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Theorem unass 3203
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 3187 . . 3 (𝑥 ∈ (𝐴 ∪ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
2 elun 3187 . . . 4 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32orbi2i 736 . . 3 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
4 elun 3187 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
54orbi1i 737 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶))
6 orass 741 . . . 4 (((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
75, 6bitr2i 184 . . 3 ((𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)) ↔ (𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶))
81, 3, 73bitrri 206 . 2 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵𝐶)))
98uneqri 3188 1 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
Colors of variables: wff set class
Syntax hints:  wo 682   = wceq 1316  wcel 1465  cun 3039
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-un 3045
This theorem is referenced by:  un12  3204  un23  3205  un4  3206  qdass  3590  qdassr  3591  rdgisucinc  6250  oasuc  6328  unfidisj  6778  undifdc  6780  djuassen  7041  fzosplitprm1  9979  hashunlem  10518
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