Metamath Proof Explorer


Theorem nfrexdw

Description: Deduction version of nfrexw . (Contributed by Mario Carneiro, 14-Oct-2016) Add disjoint variable condition to avoid ax-13 . See nfrexd for a less restrictive version requiring more axioms. (Revised by GG, 20-Jan-2024)

Ref Expression
Hypotheses nfraldw.1 ⊢ Ⅎ y φ
nfraldw.2 ⊢ φ → Ⅎ _ x A
nfraldw.3 ⊢ φ → Ⅎ x ψ
Assertion nfrexdw ⊢ φ → Ⅎ x ∃ y ∈ A ψ

Proof

Step Hyp Ref Expression
1 nfraldw.1 ⊢ Ⅎ y φ
2 nfraldw.2 ⊢ φ → Ⅎ _ x A
3 nfraldw.3 ⊢ φ → Ⅎ x ψ
4 dfrex2 ⊢ ∃ y ∈ A ψ ↔ ¬ ∀ y ∈ A ¬ ψ
5 3 nfnd ⊢ φ → Ⅎ x ¬ ψ
6 1 2 5 nfraldw ⊢ φ → Ⅎ x ∀ y ∈ A ¬ ψ
7 6 nfnd ⊢ φ → Ⅎ x ¬ ∀ y ∈ A ¬ ψ
8 4 7 nfxfrd ⊢ φ → Ⅎ x ∃ y ∈ A ψ