Metamath Proof Explorer


Theorem nfraldw

Description: Deduction version of nfralw . Version of nfrald with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 15-Feb-2013) Avoid ax-9 , ax-ext . (Revised by GG, 24-Sep-2024)

Ref Expression
Hypotheses nfraldw.1 ⊢ Ⅎ y φ
nfraldw.2 ⊢ φ → Ⅎ _ x A
nfraldw.3 ⊢ φ → Ⅎ x ψ
Assertion nfraldw ⊢ 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 df-ral ⊢ ∀ y ∈ A ψ ↔ ∀ y y ∈ A → ψ
5 2 nfcrd ⊢ φ → Ⅎ x y ∈ A
6 5 3 nfimd ⊢ φ → Ⅎ x y ∈ A → ψ
7 1 6 nfald ⊢ φ → Ⅎ x ∀ y y ∈ A → ψ
8 4 7 nfxfrd ⊢ φ → Ⅎ x ∀ y ∈ A ψ