Theorem unrab 2267

Description: Union of two restricted class abstractions.

Assertion

Ref	Expression
unrab	|- ({x e. A | ph} u. {x e. A | ps}) = {x e. A | (ph \/ ps)}

Proof of Theorem unrab

Step	Hyp	Ref	Expression
1		unab 2264	. . 3 |- ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)}) = {x | ((x e. A /\ ph) \/ (x e. A /\ ps))}
2		andi 603	. . . 4 |- ((x e. A /\ (ph \/ ps)) <-> ((x e. A /\ ph) \/ (x e. A /\ ps)))
3	2	abbii 1573	. . 3 |- {x | (x e. A /\ (ph \/ ps))} = {x | ((x e. A /\ ph) \/ (x e. A /\ ps))}
4	1, 3	eqtr4 1496	. 2 |- ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)}) = {x | (x e. A /\ (ph \/ ps))}
5		df-rab 1650	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
6		df-rab 1650	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
7	5, 6	uneq12i 2179	. 2 |- ({x e. A | ph} u. {x e. A | ps}) = ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)})
8		df-rab 1650	. 2 |- {x e. A | (ph \/ ps)} = {x | (x e. A /\ (ph \/ ps))}
9	4, 7, 8	3eqtr4 1503	1 |- ({x e. A | ph} u. {x e. A | ps}) = {x e. A | (ph \/ ps)}

Syntax hints:   \/ wo 222   /\ wa 223   = wceq 955   e. wcel 957  {cab 1462  {crab 1646   u. cun 2042

This theorem is referenced by:  kmlem3 4750

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217  ax-ext 1458

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 980  df-sb 1171  df-clab 1463  df-cleq 1468  df-clel 1471  df-rab 1650  df-v 1809  df-un 2047

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