Theorem bj-rabeqd 32108

Description: Deduction form of rabeq 3166. Note that contrary to rabeq 3166 it has no dv condition. (Contributed by BJ, 27-Apr-2019.)

Hypotheses

Ref	Expression
bj-rabeqd.nf	⊢ Ⅎ𝑥𝜑
bj-rabeqd.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
bj-rabeqd	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Proof of Theorem bj-rabeqd

Step	Hyp	Ref	Expression
1		bj-rabeqd.nf	. 2 ⊢ Ⅎ𝑥𝜑
2		bj-rabeqd.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
3		eleq2 2677	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4	3	anbi1d 737	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜓)))
5	2, 4	syl 17	. 2 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜓)))
6	1, 5	bj-rabbida2 32105	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383   = wceq 1475  Ⅎwnf 1699   ∈ wcel 1977  {crab 2900

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-rab 2905

This theorem is referenced by:  bj-rabeqbid  32109  bj-rabeqbida  32110  bj-inrab2  32116