Thanks 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 --> ... --> sn
x=0 x=0 x=1 x=2
and 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? 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.
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 --> ... --> sn
x=0 x=0 x=2 x=3
If the answer is yes, my question is: how does state s0 transit to s1? 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. You 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?
在 2018年1月16日星期二 UTC+8下午10:13:32，Stephan Merz写道：
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,
> On 16 Jan 2018, at 08:27, 杨永 <steph...@xxxxxxxxx> 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"?
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