Theorem indi 1676

Description: Distributive law for intersection over union. Exercise 10 of [TakeutiZaring] p. 17.

Assertion

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

Proof of Theorem indi

Step	Hyp	Ref	Expression
1		andi 456	. . . 4 ⊢ ((x ∈ A ∧ (x ∈ B ∨ x ∈ C)) ↔ ((x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)))
2		elun 1601	. . . . 5 ⊢ (x ∈ (B ∪ C) ↔ (x ∈ B ∨ x ∈ C))
3	2	anbi2i 367	. . . 4 ⊢ ((x ∈ A ∧ x ∈ (B ∪ C)) ↔ (x ∈ A ∧ (x ∈ B ∨ x ∈ C)))
4		elin 1635	. . . . 5 ⊢ (x ∈ (A ∩ B) ↔ (x ∈ A ∧ x ∈ B))
5		elin 1635	. . . . 5 ⊢ (x ∈ (A ∩ C) ↔ (x ∈ A ∧ x ∈ C))
6	4, 5	orbi12i 216	. . . 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		elin 1635	. . 3 ⊢ (x ∈ (A ∩ (B ∪ C)) ↔ (x ∈ A ∧ x ∈ (B ∪ C)))
9		elun 1601	. . 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:  indir 1678  difindi 1683  undisj2 1740  difdifdir 1765  resundi 2582  kmlem2 3581

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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