Theorem unass 3361

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 3345	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)))
2		elun 3345	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
3	2	orbi2i 767	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
4		elun 3345	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
5	4	orbi1i 768	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶))
6		orass 772	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
7	5, 6	bitr2i 185	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶))
8	1, 3, 7	3bitrri 207	. 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)))
9	8	uneqri 3346	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Syntax hints:   ∨ wo 713   = wceq 1395   ∈ wcel 2200   ∪ cun 3195

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-un 3201

This theorem is referenced by:  un12  3362  un23  3363  un4  3364  qdass  3763  qdassr  3764  rdgisucinc  6531  oasuc  6610  unfidisj  7084  undifdc  7086  djuassen  7399  fzosplitprm1  10440  hashunlem  11026  prdsvalstrd  13304  plyun0  15410