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) (Revised by Gino Giotto, 24-Sep-2024)

Ref Expression
Hypotheses nfraldw.1 𝑦 𝜑
nfraldw.2 ( 𝜑 𝑥 𝐴 )
nfraldw.3 ( 𝜑 → Ⅎ 𝑥 𝜓 )
Assertion nfraldw ( 𝜑 → Ⅎ 𝑥𝑦𝐴 𝜓 )

Proof

Step Hyp Ref Expression
1 nfraldw.1 𝑦 𝜑
2 nfraldw.2 ( 𝜑 𝑥 𝐴 )
3 nfraldw.3 ( 𝜑 → Ⅎ 𝑥 𝜓 )
4 df-ral ( ∀ 𝑦𝐴 𝜓 ↔ ∀ 𝑦 ( 𝑦𝐴𝜓 ) )
5 2 nfcrd ( 𝜑 → Ⅎ 𝑥 𝑦𝐴 )
6 5 3 nfimd ( 𝜑 → Ⅎ 𝑥 ( 𝑦𝐴𝜓 ) )
7 1 6 nfald ( 𝜑 → Ⅎ 𝑥𝑦 ( 𝑦𝐴𝜓 ) )
8 4 7 nfxfrd ( 𝜑 → Ⅎ 𝑥𝑦𝐴 𝜓 )