Theorem indi 3420

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 820	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
2		elin 3356	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3		elin 3356	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
4	2, 3	orbi12i 766	. . . 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 3314	. . . 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 3314	. . 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 3366	1 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))

Syntax hints:   ∧ wa 104   ∨ wo 710   = wceq 1373   ∈ wcel 2176   ∪ cun 3164   ∩ cin 3165

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-v 2774  df-un 3170  df-in 3172

This theorem is referenced by:  indir  3422  undisj2  3519  disjssun  3524  difdifdirss  3545  disjpr2  3697  diftpsn3  3774  resundi  4972  bitsinv1  12273