Theorem difab 3500

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 2225	. . 3 |- ( y e. { x | ( ph /\ -. ps ) } <-> [ y / x ] ( ph /\ -. ps ) )
2		sban 2015	. . 3 |- ( [ y / x ] ( ph /\ -. ps ) <-> ( [ y / x ] ph /\ [ y / x ] -. ps ) )
3		df-clab 2225	. . . . 5 |- ( y e. { x | ph } <-> [ y / x ] ph )
4	3	bicomi 132	. . . 4 |- ( [ y / x ] ph <-> y e. { x | ph } )
5		sbn 2012	. . . . 5 |- ( [ y / x ] -. ps <-> -. [ y / x ] ps )
6		df-clab 2225	. . . . 5 |- ( y e. { x | ps } <-> [ y / x ] ps )
7	5, 6	xchbinxr 694	. . . 4 |- ( [ y / x ] -. ps <-> -. y e. { x | ps } )
8	4, 7	anbi12i 464	. . 3 |- ( ( [ y / x ] ph /\ [ y / x ] -. ps ) <-> ( y e. { x | ph } /\ -. y e. { x | ps } ) )
9	1, 2, 8	3bitrri 207	. 2 |- ( ( y e. { x | ph } /\ -. y e. { x | ps } ) <-> y e. { x | ( ph /\ -. ps ) } )
10	9	difeqri 3349	1 |- ( { x | ph } \ { x | ps } ) = { x | ( ph /\ -. ps ) }

Syntax hints:   -. wn 3   /\ wa 104   = wceq 1402   [wsb 1815   e. wcel 2209   {cab 2224   \ cdif 3217

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222

This theorem is referenced by:  notab  3503  difrab  3507  notrab  3510  imadiflem  5455  imadif  5456