Theorem inrab 3528

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 2624	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
2		df-rab 2624	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
3	1, 2	ineq12i 3456	. 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 2624	. . 3 |- {x e. A | (ph /\ ps)} = {x | (x e. A /\ (ph /\ ps))}
5		inab 3523	. . . 4 |- ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)}) = {x | ((x e. A /\ ph) /\ (x e. A /\ ps))}
6		anandi 801	. . . . 5 |- ((x e. A /\ (ph /\ ps)) <-> ((x e. A /\ ph) /\ (x e. A /\ ps)))
7	6	abbii 2466	. . . 4 |- {x | (x e. A /\ (ph /\ ps))} = {x | ((x e. A /\ ph) /\ (x e. A /\ ps))}
8	5, 7	eqtr4i 2376	. . 3 |- ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)}) = {x | (x e. A /\ (ph /\ ps))}
9	4, 8	eqtr4i 2376	. 2 |- {x e. A | (ph /\ ps)} = ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)})
10	3, 9	eqtr4i 2376	1 |- ({x e. A | ph} i^i {x e. A | ps}) = {x e. A | (ph /\ ps)}

Syntax hints:   /\ wa 358   = wceq 1642   e. wcel 1710  {cab 2339  {crab 2619   i^i cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214

This theorem is referenced by:  rabnc  3575