Theorem inrab 3399

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 2457	. . 3 |- { x e. A | ph } = { x | ( x e. A /\ ph ) }
2		df-rab 2457	. . 3 |- { x e. A | ps } = { x | ( x e. A /\ ps ) }
3	1, 2	ineq12i 3326	. 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 2457	. . 3 |- { x e. A | ( ph /\ ps ) } = { x | ( x e. A /\ ( ph /\ ps ) ) }
5		inab 3395	. . . 4 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
6		anandi 585	. . . . 5 |- ( ( x e. A /\ ( ph /\ ps ) ) <-> ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) )
7	6	abbii 2286	. . . 4 |- { x | ( x e. A /\ ( ph /\ ps ) ) } = { x | ( ( x e. A /\ ph ) /\ ( x e. A /\ ps ) ) }
8	5, 7	eqtr4i 2194	. . 3 |- ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } ) = { x | ( x e. A /\ ( ph /\ ps ) ) }
9	4, 8	eqtr4i 2194	. 2 |- { x e. A | ( ph /\ ps ) } = ( { x | ( x e. A /\ ph ) } i^i { x | ( x e. A /\ ps ) } )
10	3, 9	eqtr4i 2194	1 |- ( { x e. A | ph } i^i { x e. A | ps } ) = { x e. A | ( ph /\ ps ) }

Syntax hints:   /\ wa 103   = wceq 1348   e. wcel 2141   {cab 2156   {crab 2452   i^i cin 3120

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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-rab 2457  df-v 2732  df-in 3127

This theorem is referenced by:  rabnc  3447  iooinsup  11240  phiprmpw  12176  unennn  12352