Metamath Proof Explorer


Theorem eleq1d

Description: Deduction from equality to equivalence of membership. (Contributed by NM, 21-Jun-1993) Allow shortening of eleq1 . (Revised by Wolf Lammen, 20-Nov-2019)

Ref Expression
Hypothesis eleq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion eleq1d ( 𝜑 → ( 𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 eleq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 1 eqeq2d ⊢ ( 𝜑 → ( 𝑥 = 𝐴 ↔ 𝑥 = 𝐵 ) )
3 2 anbi1d ⊢ ( 𝜑 → ( ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐶 ) ↔ ( 𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐶 ) ) )
4 3 exbidv ⊢ ( 𝜑 → ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐶 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐶 ) ) )
5 dfclel ⊢ ( 𝐴 ∈ 𝐶 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝑥 ∈ 𝐶 ) )
6 dfclel ⊢ ( 𝐵 ∈ 𝐶 ↔ ∃ 𝑥 ( 𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐶 ) )
7 4 5 6 3bitr4g ⊢ ( 𝜑 → ( 𝐴 ∈ 𝐶 ↔ 𝐵 ∈ 𝐶 ) )