Theorem rabswap 2791

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 437	. . 3 ⊢ ((x ∈ A ∧ x ∈ B) ↔ (x ∈ B ∧ x ∈ A))
2	1	abbii 2466	. 2 ⊢ {x ∣ (x ∈ A ∧ x ∈ B)} = {x ∣ (x ∈ B ∧ x ∈ A)}
3		df-rab 2624	. 2 ⊢ {x ∈ A ∣ x ∈ B} = {x ∣ (x ∈ A ∧ x ∈ B)}
4		df-rab 2624	. 2 ⊢ {x ∈ B ∣ x ∈ A} = {x ∣ (x ∈ B ∧ x ∈ A)}
5	2, 3, 4	3eqtr4i 2383	1 ⊢ {x ∈ A ∣ x ∈ B} = {x ∈ B ∣ x ∈ A}

Syntax hints:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339  {crab 2619

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-rab 2624

This theorem is referenced by: (None)