Theorem unass 4166

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	⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Proof of Theorem unass

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elun 4148	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)))
2		elun 4148	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
3	2	orbi2i 910	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
4		elun 4148	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
5	4	orbi1i 911	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶))
6		orass 919	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
7	5, 6	bitr2i 276	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶))
8	1, 3, 7	3bitrri 298	. 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)))
9	8	uneqri 4151	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Syntax hints:   ∨ wo 844   = wceq 1540   ∈ wcel 2105   ∪ cun 3946

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-un 3953

This theorem is referenced by:  un12  4167  un23  4168  un4  4169  dfif5  4544  qdass  4757  qdassr  4758  ssunpr  4835  oarec  8565  unfi  9175  domunfican  9323  djuassen  10176  prunioo  13463  ioojoin  13465  fzosplitpr  13746  s4prop  14866  lcmfun  16587  phlstr  17296  prdsvalstr  17403  mreexexlem2d  17594  mreexexlem4d  17596  pwmnd  18855  ordtbas  22917  reconnlem1  24563  lhop  25769  plyun0  25947  addsasslem2  27727  ex-un  29945  ex-pw  29950  indifundif  32030  subfacp1lem1  34469  poimirlem3  36795  poimirlem4  36796  poimirlem16  36808  poimirlem19  36811  dfrcl2  42728  corcltrcl  42793  cotrclrcl  42796  frege131d  42818