Metamath Proof Explorer


Theorem nfrexd

Description: Deduction version of nfrex . Usage of this theorem is discouraged because it depends on ax-13 . See nfrexdw for a version with a disjoint variable condition, but not requiring ax-13 . (Contributed by Mario Carneiro, 14-Oct-2016) (New usage is discouraged.)

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

Proof

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