Theorem rabswap 2685

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 266	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
2	1	abbii 2321	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
3		df-rab 2493	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
4		df-rab 2493	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)}
5	2, 3, 4	3eqtr4i 2236	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵} = {𝑥 ∈ 𝐵 ∣ 𝑥 ∈ 𝐴}

Syntax hints:   ∧ wa 104   = wceq 1373   ∈ wcel 2176  {cab 2191  {crab 2488

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-11 1529  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-rab 2493

This theorem is referenced by: (None)