# Re: [tlaplus] simple toy theorem

Thank you all for the clear explanation

On Wednesday, November 30, 2022 at 1:20:28 PM UTC+2 martin...@xxxxxxxxxxxxxx wrote:
I stripped your proof down to:

THEOREM T == ASSUME NEW S(_), NEW U(_), NEW M(_), NEW P(_),
NEW B(_),
\A x : U(x),
\A x : ~U(x)
PROVE
FALSE
PROOF
OBVIOUS

The equality binds stronger than the conjunction, such that the axiom is
parenthesized as \A x : (M(x) = S(x)) /\ ~U(x) allowing me to remove the other
conjunct from the universal quantification. Now it's easily visible that the

cheers,
Martin

On 11/30/22 11:26, jack malkovick 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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