0
$\begingroup$

Is this solved correctly?

Give an interpretation such that $\phi=\forall x\exists y(P(x,y)\to\lnot P(y,y))$ is not valid.

Let I be an interpretation such that the domain is $\mathbb N$ and $P(x,y) = " y\le x "$. Thus if $y=1$ always, then $\phi=F$, i.e. it's not valid for I.

$\endgroup$
1
$\begingroup$

Yes, that works!

Here is a simpler one that also works:

Domain: $\{ a \}$ (i.e. just one object $a$)

$P(x,y)$: $x=y$ (i.e. $a$ stands in relation $P$ to itself)

Given that there is just one object, the whole statement collapses to: $a = a \rightarrow \neg a = a$, and is thus false.

$\endgroup$
  • $\begingroup$ ah easier than mine $\endgroup$ – user178403 Nov 30 '17 at 0:22

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.