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 )