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Theorem unass 3155
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 3139 . . 3 (𝑥 ∈ (𝐴 ∪ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
2 elun 3139 . . . 4 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32orbi2i 714 . . 3 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
4 elun 3139 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
54orbi1i 715 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶))
6 orass 719 . . . 4 (((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
75, 6bitr2i 183 . . 3 ((𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)) ↔ (𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶))
81, 3, 73bitrri 205 . 2 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵𝐶)))
98uneqri 3140 1 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
Colors of variables: wff set class
Syntax hints:  wo 664   = wceq 1289  wcel 1438  cun 2995
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-v 2621  df-un 3001
This theorem is referenced by:  un12  3156  un23  3157  un4  3158  qdass  3534  qdassr  3535  rdgisucinc  6132  oasuc  6207  unfidisj  6612  undifdc  6614  fzosplitprm1  9610  hashunlem  10177
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