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Metamath Proof Explorer

Theorem equsalhw

Description: Version of equsalh with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 29-Nov-2015) (Proof shortened by Wolf Lammen, 8-Jul-2022)

	Ref	Expression
Hypotheses	equsalhw.1	$⊢ ψ \to \forall x ψ$
	equsalhw.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
Assertion	equsalhw	$⊢ \forall x (x = y \to φ) \leftrightarrow ψ$

Proof

Step	Hyp	Ref	Expression
1		equsalhw.1	$⊢ ψ \to \forall x ψ$
2		equsalhw.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
3	1	nf5i	$⊢ Ⅎ x ψ$
4	3 2	equsalv	$⊢ \forall x (x = y \to φ) \leftrightarrow ψ$