Theorem rabswap 1774

Description: Swap with a membership relation in a restricted class abstraction.

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 1578	. 2 ⊢ {x∣(x ∈ A ⋀ x ∈ B)} = {x∣(x ∈ B ⋀ x ∈ A)}
3		df-rab 1655	. 2 ⊢ {x ∈ A∣x ∈ B} = {x∣(x ∈ A ⋀ x ∈ B)}
4		df-rab 1655	. 2 ⊢ {x ∈ B∣x ∈ A} = {x∣(x ∈ B ⋀ x ∈ A)}
5	2, 3, 4	3eqtr4 1508	1 ⊢ {x ∈ A∣x ∈ B} = {x ∈ B∣x ∈ A}

Syntax hints:   ⋀ wa 223   = wceq 958   ∈ wcel 960  {cab 1466  {crab 1651

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-rab 1655

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