Metamath Proof Explorer


Theorem nfrexd

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

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

Proof

Step Hyp Ref Expression
1 nfrexd.1 𝑦 𝜑
2 nfrexd.2 ( 𝜑 𝑥 𝐴 )
3 nfrexd.3 ( 𝜑 → Ⅎ 𝑥 𝜓 )
4 dfrex2 ( ∃ 𝑦𝐴 𝜓 ↔ ¬ ∀ 𝑦𝐴 ¬ 𝜓 )
5 3 nfnd ( 𝜑 → Ⅎ 𝑥 ¬ 𝜓 )
6 1 2 5 nfraldw ( 𝜑 → Ⅎ 𝑥𝑦𝐴 ¬ 𝜓 )
7 6 nfnd ( 𝜑 → Ⅎ 𝑥 ¬ ∀ 𝑦𝐴 ¬ 𝜓 )
8 4 7 nfxfrd ( 𝜑 → Ⅎ 𝑥𝑦𝐴 𝜓 )