Theorem undisj1 3466

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 3455	. 2 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
2		indir 3371	. . 3 ⊢ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶))
3	2	eqeq1i 2173	. 2 ⊢ (((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 186	1 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

Syntax hints:   ∧ wa 103   ↔ wb 104   = wceq 1343   ∪ cun 3114   ∩ cin 3115  ∅c0 3409

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-dif 3118  df-un 3120  df-in 3122  df-ss 3129  df-nul 3410

This theorem is referenced by:  funtp  5241