Theorem undi 1677

Description: Distributive law for union over intersection. Exercise 11 of [TakeutiZaring] p. 17.

Assertion

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

Proof of Theorem undi

Step	Hyp	Ref	Expression
1		ordi 452	. . . 4 ⊢ ((x ∈ A ∨ (x ∈ B ∧ x ∈ C)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C)))
2		elin 1635	. . . . 5 ⊢ (x ∈ (B ∩ C) ↔ (x ∈ B ∧ x ∈ C))
3	2	orbi2i 214	. . . 4 ⊢ ((x ∈ A ∨ x ∈ (B ∩ C)) ↔ (x ∈ A ∨ (x ∈ B ∧ x ∈ C)))
4		elun 1601	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
5		elun 1601	. . . . 5 ⊢ (x ∈ (A ∪ C) ↔ (x ∈ A ∨ x ∈ C))
6	4, 5	anbi12i 369	. . . 4 ⊢ ((x ∈ (A ∪ B) ∧ x ∈ (A ∪ C)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C)))
7	1, 3, 6	3bitr4 158	. . 3 ⊢ ((x ∈ A ∨ x ∈ (B ∩ C)) ↔ (x ∈ (A ∪ B) ∧ x ∈ (A ∪ C)))
8		elun 1601	. . 3 ⊢ (x ∈ (A ∪ (B ∩ C)) ↔ (x ∈ A ∨ x ∈ (B ∩ C)))
9		elin 1635	. . 3 ⊢ (x ∈ ((A ∪ B) ∩ (A ∪ C)) ↔ (x ∈ (A ∪ B) ∧ x ∈ (A ∪ C)))
10	7, 8, 9	3bitr4 158	. 2 ⊢ (x ∈ (A ∪ (B ∩ C)) ↔ x ∈ ((A ∪ B) ∩ (A ∪ C)))
11	10	cleqri 1101	1 ⊢ (A ∪ (B ∩ C)) = ((A ∪ B) ∩ (A ∪ C))

Syntax hints:   ∨ wo 195   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ∪ cun 1485   ∩ cin 1486

This theorem is referenced by:  undir 1679

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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