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Metamath Proof Explorer

Theorem bj-cbval2vv

Description: Version of cbval2vv 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-cbval2vv.1	⊢ ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → ( 𝜑 ↔ 𝜓 ) )
	Assertion	bj-cbval2vv	⊢ ( ∀ 𝑥 ∀ 𝑦 𝜑 ↔ ∀ 𝑧 ∀ 𝑤 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		bj-cbval2vv.1	⊢ ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → ( 𝜑 ↔ 𝜓 ) )
2		nfv	⊢ Ⅎ 𝑧 𝜑
3		nfv	⊢ Ⅎ 𝑤 𝜑
4		nfv	⊢ Ⅎ 𝑥 𝜓
5		nfv	⊢ Ⅎ 𝑦 𝜓
6	2 3 4 5 1	cbval2v	⊢ ( ∀ 𝑥 ∀ 𝑦 𝜑 ↔ ∀ 𝑧 ∀ 𝑤 𝜓 )