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Metamath Proof Explorer

Theorem pm2.61ne

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 25-Nov-2019)

	Ref	Expression
Hypotheses	pm2.61ne.1	$⊢ A = B \to (ψ \leftrightarrow χ)$
	pm2.61ne.2	$⊢ (φ \land A \neq B) \to ψ$
	pm2.61ne.3	$⊢ φ \to χ$
Assertion	pm2.61ne	$⊢ φ \to ψ$

Proof

Step	Hyp	Ref	Expression
1		pm2.61ne.1	$⊢ A = B \to (ψ \leftrightarrow χ)$
2		pm2.61ne.2	$⊢ (φ \land A \neq B) \to ψ$
3		pm2.61ne.3	$⊢ φ \to χ$
4	3 1	imbitrrid	$⊢ A = B \to (φ \to ψ)$
5	2	expcom	$⊢ A \neq B \to (φ \to ψ)$
6	4 5	pm2.61ine	$⊢ φ \to ψ$