Metamath Proof Explorer


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