Theorem undi 4235

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 3915	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
2	1	orbi2i 912	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
3		ordi 1007	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
4		elin 3915	. . . 4 ⊢ (𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)))
5		elun 4103	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6		elun 4103	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
7	5, 6	anbi12i 628	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
8	4, 7	bitr2i 276	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
9	2, 3, 8	3bitri 297	. 2 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
10	9	uneqri 4106	1 ⊢ (𝐴 ∪ (𝐵 ∩ 𝐶)) = ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶))

Syntax hints:   ∧ wa 395   ∨ wo 847   = wceq 1541   ∈ wcel 2113   ∪ cun 3897   ∩ cin 3898

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2713  df-cleq 2726  df-clel 2809  df-v 3440  df-un 3904  df-in 3906

This theorem is referenced by:  undir  4237  dfif4  4493  dfif5  4494