Theorem undisj2 4393

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 4373	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
2		indi 4204	. . 3 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
3	2	eqeq1i 2743	. 2 ⊢ ((𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅ ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 277	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)

Syntax hints:   ↔ wb 205   ∧ wa 395   = wceq 1539   ∪ cun 3881   ∩ cin 3882  ∅c0 4253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254

This theorem is referenced by:  disjtp2  4649  f1oun2prg  14558  cnfldfun  20522