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

Theorem symdifv 5015
Description: The symmetric difference with the universal class is the complement. (Contributed by Scott Fenton, 24-Apr-2012.)
Assertion
Ref Expression
symdifv (𝐴 △ V) = (V ∖ 𝐴)

Proof of Theorem symdifv
StepHypRef Expression
1 df-symdif 4181 . 2 (𝐴 △ V) = ((𝐴 ∖ V) ∪ (V ∖ 𝐴))
2 ssv 3939 . . . 4 𝐴 ⊆ V
3 ssdif0 4294 . . . 4 (𝐴 ⊆ V ↔ (𝐴 ∖ V) = ∅)
42, 3mpbi 231 . . 3 (𝐴 ∖ V) = ∅
54uneq1i 4094 . 2 ((𝐴 ∖ V) ∪ (V ∖ 𝐴)) = (∅ ∪ (V ∖ 𝐴))
6 0un 4324 . 2 (∅ ∪ (V ∖ 𝐴)) = (V ∖ 𝐴)
71, 5, 63eqtri 2766 1 (𝐴 △ V) = (V ∖ 𝐴)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1547  Vcvv 3431  cdif 3880  cun 3881  wss 3883  csymdif 4180  c0 4261
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-v 3433  df-dif 3886  df-un 3888  df-ss 3900  df-symdif 4181  df-nul 4262
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator