Theorem un4 1618

Description: A rearrangement of the union of 4 classes.

Assertion

Ref	Expression
un4	⊢ ((A ∪ B) ∪ (C ∪ D)) = ((A ∪ C) ∪ (B ∪ D))

Proof of Theorem un4

Step	Hyp	Ref	Expression
1		un12 1616	. . 3 ⊢ (B ∪ (C ∪ D)) = (C ∪ (B ∪ D))
2	1	uneq2i 1608	. 2 ⊢ (A ∪ (B ∪ (C ∪ D))) = (A ∪ (C ∪ (B ∪ D)))
3		unass 1615	. 2 ⊢ ((A ∪ B) ∪ (C ∪ D)) = (A ∪ (B ∪ (C ∪ D)))
4		unass 1615	. 2 ⊢ ((A ∪ C) ∪ (B ∪ D)) = (A ∪ (C ∪ (B ∪ D)))
5	2, 3, 4	3eqtr4 1126	1 ⊢ ((A ∪ B) ∪ (C ∪ D)) = ((A ∪ C) ∪ (B ∪ D))

Syntax hints:   = wceq 1091   ∪ cun 1485

This theorem is referenced by:  unundi 1619  unundir 1620  xpun 2463

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-v 1349  df-un 1490

metamath.org