red is not on in the current (first) state.
- No xeec Red is not on in the first state
- No xqZA Red is not on in the first state, which is the only one checked.
- No ydkm Red is not in the first state's lit colors.
- No yoxy State 1 doesn't have Red lit.
- No yypq Red in Panel.lit only applies to the first state in the trace, where R is not present.
- No zf66 In the first state, the red light is not lit.
correct, implicit G, N = 1
- No uOG8 There is a trace where Red is not lit
correct, guessed, N = 1
- No j9mq I'm not sure how statements that don't use any temporal operators should be interpreted.
incorrect, implicit F + misunderstood question format?, N = 8
- Yes 3r0 Red is on in one states.
- Yes 55oa Even though the first state does not have Red in it, the formula never specifies that Red has to be lit in every state, so the fact that it is lit in some states is enough to satisfy.
- Yes 8E1h The Red light was present in lit in at least one trace.
- Yes N3Cw I think the formula says that the red light has to be on in at least one state, which is satisfied by the {RGB} states.
- Yes XLuy Red is in the last four traces.
- Yes able Because it doesn't talk about red always lit
- Yes rahc There is at least one state in which the red light is on.
- Yes tyos It doesn't say always so I just looked to find a single trace with `Red in Panel.lit` - e. g. {RGB}
Q6.5
formula: | after { after { after { Red in Panel.lit } } } |
trace: | {R} {} {} {R} {} |
answer | Yes |
54 / 57 correct (95%)
no student thinks that X spans
correct, N = 46
- Yes 3r0 After the first three states, red light is on.
- Yes 7jm The fourth part of the trace is { R }, which contains a R.
- Yes w4t It is stating that in the fourth state the Red light is on, which is true of the trace.
- Yes 1vwv Red is on in the 3rd state in the future (the 4th state).
- Yes 31qw Red is on in the fourth state
- Yes 7opy The Red light is on in the forth state (after three steps from the first state)
- Yes DzoD Because R is lit in 4th state and there are 3 "after" statements
- Yes N3Cw In state 1, the formula is satisfied because three states after, the Red light turns on.
- Yes XLuy Red is in the third state after the first.
- Yes XU9x after {after { after { x }}} is checking for x three states after the initial, which is the 4th state. The red light is lit in the 4th state.
- Yes able The red should be there after the first three states which it is. voila!
- Yes bx1r Red light is lit in the fourth state, which is 3 states after the initial state
- Yes bxd4 Because the after after after state of the current state, the red light is also turned on.
- Yes c1fn The use of 3 'after' keywords makes the formula imply that the red light should be on 3 states after the present one. The state 3 states after the first, according to the trace, is { R }, where the red light is on, thereby satisfying this requirement.
- Yes c9jy The red light is on in the fourth state (indicated by three "afters").
- Yes cirf This formula requires that the red light be turned on in the fourth state, which is satisfied by this trace.
- Yes clU0 The Red light is lit in the fourth state, i.e., three states after the first state.
- Yes duq8 The statement never said anything about the first state, so whatever value in it would suffice, the same applies to 2nd, 3rd, and 5th. It only specified that the 4th state must have red.
- Yes dznh Because the Red light is on in the 4th state of the trace (3 states after the initial state).
- Yes ekny 3 afters is the 4th state, in which R is lit.
- Yes eolW Current trace + 3 after = 4th Trace, Red is on
- Yes fojf The Red light is on the third state.
- Yes fsqw The red light is on in the fourth state.
- Yes hDZG Red is lit in the state after the state after the state after the first state (i.e., the fourth state).
- Yes iVk5 This formula is satisfied by the fourth state. There are three states before the first state so the fourth state is (after after after) the first state and Red is in Panel.lit in the fourth state.
- Yes j7t9 Because Red light is on in the 4th state
- Yes m0p6 Red is in the 4th state
- Yes mxad Regardless of what happens in the first three states, as long as Red is lit in the fourth state, the trace should satisfy the formula. As such, this trace is valid.
- Yes n4vd The red light is lit in instance 4
- Yes ndij Red light is on in the 4th state.
- Yes nieh Because the Red light is on on the third state.
- Yes nr65 Red is lit in 3 states (so 4th state beginning on 1st state).
- Yes pfdd Red it lit in the 4th state.
- Yes ptyk Given the 3 afters in this formula, the statement Red in Panel.lit is being evaluated at state 4. In state 4, the Red light is on, so this formula is satisfied.
- Yes qjpx Red light is lit in the fourth state
- Yes rnC9 The red light is on three states after the initial state.
- Yes s0nv This formula says that three states after the first state, the red light will be on, which is represented in the trace, because the fourth state (3 states after the first) has the red light on.
- Yes sain Red in 4th state
- Yes uOG8 the fourth trace has red present
- Yes vz2g after is the 2nd, after after is the 3rd, after after after is the 4th. Red is lit in the 4th.
- Yes xeec Red is on in the fourth state
- Yes xqZA First state is 1, after 3 time steps we are in state 4 with Red on.
- Yes ydkm In the 4th state, 3 states after state 0, the Red is lit.
- Yes yoxy Red is lit in state 4.
- Yes yypq The formula states that Red will be lit after 3 time steps, which is the 4th state.
- Yes zf66 The red light is lit in the fourth state.
correct, implicit F, 4
- Yes 55oa Again, since this formula never specifies that it should hold in all states, the fact that it holds in states 2-4 is sufficient
- Yes 8E1h Red was lit in one state that is 3 states after some state.
- Yes kuaa It can be! If we are starting on the first trace, then yes, because it appears three traces later (which is the number of afters)
- Yes rahc There is a state whose third state afterwards has the red light on in that state.
correct, miscount X, N = 4
- Yes 917u The red light is on three states from the initial state.
- Yes j9mq The state 2 states after the initial state has red lit.
- Yes qMyf We're looking at the 4th time step (3 afters); at which point, Red is lit.
- Yes tyos {R} plus 3 jumps gets you back to {R}
incorrect, implicit G, N = 1
- No gbrV For the second step in the trace this doesn't hold as after after after is none
incorrect, mis-count X + confused off / on?, N = 2
- No KXjz The red light is not lit in the fourth state.
- No kkzx I think the formula defines that the red light to be on in the third state, but the trace didn't have the red light on in the third state.
Q6.7
formula: | always { Red in Panel.lit implies after { after { after { Red in Panel.lit } } } } |
trace: | {} {RGB} {RGB} {} {RGB} |
answer | Yes |
50 / 57 correct (88%)
12 students (21%) do not understand that the final trace state is the lasso
correct, N = 43
- Yes w4t Since the formula uses implies, it is true when Red is not lit, so States 1 and 4 of the trace matches the formula. For State 2 and 3 (where the Red light is on), the formula checks if the Red light is on 3 states later. Since the last state is a lasso state, it will always be Red afterwards, so both State 2 and 3 will have a Red light on 3 states later.
- Yes 1vwv Red is on in the 2nd state, so it must be on 3 states in the future (the 5th state), and it is. Red is on in the 3rd and 5th states, so it must be on in the 6th and 8th states, respectively. It is on in these states because the last state, which has Red on, repeats forever. Red is on in all states after the 5th state so the formula is always satisfied.
- Yes 31qw All of the states that have R on have R on also in the next of next of next state
- Yes 55oa The formula is saying that whenever there is a Red light, there will be another one 3 states away. The first Red light is trace 2 and 3 states away there is also one. There is also a Red light 3 traces away from state 3. Finally, we lasso on a state with the Red light lit so this will always hold.
- Yes 7opy Whenever Red is on in state i, the Red light is on again in state i+3.
- Yes 8E1h It is always true that Red is lit in a state 3 states after a state with Red lit. This is because for state 2, Red is lit in state 5, and for state 3 and 5, since state 5 repeats forever, Red is lit in all those states.
- Yes 917u Whenever the red light is on the red light is also on three states from now.
- Yes XLuy Red is in the second trace and in the fifth trace it is also turned on.
- Yes XU9x Because the last state lassos, even state 2 and 5 will also be satisfy this predicte even though their "3 states away" state will be {RGB}
- Yes bx1r Once the red light turns on, then it must be on again three states after that
- Yes bxd4 The second state has Red light on, and the fifth state has a red light on. The third state has the REd light on, and since the fifth state will repeat, it will also get a state with a red light on. Same for the fifth state and so on.
- Yes c1fn The formula implies that whenever the red light is on in a given state, then it will also be on in the state that occurs 3 states away from this original state. Consider the cases in which red is on in the given trace. 1. Second state: R is present in the second state as well as the fifth state, which occurs 3 states afterwards, thereby satisfying the formula. 2. Third state: The last state { RGB } is repeats indefinitely, which implies the sixth state (occurring 3 states after the third) also contains R. 3. Fifth state: Again, since { RGB } repeats indefinitely at the end, the eighth state, occurring 3 states after the fifth, also contains R. Therefore, this formula is always satisfied.
- Yes c9jy The always indicates that the formula holds at any state. In this case, the Red in Panel.lit implies... part of the formula applies at the second state, at which R is lit three states later (in the fifth state).
- Yes cirf Whenever the red light is turned on, it is on again after three transitions, and the infinite repetition at the end ensures that the red light will always be on for all states after the fourth state of this trace, which is all that is needed since the red light is not on in the first state.
- Yes clU0 Whenever the red light is on, it is also on three states in the future. The red light is on in the second state and also in the fifth state, in the third state and also in the sixth state (as part of the lasso), and whenever it is on in the lasso it will also be on three steps later in the lasso.
- Yes duq8 In the 2nd state, the color is red, so the 5th state must be red, which is true. In the 3rdstate, the color is red, so the 6th state must be red, which is true because the final state repeats forever. Since the repeated state has red, this applies to the 5th state too and so on.
- Yes dznh because for each element of the trace (including the final lasso state) : 1. the Red light is off 2. Or the red light is on and the red light 3 states 'later' is on.
- Yes ekny This formula states that for every state where Red is lit, 3 states after, Red will still be lit. In the 2nd state, Red is lit, and is satisfied by the 5th state where it is still lit. As the last state repeats forever, all states after are satisfied too.
- Yes eolW Whenever Red is on, the 3rd state after that should have Red on. Since the last state that repeats forever has Red on, the 3rd state for every trace with red on is RGB.
- Yes fojf For each state in which the Red light is on, the Red light is on in the state that is three states in the future. Additionally, the last state repeats forever with the Red light on.
- Yes fsqw The red light is on in the second state, and correspondingly, it is on in the fifth state, three states later. The red light is on in the third state, and since the final state repeats, it is also on three states later. Since the final state repeats, whenever the red light is on in a state after and including the fifth state, it will be on three states later.
- Yes gbrV In the first state the LHS of the implies is false so the expression is true. In the second state after after after is RGB so true implies true = true. The last state keeps repeating so the RHS will always be true from this point on
- Yes hDZG In every state where Red is on, Red is also on three states in the future (repeating the end state as many times as necessary to get there).
- Yes kkzx The formula defines that when the red light is on in (i)th state, it is on again in (i+3)th state. State i+1 and i+2 doesn't have any restrictions. From the trace we can see that the red light was on in the second state and then on again in the fifth state.
- Yes m0p6 When there is red, it is on 3 states after
- Yes mxad The formula shows that the first time Red is lit (the second state) means that in the sixth state, Red must also be lit. This also means that every four states, starting from the second, Red will be lit. Since the fifth state shown, {RGB}, continues to infinity, Red is lit in every state to infinity, and as such it will necessarily be lit in every fourth state, starting with the second.
- Yes n4vd Since the last instance repeats forever, all instances with a red light also have a red light lit three instances later.
- Yes ndij If the red light is on, then the red light should be on again after three states. The red light is on in the 2nd state, and is on in the 5th state. Because the final state repeats forever, the red light is on from the 5th state forward. So the red light is on in the 3rd state and again in the 6th state; on in the 5th state and again in the 8th state.
- Yes nieh Because when the Red light is on, the Red light is on again in every 3 other state.
- Yes nr65 Whenever Red is lit, it will be lit in 3 states. Red lit in state 2 requires Red to be lit in state 5 (yes). Red lit in state 3 requires Red to be lit in state 6 (yes - a lasso state). After state 5, Red is always lit anyways.
- Yes ptyk Because this formula begins with always, we must evaluate it for every state. Red in Panel.lit is true in states 2, 3, 5, and beyond. At state 2, after after after refers to state 5. In state 5, the Red light is on, so this is satisfied. In state 3, after after after refers to state 6, which is equivalent to state 5, so again this is satisfied. This is also true in state 5. In state 5, after after after refers to state 8, which is a lasso of state 5, so it is still satisfied. Red in Panel.lit will also be true in all states beyond state 5. Three states after all of these states will be replicas of state 5, so will contain Red in Panel.lit, meaning that this will always be satisfied.
- Yes qMyf We note that the last state repeats forever. In the 2nd step, we hope that In the 5th step, R will be lit, and it is. In the 3rd, we hope R is lit in the 6th, and it is. For all steps i, i >= 5, we hope that it is lit in step i+3; since it is always lit after step 5, this holds.
- Yes qjpx Any Red state is followed by a Red state three states after, including the final lasso state
- Yes rahc It is always the case that if red light is on in one state, the red light is also on in the third following state.
- Yes rnC9 For all states with the red light on, three states later it is also on. For all of the states this is covered by the lasso.
- Yes s0nv Now, we generalize to saying that at any point in time, if the red light is on, then three states afterward, the red light will be on. We see that in the second state, the red light is on, and we get that it is also on in the fifth state. Since the fifth state repeats forever, we see that any state i with a red light after the second state will have state i + 3 also having the red light on.
- Yes sain if a state has red in it, the state 3 changes after also has red because the final state repeats forever
- Yes uOG8 when red was present it was present three traces later
- Yes vz2g Assuming the last state is the one that repeats forever as stated above in the constraints, then any state with Red lit will always have a 3rd following state that also has Red lit.
- Yes xeec Red is on in the second state, so it must be on in the state three later which it is. It is also on in the third state, and since the final state repeats forever, then this is satisfied.
- Yes xqZA Each Red being on state has Red on in the state 3 time steps later.
- Yes yoxy Red is lit in the second state, and also lit in the third state from the second (state 5).
- Yes zf66 In the three states (second, third, and last) that have the red light lit, the state that comes three states later also has the red light lit.
correct, misunderstood question?, N = 2
- Yes kuaa Red is not always in Panel.lit; therefore, any set of conditions can be true
- Yes 3r0 For the given states, if in the second state red light is on, in the fifth state it must on.
correct, wrong lasso, N = 5
- Yes DzoD State 2 has it's "afters" satisfied, the other 2 where red is lit don't need to since there are only 5 total states.
- Yes KXjz For all traces where red is lit, red is lit three states later unless a jump of three states exceeds the number of states checked.
- Yes N3Cw In any state where the Red light is on, the state three states after also has the red light on (if the final state loops).
- Yes tyos I'm not sure cause I can't tell where this trace is lasso'ing. I'm going to guess it lassos at the last {RGB}, hence this is true?
- Yes ydkm Since the traces repeat the lasso infinitely, the last two states that contain Red will meet this definition, and 3 states after the second state is the fifth state, which contains Red.
incorrect, wrong lasso, N = 7
- No 7jm Red is in the 3rd stage, but not in the 6th stage (which is equivalent to the 1st stage). Alternatively, how I first thought of it is gcd(3, 5) = 1, so the only trace this is true is the one where Red is in every trace.
- No able Because although there is a red for the first RGB but not for the second RGB
- No iVk5 the state that is three states after the third state (which has Red in Panel.lit) is the first state, which does not have Red in Panel.lit
- No j7t9 In the 3rd state, Red light is on, but 3 states after, it is not on
- No j9mq This doesn't hold when you consider the trace as a loop.
- No pfdd The 3rd state does not satisfy this, as going three states after the 3rd state gets us back to the 1st, where Red is off.
- No yypq The third state has the Red light lit up, but if we travel 3 time steps ahead of that state, we end up at the first state, which does not have the Red light lit up.
Q6.9
formula: | after { Red in Panel.lit } until after { Green in Panel.lit } |
trace: | {RB} {RB} {RB} {RGB} {B} |
answer | Yes |
44 / 57 correct (77%)
9 think until negates rhs
correct, N = 39
- Yes 3r0 In the thrid state, the red is on in fourth state and green is on in the fourth state, so that in the following state, the next state of current state red light will never on.
- Yes w4t For State 1 and 2, State 2 and 3 both have the Red light on, so it satisfies the requirement for formula-a. This requirement ceases to hold for State 3, where the state afterwards (State 4) has the Green light turned on. However, the trace can still meet this condition (just not required to), so the formula is still true for State 3. For State 4 and onwards, it no longer needs to meet any conditions, so anything afterwards is valid.
- Yes 1vwv Red is on in the 2nd state (`after` of the current state) and the 3rd state (the state whose `after` has Green on) so the `until` is satisfied. (I would change my answer to No if these statements are actually meant to be on separate lines).
- Yes 31qw Red is always on in the next state until green is on in the next state
- Yes 7opy The third state triggers the until because Green is on in the forth state {RGB}. And Red is on in all of the first four states.
- Yes KXjz yes, state 3 is when the "after { Green in Panel.lit }" predicate holds, and in all previous states the "after { Red in Panel.lit }" predicate holds.
- Yes N3Cw For all the states before state 3 (where the green light is on in the next state), each next state has red in it. Or, for 1 <= i < 3, state_{i + 1} has the red light on.
- Yes XU9x Even though {RGB} breaks the pattern (because {B} doesn't have Red), {RGB} contains G so the requirement to see R in subsequent states is over.
- Yes able because after the second R there is always are until G in the fourth state.
- Yes bx1r There is always a red light in the next state until there is also a green light in the next state, at which point there is neither in the next state.
- Yes c1fn The red light is on until the fourth state, when the green light also turns on. This implies the third state { RB } was the last time the condition before the 'until' keyword had to hold. As a consequence, the indefinite repetition of { B } afterwards satisfies the formula at hand.
- Yes c9jy The red light is on and holds for all states up until the one which "after {Green in Panel.lit}", which does not specify that the Red light has to turn off.
- Yes cirf The red light is required to be on in the next state until the green light is turned on in the next state. Given the technical way "until" works, I don't think red even needs to be on in the fourth state, but since the green light is on in the next state when we are at the third state, the red light does not have to be on for the rest of the trace, so it is fine that only the blue light is on after the fourth state. "Until" doesn't require fmla-a to be false once fmla-b is true, just that it holds as long as fmla-b has not yet become true.
- Yes clU0 For states 1 and 2, it is true that Red is lit in the next state, and in state 3, it is true that Green is lit in the next state. Together, these fulfill the obligations of the until statement.
- Yes duq8 The first part specifies that red should be in the state after the first state, which happens. And each state should have red until green shows up in the state after the until state. Since green showed up in the 4th state, this means that the 2nd and 3rd must have red, which they do.
- Yes dznh The Red light stays on in the trace from the 2nd state until the Green light comes on.
- Yes ekny The formula is saying that Red will be lit until the state after Green is lit. Red is lit in all the states except the fifth one, which is okay because in the 4th state Green is lit.
- Yes fojf At every state before the Green light is lit, Red is lit in the following state.
- Yes fsqw The red light is on in the second state and third state, which means "after { Red in Panel.lit }" is true in the first and second state. In the fourth state, the green light is lit, which means "after { Green in Panel.lit }" is true in the third state.
- Yes gbrV Red is always on until Green is turned on
- Yes hDZG Yes. The second part, “after { Green in Panel.lit }”, is satisfied by the third state (since Green is lit in the fourth state). In both of the states before this, “after { Red in Panel.lit }” is satisfied, because Red is lit in the second and third states (the “after”s of the first and second states). These two facts make the “until” true.
- Yes iVk5 (after Red in Panel.lit) holds in all states where the following state has Red in Panel.lit. This is true for the first 3 states. (after Green in Panel.lit) holds in the third state because Green is in Panel.lit in the 4th state. So, the LHS of the until is true until the RHS holds.
- Yes j9mq When green is in the next state, the predicate on red is released. That predicate also holds up to this point.
- Yes kkzx If initial state is i, the red light should be on from (i+1)th state to the (j)th state where the green light appears, and it should be off from the (j+1)th state. Other states are not specified. From the trace we see the red light was on until the last state - which is the (j+1)th state. So the formula is satisfied by this trace.
- Yes kuaa Because red must be in Panel.lit in the following state until Green is in Panel.lit in the next state. Since both conditions are met, it satisfies the trace
- Yes n4vd The red light is always lit in the next trace until the green light is lit in the next trace.
- Yes ndij The red light is on in the 2nd state, so the part before until (after { Red in Panel.lit }) is satisfied. The green light is on in the 4th state, and the red light is on in the 2nd, 3rd and 4th states.
- Yes pfdd It is true that from the first state until the third state (after which Green is lit) Red will be lit in the next state.
- Yes qMyf The formula basically says "Red will be lit in the next step until Green is lit in the next step". Green is lit in the 4th step; every step up to and including it, Red is lit, so this holds.
- Yes qjpx Red stay lit until the green light turns on
- Yes rnC9 State 3 satisfies the right side as in state 4 the green light is on, and for states 1 and 2 the red light is on in the next state. Thus the red light is on in the next state until the green light is on in the next state.
- Yes s0nv The red light is continually on until the state after the green light turns on, and once this event occurs, the formula expires and we are free to repeat the last state infinitely.
- Yes sain the two first states satisfy the condition that the next state has red in it
- Yes uOG8 until does not guarantee that statement A does not hold in traces after
- Yes xeec Because red is on in every state until the state after the first time green is on.
- Yes xqZA The requirement for Red to be on in the next time step is lifted in state 3, when Green is on in the next time step (state 4).
- Yes ydkm Yes, because for states 1 and 2, the next state has red as lit, and then in state 3 the after state (state 4 has green) so the obiligation ceases to hold. Red is still lit in state 4, but that is allowed as until does not care about the value of the first expression after the state where the second expression is met.
- Yes yypq The "after { Green in Panel.lit }" condition is fulfilled in the 3rd state and all states from the 1st state to the 3rd state fulfill the condition "after { Red in Panel.lit }"
- Yes zf66 The first three states satisfy the left side of the until. The fourth state satisfies the right side of the until, freeing the fifth and later states from any obligations.
correct, N = 5
- Yes DzoD because of the after statements, Red and Green can both be on one time
- Yes XLuy Red is not turned on in the fifth state once green turns on in the fourth.
- Yes bxd4 Because of the Green in the fourth state, the fifth state(repeated state) has no Red in it anymore.
- Yes mxad The formula writes that Red will be lit until the state after Green is lit. Since Red is lit in states 1-4, and Green is lit in state 4, Red must not be lit in any state after the 4th. Indeed, the 5th and lasso state continues as {B} forever, meaning Red is never lit after the 4th state, so the trace is satisfied.
- Yes tyos This might be a tautology.
incorrect, U forces lhs to become false, N = 9
- No 7jm Red is in the 4th state, but Green is too, but the statement says Red should dissappear when Green shows up.
- No 55oa I believe this formula is saying that red will continually be in lit until the green one is. As I interpret it, the red and the green should never be lit at the same time, which is exactly what's happening in trace 4.
- No 8E1h I don't really understand this one, but I think that since Red is lit in the after the state of a state for which Green is lit in the after state, then the "until" part doesn't hold.
- No eolW Red and Green shouldn't be on at the same time, the until will trigger at index 2 which means in index 3, Red should not be on.
- No j7t9 In the 4th state, Green light is already on but Red light is not off
- No m0p6 The 4th state has a red in it, but it should not because after {green in panel.lit} is true in the 3rd state, so the 4th state should not have a red
- No ptyk after {Green in Panel.lit} is true in state 3 because in state 4, the Green light is on. This means, in state 3, after {Red in Panel.lit} cannot be true if this statement were to be satisfiable. Since, however, we see that in state 4, the Red light is on, after {Red in Panel.lit} is true in state 3. Thus this formula is not satisfied by the trace.
- No vz2g State 3 has Red lit, but State 4 has Green lit. Red is supposed to turn off once the state after has Green lit.
- No yoxy Red should not be lit in states 3 or 4.
incorrect, missing U, N = 1
- No 917u The red light is on in the next state without the green light being on. I read the formula as after{Red in Panel.lit and Green in Panel.lit}
incorrect, implicit G, N = 2
- No nr65 At state 4, there is a state where neither are true. until requires that one remains true until the other becomes true. The next state is solely a Blue state.
- No rahc I'm not sure if I understand the formula... But I think it's saying that, a state whose following state has the red light on happens until a state whose following state has the green light on. The fourth state has the red light on, but there is no state after it (including itself) that has the green light on.
incorrect, participant guessed, N = 1
- No nieh I'm not sure of either answer!
Q6.11
formula: | eventually { Red in Panel.lit} and eventually { Green in Panel.lit } |
trace: | {} {G} {} {} {R} |
answer | Yes |
57 / 57 correct (100%)
no students confused by scope here
correct, N = 56
- Yes 7jm R and G are in the trace.
- Yes w4t From the first state, the green light is eventually on and the red light is eventually on.
- Yes 1vwv From the current state, Red and Green are both eventually on.
- Yes 31qw at some point green goes on, and at some point red goes on.
- Yes 55oa Eventually means that it needs to happen in at least one state in the future. Since Green is lit in state 2 and Red is lit in the lasso, this holds.
- Yes 7opy Red is on in some states. So is Green.
- Yes 8E1h There is a state for which eventually in the future Green is lit and Red is also lit at some point (not necessarily both lit in the same state)
- Yes 917u There is some future state where the green light is on and some future state where the red light is on.
- Yes DzoD Both lights turn on eventually
- Yes KXjz both red and green are eventually lit
- Yes N3Cw The red light and the green light are on at some point.
- Yes XLuy Both green and red eventually become turned on in the second and fifth states respectively.
- Yes XU9x They are both eventually lit, the formula never said it had to be in the same places.
- Yes able because after the start one red and one green is present.
- Yes bx1r Both eventually operators don't have to happen at once, they just both have to eventually happen at some point.
- Yes bxd4 Green and Red should have happened eventually during the states. They both turned on eventually! :) 2nd and fifth
- Yes c1fn The absence of the 'always' and 'until' keywords implies that the conditions within either 'eventually' block need only hold once. Since both the green and red lights turn on at different points within this trace, this formula is satisfied.
- Yes c9jy Both the red and green lights are eventually on; the order doesn't matter due to the bracket ordering and usage of the keyword "and".
- Yes cirf This only requires that the eventually statements hold in the first state, so as long as there is at least one state in which the green light is on and at least one state in which the red light is on, the formula should be satisfied. Since this is the case in the given trace, the formula should be satisfied.
- Yes clU0 The Red light is eventually lit (e.g. in the fifth state), and the Green light is also eventually lit (in the second state).
- Yes duq8 The problem wants it so that eventually the panel has red and eventually the panel has green, but not necessarily that they should happen at the same time. The trace has green and red, so this is satisfied.
- Yes dznh Both the Green and Red lights turn on at some point.
- Yes ekny The last state satisfies the first condition and the second state satisfies the second condition.
- Yes eolW Green and Red both turn on eventually.
- Yes fojf The Red light is lit at some point on or after the first state, and so is the Green light.
- Yes fsqw The red light is on in some state, and the green light is on in some state.
- Yes gbrV Green is turned on in the seconds state and red in the last, satisfying the eventually construct.
- Yes hDZG Red is eventually lit (in the fifth state), and, separately, Green is eventually lit (in the second state).
- Yes iVk5 There are traces where Red and Green are in Panel.lit
- Yes j7t9 Red light is on at some state, so is the Green light
- Yes j9mq Both "eventually" predicates hold at some point
- Yes kkzx Both the red light and the green light are on at some point in the trace.
- Yes kuaa Both Green and Red eventually appear in the traces
- Yes m0p6 Eventually both red and green appeared
- Yes mxad Yes, we must have Green and Red appear at some point in the trace. Since they both appear, the formula is valid.
- Yes n4vd Both red and green are eventually lit
- Yes ndij At some point the red light is on, and at some point the green light is on. Both conditions are satisfied.
- Yes nieh Eventually both conditions are handled, doesn't have to be at the same time.
- Yes nr65 Eventually, there are states for which Red is lit and independently for which Green is lit (state 5 and 2 respectively). It doesn't require that they are lit in perpetuity.
- Yes pfdd Eventually red is lit (5th state) and eventually green is lit (2nd state)
- Yes ptyk Eventually {Red in Panel.lit} means that at some point in the trace, Red will be lit. We see in state 5 (and beyond) that this is true. Similarly, eventually {Green in Panel.lit} means that at some point in the trace Green will be lit. This is true in state 2. Thus these two formulas are both true and the entire formula is also true.
- Yes qMyf Red and Green are both lit at some point during the trace, so they are both eventually lit.
- Yes qjpx Red and Green eventually lit in some states
- Yes rahc The formula requires that eventually red light is lit and eventually green light is lit.
- Yes rnC9 The green and red lights both turn on
- Yes s0nv The formula states that at some point, the green light must be on and the right light must be on. It does not enforce that this must ALWAYS be the case. So, we see that the second state has the green light, and all states after and including the fifth state have the red light, so our formula is satisfied.
- Yes sain both the red and green lights are lit at some point
- Yes tyos {G} and {R}
- Yes uOG8 eventually both were lit
- Yes vz2g Both Green and Red are lit at some point.
- Yes xeec In the first state, both green and red lights are eventually on in subsequent states
- Yes xqZA Red and Green are both on at some point.
- Yes ydkm Both red and green are eventually lit
- Yes yoxy Green and Red are both lit at some point.
- Yes yypq The formula tells us that the Red light must eventually be lit and the Green light must also eventually be lit. The Red light is lit in the 5th state and the Green light is lit in the 2nd state.
- Yes zf66 The red light is lit in at least one state, as is the green light.
correct, N = 1
- Yes 3r0 Red and green in on in the last four states.
Q6.13
formula: | after { after { eventually { Red in Panel.lit } } } |
trace: | {RGB} {RGB} {RGB} {RGB} {RGB} |
answer | Yes |
56 / 57 correct (98%)
only one student thinks: "X negates current state"
correct, N = 48
- Yes 3r0 red light is on in all states. The formula doesn't constran the first two states.
- Yes 7jm R is in the 3rd time unit (and the 4th and 5th)
- Yes w4t In the third state, the Red light is already on, so it satisfies that it is eventually turned on.
- Yes 1vwv After the 2nd state, Red is eventually on (it is on in the 3rd state, in fact).
- Yes 31qw Looking at the third state onwards, it's true that red comes on at some point.
- Yes 55oa Since Red is in every trace, it will always satisfy this
- Yes 7opy Red is always on.
- Yes 917u there is some state future to two states from now where the red light is on
- Yes KXjz the red light is "eventually" lit after the 3rd state, since it's lit in the 4th state.
- Yes N3Cw In state 1 the formula holds, because in state 3, the red light is eventually on (in fact, the red light is on in state 3).
- Yes XU9x Well R is lit in all the traces, so this is pretty trivial.
- Yes able Because after two states red is present always although its required to appear once.
- Yes bx1r Two states after the first, there is some state after that at some point where the Red panel is lit
- Yes c1fn The formula necessitates that there is a state at or beyond the third state (i.e. second state after the first) where the red light turns on. Since, in this trace, the red light is always on, this is automatically satisfied.
- Yes c9jy The red light is on at some point at/after the third state (indicated by the two "after").
- Yes cirf Since the red light is always on in this trace, any requirement that the red light be on at some point in time (without untils or requirements that the red light be off at some point) is satisfied by it.
- Yes clU0 In the third state (i.e., 2 states after the first state), it is the case that the Red light is eventually lit (e.g. in the third state itself).
- Yes duq8 The problem wants it so that after the first two states, there exists a state where red is in the panel, which happens in the third state.
- Yes dznh Because, at some point after the 2nd state in the trace, the Red light is on.
- Yes ekny Red is lit after the 2nd state.
- Yes eolW Red is on in 4th trace.
- Yes fojf Starting in the third state, there is some state there or in the future in which the Red light is on.
- Yes fsqw The red light is on in some state after and including the third state.
- Yes hDZG Red is lit in the third state, so “eventually { Red in Panel.lit }” is true in the third state. Therefore, the original formula is true in the first state.
- Yes iVk5 Red is always in Panel.lit, so regardless of the number of "after"s, Red will always be in Panel.lit in the next state. So, the eventually always holds.
- Yes j7t9 starting from the third state, Red light is on at some state
- Yes j9mq Red is always lit, so it's necessarily lit eventually regardless of where you start in the trace.
- Yes kuaa Red eventually appears in the lit panel
- Yes mxad Yes, the formula requires that Red is lit at some point from the 3rd state onwards, but every state in this trace (including the lasso) lights all three colors, so Red will necessarily always be lit.
- Yes n4vd Yes, since red is lit in all isntances
- Yes ndij Two afters mean that we only need to look at traces starting from the 3rd state, and we see that the red light is on at some point (in the 3rd state).
- Yes nieh Red light turns on eventually after 2 turns.
- Yes nr65 In 2 state, Red is eventually lit. It is lit in state 3.
- Yes pfdd Red is lit in the 5th state, which occurs after the 3rd state.
- Yes ptyk This statement is saying that beginning in state 3, there will be some state greater than or equal to 3 in which the Red light is on. We see that in states 3, 4, 5, and beyond the Red light is on, so this formula is satisfiable.
- Yes qMyf Red is lit in every step in the trace, so it is lit after the 3rd step.
- Yes qjpx Red is always lit, so it will eventually lit from the third state
- Yes s0nv Since the red light is on in every state, we see that every state satisfies the `eventually` clause. `After after` is referring to the third state in our trace, and we see that the third state itself satisfies the `eventually` clause.
- Yes sain red is always lit
- Yes tyos Theres several states where Red panel is lit following the `after after`.
- Yes uOG8 red was eventually lit
- Yes vz2g Red is lit after the second state.
- Yes xeec Red is on in states after the third state
- Yes xqZA Red is on at some point after/including state 3.
- Yes ydkm Red is lit in at least one state after state 2, which satisfies eventually.
- Yes yoxy Starting from the third state, red is lit at least once.
- Yes yypq All of the states have the red light lit up.
- Yes zf66 the red light is always lit
correct, N = 8
- Yes 8E1h There is a trace for which two traces later, eventually Red is lit
- Yes DzoD Red is on in state 5
- Yes bxd4 For every state the Red light is turned on at some point after after that state (since Red is already turned on in every state)
- Yes gbrV Red is always on so basically any surrounding condition is irrelevant here
- Yes kkzx The formula defines that the red light should be on at some point. The trace given has the red light on in every state and therefore the formula is sat.
- Yes m0p6 Red is in the 4th and 5th state, which makes this true
- Yes rahc For all the states, its second following state is such that eventually the red light is on.
- Yes rnC9 Red is on in the lasso thus the eventually is always satisfied
incorrect, X requires negation in current state, N = 1
- No XLuy Red is turned on in the first two states.
Q6.15
formula: | Red in Panel.lit until Blue in Panel.lit |
trace: | {R} {R} {R} {R} {R} |
answer | No |
28 / 57 correct (49%)
51% think until is weak
correct, N = 28
- No w4t Formula b must hold in some state afterwards in the trace.
- No 1vwv Blue is not eventually in Panel.lit, so `until` is not satisfied.
- No 31qw Blue never turns on.
- No 7opy Blue is never on. But the until formula claims that it should be on in some state.
- No N3Cw I think using "until" means that the "Blue in Panel.lit" has to hold at some point, but the blue light never turns on.
- No able Because blue is not there in any state
- No bx1r The blue light being on is required for the until operator to be satisfied
- No bxd4 There is no Blue in the traces. Blue should happen in the state but since fmla-b doesn't hold, the formula cannot be satisfied by the tracee.
- No cirf The documentation page states that until requires that the second formula be true at some point (in other words, this formula implies that eventually Blue in Panel.lit). Since the blue light never turns on this trace does not satisfy the formula (even though I think the intuitive reading of until would indicate that it should).
- No clU0 A statement of the form until requires to eventually hold, but it is never the case that the Blue light is on.
- No duq8 Until assumes the 'Red in Panel.lit' is only held for a finite number of steps, but since trace will make Red with infinite steps.
- No dznh Because the Blue light never turns on.
- No fojf Until requires here that the Blue light is eventually lit, but it is never lit.
- No fsqw The blue light is never on in any state.
- No hDZG Blue is never lit, so the second part of the “until” is not satisfied.
- No iVk5 Until specifies that the RHS must hold at some point, and Blue is never in Panel.lit in this trace
- No j9mq Blue is never lit in the trace, which falsifies the 'until'.
- No nr65 fmla-b (Blue in Panel.lit) must hold in some finite state after the current state.
- No pfdd Blue is not lit in any state
- No ptyk This is not satisfiable because the until operator guarantees that Blue in Panel.lit will hold at some point in the trace, but we see here that it never does.
- No qjpx Blue never lits
- No rahc The formula requires that the blue light is on at some point, and at this point onwards the red light is never on. This trace doesn't have any state where blue light is on.
- No rnC9 As blue is never on, the right side of the until can never be true.
- No s0nv As per the ANON documentation: " until is true in a state i if and only if: fmla-b holds in some state j>=i and fmla-a holds in all states k where i <= k < j." In this case, is {Blue in Panel.lit}, but we see that there is no state in this trace (including the lasso step) in which fmla-b holds, since the light is never blue. Therefore, fmla-b does not hold in some state j>=i, which means that this trace does not satisfy the formula.
- No xqZA Blue is never on.
- No ydkm For until to work there must be some state where Blue is lit.
- No yypq In order to use "until" and return true, the second condition must be satisfied somewhere. The Blue light is never true, so we can't be certain that the Red light is on until the Blue light is on.
- No zf66 There is no state where the blue light is lit.
incorrect, weak U semantics, N = 29
- Yes 3r0 Blue light is never on, so red can always on.
- Yes 7jm Red is always lit, and Blue is never, so this is always true.
- Yes 55oa The only time Red should stop being lit is if Blue is. Seeing as Blue is never lit, that means that Red should always be lit (which it is)
- Yes 8E1h I'm not as sure about this one, but I think that the idea that Red is always lit until Blue is lit, and since Blue isn't lit, it's true.
- Yes 917u The red light is always on because the blue light is never turned on. So the formula is not violated.
- Yes DzoD Blue never turns on, so red never needs to turn off
- Yes KXjz the first predicate always can hold since the second never does.
- Yes XLuy Blue is never turned on so Red can stay on for all 5 traces.
- Yes XU9x Blue is never lit, so Red must stay lit.
- Yes c1fn The 'until' keyword implies the red light should always stay on until the blue light turns on for the first time. Since the blue light never turns on and the red light always stays on, this formula is satisfied.
- Yes c9jy "Red in Panel.lit" always holds, since the "Blue in Panel.lit" (blue light on) state never occurs.
- Yes ekny The first condition is always true.
- Yes eolW Blue is never on
- Yes gbrV Blue is never turned on so red always stays on
- Yes j7t9 Blue light is never on and Red light is always on
- Yes kkzx The formula says the red light should be on until the blue light is on. In the trace the blue light is never on and the red light is always on - it satisfies the formula since the formula doesn't say blue light must be on at some point. Since the until condition is never hit it makes sense that the red light is on throughout the trace.
- Yes kuaa Blue never appears in the panel; therefore, red must always appear
- Yes m0p6 There is no blue in panel.lit, so it's fine that red is always in panel.lit
- Yes mxad This is true because Blue never appears lit in the trace. As shown, Red will stay lit to infinity, and so Blue will never light up and thus Red should never shut off.
- Yes n4vd Yes, since blue is never lit the until statement is equivalent to just the antecedent.
- Yes ndij Because the blue light is never on, it's vacuously true.
- Yes nieh The second clause of the until statement is never satisfied so it can keep being red.
- Yes qMyf Since Blue is never lit, Red can continue to be lit forever.
- Yes sain red is always lit
- Yes tyos Red in Panel.lit was enough to satisfy the formula
- Yes uOG8 just cause we dont see the blue doesnt mean it wont be there in the future
- Yes vz2g Blue is never lit, so Red is always lit.
- Yes xeec Blue is never lit, so red should be always lit
- Yes yoxy The until clause doesn't ever have to become true.
Q6.17
formula: | eventually { always { Red in Panel.lit } } |
trace: | {} {RGB} {} {RGB} {} |
answer | No |
57 / 57 correct (100%)
3 think F disallows flickers
correct, N = 53
- No 3r0 In the fifth state, red light should turn on.
- No 7jm There is never a time where Red is always lit, since at least every other time has Red not being lit.
- No w4t At no point in any of the states is the Red light always on.
- No 1vwv There is no point beyond which Red is always on because the final empty state repeats forever.
- No 31qw All states not shown are {}, so there's no point after which red is always on.
- No 55oa This is saying that at some point in time, Red will become lit and stay that way forever. While Red does become lit twice, in the next states it immediately turns off. If the lasso state had the Red light lit this would be satisfied, but seeing as it doesn't, this is definitely not satisfied.
- No 7opy The trace ends with a repetition of {}, which makes it impossible to have Red always on.
- No 8E1h There is no state for which every state after that state has Red lit.
- No 917u There is not state from where on the Red light is always on.
- No DzoD Red eventually turns off, which means the "always" statement is not satisfied.
- No KXjz it is never the case that red is always lit since in the two instances it is, it isn't in the following state.
- No N3Cw The formula says that eventually the Red light will turn on and stay on forever, which doesn't happen.
- No XLuy Red is never on continuously.
- No XU9x The lasso state doesn't contain R
- No able Red is not always on.
- No bx1r eventually always forces at least the last state to satisfy the condition
- No bxd4 There isn't a trace of states where all the traces have Red in the Panel.lit. It would've been true if the fifth state had Red in it because it repeats.
- No c1fn The final lasso state, { }, repeats indefinitely, and the red light is not on in that state. Therefore, we never reach a state, at and after which the red light always stays on, so the formula is not satisfied.
- No c9jy No state occurs in which the Red light is permanently turned on.
- No cirf This formula requires that at some point the red light turns on and stays on forever, but since this trace ends with an infinite sequence of states where all the lights are off, this trace does not satisfy the formula.
- No clU0 There is no point in the trace at which the Red light stays on forever.
- No duq8 The formula statements that eventually there exists a state where all states including that one has red in it. This clearly does not happen since the last state getting repeated has no colors.
- No dznh because there is no state after which the Red light is always on.
- No ekny There is never a point after which Red is always lit (it loops on a state where no lights are lit).
- No eolW Lasso around all lights off - red is not on.
- No fojf Since the empty state (in which no light, including Red, is lit) cycles forever at the end, there is no state after or on the first state such that the Red light will be on at every state in the future.
- No fsqw The red light is never always on after any of the states.
- No gbrV There is no state after which red is always on
- No hDZG The last state does not have the red light lit, and since that repeats forever, that means that Red is off forever. Therefore, there is no state after which Red is always lit.
- No iVk5 There is no state after which all of the following states have Red in Panel.lit
- No j7t9 There is no state after which Red light is always on
- No j9mq There is no time step such that red is always lit from that point forward
- No kkzx The red light is not on in the last state, but the formula states that at some point the red light will always be on.
- No mxad After Red appears at some point, it must be the case the Red will always appear after that. There are two opportunities here to satisfy this formula, but Red doesn't stay lit for even two states in a row. Since the final and lasso state has nothing lit, there are no more opportunities to satisfy the formula, so this trace does not satisfy it.
- No n4vd No, because there is no instance where the red panel is always lit from then on out.
- No ndij Because the last state (all lights off) repeats forever, we are never getting to a point where the red light stays on.
- No nieh The last state does not contain the red light therefore the always statement is never eventually satisfied.
- No nr65 There is no point at which every state after that state has Red lit. The trace lassos on nothing being lit.
- No pfdd Whenever red is lit, it turns off immediately, so it is never the case that red is always lit.
- No ptyk This trace is looking for a state after which Red is always lit. If Red were lit in state 5 (and therefore in all future states), this would be true. Since it is not, we see that this formula is not satisfiable.
- No qMyf The empty state repeats forever, so Red is only lit twice. Thus, Red is never always lit.
- No qjpx The final state does not have Red in it. Thus it cannot be forever lit.
- No rahc There is no state such that eventually there is a state following it where all the subsequent states have the red light on.
- No rnC9 As red is not on in the lasso, it is never true that the red light will always be on after a given state
- No s0nv This states that there exists a state such that all states afterward will have the Red light on. However, we see that this does not occur in the first four states, and in the last repeating state, no lights are on, so this formula will never hold in the future.
- No sain there is no state after which all states have the red light
- No tyos The last trace does not have Red in Panel.lit.
- No uOG8 if there were more traces in the future, this could hold, but in this set of traces there is no point in which red is always lit
- No xeec There is no state where red is always lit in all subsequent states
- No xqZA There is no point at which Red turns on and stays on.
- No ydkm There is no state such that every state after has red lit, as the last state does not have red lit.
- No yoxy Red becomes lit but doesn't permanently stay on once it is.
- No zf66 There is no state after which the red light is always lit.
correct, F disallows flickers, N = 3 students
- No kuaa According to these set of conditions, once red appears, it must always appear.
- No m0p6 3 and 5th state need to have red in it to make it true
- No vz2g Once Red is lit, it should stay lit. (eventually always)
correct, wrong lasso, N = 1
- No yypq There is never a point where Red is always lit afterwards. The Red light will always flicker on and off based on the given states.
Q6.19
formula: | always { Red in Panel.lit implies Green in Panel.lit } |
trace: | {} {} {} {} {} |
answer | Yes |
57 / 57 correct (100%)
no evidence of "=> jumps to satisfying state" misconception
57, N = 57
- Yes 3r0 Red light is never turned on, and green lit can be turn on or not.
- Yes 7jm Red is never lit, so this is vacuously true.
- Yes w4t The implies statement returns true when formula-a is not met. Since the Red light is never on, the implies statement always returns true.
- Yes 1vwv Red is never on, so the implication is always satisfied.
- Yes 31qw Conditional is true when antecedent is not satisfied - red never on.
- Yes 55oa Because of how "implies" works, this formula is saying that the Green light will be lit only if the Red light is. It also says that if the Red light isn't lit, nothing happens. Therefore, this trace holds because False implies is always True.
- Yes 7opy Red is never on. So the implication is trivially true.
- Yes 8E1h Since Red is never lit, the statement within the always {} is always true.
- Yes 917u The red light is never turned on so the green light also does not need to be turned on. The formula is never violated.
- Yes DzoD Because red never turns on, there is never an opportunity to violate the statement.
- Yes KXjz no lights are ever lit, so there's no obligation to satisfy that implication, and thus it always holds.
- Yes N3Cw Implies statements are true if the left hand is always false, which is the case here (red light never turns on).
- Yes XLuy Red is never lit so Green also never has to be lit.
- Yes XU9x Green has no obligation to be lit if Red isn't (and it never gets lit)
- Yes able Because red is never lit itself
- Yes bx1r The red light is never on, so the green light never has to be on either. The implication is always vacuously true
- Yes bxd4 There was no Red in the trace, so whether Green is in the trace or not doesn't matter. The trace doesn't 'unsatisfy' the formula.
- Yes c1fn The given trace shows that in all states, all lights are off. Therefore, the precondition for the implies statement (i.e. the red light being on) is never satisfied, so the implies statement evaluates to true. Since this is always the case, the formula is satisfied.
- Yes c9jy The red light is never turned on, so it doesn't matter whether the green light is turned on or not.
- Yes cirf Since the red light is never on, the implication always holds true, so the trace satisfies the formula.
- Yes clU0 In every state, the Red light is not lit, so no obligation is incurred on the Green light, meaning the implication always (vacuously) holds.
- Yes duq8 The formula only needs to ensure that whenever red is in the traces, green should be in it, but red doesn't have to be in the traces, so all empty colors would suffice.
- Yes dznh Because the Right light never turns on. As a result, the implication is always true.
- Yes ekny The formula is vacuously true because the first part of the implies is never satisfied.
- Yes eolW Red never lit so implies never invoked,
- Yes fojf The Red light is never lit. When the first part of an implies statement is false, the implies statement is true. Sine the first part of this implies statement is always false, the implies statement is always true.
- Yes fsqw The red light is never on, so there's no condition that the trace needs to satisfy.
- Yes gbrV The LHS is always false so the expression is always true
- Yes hDZG Red is never lit, so the implication is vacuously true.
- Yes iVk5 The LHS of the implies never holds since Red is never in Panel.lit. So, the RHS never has to hold for the formula to be true.
- Yes j7t9 The implication is satisfied vacuously
- Yes j9mq The first half of the implication is always false, so the expression as a whole is always true.
- Yes kkzx The formula says whenever the red light is on, the green light is also on. Since in the trace the red light is never on, the behavior is unspecified and therefore all lights off is valid.
- Yes kuaa The condition above only holds true if red appears in the panel. If there is no red, then any set of traces can appear without red
- Yes m0p6 Red is never in panel.lit, so therefore there is nothing to worry about
- Yes mxad This is an infinitely empty trace. Since Red is never in Panel.lit, it doesn't constrain Green at all. The implies statement is still true because the left-side of the expression is false.
- Yes n4vd Yes, the implication is satisfied vacuously.
- Yes ndij Because the red light is never on, it's vacuously true.
- Yes nieh Red light is never turned on therefore we don't ever have to satisfy the latter part of the implication.
- Yes nr65 Red is never lit, so this is vacuously satisfied.
- Yes pfdd Red is never lit, so the implication is always true.
- Yes ptyk In all states, Red is not lit. Therefore the first portion of this implies statement is not true. That means that the entire statement is true. An implies is only false if the first portion holds but the second does not.
- Yes qMyf Since Red is never lit, Green never has an obligation to be lit, so this formula isn't violated.
- Yes qjpx Red is never lit, so the implies is always true
- Yes rahc Red light is never on, so the part inside { } is always satisfied.
- Yes rnC9 as Red is never on, this is trivially satisfied.
- Yes s0nv Red is never in Panel.lit, so the implication is vacuously true for all states, thus satisfying our formula for all states.
- Yes sain first condition never holds so its always true
- Yes tyos LHS of implies is false. Therefore this is true.
- Yes uOG8 red was never there so green was never there
- Yes vz2g It's an implies statement. Nothing requires Red to be lit. Red and Green both never being lit is therefore satisfactory.
- Yes xeec Red is not lit, so the implies statement is always true
- Yes xqZA The implication is vacuously true since Red is never on.
- Yes ydkm Because red is never lit, False implies False results in True.
- Yes yoxy Red doesn't necessarily have to be lit.
- Yes yypq The Red light is never lit, so the implies statement is always satisfied. Since, the implies is always satisfied, the formula is satisfied.
- Yes zf66 The red light is never lit, so the `implies` statement is always true by default.