Theorem inrab 3236

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 2357	. . 3 |- { x e. A | ph } = { x | ( x e. A /\ ph ) }
2		df-rab 2357	. . 3 |- { x e. A | ps } = { x | ( x e. A /\ ps ) }
3	1, 2	ineq12i 3165	. 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 2357	. . 3 |- { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
5		inab 3232	. . . 4 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
6		anandi 554	. . . . 5 |- ( ( x e. A /\ ( ph /\ ps ) ) <-> ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) )
7	6	abbii 2194	. . . 4 |- { x | ( x e. A /\ ( ph /\ ps ) ) } = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
8	5, 7	eqtr4i 2104	. . 3 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( x e. A /\ ( ph /\ ps ) ) }
9	4, 8	eqtr4i 2104	. 2 |- { x e. A | ( ph /\ ps ) } = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
10	3, 9	eqtr4i 2104	1 |- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Syntax hints:   /\ wa 102   = wceq 1284   e. wcel 1433   {cab 2067   {crab 2352   i^i cin 2972

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-rab 2357  df-v 2603  df-in 2979

This theorem is referenced by:  rabnc  3277