Theorem rabeqbii 37160

Description: Equality theorem for restricted class abstractions. Inference version. (Contributed by GG, 1-Sep-2025.)

Hypotheses

Ref	Expression
rabeqbii.1	⊢ 𝐴 = 𝐵
rabeqbii.2	⊢ (𝜑 ↔ 𝜓)

Assertion

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

Proof of Theorem rabeqbii

Step	Hyp	Ref	Expression
1		rabeqbii.1	. . . . 5 ⊢ 𝐴 = 𝐵
2	1	eleq2i 2856	. . . 4 ⊢ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		rabeqbii.2	. . . 4 ⊢ (𝜑 ↔ 𝜓)
4	2, 3	anbi12i 640	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ 𝜓))
5	4	abbii 2830	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}
6		df-rab 3417	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7		df-rab 3417	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}
8	5, 6, 7	3eqtr4i 2796	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐵 ∣ 𝜓}

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147  {cab 2741  {crab 3416

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417

This theorem is used by: (None)