Metamath Proof Explorer


Theorem nfceqdf

Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 14-Oct-2016) Avoid ax-8 and df-clel . (Revised by WL and SN, 23-Aug-2024)

Ref Expression
Hypotheses nfceqdf.1 ⊢ Ⅎ 𝑥 𝜑
nfceqdf.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion nfceqdf ( 𝜑 → ( Ⅎ 𝑥 𝐴 ↔ Ⅎ 𝑥 𝐵 ) )

Proof

Step Hyp Ref Expression
1 nfceqdf.1 ⊢ Ⅎ 𝑥 𝜑
2 nfceqdf.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
3 eleq2w2 ⊢ ( 𝐴 = 𝐵 → ( 𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵 ) )
4 2 3 syl ⊢ ( 𝜑 → ( 𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵 ) )
5 1 4 nfbidf ⊢ ( 𝜑 → ( Ⅎ 𝑥 𝑦 ∈ 𝐴 ↔ Ⅎ 𝑥 𝑦 ∈ 𝐵 ) )
6 5 albidv ⊢ ( 𝜑 → ( ∀ 𝑦 Ⅎ 𝑥 𝑦 ∈ 𝐴 ↔ ∀ 𝑦 Ⅎ 𝑥 𝑦 ∈ 𝐵 ) )
7 df-nfc ⊢ ( Ⅎ 𝑥 𝐴 ↔ ∀ 𝑦 Ⅎ 𝑥 𝑦 ∈ 𝐴 )
8 df-nfc ⊢ ( Ⅎ 𝑥 𝐵 ↔ ∀ 𝑦 Ⅎ 𝑥 𝑦 ∈ 𝐵 )
9 6 7 8 3bitr4g ⊢ ( 𝜑 → ( Ⅎ 𝑥 𝐴 ↔ Ⅎ 𝑥 𝐵 ) )