Theorem undisj2 3510

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 3498	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
2		indi 3411	. . 3 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
3	2	eqeq1i 2204	. 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 1364   ∪ cun 3155   ∩ cin 3156  ∅c0 3451

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 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-dif 3159  df-un 3161  df-in 3163  df-ss 3170  df-nul 3452

This theorem is referenced by: (None)