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Theorem unass 3228
 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 3212 . . 3 (𝑥 ∈ (𝐴 ∪ (𝐵𝐶)) ↔ (𝑥𝐴𝑥 ∈ (𝐵𝐶)))
2 elun 3212 . . . 4 (𝑥 ∈ (𝐵𝐶) ↔ (𝑥𝐵𝑥𝐶))
32orbi2i 751 . . 3 ((𝑥𝐴𝑥 ∈ (𝐵𝐶)) ↔ (𝑥𝐴 ∨ (𝑥𝐵𝑥𝐶)))
4 elun 3212 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
54orbi1i 752 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∨ 𝑥𝐶) ↔ ((𝑥𝐴𝑥𝐵) ∨ 𝑥𝐶))
6 orass 756 . . . 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 3213 1 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
 Colors of variables: wff set class Syntax hints:   ∨ wo 697   = wceq 1331   ∈ wcel 1480   ∪ cun 3064 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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119 This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-un 3070 This theorem is referenced by:  un12  3229  un23  3230  un4  3231  qdass  3615  qdassr  3616  rdgisucinc  6275  oasuc  6353  unfidisj  6803  undifdc  6805  djuassen  7066  fzosplitprm1  10004  hashunlem  10543
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