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

Theorem nannot 1453
Description: Show equivalence between negation and the Nicod version. To derive nic-dfneg 1595, apply nanbi 1454. (Contributed by Jeff Hoffman, 19-Nov-2007.)
Assertion
Ref Expression
nannot 𝜓 ↔ (𝜓𝜓))

Proof of Theorem nannot
StepHypRef Expression
1 df-nan 1448 . . 3 ((𝜓𝜓) ↔ ¬ (𝜓𝜓))
2 anidm 676 . . 3 ((𝜓𝜓) ↔ 𝜓)
31, 2xchbinx 324 . 2 ((𝜓𝜓) ↔ ¬ 𝜓)
43bicomi 214 1 𝜓 ↔ (𝜓𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196  wa 384  wnan 1447
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 197  df-an 386  df-nan 1448
This theorem is referenced by:  nanbi  1454  trunantru  1524  falnanfal  1527  nic-dfneg  1595  andnand1  32398  imnand2  32399
  Copyright terms: Public domain W3C validator