Theorem un23 3423

Description: A rearrangement of union. (Contributed by NM, 12-Aug-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
un23	⊢ ((A ∪ B) ∪ C) = ((A ∪ C) ∪ B)

Proof of Theorem un23

Step	Hyp	Ref	Expression
1		unass 3421	. 2 ⊢ ((A ∪ B) ∪ C) = (A ∪ (B ∪ C))
2		un12 3422	. 2 ⊢ (A ∪ (B ∪ C)) = (B ∪ (A ∪ C))
3		uncom 3409	. 2 ⊢ (B ∪ (A ∪ C)) = ((A ∪ C) ∪ B)
4	1, 2, 3	3eqtri 2377	1 ⊢ ((A ∪ B) ∪ C) = ((A ∪ C) ∪ B)

Syntax hints:   = wceq 1642   ∪ 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:  ssunpr  3869