Theorem rabeqd 3442

Description: Deduction form of rabeq 3428. Note that contrary to rabeq 3428 it has no disjoint variable condition. (Contributed by BJ, 27-Apr-2019.)

Hypotheses

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

Assertion

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

Proof of Theorem rabeqd

Step	Hyp	Ref	Expression
1		rabeqd.nf	. 2 ⊢ Ⅎ𝑥𝜑
2		rabeqd.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
3		eleq2 2850	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4	3	anbi1d 642	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜓)))
5	2, 4	syl 18	. 2 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜓)))
6	1, 5	rabbida4 3439	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1568  Ⅎwnf 1811   ∈ wcel 2141  {crab 3414

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143  ax-9 2151  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1808  df-nf 1812  df-sb 2095  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3415

This theorem is referenced by:  rabeqbida  3443  bj-rabeqbid  37522  bj-inrab2  37530  smfinfmpt  47503