Theorem inrab 2267

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 2264	. . 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 1572	. . 3 |- {x | (x e. A /\ (ph /\ ps))} = {x | ((x e. A /\ ph) /\ (x e. A /\ ps))}
4	1, 3	eqtr4 1495	. 2 |- ({x | (x e. A /\ ph)} i^i {x | (x e. A /\ ps)}) = {x | (x e. A /\ (ph /\ ps))}
5		df-rab 1649	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
6		df-rab 1649	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
7	5, 6	ineq12i 2211	. 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 1649	. 2 |- {x e. A | (ph /\ ps)} = {x | (x e. A /\ (ph /\ ps))}
9	4, 7, 8	3eqtr4 1502	1 |- ({x e. A | ph} i^i {x e. A | ps}) = {x e. A | (ph /\ ps)}

Syntax hints:   /\ wa 223   = wceq 954   e. wcel 956  {cab 1461  {crab 1645   i^i cin 2042

This theorem is referenced by:  iooint 6317  blin 7804

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-rab 1649  df-v 1808  df-in 2047

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