Theorem undi 3179

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 3120	. . . 4 ⊢ (x ∈ (B ∩ 𝐶) ↔ (x ∈ B ∧ x ∈ 𝐶))
2	1	orbi2i 678	. . 3 ⊢ ((x ∈ A ∨ x ∈ (B ∩ 𝐶)) ↔ (x ∈ A ∨ (x ∈ B ∧ x ∈ 𝐶)))
3		ordi 728	. . 3 ⊢ ((x ∈ A ∨ (x ∈ B ∧ x ∈ 𝐶)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶)))
4		elin 3120	. . . 4 ⊢ (x ∈ ((A ∪ B) ∩ (A ∪ 𝐶)) ↔ (x ∈ (A ∪ B) ∧ x ∈ (A ∪ 𝐶)))
5		elun 3078	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
6		elun 3078	. . . . 5 ⊢ (x ∈ (A ∪ 𝐶) ↔ (x ∈ A ∨ x ∈ 𝐶))
7	5, 6	anbi12i 433	. . . 4 ⊢ ((x ∈ (A ∪ B) ∧ x ∈ (A ∪ 𝐶)) ↔ ((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶)))
8	4, 7	bitr2i 174	. . 3 ⊢ (((x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ 𝐶)) ↔ x ∈ ((A ∪ B) ∩ (A ∪ 𝐶)))
9	2, 3, 8	3bitri 195	. 2 ⊢ ((x ∈ A ∨ x ∈ (B ∩ 𝐶)) ↔ x ∈ ((A ∪ B) ∩ (A ∪ 𝐶)))
10	9	uneqri 3079	1 ⊢ (A ∪ (B ∩ 𝐶)) = ((A ∪ B) ∩ (A ∪ 𝐶))

Syntax hints:   ∧ wa 97   ∨ wo 628   = wceq 1242   ∈ wcel 1390   ∪ cun 2909   ∩ cin 2910

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-in 2918

This theorem is referenced by:  undir  3181