Theorem unass 3125

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 3109	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)))
2		elun 3109	. . . 4 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
3	2	orbi2i 687	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ (𝐵 ∪ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
4		elun 3109	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
5	4	orbi1i 688	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶))
6		orass 692	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
7	5, 6	bitr2i 178	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∨ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶))
8	1, 3, 7	3bitrri 200	. 2 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∨ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∪ (𝐵 ∪ 𝐶)))
9	8	uneqri 3110	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))

Syntax hints:   ∨ wo 637   = wceq 1257   ∈ wcel 1407   ∪ cun 2940

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036

This theorem depends on definitions:  df-bi 114  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-v 2574  df-un 2947

This theorem is referenced by:  un12  3126  un23  3127  un4  3128  qdass  3492  qdassr  3493  rdgisucinc  6000  oasuc  6072  fzosplitprm1  9162