Theorem difab 3896

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

Assertion

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

Proof of Theorem difab

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2609	. . 3 |- ( y e. { x | ( ph /\ -. ps ) } <-> [ y / x ] ( ph /\ -. ps ) )
2		sban 2399	. . 3 |- ( [ y / x ] ( ph /\ -. ps ) <-> ( [ y / x ] ph /\ [ y / x ] -. ps ) )
3		df-clab 2609	. . . . 5 |- ( y e. { x | ph } <-> [ y / x ] ph )
4	3	bicomi 214	. . . 4 |- ( [ y / x ] ph <-> y e. { x | ph } )
5		sbn 2391	. . . . 5 |- ( [ y / x ] -. ps <-> -. [ y / x ] ps )
6		df-clab 2609	. . . . 5 |- ( y e. { x | ps } <-> [ y / x ] ps )
7	5, 6	xchbinxr 325	. . . 4 |- ( [ y / x ] -. ps <-> -. y e. { x | ps } )
8	4, 7	anbi12i 733	. . 3 |- ( ( [ y / x ] ph /\ [ y / x ] -. ps ) <-> ( y e. { x | ph } /\ -. y e. { x | ps } ) )
9	1, 2, 8	3bitrri 287	. 2 |- ( ( y e. { x | ph } /\ -. y e. { x | ps } ) <-> y e. { x | ( ph /\ -. ps ) } )
10	9	difeqri 3730	1 |- ( { x | ph } \ { x | ps } ) = { x | ( ph /\ -. ps ) }

Syntax hints:   -. wn 3   /\ wa 384   = wceq 1483   [wsb 1880   e. wcel 1990   {cab 2608   \ cdif 3571

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577

This theorem is referenced by:  notab  3897  difrab  3901  notrab  3904