Theorem undisj2 3473

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 3461	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
2		indi 3374	. . 3 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
3	2	eqeq1i 2178	. 2 ⊢ ((𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅ ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 186	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)

Syntax hints:   ∧ wa 103   ↔ wb 104   = wceq 1348   ∪ cun 3119   ∩ cin 3120  ∅c0 3414

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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-un 3125  df-in 3127  df-ss 3134  df-nul 3415

This theorem is referenced by: (None)