Theorem undi 4201

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 3897	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶))
2	1	orbi2i 910	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)))
3		ordi 1003	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
4		elin 3897	. . . 4 ⊢ (𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)))
5		elun 4076	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6		elun 4076	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
7	5, 6	anbi12i 629	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝑥 ∈ (𝐴 ∪ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)))
8	4, 7	bitr2i 279	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
9	2, 3, 8	3bitri 300	. 2 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∩ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)))
10	9	uneqri 4078	1 ⊢ (𝐴 ∪ (𝐵 ∩ 𝐶)) = ((𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶))

Syntax hints:   ∧ wa 399   ∨ wo 844   = wceq 1538   ∈ wcel 2111   ∪ cun 3879   ∩ cin 3880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-v 3443  df-un 3886  df-in 3888

This theorem is referenced by:  undir  4203  dfif4  4440  dfif5  4441