1
$\begingroup$

I came across a question where i had to find the necessary and sufficient condition for a property $p$. So i thought of finding a necessary condition $n$ and a sufficient condition $s$ separately and then putting a conjunction between them. But I'm not sure whether that would give me the necessary and sufficient condition. I'm not able to think of a counter example to this at the moment. Formally my query is,

Are the sequents $(p\Rightarrow n) \land (s\Rightarrow p) \vdash (n\land s) \Leftrightarrow p$ and $(n\land s) \Leftrightarrow p\vdash (p\Rightarrow n) \land (s\Rightarrow p)$ valid?

$\endgroup$
1
$\begingroup$

No, because the sufficient property could be false: if $n=p=\top,s=\bot$, then $p⇒ n$ and $s⇒p$ are both true, but $n\wedge s=\bot$ while $p$ remains true, so the equivalence certainly cannot hold.

One could see that on an intuitive level by noting that since $s\implies p\implies n$, the statement $s\wedge n$ only depends on the statement of the stronger of both, namely the sufficient condition $s$. But that would reduce the right side to $s\Leftrightarrow p$, which is absurd in general.

$\endgroup$
1
  • 1
    $\begingroup$ Oh i see now, any false statement will be a sufficient condition for my property! $\endgroup$
    – Jamāl
    Sep 3 '20 at 4:30

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.