Theorem rabeqd 3473

Description: Deduction form of rabeq 3458. Note that contrary to rabeq 3458 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 2833	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4	3	anbi1d 630	. . 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	rabbida4 3470	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537  Ⅎwnf 1781   ∈ wcel 2108  {crab 3443

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444

This theorem is referenced by:  rabeqbida  3474  bj-rabeqbid  36887  bj-inrab2  36894