Theorem indi 3369

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 808	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
2		elin 3305	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3		elin 3305	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
4	2, 3	orbi12i 754	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
5	1, 4	bitr4i 186	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)))
6		elun 3263	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
7	6	anbi2i 453	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
8		elun 3263	. . 3 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)))
9	5, 7, 8	3bitr4i 211	. 2 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)))
10	9	ineqri 3315	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:  indir  3371  undisj2  3467  disjssun  3472  difdifdirss  3493  disjpr2  3640  diftpsn3  3714  resundi  4897