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Metamath Proof Explorer

Theorem bj-cbvexdvav

Description: Version of cbvexdva with a disjoint variable condition, which does not require ax-13 . (Contributed by BJ, 16-Jun-2019) (Proof modification is discouraged.)

		Ref	Expression
	Hypothesis	bj-cbvaldvav.1	$⊢ (φ \land x = y) \to (ψ \leftrightarrow χ)$
	Assertion	bj-cbvexdvav	$⊢ φ \to (\exists x ψ \leftrightarrow \exists y χ)$

Proof

Step	Hyp	Ref	Expression
1		bj-cbvaldvav.1	$⊢ (φ \land x = y) \to (ψ \leftrightarrow χ)$
2		nfv	$⊢ Ⅎ y φ$
3		nfvd	$⊢ φ \to Ⅎ y ψ$
4	1	ex	$⊢ φ \to (x = y \to (ψ \leftrightarrow χ))$
5	2 3 4	bj-cbvexdv	$⊢ φ \to (\exists x ψ \leftrightarrow \exists y χ)$