Metamath Proof Explorer


Theorem incom

Description: Commutative law for intersection of classes. Exercise 7 of TakeutiZaring p. 17. (Contributed by NM, 21-Jun-1993) (Proof shortened by SN, 12-Dec-2023)

Ref Expression
Assertion incom
|- ( A i^i B ) = ( B i^i A )

Proof

Step Hyp Ref Expression
1 rabswap
 |-  { x e. A | x e. B } = { x e. B | x e. A }
2 dfin5
 |-  ( A i^i B ) = { x e. A | x e. B }
3 dfin5
 |-  ( B i^i A ) = { x e. B | x e. A }
4 1 2 3 3eqtr4i
 |-  ( A i^i B ) = ( B i^i A )