Theorem unrab 1694

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 1691	. . 3 |- ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)}) = {x | ((x e. A /\ ph) \/ (x e. A /\ ps))}
2		andi 456	. . . . 5 |- ((x e. A /\ (ph \/ ps)) <-> ((x e. A /\ ph) \/ (x e. A /\ ps)))
3	2	biabi 1181	. . . 4 |- {x | (x e. A /\ (ph \/ ps))} = {x | ((x e. A /\ ph) \/ (x e. A /\ ps))}
4	3	cleqcomi 1105	. . 3 |- {x | ((x e. A /\ ph) \/ (x e. A /\ ps))} = {x | (x e. A /\ (ph \/ ps))}
5	1, 4	eqtr 1119	. 2 |- ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)}) = {x | (x e. A /\ (ph \/ ps))}
6		df-rab 1208	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
7		df-rab 1208	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
8	6, 7	uneq12i 1609	. 2 |- ({x e. A | ph} u. {x e. A | ps}) = ({x | (x e. A /\ ph)} u. {x | (x e. A /\ ps)})
9		df-rab 1208	. 2 |- {x e. A | (ph \/ ps)} = {x | (x e. A /\ (ph \/ ps))}
10	5, 8, 9	3eqtr4 1126	1 |- ({x e. A | ph} u. {x e. A | ps}) = {x e. A | (ph \/ ps)}

Syntax hints:   \/ wo 195   /\ wa 196  {cab 1090   = wceq 1091   e. wcel 1092  {crab 1204   u. cun 1485

This theorem is referenced by:  kmlem3 3582

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rab 1208  df-v 1349  df-un 1490

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