Metamath Proof Explorer


Theorem bj-cbv1hv

Description: Version of cbv1h 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-cbv1hv.1 ⊢ φ → ψ → ∀ y ψ
bj-cbv1hv.2 ⊢ φ → χ → ∀ x χ
bj-cbv1hv.3 ⊢ φ → x = y → ψ → χ
Assertion bj-cbv1hv ⊢ ∀ x ∀ y φ → ∀ x ψ → ∀ y χ

Proof

Step Hyp Ref Expression
1 bj-cbv1hv.1 ⊢ φ → ψ → ∀ y ψ
2 bj-cbv1hv.2 ⊢ φ → χ → ∀ x χ
3 bj-cbv1hv.3 ⊢ φ → x = y → ψ → χ
4 nfa1 ⊢ Ⅎ x ∀ x ∀ y φ
5 nfa2 ⊢ Ⅎ y ∀ x ∀ y φ
6 2sp ⊢ ∀ x ∀ y φ → φ
7 6 1 syl ⊢ ∀ x ∀ y φ → ψ → ∀ y ψ
8 5 7 nf5d ⊢ ∀ x ∀ y φ → Ⅎ y ψ
9 6 2 syl ⊢ ∀ x ∀ y φ → χ → ∀ x χ
10 4 9 nf5d ⊢ ∀ x ∀ y φ → Ⅎ x χ
11 6 3 syl ⊢ ∀ x ∀ y φ → x = y → ψ → χ
12 4 5 8 10 11 cbv1v ⊢ ∀ x ∀ y φ → ∀ x ψ → ∀ y χ