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# Re: [tlaplus] Confusion on "invariant under stuttering"

 Temporal formulas need to be evaluated over behaviors (infinite sequences of states), not states, and s^(+n) denotes the suffix of the behavior s starting at the n-th state (so that s^(+0) is just s). Of course, if F is a state predicate, evaluation of s |= F actually only considers the first state of s.I don't know what you mean by the notation "[]s^(+n) |= F", it does not appear in the book. In order to evaluate  s^(+n) |= []Fyou consider all suffixes, i.e. you check whether  (s^(+n))^(+m) |= F     i.e.   s^(n+m) |= Fholds for all non-negative integers m.StephanOn 19 Jan 2018, at 00:37, 杨永 wrote:Hi, StephanThanks a lot for your patient explanation. I think I seemed to understand a little bit. Can I understand from s^(+n) |= F that we just consider the state at time n for temporal formula F without considering the state of other states at any other time? and,  as for []s^(+n) |= F we should consider all states of the behavior s?Thanks again for your help!best regards,Yong在 2018年1月18日星期四 UTC+8下午3:48:40，Stephan Merz写道：Hello again,I'll try to be clearer this time.On 17 Jan 2018, at 03:13, 杨永 wrote:Hello, StephanThanks for your help, but, I'm still confused with the problem.You said that action formulas are evaluated over two states, that is, I should consider two states if I want to evaluate the Boolean value of an action. For example, there is a behavior "s" with the state sequence as s0     -->   s1      -->     s2    -->    s3    -->   ...   -->   snx=0         x=0             x=1          x=2and an action [x'=x+1]_x.Does it mean I should judge s0 and s1 in "s" if I want to determine the value of s |= [x'=x+1]_x? First of all, an action formula is not a temporal formula, so TLA does not really define the semantics of s |= [x' = x+1]_x. But for the sake of the argument, let's pretend it was and let's take s |= A (for an action A) to mean that A is true of the first two states of s, just as the meaning of s |= P for a state predicate P is that P is true of the first state of s. So, indeed, we look at states s0 and s1 in order to find out if the formula holds.As in this example, s0 --> s1 is a stuttering step, because this step satisfies [x'=x+1]_x that the value of "x" is UNCHANGED. So, s |= [x'=x+1]_x equals TRUE in this example.Exactly.However, the example in the book (Page 90) says "the action [ x' = x + 1]_x is satisfied by a behavior "s" in which x is left unchanged in the first step and incremented by 2 in the second step". Does the example means action [x'=x+1]_x also satisfies a behavior "s" as s0     -->   s1      -->     s2    -->    s3    -->   ...   -->   snx=0         x=0             x=2          x=3Again, we look at states s0 and s1, and since the variable x has the same value in both states, the formula again evaluates to true for this behavior. However the formula is false for the behavior   s1  -->  s2  --> s3 --> ...which results from s by removing a stuttering transition, and this shows that the formula [x' = x+1]_x is not invariant under (addition or removal of) stuttering.If the answer is yes, my question is: how does state s0 transit to s1? I think here is the root of your misunderstanding. When we evaluate the semantics s |= F for a behavior s and a formula F, both are given to us and we don't ask how the behavior or the formula are produced. Your sequence s above is a behavior, so it makes sense to ask if some formula, such as [x' = x+1]_x, is true or false of it.I think the value of x in state s2 should be 0 or 1 if the step is a [ x' = x + 1]_x step while x=0 in state s1, because [x'=x+1]_x == (x'=x+1) \/ (x'=x), that is, the value of x should be UNCHANGED or increased by 1 in each step of the behavior. The action formula [x' = x+1]_x only considers states s0 and s1, therefore it does not restrict in any way the value of x in state s2. On the contrary, the temporal formula [][x' = x+1]_x considers all states of s, and that formula is false of the behavior s, and of the suffix starting at s1.Hope this helps,StephanYou said "Any formula [A]_f is true for a stuttering transition (two identical states), independently of the actual formula A.". As in this example, the value of x in state s1 is 0, if [x'=x+1]_x is TRUE, either (x'=x+1) or (x'=x) should be executed which leads to the value of x in state s2 equals 1 or UNCHANGED (i.e., 0). x'=x should be executes if we do not consider x'=x+1, so, the value of x should be 0 in state s2 in this example. I'm confused how does the value of x become 2 in state s2 while it is 0 in state s1.If the answer is no, could you please list the transition of state under action [ x' = x + 1]_x satisfies that x is left unchanged in the first step and incremented by 2 in the second step?Or, is there any misunderstanding?Kind regards,Yong在 2018年1月16日星期二 UTC+8下午10:13:32，Stephan Merz写道：Hello, action formulas are evaluated over two states. The action formula [x'=x+1]_x is true for two states s and t if either the two states assign the same value to x or if the value of x in t is the value of x in s plus one. Any formula [A]_f is true for a stuttering transition (two identical states), independently of the actual formula A. As an extreme example, [FALSE]_f is true for a stuttering transition, but false for any non-stuttering transition. These examples show that [A]_f is in general not invariant under stuttering: assume for a moment that [A]_f were a temporal formula (similarly as state predicates are lifted to temporal formulas) and suppose that you evaluate that formula over a behavior that starts with a stuttering transition, then that behavior satisfies [A]_f. If you remove the initial stuttering transition, there is no reason why the resulting behavior should still satisfy [A]_f. This is exactly the point of the example given in the book. In contrast, it is not hard to convince yourself that the temporal formula [][A]_f is invariant under addition or removal of (finitely many) stuttering transitions. Hope this helps, Stephan > On 16 Jan 2018, at 08:27, 杨永 wrote: > > Hi, guys > > I am reading "TLA+ Specifying Systems", and I'm confusing on the problem of invariant under stuttering(Page 90). The 3rd paragraph is as follows: > > A state predicate (viewed as a temporal formula) is invariant under stuttering, since its truth depends only on the first state of a behavior, and adding a stuttering step doesn't change the first state. An arbitrary action is not invariant under stuttering. For example, the action [ x' = x + 1]_x is satisfied by a behavior \$a\$ in which x is left unchanged in the first step and incremented by 2 in the second step; it isn't satisfied by the behavior obtained by removing the initial stuttering step from \$a\$ . However, the formula [][ x' = x + 1]_x is invariant under stuttering, since it is satisfied by a behavior iff every step that changes x is an x' = x + 1 step -- a condition not a affected by adding or deleting stuttering steps. > > My question is, > > since [A]_f = A \/ (f'=f), I think, in action [ x' = x + 1]_x, x is left unchanged in the first step, the second step should be increased x by 1 after execute the action. How to understand that "x is incremented by 2 in the second step"? > > > Thanks, > > Yong > > -- > You received this message because you are subscribed to the Google Groups "tlaplus" group. > To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+u...@googlegroups.com. > To post to this group, send email to tla...@googlegroups.com. > Visit this group at https://groups.google.com/group/tlaplus. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "tlaplus" group.To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+u...@googlegroups.com.To post to this group, send email to tla...@xxxxxxxxxxxxxxxx.Visit this group at https://groups.google.com/group/tlaplus.For more options, visit https://groups.google.com/d/optout.-- You received this message because you are subscribed to the Google Groups "tlaplus" group.To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+u...@xxxxxxxxxxxxxxxx.To post to this group, send email to tla...@xxxxxxxxxxxxxxxx.Visit this group at https://groups.google.com/group/tlaplus.For more options, visit https://groups.google.com/d/optout.

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