Theorem undisj2 3530

Description: The union of disjoint classes is disjoint. (Contributed by NM, 13-Sep-2004.)

Assertion

Ref	Expression
undisj2	⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)

Proof of Theorem undisj2

Step	Hyp	Ref	Expression
1		un00 3518	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
2		indi 3431	. . 3 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
3	2	eqeq1i 2217	. 2 ⊢ ((𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅ ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 187	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)

Syntax hints:   ∧ wa 104   ↔ wb 105   = wceq 1375   ∪ cun 3175   ∩ cin 3176  ∅c0 3471

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 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-v 2781  df-dif 3179  df-un 3181  df-in 3183  df-ss 3190  df-nul 3472

This theorem is referenced by: (None)