another simple theorem

[jack malkovick]

How could I help TLAPS to prove this simple theorem?

THEOREM TT ==
    ASSUME
        NEW P(_), NEW Q(_)
    PROVE
        \A x : (P(x) # Q(x)) => (P(x) = ~Q(x))
    PROOF
        OBVIOUS

PS. it can prove with no problem the reverse implication

       \A x : (P(x) = ~Q(x)) => (P(x) # Q(x))

[Stephan Merz]

Hi,

I am glad that TLAPS doesn’t prove this because it’s wrong. As a simple counter-example consider

P(x) == 1

Q(x) == 2

Clearly, you have P(x) # Q(x) for any x but we do not expect to prove 1 = ~2.

What TLAPS should prove is

ASSUME NEW P(), NEW Q()

PROVE \A x : ~(P(x) <=> Q(x)) => (P(x) <=> ~Q(x))

Stephan

On 1 Dec 2022, at 19:58, jack malkovick <sillym...@gmail.com> wrote:



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[jack malkovick]

Ah... I supposed that P and Q will be predicates.

It this the canonical way to specify that P and Q are predicates?  

THEOREM TT ==
    ASSUME
        NEW P(_), NEW Q(_),
        \A x : P(x) \in {TRUE, FALSE} /\ Q(x) \in {TRUE, FALSE}

    PROVE
        \A x : (P(x) # Q(x)) => (P(x) = ~Q(x))
    PROOF
        OBVIOUS

[Stephan Merz]

Yes, you can also write … \in BOOLEAN. -s

On 1 Dec 2022, at 20:44, jack malkovick <sillym...@gmail.com> wrote:

Ah... I supposed that P and Q will be predicates.

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[jack malkovick]

Right... silly me. Thank you!