Metamath Proof Explorer


Theorem bj-cbv2v

Description: Version of cbv2 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-cbv2v.1 ⊢ Ⅎ 𝑥 𝜑
bj-cbv2v.2 ⊢ Ⅎ 𝑦 𝜑
bj-cbv2v.3 ⊢ ( 𝜑 → Ⅎ 𝑦 𝜓 )
bj-cbv2v.4 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
bj-cbv2v.5 ⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
Assertion bj-cbv2v ( 𝜑 → ( ∀ 𝑥 𝜓 ↔ ∀ 𝑦 𝜒 ) )

Proof

Step Hyp Ref Expression
1 bj-cbv2v.1 ⊢ Ⅎ 𝑥 𝜑
2 bj-cbv2v.2 ⊢ Ⅎ 𝑦 𝜑
3 bj-cbv2v.3 ⊢ ( 𝜑 → Ⅎ 𝑦 𝜓 )
4 bj-cbv2v.4 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
5 bj-cbv2v.5 ⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
6 2 nf5ri ⊢ ( 𝜑 → ∀ 𝑦 𝜑 )
7 1 nfal ⊢ Ⅎ 𝑥 ∀ 𝑦 𝜑
8 7 nf5ri ⊢ ( ∀ 𝑦 𝜑 → ∀ 𝑥 ∀ 𝑦 𝜑 )
9 6 8 syl ⊢ ( 𝜑 → ∀ 𝑥 ∀ 𝑦 𝜑 )
10 3 nf5rd ⊢ ( 𝜑 → ( 𝜓 → ∀ 𝑦 𝜓 ) )
11 4 nf5rd ⊢ ( 𝜑 → ( 𝜒 → ∀ 𝑥 𝜒 ) )
12 10 11 5 bj-cbv2hv ⊢ ( ∀ 𝑥 ∀ 𝑦 𝜑 → ( ∀ 𝑥 𝜓 ↔ ∀ 𝑦 𝜒 ) )
13 9 12 syl ⊢ ( 𝜑 → ( ∀ 𝑥 𝜓 ↔ ∀ 𝑦 𝜒 ) )