Theorem inrab 3445

Description: Intersection of two restricted class abstractions. (Contributed by NM, 1-Sep-2006.)

Assertion

Ref	Expression
inrab	|- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Proof of Theorem inrab

Step	Hyp	Ref	Expression
1		df-rab 2493	. . 3 |- { x e. A | ph } = { x | ( x e. A /\ ph ) }
2		df-rab 2493	. . 3 |- { x e. A | ps } = { x | ( x e. A /\ ps ) }
3	1, 2	ineq12i 3372	. 2 |- ( { x e. A | ph } i^i { x e. A | ps } ) = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
4		df-rab 2493	. . 3 |- { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
5		inab 3441	. . . 4 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
6		anandi 590	. . . . 5 |- ( ( x e. A /\ ( ph /\ ps ) ) <-> ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) )
7	6	abbii 2321	. . . 4 |- { x | ( x e. A /\ ( ph /\ ps ) ) } = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
8	5, 7	eqtr4i 2229	. . 3 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( x e. A /\ ( ph /\ ps ) ) }
9	4, 8	eqtr4i 2229	. 2 |- { x e. A | ( ph /\ ps ) } = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
10	3, 9	eqtr4i 2229	1 |- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Syntax hints:   /\ wa 104   = wceq 1373   e. wcel 2176   {cab 2191   {crab 2488   i^i cin 3165

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-io 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-rab 2493  df-v 2774  df-in 3172

This theorem is referenced by:  rabnc  3493  iooinsup  11588  phiprmpw  12544  unennn  12768  dfrhm2  13916  lgsquadlem2  15555