Theorem rabswap 2546

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 263	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
2	1	abbii 2204	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
3		df-rab 2369	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
4		df-rab 2369	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
5	2, 3, 4	3eqtr4i 2119	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴}

Syntax hints:   ∧ wa 103   = wceq 1290   ∈ wcel 1439  {cab 2075  {crab 2364

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-11 1443  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-tru 1293  df-nf 1396  df-sb 1694  df-clab 2076  df-cleq 2082  df-rab 2369

This theorem is referenced by: (None)