Theorem cbvrabdavw2 36856

Description: Change bound variable and domain in restricted class abstractions. Deduction form. (Contributed by GG, 14-Aug-2025.)

Hypotheses

Ref	Expression
cbvrabdavw2.1	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
cbvrabdavw2.2	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐴 = 𝐵)

Assertion

Ref	Expression
cbvrabdavw2	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑦 ∈ 𝐵 ∣ 𝜒})

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

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

Proof of Theorem cbvrabdavw2

Step	Hyp	Ref	Expression
1		eleq1w 2848	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2	1	adantl 487	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
3		cbvrabdavw2.2	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐴 = 𝐵)
4	3	eleq2d 2851	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
5	2, 4	bitrd 282	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
6		cbvrabdavw2.1	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
7	5, 6	anbi12d 644	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑦 ∈ 𝐵 ∧ 𝜒)))
8	7	cbvabdavw 36827	. 2 ⊢ (𝜑 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜒)})
9		df-rab 3419	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
10		df-rab 3419	. 2 ⊢ {𝑦 ∈ 𝐵 ∣ 𝜒} = {𝑦 ∣ (𝑦 ∈ 𝐵 ∧ 𝜒)}
11	8, 9, 10	3eqtr4g 2825	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑦 ∈ 𝐵 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146  {cab 2743  {crab 3418

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419

This theorem is used by: (None)