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Theorem indi 3178
 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 ∩ (B𝐶)) = ((AB) ∪ (A𝐶))

Proof of Theorem indi
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 andi 730 . . . 4 ((x A (x B x 𝐶)) ↔ ((x A x B) (x A x 𝐶)))
2 elin 3120 . . . . 5 (x (AB) ↔ (x A x B))
3 elin 3120 . . . . 5 (x (A𝐶) ↔ (x A x 𝐶))
42, 3orbi12i 680 . . . 4 ((x (AB) x (A𝐶)) ↔ ((x A x B) (x A x 𝐶)))
51, 4bitr4i 176 . . 3 ((x A (x B x 𝐶)) ↔ (x (AB) x (A𝐶)))
6 elun 3078 . . . 4 (x (B𝐶) ↔ (x B x 𝐶))
76anbi2i 430 . . 3 ((x A x (B𝐶)) ↔ (x A (x B x 𝐶)))
8 elun 3078 . . 3 (x ((AB) ∪ (A𝐶)) ↔ (x (AB) x (A𝐶)))
95, 7, 83bitr4i 201 . 2 ((x A x (B𝐶)) ↔ x ((AB) ∪ (A𝐶)))
109ineqri 3124 1 (A ∩ (B𝐶)) = ((AB) ∪ (A𝐶))
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ∨ wo 628   = wceq 1242   ∈ wcel 1390   ∪ cun 2909   ∩ cin 2910 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-in 2918 This theorem is referenced by:  indir  3180  undisj2  3274  disjssun  3279  difdifdirss  3301  disjpr2  3425  diftpsn3  3496  resundi  4568
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