Theorem undisj2 4174

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 4154	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
2		indi 4016	. . 3 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
3	2	eqeq1i 2765	. 2 ⊢ ((𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅ ↔ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 267	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)

Syntax hints:   ↔ wb 196   ∧ wa 383   = wceq 1632   ∪ cun 3713   ∩ cin 3714  ∅c0 4058

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-v 3342  df-dif 3718  df-un 3720  df-in 3722  df-ss 3729  df-nul 4059

This theorem is referenced by:  disjtp2  4396  f1oun2prg  13862  cnfldfun  19960