Theorem indi 3456

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 826	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
2		elin 3392	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3		elin 3392	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
4	2, 3	orbi12i 772	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
5	1, 4	bitr4i 187	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)))
6		elun 3350	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
7	6	anbi2i 457	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
8		elun 3350	. . 3 ⊢ (𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∨ 𝑥 ∈ (𝐴 ∩ 𝐶)))
9	5, 7, 8	3bitr4i 212	. 2 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ 𝑥 ∈ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)))
10	9	ineqri 3402	1 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))

Syntax hints:   ∧ wa 104   ∨ wo 716   = wceq 1398   ∈ wcel 2202   ∪ cun 3199   ∩ cin 3200

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-un 3205  df-in 3207

This theorem is referenced by:  indir  3458  undisj2  3555  disjssun  3560  difdifdirss  3581  disjpr2  3737  diftpsn3  3819  resundi  5032  bitsinv1  12603