Theorem undisj1 3603

Description: The union of disjoint classes is disjoint. (Contributed by NM, 26-Sep-2004.)

Assertion

Ref	Expression
undisj1	⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∪ B) ∩ C) = ∅)

Proof of Theorem undisj1

Step	Hyp	Ref	Expression
1		un00 3587	. 2 ⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∩ C) ∪ (B ∩ C)) = ∅)
2		indir 3504	. . 3 ⊢ ((A ∪ B) ∩ C) = ((A ∩ C) ∪ (B ∩ C))
3	2	eqeq1i 2360	. 2 ⊢ (((A ∪ B) ∩ C) = ∅ ↔ ((A ∩ C) ∪ (B ∩ C)) = ∅)
4	1, 3	bitr4i 243	1 ⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∪ B) ∩ C) = ∅)

Syntax hints:   ↔ wb 176   ∧ wa 358   = wceq 1642   ∪ cun 3208   ∩ cin 3209  ∅c0 3551

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-in 3214  df-un 3215  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by: (None)