Theorem unass 3284

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 3268	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)))
2		elun 3268	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
3	2	orbi2i 757	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
4		elun 3268	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
5	4	orbi1i 758	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶))
6		orass 762	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
7	5, 6	bitr2i 184	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶))
8	1, 3, 7	3bitrri 206	. 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)))
9	8	uneqri 3269	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Syntax hints:   ∨ wo 703   = wceq 1348   ∈ wcel 2141   ∪ cun 3119

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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125

This theorem is referenced by:  un12  3285  un23  3286  un4  3287  qdass  3680  qdassr  3681  rdgisucinc  6364  oasuc  6443  unfidisj  6899  undifdc  6901  djuassen  7194  fzosplitprm1  10190  hashunlem  10739