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Metamath Proof Explorer

Theorem bj-cbvaldv

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

	Ref	Expression
Hypotheses	bj-cbvaldv.1	$⊢ Ⅎ y φ$
	bj-cbvaldv.2	$⊢ φ \to Ⅎ y ψ$
	bj-cbvaldv.3	$⊢ φ \to (x = y \to (ψ \leftrightarrow χ))$
Assertion	bj-cbvaldv	$⊢ φ \to (\forall x ψ \leftrightarrow \forall y χ)$

Proof

Step	Hyp	Ref	Expression
1		bj-cbvaldv.1	$⊢ Ⅎ y φ$
2		bj-cbvaldv.2	$⊢ φ \to Ⅎ y ψ$
3		bj-cbvaldv.3	$⊢ φ \to (x = y \to (ψ \leftrightarrow χ))$
4		nfv	$⊢ Ⅎ x φ$
5		nfv	$⊢ Ⅎ x χ$
6	5	a1i	$⊢ φ \to Ⅎ x χ$
7	4 1 2 6 3	bj-cbv2v	$⊢ φ \to (\forall x ψ \leftrightarrow \forall y χ)$