Theorem indi 3502

Description: Distributive law for intersection over union. Exercise 10 of [TakeutiZaring] p. 17. (Contributed by NM, 30-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem indi

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		andi 837	. . . 4 ⊢ ((x ∈ A ∧ (x ∈ B ∨ x ∈ C)) ↔ ((x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)))
2		elin 3220	. . . . 5 ⊢ (x ∈ (A ∩ B) ↔ (x ∈ A ∧ x ∈ B))
3		elin 3220	. . . . 5 ⊢ (x ∈ (A ∩ C) ↔ (x ∈ A ∧ x ∈ C))
4	2, 3	orbi12i 507	. . . 4 ⊢ ((x ∈ (A ∩ B) ∨ x ∈ (A ∩ C)) ↔ ((x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)))
5	1, 4	bitr4i 243	. . 3 ⊢ ((x ∈ A ∧ (x ∈ B ∨ x ∈ C)) ↔ (x ∈ (A ∩ B) ∨ x ∈ (A ∩ C)))
6		elun 3221	. . . 4 ⊢ (x ∈ (B ∪ C) ↔ (x ∈ B ∨ x ∈ C))
7	6	anbi2i 675	. . 3 ⊢ ((x ∈ A ∧ x ∈ (B ∪ C)) ↔ (x ∈ A ∧ (x ∈ B ∨ x ∈ C)))
8		elun 3221	. . 3 ⊢ (x ∈ ((A ∩ B) ∪ (A ∩ C)) ↔ (x ∈ (A ∩ B) ∨ x ∈ (A ∩ C)))
9	5, 7, 8	3bitr4i 268	. 2 ⊢ ((x ∈ A ∧ x ∈ (B ∪ C)) ↔ x ∈ ((A ∩ B) ∪ (A ∩ C)))
10	9	ineqri 3450	1 ⊢ (A ∩ (B ∪ C)) = ((A ∩ B) ∪ (A ∩ C))

Syntax hints:   ∨ wo 357   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ∪ cun 3208   ∩ cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215

This theorem is referenced by:  indir  3504  difindi  3510  undisj2  3604  disjssun  3609  difdifdir  3638  diftpsn3  3850  addcass  4416  resundi  4982  sbthlem1  6204