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

Theorem disjssun 4431
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 4125 . . . 4 ((𝐴𝐵) = ∅ → ((𝐴𝐶) ∪ (𝐴𝐵)) = ((𝐴𝐶) ∪ ∅))
2 indi 4247 . . . . 5 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
32equncomi 4123 . . . 4 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐶) ∪ (𝐴𝐵))
4 un0 4357 . . . . 5 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
54eqcomi 2738 . . . 4 (𝐴𝐶) = ((𝐴𝐶) ∪ ∅)
61, 3, 53eqtr4g 2789 . . 3 ((𝐴𝐵) = ∅ → (𝐴 ∩ (𝐵𝐶)) = (𝐴𝐶))
76eqeq1d 2731 . 2 ((𝐴𝐵) = ∅ → ((𝐴 ∩ (𝐵𝐶)) = 𝐴 ↔ (𝐴𝐶) = 𝐴))
8 dfss2 3932 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴 ∩ (𝐵𝐶)) = 𝐴)
9 dfss2 3932 . 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 1540  cun 3912  cin 3913  wss 3914  c0 4296
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3449  df-dif 3917  df-un 3919  df-in 3921  df-ss 3931  df-nul 4297
This theorem is referenced by:  ssfi  9137  hashbclem  14417  alexsubALTlem2  23935  iccntr  24710  reconnlem1  24715  dvne0  25916
  Copyright terms: Public domain W3C validator