Metamath Proof Explorer


Theorem nfbidf

Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 4-Oct-2016) df-nf changed. (Revised by Wolf Lammen, 18-Sep-2021)

Ref Expression
Hypotheses albid.1 𝑥 𝜑
albid.2 ( 𝜑 → ( 𝜓𝜒 ) )
Assertion nfbidf ( 𝜑 → ( Ⅎ 𝑥 𝜓 ↔ Ⅎ 𝑥 𝜒 ) )

Proof

Step Hyp Ref Expression
1 albid.1 𝑥 𝜑
2 albid.2 ( 𝜑 → ( 𝜓𝜒 ) )
3 1 2 exbid ( 𝜑 → ( ∃ 𝑥 𝜓 ↔ ∃ 𝑥 𝜒 ) )
4 1 2 albid ( 𝜑 → ( ∀ 𝑥 𝜓 ↔ ∀ 𝑥 𝜒 ) )
5 3 4 imbi12d ( 𝜑 → ( ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) ↔ ( ∃ 𝑥 𝜒 → ∀ 𝑥 𝜒 ) ) )
6 df-nf ( Ⅎ 𝑥 𝜓 ↔ ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) )
7 df-nf ( Ⅎ 𝑥 𝜒 ↔ ( ∃ 𝑥 𝜒 → ∀ 𝑥 𝜒 ) )
8 5 6 7 3bitr4g ( 𝜑 → ( Ⅎ 𝑥 𝜓 ↔ Ⅎ 𝑥 𝜒 ) )