Theorem undisj1 1739

Description: The union of disjoint classes is disjoint.

Assertion

Ref	Expression
undisj1	⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∪ B) ∩ C) = ∅)

Proof of Theorem undisj1

Step	Hyp	Ref	Expression
1		un00 1728	. 2 ⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∩ C) ∪ (B ∩ C)) = ∅)
2		indir 1678	. . 3 ⊢ ((A ∪ B) ∩ C) = ((A ∩ C) ∪ (B ∩ C))
3	2	cleq1i 1108	. 2 ⊢ (((A ∪ B) ∩ C) = ∅ ↔ ((A ∩ C) ∪ (B ∩ C)) = ∅)
4	1, 3	bitr4 154	1 ⊢ (((A ∩ C) = ∅ ∧ (B ∩ C) = ∅) ↔ ((A ∪ B) ∩ C) = ∅)

Syntax hints:   ↔ wb 127   ∧ wa 196   = wceq 1091   ∪ cun 1485   ∩ cin 1486  ∅c0 1707

This theorem is referenced by:  cdaassen 3725

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708

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