Theorem cbvrabv2w 46282

Description: A more general version of cbvrabv 3425. Version of cbvrabv2 46281 with a disjoint variable condition, which does not require ax-13 2404. (Contributed by Glauco Siliprandi, 23-Oct-2021.) (Revised by GG, 14-Aug-2025.)

Hypotheses

Ref	Expression
cbvrabv2w.1	⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
cbvrabv2w.2	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
cbvrabv2w	⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ 𝜓}

Distinct variable groups:   𝑦,𝐴   𝑥,𝐵   𝜑,𝑦   𝜓,𝑥   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)   𝐴(𝑥)   𝐵(𝑦)

Proof of Theorem cbvrabv2w

Step	Hyp	Ref	Expression
1		id 23	. . . . 5 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑦)
2		cbvrabv2w.1	. . . . 5 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3	1, 2	eleq12d 2857	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
4		cbvrabv2w.2	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	3, 4	anbi12d 644	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ 𝜓)))
6	5	cbvabv 2833	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜓)}
7		df-rab 3416	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
8		df-rab 3416	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ 𝜓} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜓)}
9	6, 7, 8	3eqtr4i 2796	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ 𝜓}

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2146  {cab 2741  {crab 3415

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3416

This theorem is used by:  smfsuplem2  47961