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Re: [tlaplus] What do formulas inside "[]" operate on?

Hello Pascal,

you are completely right: P ~> Q is shorthand for [](P => <>Q), and your semantic intuitions are correct. For more details on the semantics of TLA, you may want to look at [1]. In fact, the equivalence holds not just for state predicates P and Q, but even if P and Q are arbitrary temporal formulas.

Equivalences such as this one are not typically checked using TLC, although you could create a spec in which P and Q are just Boolean variables that can change values non-deterministically and then verify the two implications separately.


[1] https://members.loria.fr/SMerz/papers/tla+logic2008.html

> On 27 Jul 2018, at 14:47, pascal.s...@xxxxxxxxx wrote:
> Hi, newbie here.
> I was playing around with "~>" and wondering if it can be defined in terms of "[]" and "<>".
> So I came up with this: I think "P ~> Q" is equivalent to "[](P => <>Q)". Unfortunately, I can't compare the two above Temporal Formulas with <=>, as TLC shouts "cannot handle" at me when I try..
> When Lamport mentioned that applying a state formula to a behaviour is the same as applying it to the first state of that behaviour, similarly for action formulas, I came up with the intuition that "[]" and "<>" could therefore probably be seen as quantifiers over every possible suffix of a behaviour.
> (I'm also assuming that what Lamport calls a behaviour is just a sequence of states. Please correct me if I'm wrong.)
> This now leads me to believe that the above equivalence holds, because of the following argument:
> Let P and Q be state formulas, and let B be a behaviour, then []P applied to B (as I understand it) means "(\A B_s \in suffixes(B) : P applied to B_s)". And since P applied to a behaviour is the same a P applied to the first state of that behaviour, then "[](P => <>Q)" means that for every state s in B where P is true, Q must hold true in some state t in the suffix of B, starting at s, which is equivalent to the definition of "~>".
> Can anyone confirm or refute my assumptions, such as the assumption that everything inside of "[]" operates on a suffix? Thank you very much for your time and sorry for the long post.
> Best regards
> Pascal
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