Theorem inrab 3353

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 2426	. . 3 |- { x e. A | ph } = { x | ( x e. A /\ ph ) }
2		df-rab 2426	. . 3 |- { x e. A | ps } = { x | ( x e. A /\ ps ) }
3	1, 2	ineq12i 3280	. 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 2426	. . 3 |- { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
5		inab 3349	. . . 4 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
6		anandi 580	. . . . 5 |- ( ( x e. A /\ ( ph /\ ps ) ) <-> ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) )
7	6	abbii 2256	. . . 4 |- { x | ( x e. A /\ ( ph /\ ps ) ) } = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
8	5, 7	eqtr4i 2164	. . 3 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( x e. A /\ ( ph /\ ps ) ) }
9	4, 8	eqtr4i 2164	. 2 |- { x e. A | ( ph /\ ps ) } = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
10	3, 9	eqtr4i 2164	1 |- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Syntax hints:   /\ wa 103   = wceq 1332   e. wcel 1481   {cab 2126   {crab 2421   i^i cin 3075

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-rab 2426  df-v 2691  df-in 3082

This theorem is referenced by:  rabnc  3400  iooinsup  11078  phiprmpw  11934  unennn  11946