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 2770	. 2 ⊢ (((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 281	1 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∪ cun 3904   ∩ cin 3905  ∅c0 4286

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287

This theorem is used by:  disjtpsn  4683  disjtp2  4684  funtp  6597  prinfzo0  13744  hash7g  14541  f1oun2prg  14978  cnfldfunALT  21587