Theorem eqneqall

Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018)

	Ref	Expression
Assertion	eqneqall	$$⊢A=B\to \left(A\ne B\to \phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		df-ne	$$⊢A\ne B\leftrightarrow \neg A=B$$
2		pm2.24	$$⊢A=B\to \left(\neg A=B\to \phi \right)$$
3	1 2	syl5bi	$$⊢A=B\to \left(A\ne B\to \phi \right)$$