Theorem rabeqbidva 3411

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 3401	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . . . 5 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	eleq2d 2817	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	4	anbi1d 631	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜒) ↔ (𝑥 ∈ 𝐵 ∧ 𝜒)))
6	5	rabbidva2 3397	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
7	2, 6	eqtrd 2766	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1541   ∈ wcel 2111  {crab 3395

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396

This theorem is referenced by:  rabeqbidv  3413  natpropd  17886  gsumpropd2lem  18587  elntg  28962  rmfsupp2  33205  poimirlem28  37687  scotteqd  44329  uspgrlimlem1  48087  domnmsuppn0  48468  eenglngeehlnm  48839