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*From*: jack malkovick <sillymouse333@xxxxxxxxx>*Date*: Thu, 1 Dec 2022 11:58:15 -0800 (PST)*References*: <9f7efea9-a54a-47fa-b70f-f29a522449ecn@googlegroups.com> <567A33A9-D867-4DB8-8A9B-D6E7E8D31E86@gmail.com>

Right... silly me. Thank you!

On Thursday, December 1, 2022 at 9:53:54 PM UTC+2 Stephan Merz wrote:

Yes, you can also write … \in BOOLEAN. -sOn 1 Dec 2022, at 20:44, jack malkovick <sillym...@xxxxxxxxx> wrote: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--On Thursday, December 1, 2022 at 9:29:02 PM UTC+2 Stephan Merz wrote:Hi,I am glad that TLAPS doesn’t prove this because it’s wrong. As a simple counter-example considerP(x) == 1Q(x) == 2Clearly, you have P(x) # Q(x) for any x but we do not expect to prove 1 = ~2.What TLAPS should prove isASSUME NEW P(), NEW Q()PROVE \A x : ~(P(x) <=> Q(x)) => (P(x) <=> ~Q(x))StephanOn 1 Dec 2022, at 19:58, jack malkovick <sillym...@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

OBVIOUSPS. it can prove with no problem the reverse implication\A x : (P(x) = ~Q(x)) => (P(x) # Q(x))--

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**References**:**Re: [tlaplus] another simple theorem***From:*jack malkovick

**Re: [tlaplus] another simple theorem***From:*Stephan Merz

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