Theorem rabswap 3318

Description: Swap with a membership relation in a restricted class abstraction. (Contributed by NM, 4-Jul-2005.)

Assertion

Ref	Expression
rabswap	⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴}

Proof of Theorem rabswap

Step	Hyp	Ref	Expression
1		ancom 453	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
2	1	abbii 2839	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
3		df-rab 3092	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
4		df-rab 3092	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
5	2, 3, 4	3eqtr4i 2807	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴}

Syntax hints:   ∧ wa 387   = wceq 1508   ∈ wcel 2051  {cab 2753  {crab 3087

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1759  ax-4 1773  ax-5 1870  ax-6 1929  ax-7 1966  ax-9 2060  ax-ext 2745

This theorem depends on definitions:  df-bi 199  df-an 388  df-ex 1744  df-sb 2017  df-clab 2754  df-cleq 2766  df-rab 3092

This theorem is referenced by: (None)