Theorem rabeqd 3428

Description: Deduction form of rabeq 3414. Note that contrary to rabeq 3414 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 2826	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4	3	anbi1d 632	. . 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 3425	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1542  Ⅎwnf 1785   ∈ wcel 2114  {crab 3400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3401

This theorem is referenced by:  rabeqbida  3429  bj-rabeqbid  37097  bj-inrab2  37104  smfinfmpt  47099