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

Theorem disjssun 4413
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 4130 . . . 4 ((𝐴𝐵) = ∅ → ((𝐴𝐶) ∪ (𝐴𝐵)) = ((𝐴𝐶) ∪ ∅))
2 indi 4247 . . . . 5 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐵) ∪ (𝐴𝐶))
32equncomi 4128 . . . 4 (𝐴 ∩ (𝐵𝐶)) = ((𝐴𝐶) ∪ (𝐴𝐵))
4 un0 4341 . . . . 5 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
54eqcomi 2827 . . . 4 (𝐴𝐶) = ((𝐴𝐶) ∪ ∅)
61, 3, 53eqtr4g 2878 . . 3 ((𝐴𝐵) = ∅ → (𝐴 ∩ (𝐵𝐶)) = (𝐴𝐶))
76eqeq1d 2820 . 2 ((𝐴𝐵) = ∅ → ((𝐴 ∩ (𝐵𝐶)) = 𝐴 ↔ (𝐴𝐶) = 𝐴))
8 df-ss 3949 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴 ∩ (𝐵𝐶)) = 𝐴)
9 df-ss 3949 . 2 (𝐴𝐶 ↔ (𝐴𝐶) = 𝐴)
107, 8, 93bitr4g 315 1 ((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207   = wceq 1528  cun 3931  cin 3932  wss 3933  c0 4288
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-v 3494  df-dif 3936  df-un 3938  df-in 3940  df-ss 3949  df-nul 4289
This theorem is referenced by:  hashbclem  13798  alexsubALTlem2  22584  iccntr  23356  reconnlem1  23361  dvne0  24535
  Copyright terms: Public domain W3C validator