Metamath Proof Explorer


Theorem rabeqbidva

Description: Equality of restricted class abstractions. (Contributed by Mario Carneiro, 26-Jan-2017) Remove DV conditions. (Revised by GG, 1-Sep-2025)

Ref Expression
Hypotheses rabeqbidva.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
rabeqbidva.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion rabeqbidva ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )

Proof

Step Hyp Ref Expression
1 rabeqbidva.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 rabeqbidva.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
3 2 rabbidva ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ 𝜒 } )
4 1 eleq2d ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
5 4 anbi1d ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜒 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜒 ) ) )
6 5 rabbidva2 ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜒 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )
7 3 6 eqtrd ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )