Theorem unass 4093

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 4076	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)))
2		elun 4076	. . . 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 4076	. . . . 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 279	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶))
8	1, 3, 7	3bitrri 301	. 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)))
9	8	uneqri 4078	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Syntax hints:   ∨ wo 844   = wceq 1538   ∈ wcel 2111   ∪ cun 3879

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-v 3443  df-un 3886

This theorem is referenced by:  un12  4094  un23  4095  un4  4096  dfif5  4441  qdass  4649  qdassr  4650  ssunpr  4725  oarec  8171  domunfican  8775  djuassen  9589  prunioo  12859  ioojoin  12861  fzosplitpr  13141  s4prop  14263  lcmfun  15979  phlstr  16645  prdsvalstr  16718  mreexexlem2d  16908  mreexexlem4d  16910  pwmnd  18094  ordtbas  21797  reconnlem1  23431  lhop  24619  plyun0  24794  ex-un  28209  ex-pw  28214  indifundif  30297  subfacp1lem1  32539  poimirlem3  35060  poimirlem4  35061  poimirlem16  35073  poimirlem19  35076  dfrcl2  40375  corcltrcl  40440  cotrclrcl  40443  frege131d  40465