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

Theorem vss 4155
Description: Only the universal class has the universal class as a subclass. (Contributed by NM, 17-Sep-2003.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
vss (V ⊆ 𝐴𝐴 = V)

Proof of Theorem vss
StepHypRef Expression
1 ssv 3766 . . 3 𝐴 ⊆ V
21biantrur 528 . 2 (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
3 eqss 3759 . 2 (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
42, 3bitr4i 267 1 (V ⊆ 𝐴𝐴 = V)
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 383   = wceq 1632  Vcvv 3340  wss 3715
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-v 3342  df-in 3722  df-ss 3729
This theorem is referenced by:  vdif0  4181
  Copyright terms: Public domain W3C validator