Theorem undir 1679

Description: Distributive law for union over intersection. Theorem 29 of [Suppes] p. 27.

Assertion

Ref	Expression
undir	⊢ ((A ∩ B) ∪ C) = ((A ∪ C) ∩ (B ∪ C))

Proof of Theorem undir

Step	Hyp	Ref	Expression
1		undi 1677	. 2 ⊢ (C ∪ (A ∩ B)) = ((C ∪ A) ∩ (C ∪ B))
2		uncom 1604	. 2 ⊢ ((A ∩ B) ∪ C) = (C ∪ (A ∩ B))
3		uncom 1604	. . 3 ⊢ (A ∪ C) = (C ∪ A)
4		uncom 1604	. . 3 ⊢ (B ∪ C) = (C ∪ B)
5		ineq12 1640	. . 3 ⊢ (((A ∪ C) = (C ∪ A) ∧ (B ∪ C) = (C ∪ B)) → ((A ∪ C) ∩ (B ∪ C)) = ((C ∪ A) ∩ (C ∪ B)))
6	3, 4, 5	mp2an 520	. 2 ⊢ ((A ∪ C) ∩ (B ∪ C)) = ((C ∪ A) ∩ (C ∪ B))
7	1, 2, 6	3eqtr4 1126	1 ⊢ ((A ∩ B) ∪ C) = ((A ∪ C) ∩ (B ∪ C))

Syntax hints:   = wceq 1091   ∪ cun 1485   ∩ cin 1486

This theorem is referenced by:  undif1 1761

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  df-in 1491

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