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# Re: [tlaplus] Q about the existential quantifier

I misspoke: even the implication

[](\E x : F) => \EE x : []F

is not valid in general, but it is valid if F is a state predicate, or somewhat more generally, if x does not occur in the scope of temporal operators. For example, assume that v is a state variable, then

[](\E x : x = v /\ \E y : [](x=y))

is valid (it suffices at every instant to take x = v, y=x and since both x and y are rigid variables their values never change), but the formula

\EE x : [](x = v /\ \E y : [](x=y))

will not hold in general: the first conjunct requires that x and v always have the same value, but the second conjunct requires that x remains constant, which is impossible if v changes during the behavior.

For an arbitrary formula F(x), we only have

[](\E x : F(x)) => \EE x : [](\E y : y=x /\ F(y))

but this formula does not appear to be very useful.

Sincere apologies: temporal quantifiers are really delicate!

Stephan

On Saturday, May 29, 2021 at 8:42:23 AM UTC+2 Stephan Merz wrote:
Hello,

the implication

[](\E x : F) => \EE x : []F

is indeed valid. The reverse implication need not hold. For example,

\EE x : []<> << x' # x >>_x

is valid, but

[] \E x : <> << x' # x >>_x

is a contradiction. If F is a state predicate then

(\EE x : []F) => [](\E x : F)

is valid.

Stephan

On 29 May 2021, at 05:51, ns <nedsr...@xxxxxxxxx> wrote:

hello, the Specifying Systems book says that a regular existentially quantified variable functions as a constant. And from the definition this does seem to be the case
\sigma |= (\E x: F) == \E x: (\sigma |= F)
where \sigma is a trace. However, if I have
[] (\E x: F)
its now possible for each state of \sigma to have a different value of x and satisfy this formula. But this now seems closer to a temporal existential. So would it be equivalent to
\EE x: [] F
?
thanks

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