Theorem undisj1 3549

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 3538	. 2 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
2		indir 3453	. . 3 ⊢ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶))
3	2	eqeq1i 2237	. 2 ⊢ (((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 187	1 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

Syntax hints:   ∧ wa 104   ↔ wb 105   = wceq 1395   ∪ cun 3195   ∩ cin 3196  ∅c0 3491

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-dif 3199  df-un 3201  df-in 3203  df-ss 3210  df-nul 3492

This theorem is referenced by:  funtp  5373