MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  disjssun Structured version   Visualization version   GIF version

Theorem disjssun 4228
Description: Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
disjssun ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))

Proof of Theorem disjssun
StepHypRef Expression
1 uneq2 3957 . . . 4 ((𝐴𝐵) = ∅ → ((𝐴𝐶) ∪ (𝐴𝐵)) = ((𝐴𝐶) ∪ ∅))
2 indi 4072 . . . . 5 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
32equncomi 3955 . . . 4 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐶) ∪ (𝐴𝐵))
4 un0 4161 . . . . 5 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
54eqcomi 2806 . . . 4 (𝐴𝐶) = ((𝐴𝐶) ∪ ∅)
61, 3, 53eqtr4g 2856 . . 3 ((𝐴𝐵) = ∅ → (𝐴 ∩ (𝐵𝐶)) = (𝐴𝐶))
76eqeq1d 2799 . 2 ((𝐴𝐵) = ∅ → ((𝐴 ∩ (𝐵𝐶)) = 𝐴 ↔ (𝐴𝐶) = 𝐴))
8 df-ss 3781 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴 ∩ (𝐵𝐶)) = 𝐴)
9 df-ss 3781 . 2 (𝐴𝐶 ↔ (𝐴𝐶) = 𝐴)
107, 8, 93bitr4g 306 1 ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198   = wceq 1653  cun 3765  cin 3766  wss 3767  c0 4113
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-ext 2775
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2784  df-cleq 2790  df-clel 2793  df-nfc 2928  df-v 3385  df-dif 3770  df-un 3772  df-in 3774  df-ss 3781  df-nul 4114
This theorem is referenced by:  hashbclem  13481  alexsubALTlem2  22177  iccntr  22949  reconnlem1  22954  dvne0  24112
  Copyright terms: Public domain W3C validator