Re: [tlaplus] another simple theorem

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 <sillymouse333@xxxxxxxxx> wrote:

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))

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