Theorem undisj1 4422

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 4364	. 2 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
2		indir 4239	. . 3 ⊢ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶))
3	2	eqeq1i 2768	. 2 ⊢ (((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 281	1 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

Syntax hints:   ↔ wb 209   ∧ wa 400   = wceq 1570   ∪ cun 3903   ∩ cin 3904  ∅c0 4286

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287

This theorem is referenced by:  disjtpsn  4681  disjtp2  4682  funtp  6593  prinfzo0  13723  hash7g  14519  f1oun2prg  14950  cnfldfunALT  21537