Theorem unundi 3425

Description: Union distributes over itself. (Contributed by NM, 17-Aug-2004.)

Assertion

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

Proof of Theorem unundi

Step	Hyp	Ref	Expression
1		unidm 3408	. . 3 ⊢ (A ∪ A) = A
2	1	uneq1i 3415	. 2 ⊢ ((A ∪ A) ∪ (B ∪ C)) = (A ∪ (B ∪ C))
3		un4 3424	. 2 ⊢ ((A ∪ A) ∪ (B ∪ C)) = ((A ∪ B) ∪ (A ∪ C))
4	2, 3	eqtr3i 2375	1 ⊢ (A ∪ (B ∪ C)) = ((A ∪ B) ∪ (A ∪ C))

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:  dfif5  3675