Theorem rabeqbidva 3428

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 3418	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . . . 5 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	eleq2d 2846	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	4	anbi1d 643	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜒) ↔ (𝑥 ∈ 𝐵 ∧ 𝜒)))
6	5	rabbidva2 3414	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
7	2, 6	eqtrd 2795	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {crab 3412

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 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413

This theorem is used by:  rabeqbidv  3429  scotteqd  9887  natpropd  18101  gsumpropd2lem  18815  elntg  29481  rmfsupp2  33717  poimirlem28  38480  uspgrlimlem1  49002  domnmsuppn0  49397  eenglngeehlnm  49767