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 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion	bj-cbvexdvav	⊢ ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑦 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		bj-cbvaldvav.1	⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
2		nfv	⊢ Ⅎ 𝑦 𝜑
3		nfvd	⊢ ( 𝜑 → Ⅎ 𝑦 𝜓 )
4	1	ex	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
5	2 3 4	bj-cbvexdv	⊢ ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑦 𝜒 ) )