Theorem unass 3421

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 3221	. . 3 ⊢ (x ∈ (A ∪ (B ∪ C)) ↔ (x ∈ A ∨ x ∈ (B ∪ C)))
2		elun 3221	. . . 4 ⊢ (x ∈ (B ∪ C) ↔ (x ∈ B ∨ x ∈ C))
3	2	orbi2i 505	. . 3 ⊢ ((x ∈ A ∨ x ∈ (B ∪ C)) ↔ (x ∈ A ∨ (x ∈ B ∨ x ∈ C)))
4		elun 3221	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
5	4	orbi1i 506	. . . 4 ⊢ ((x ∈ (A ∪ B) ∨ x ∈ C) ↔ ((x ∈ A ∨ x ∈ B) ∨ x ∈ C))
6		orass 510	. . . 4 ⊢ (((x ∈ A ∨ x ∈ B) ∨ x ∈ C) ↔ (x ∈ A ∨ (x ∈ B ∨ x ∈ C)))
7	5, 6	bitr2i 241	. . 3 ⊢ ((x ∈ A ∨ (x ∈ B ∨ x ∈ C)) ↔ (x ∈ (A ∪ B) ∨ x ∈ C))
8	1, 3, 7	3bitrri 263	. 2 ⊢ ((x ∈ (A ∪ B) ∨ x ∈ C) ↔ x ∈ (A ∪ (B ∪ C)))
9	8	uneqri 3407	1 ⊢ ((A ∪ B) ∪ C) = (A ∪ (B ∪ C))

Syntax hints:   ∨ wo 357   = wceq 1642   ∈ wcel 1710   ∪ cun 3208

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215

This theorem is referenced by:  un12  3422  un23  3423  un4  3424  dfif5  3675  qdass  3820  qdassr  3821  ssunpr  3869  addcass  4416