Metamath Proof Explorer


Theorem nf5di

Description: Since the converse holds by a1i , this inference shows that we can represent a not-free hypothesis with either F/ x ph (inference form) or ( ph -> F/ x ph ) (deduction form). (Contributed by NM, 17-Aug-2018) (Proof shortened by Wolf Lammen, 10-Jul-2019) Reduce axiom dependencies. (Revised by SN and Sky Winter, 23-Oct-2026)

Ref Expression
Hypothesis nf5di.1 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜑 )
Assertion nf5di Ⅎ 𝑥 𝜑

Proof

Step Hyp Ref Expression
1 nf5di.1 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜑 )
2 hbe1 ⊢ ( ∃ 𝑥 𝜑 → ∀ 𝑥 ∃ 𝑥 𝜑 )
3 1 nfrd ⊢ ( 𝜑 → ( ∃ 𝑥 𝜑 → ∀ 𝑥 𝜑 ) )
4 3 com12 ⊢ ( ∃ 𝑥 𝜑 → ( 𝜑 → ∀ 𝑥 𝜑 ) )
5 2 4 eximdh ⊢ ( ∃ 𝑥 𝜑 → ( ∃ 𝑥 𝜑 → ∃ 𝑥 ∀ 𝑥 𝜑 ) )
6 5 pm2.43i ⊢ ( ∃ 𝑥 𝜑 → ∃ 𝑥 ∀ 𝑥 𝜑 )
7 hbe1a ⊢ ( ∃ 𝑥 ∀ 𝑥 𝜑 → ∀ 𝑥 𝜑 )
8 6 7 syl ⊢ ( ∃ 𝑥 𝜑 → ∀ 𝑥 𝜑 )
9 8 nfi ⊢ Ⅎ 𝑥 𝜑