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Theorem unass 3125
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 3109 . . 3 (𝑥 ∈ (𝐴 ∪ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
2 elun 3109 . . . 4 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32orbi2i 687 . . 3 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
4 elun 3109 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
54orbi1i 688 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶))
6 orass 692 . . . 4 (((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
75, 6bitr2i 178 . . 3 ((𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)) ↔ (𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶))
81, 3, 73bitrri 200 . 2 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵𝐶)))
98uneqri 3110 1 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
Colors of variables: wff set class
Syntax hints:  wo 637   = wceq 1257  wcel 1407  cun 2940
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036
This theorem depends on definitions:  df-bi 114  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-v 2574  df-un 2947
This theorem is referenced by:  un12  3126  un23  3127  un4  3128  qdass  3492  qdassr  3493  rdgisucinc  6000  oasuc  6072  fzosplitprm1  9162
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