Metamath Proof Explorer


Theorem inteq

Description: Equality law for intersection. (Contributed by NM, 13-Sep-1999)

Ref Expression
Assertion inteq
|- ( A = B -> |^| A = |^| B )

Proof

Step Hyp Ref Expression
1 raleq
 |-  ( A = B -> ( A. y e. A x e. y <-> A. y e. B x e. y ) )
2 1 abbidv
 |-  ( A = B -> { x | A. y e. A x e. y } = { x | A. y e. B x e. y } )
3 dfint2
 |-  |^| A = { x | A. y e. A x e. y }
4 dfint2
 |-  |^| B = { x | A. y e. B x e. y }
5 2 3 4 3eqtr4g
 |-  ( A = B -> |^| A = |^| B )