# Re: [tlaplus] simple toy theorem

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 assumptions are a contradiction :)

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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