0
$\begingroup$

I'm trying to prove $\neg\neg\bullet\varphi$

in system $L(\neg, \to, \bullet)$, where $\bullet$ is constant truth, i.e. $\bullet \varphi \approx (\varphi \to \varphi)$

Using modus ponens with axiomas:

A1) $\neg\neg\bullet\bullet\varphi$

A2) $(\neg\bullet\varphi \to \neg \psi)$

A3) $((\varphi\to\psi) \to (\neg\psi\to\neg\varphi))$

but still with no success. Could anybody suggest me some flow of proof?

$\endgroup$
2
$\begingroup$

Hint Take $\psi=\bullet\bullet\varphi$ in A2 and apply A3 and modus ponens.

$\endgroup$
  • $\begingroup$ Probably I don't understand something deeply.. $\endgroup$ – Levitan Nov 2 '14 at 17:06
  • $\begingroup$ We have then $\psi \approx \bullet\bullet\varphi$, $\varphi \approx \varphi$, then using a2 $(\neg\bullet\varphi \to \neg\bullet\bullet\varphi)$ and a3 $((\varphi \to \bullet\bullet\varphi) \to (\neg\bullet\bullet\varphi\to\neg\varphi) )$ and using modus ponens if $a$, and $a \to b$ then $b$, but I don't see what can be $a$, and $b$ here. $\endgroup$ – Levitan Nov 2 '14 at 17:16
  • 2
    $\begingroup$ @Levitan You're not using (A3) in the intended way. Allow me to rewrite (A3) as $((A\to B)\to (\neg B\to \neg A))$. You want to use (A3) with $A=\neg \bullet \varphi$ and $B=\neg \bullet \bullet \varphi$, so that you can use (A2). $\endgroup$ – Git Gud Nov 2 '14 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.