Theorem inab

Description: Intersection of two class abstractions. (Contributed by NM, 29-Sep-2002) (Proof shortened by Andrew Salmon, 26-Jun-2011)

	Ref	Expression
Assertion	inab	|- ( { x | ph } i^i { x | ps } ) = { x | ( ph /\ ps ) }

Proof

Step	Hyp	Ref	Expression
1		sban	|- ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
2		df-clab	|- ( y e. { x | ( ph /\ ps ) } <-> [ y / x ] ( ph /\ ps ) )
3		df-clab	|- ( y e. { x | ph } <-> [ y / x ] ph )
4		df-clab	|- ( y e. { x | ps } <-> [ y / x ] ps )
5	3 4	anbi12i	|- ( ( y e. { x | ph } /\ y e. { x | ps } ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
6	1 2 5	3bitr4ri	|- ( ( y e. { x | ph } /\ y e. { x | ps } ) <-> y e. { x | ( ph /\ ps ) } )
7	6	ineqri	|- ( { x | ph } i^i { x | ps } ) = { x | ( ph /\ ps ) }