Theorem indi 3406

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 819	. . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∨ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
2		elin 3342	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3		elin 3342	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶))
4	2, 3	orbi12i 765	. . . 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 3300	. . . 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 3300	. . 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 3352	1 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))

Syntax hints:   ∧ wa 104   ∨ wo 709   = wceq 1364   ∈ wcel 2164   ∪ cun 3151   ∩ cin 3152

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-un 3157  df-in 3159

This theorem is referenced by:  indir  3408  undisj2  3505  disjssun  3510  difdifdirss  3531  disjpr2  3682  diftpsn3  3759  resundi  4955