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

Theorem disjssun 4475
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 4173 . . . 4 ((𝐴𝐵) = ∅ → ((𝐴𝐶) ∪ (𝐴𝐵)) = ((𝐴𝐶) ∪ ∅))
2 indi 4291 . . . . 5 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
32equncomi 4171 . . . 4 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐶) ∪ (𝐴𝐵))
4 un0 4401 . . . . 5 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
54eqcomi 2745 . . . 4 (𝐴𝐶) = ((𝐴𝐶) ∪ ∅)
61, 3, 53eqtr4g 2801 . . 3 ((𝐴𝐵) = ∅ → (𝐴 ∩ (𝐵𝐶)) = (𝐴𝐶))
76eqeq1d 2738 . 2 ((𝐴𝐵) = ∅ → ((𝐴 ∩ (𝐵𝐶)) = 𝐴 ↔ (𝐴𝐶) = 𝐴))
8 dfss2 3982 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴 ∩ (𝐵𝐶)) = 𝐴)
9 dfss2 3982 . 2 (𝐴𝐶 ↔ (𝐴𝐶) = 𝐴)
107, 8, 93bitr4g 314 1 ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206   = wceq 1538  cun 3962  cin 3963  wss 3964  c0 4340
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1541  df-fal 1551  df-ex 1778  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-v 3481  df-dif 3967  df-un 3969  df-in 3971  df-ss 3981  df-nul 4341
This theorem is referenced by:  ssfi  9218  hashbclem  14494  alexsubALTlem2  24078  iccntr  24865  reconnlem1  24870  dvne0  26073
  Copyright terms: Public domain W3C validator