Theorem bj-cbvexdv

Description: Version of cbvexd 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 → Ⅎ 𝑦 𝜓 )
	bj-cbvaldv.3	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
Assertion	bj-cbvexdv	⊢ ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑦 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		bj-cbvaldv.1	⊢ Ⅎ 𝑦 𝜑
2		bj-cbvaldv.2	⊢ ( 𝜑 → Ⅎ 𝑦 𝜓 )
3		bj-cbvaldv.3	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
4	2	nfnd	⊢ ( 𝜑 → Ⅎ 𝑦 ¬ 𝜓 )
5		notbi	⊢ ( ( 𝜓 ↔ 𝜒 ) ↔ ( ¬ 𝜓 ↔ ¬ 𝜒 ) )
6	3 5	imbitrdi	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( ¬ 𝜓 ↔ ¬ 𝜒 ) ) )
7	1 4 6	bj-cbvaldv	⊢ ( 𝜑 → ( ∀ 𝑥 ¬ 𝜓 ↔ ∀ 𝑦 ¬ 𝜒 ) )
8	7	notbid	⊢ ( 𝜑 → ( ¬ ∀ 𝑥 ¬ 𝜓 ↔ ¬ ∀ 𝑦 ¬ 𝜒 ) )
9		df-ex	⊢ ( ∃ 𝑥 𝜓 ↔ ¬ ∀ 𝑥 ¬ 𝜓 )
10		df-ex	⊢ ( ∃ 𝑦 𝜒 ↔ ¬ ∀ 𝑦 ¬ 𝜒 )
11	8 9 10	3bitr4g	⊢ ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑦 𝜒 ) )