Re: [tlaplus] simple toy theorem

[Stephan Merz <stephan.merz@xxxxxxxxx>]

I think this is a matter of parsing precedence.

  \A x : M(x) = S(x) /\ ~U(x)

is parsed as

  \A x : (M(x) = S(x)) /\ ~U(x)

and therefore your hypotheses imply

   \A x : ~ U(x)

For the analogous reason, the hypothesis

  NEW B(_), \A x : B(x) = S(x) /\ U(x)

asserts

  \A x : U(x)

and now you get an inconsistent set of assumptions. Had you used "<=>" instead of "=", the result would have been different because conjunction binds more tightly than equivalence.

Stephan

On 30 Nov 2022, at 11:26, jack malkovick <sillymouse333@xxxxxxxxx> wrote:

Let's say we have this simple theorem

THEOREM T ==
    ASSUME
        NEW NEW S(_), NEW U(_), NEW M(_), NEW P(_),
        \A x : M(x) = S(x) /\ ~U(x),
        \A x : P(x) = M(x)
   PROVE
        \A x : P(x) => S(x)

   PROOF
        OBVIOUS

It is true.

If I negate the goal to \E x : ~(P(x) => S(x)) same as \E x : P(x) /\ ~S(x) it becomes red.

However, if I add another assumption
NEW B(_), \A x : B(x) = S(x) /\ U(x),

The theorem turns green! How can this new assumption make the theorem true?

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