Theorem rabeqbidva 3405

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

Hypotheses

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

Assertion

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

Distinct variable group:   𝜑,𝑥

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

Proof of Theorem rabeqbidva

Step	Hyp	Ref	Expression
1		rabeqbidva.2	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
2	1	rabbidva 3395	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . . . 5 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	eleq2d 2822	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	4	anbi1d 632	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜒) ↔ (𝑥 ∈ 𝐵 ∧ 𝜒)))
6	5	rabbidva2 3391	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
7	2, 6	eqtrd 2771	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1542   ∈ wcel 2114  {crab 3389

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-rab 3390

This theorem is referenced by:  rabeqbidv  3407  natpropd  17946  gsumpropd2lem  18647  elntg  29053  rmfsupp2  33299  poimirlem28  37969  scotteqd  44664  uspgrlimlem1  48464  domnmsuppn0  48845  eenglngeehlnm  49215