Intuitionistic Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >  unass GIF version

Theorem unass 3100
 Description: Associative law for union of classes. Exercise 8 of [TakeutiZaring] p. 17. (Contributed by NM, 3-May-1994.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
unass ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))

Proof of Theorem unass
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 elun 3084 . . 3 (𝑥 ∈ (𝐴 ∪ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
2 elun 3084 . . . 4 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32orbi2i 679 . . 3 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
4 elun 3084 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
54orbi1i 680 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶))
6 orass 684 . . . 4 (((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
75, 6bitr2i 174 . . 3 ((𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)) ↔ (𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶))
81, 3, 73bitrri 196 . 2 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵𝐶)))
98uneqri 3085 1 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
 Colors of variables: wff set class Syntax hints:   ∨ wo 629   = wceq 1243   ∈ wcel 1393   ∪ cun 2915 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 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-un 2922 This theorem is referenced by:  un12  3101  un23  3102  un4  3103  qdass  3467  qdassr  3468  rdgisucinc  5972  oasuc  6044  fzosplitprm1  9090
 Copyright terms: Public domain W3C validator