Theorem inrab 3497

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 2531	. . 3 |- { x e. A | ph } = { x | ( x e. A /\ ph ) }
2		df-rab 2531	. . 3 |- { x e. A | ps } = { x | ( x e. A /\ ps ) }
3	1, 2	ineq12i 3424	. 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 2531	. . 3 |- { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
5		inab 3493	. . . 4 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
6		anandi 594	. . . . 5 |- ( ( x e. A /\ ( ph /\ ps ) ) <-> ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) )
7	6	abbii 2350	. . . 4 |- { x | ( x e. A /\ ( ph /\ ps ) ) } = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
8	5, 7	eqtr4i 2258	. . 3 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( x e. A /\ ( ph /\ ps ) ) }
9	4, 8	eqtr4i 2258	. 2 |- { x e. A | ( ph /\ ps ) } = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
10	3, 9	eqtr4i 2258	1 |- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Syntax hints:   /\ wa 104   = wceq 1398   e. wcel 2205   {cab 2220   {crab 2526   i^i cin 3213

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-rab 2531  df-v 2817  df-in 3220

This theorem is referenced by:  rabnc  3545  hashfibclem  11231  iooinsup  11987  phiprmpw  12944  unennn  13232  dfrhm2  14399  lgsquadlem2  16077  umgrislfupgrenlem  16251