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.


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.

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

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