Theorem rabswap 2609

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 264	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
2	1	abbii 2255	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
3		df-rab 2425	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
4		df-rab 2425	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
5	2, 3, 4	3eqtr4i 2170	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴}

Syntax hints:   ∧ wa 103   = wceq 1331   ∈ wcel 1480  {cab 2125  {crab 2420

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-rab 2425

This theorem is referenced by: (None)