Theorem undi 3370

Description: Distributive law for union over intersection. Exercise 11 of [TakeutiZaring] p. 17. (Contributed by NM, 30-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem undi

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elin 3305	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
2	1	orbi2i 752	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
3		ordi 806	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
4		elin 3305	. . . 4 ⊢ (𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)))
5		elun 3263	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6		elun 3263	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
7	5, 6	anbi12i 456	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
8	4, 7	bitr2i 184	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
9	2, 3, 8	3bitri 205	. 2 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
10	9	uneqri 3264	1 ⊢ (𝐴 ∪ (𝐵 ∩ 𝐶)) = ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶))

Syntax hints:   ∧ wa 103   ∨ wo 698   = wceq 1343   ∈ wcel 2136   ∪ cun 3114   ∩ cin 3115

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-un 3120  df-in 3122

This theorem is referenced by:  undir  3372  undifdc  6889