Theorem inrab 2271

Description: Intersection of two restricted class abstractions.

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		inab 2268	. . 3 |- ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)}) = {x | ((x e. A /\ ph) /\ (x e. A /\ ps))}
2		anandi 510	. . . 4 |- ((x e. A /\ (ph /\ ps)) <-> ((x e. A /\ ph) /\ (x e. A /\ ps)))
3	2	abbii 1575	. . 3 |- {x | (x e. A /\ (ph /\ ps))} = {x | ((x e. A /\ ph) /\ (x e. A /\ ps))}
4	1, 3	eqtr4 1498	. 2 |- ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)}) = {x | (x e. A /\ (ph /\ ps))}
5		df-rab 1652	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
6		df-rab 1652	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
7	5, 6	ineq12i 2215	. 2 |- ({x e. A | ph} i^i {x e. A | ps}) = ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)})
8		df-rab 1652	. 2 |- {x e. A | (ph /\ ps)} = {x | (x e. A /\ (ph /\ ps))}
9	4, 7, 8	3eqtr4 1505	1 |- ({x e. A | ph} i^i {x e. A | ps}) = {x e. A | (ph /\ ps)}

Syntax hints:   /\ wa 223   = wceq 956   e. wcel 958  {cab 1463  {crab 1648   i^i cin 2046

This theorem is referenced by:  iooint 6372  blin 7852

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-rab 1652  df-v 1812  df-in 2051

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