Metamath Proof Explorer


Theorem cbvrabv2w

Description: A more general version of cbvrabv . Version of cbvrabv2 with a disjoint variable condition, which does not require ax-13 . (Contributed by Glauco Siliprandi, 23-Oct-2021) (Revised by GG, 14-Aug-2025)

Ref Expression
Hypotheses cbvrabv2w.1 ⊢ ( 𝑥 = 𝑦 → 𝐴 = 𝐵 )
cbvrabv2w.2 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion cbvrabv2w { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑦 ∈ 𝐵 ∣ 𝜓 }

Proof

Step Hyp Ref Expression
1 cbvrabv2w.1 ⊢ ( 𝑥 = 𝑦 → 𝐴 = 𝐵 )
2 cbvrabv2w.2 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
3 id ⊢ ( 𝑥 = 𝑦 → 𝑥 = 𝑦 )
4 3 1 eleq12d ⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵 ) )
5 4 2 anbi12d ⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝜓 ) ) )
6 5 cbvabv ⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } = { 𝑦 ∣ ( 𝑦 ∈ 𝐵 ∧ 𝜓 ) }
7 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
8 df-rab ⊢ { 𝑦 ∈ 𝐵 ∣ 𝜓 } = { 𝑦 ∣ ( 𝑦 ∈ 𝐵 ∧ 𝜓 ) }
9 6 7 8 3eqtr4i ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑦 ∈ 𝐵 ∣ 𝜓 }