Metamath Proof Explorer


Theorem nfbidv

Description: An equality theorem for nonfreeness. See nfbidf for a version without disjoint variable condition but requiring more axioms. (Contributed by Mario Carneiro, 4-Oct-2016) Remove dependency on ax-6 , ax-7 , ax-12 by adapting proof of nfbidf . (Revised by BJ, 25-Sep-2022)

Ref Expression
Hypothesis albidv.1 ( 𝜑 → ( 𝜓𝜒 ) )
Assertion nfbidv ( 𝜑 → ( Ⅎ 𝑥 𝜓 ↔ Ⅎ 𝑥 𝜒 ) )

Proof

Step Hyp Ref Expression
1 albidv.1 ( 𝜑 → ( 𝜓𝜒 ) )
2 1 exbidv ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑥 𝜒 ) )
3 1 albidv ( 𝜑 → ( ∀ 𝑥 𝜓 ↔ ∀ 𝑥 𝜒 ) )
4 2 3 imbi12d ( 𝜑 → ( ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) ↔ ( ∃ 𝑥 𝜒 → ∀ 𝑥 𝜒 ) ) )
5 df-nf ( Ⅎ 𝑥 𝜓 ↔ ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) )
6 df-nf ( Ⅎ 𝑥 𝜒 ↔ ( ∃ 𝑥 𝜒 → ∀ 𝑥 𝜒 ) )
7 4 5 6 3bitr4g ( 𝜑 → ( Ⅎ 𝑥 𝜓 ↔ Ⅎ 𝑥 𝜒 ) )