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Metamath Proof Explorer

Theorem eqneqall

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

		Ref	Expression
	Assertion	eqneqall	$⊢ A = B \to (A \neq B \to φ)$

Step	Hyp	Ref	Expression
1		df-ne	$⊢ A \neq B \leftrightarrow \neg A = B$
2		pm2.24	$⊢ A = B \to (\neg A = B \to φ)$
3	1 2	biimtrid	$⊢ A = B \to (A \neq B \to φ)$