Theorem difab

Description: Difference of two class abstractions. (Contributed by NM, 23-Oct-2004) (Proof shortened by Andrew Salmon, 26-Jun-2011)

	Ref	Expression
Assertion	difab	|- ( { x | ph } \ { x | ps } ) = { x | ( ph /\ -. ps ) }

Proof

Step	Hyp	Ref	Expression
1		df-clab	|- ( y e. { x | ( ph /\ -. ps ) } <-> [ y / x ] ( ph /\ -. ps ) )
2		sban	|- ( [ y / x ] ( ph /\ -. ps ) <-> ( [ y / x ] ph /\ [ y / x ] -. ps ) )
3		df-clab	|- ( y e. { x | ph } <-> [ y / x ] ph )
4	3	bicomi	|- ( [ y / x ] ph <-> y e. { x | ph } )
5		sbn	|- ( [ y / x ] -. ps <-> -. [ y / x ] ps )
6		df-clab	|- ( y e. { x | ps } <-> [ y / x ] ps )
7	5 6	xchbinxr	|- ( [ y / x ] -. ps <-> -. y e. { x | ps } )
8	4 7	anbi12i	|- ( ( [ y / x ] ph /\ [ y / x ] -. ps ) <-> ( y e. { x | ph } /\ -. y e. { x | ps } ) )
9	1 2 8	3bitrri	|- ( ( y e. { x | ph } /\ -. y e. { x | ps } ) <-> y e. { x | ( ph /\ -. ps ) } )
10	9	difeqri	|- ( { x | ph } \ { x | ps } ) = { x | ( ph /\ -. ps ) }