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 )