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Metamath Proof Explorer

Theorem equsalvw

Description: Version of equsalv with a disjoint variable condition, and of equsal with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw . (Contributed by BJ, 31-May-2019)

		Ref	Expression
	Hypothesis	equsalvw.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
	Assertion	equsalvw	$⊢ \forall x (x = y \to φ) \leftrightarrow ψ$

Proof

Step	Hyp	Ref	Expression
1		equsalvw.1	$⊢ x = y \to (φ \leftrightarrow ψ)$
2	1	pm5.74i	$⊢ (x = y \to φ) \leftrightarrow (x = y \to ψ)$
3	2	albii	$⊢ \forall x (x = y \to φ) \leftrightarrow \forall x (x = y \to ψ)$
4		equsv	$⊢ \forall x (x = y \to ψ) \leftrightarrow ψ$
5	3 4	bitri	$⊢ \forall x (x = y \to φ) \leftrightarrow ψ$