Team,
I need to give a counter model for $\exists x \forall y ((Rxy \land \neg Ryx) \rightarrow (Rxx \leftrightarrow Ryy))$.
To do that, though, I'm trying to paraphrase this into a sentence in English. I'm having trouble. Can anyone help me with the paraphrase so that I can get thinking about the counter model?
Ugh .. there is really no fluent reading of this in English that will actually help you further understand what this statement is claiming.
So, instead, just work with the formula and see how we can make it false.
So, how can we make $\exists x \forall y ((Rxy \land \neg Ryx) \rightarrow (Rxx \leftrightarrow Ryy))$ false? Well, that would be the same as making $\neg \exists x \forall y ((Rxy \land \neg Ryx) \rightarrow (Rxx \leftrightarrow Ryy))$ true, and bringing the negation inside, that statement is equivalent to:
$\forall x \exists y ((Rxy \land \neg Ryx) \land \neg (Rxx \leftrightarrow Ryy))$
OK, so we clearly need at least two different objects, otherwise we cannot have both $Rxy$ and $\neg Ryx$, so let's say we have $a$ and $b$, and so when we let $x=a$, we need $Rab$ and $\neg Rba$. OK, so now we also need $\neg (Raa \leftrightarrow Rbb))$, meaning we need exactly one of $Raa$ and $Rbb$, so let's say we have $Raa$ but not $Rbb$.
Are we there? No, because while the $\exists y ((Rxy \land \neg Ryx) \land \neg (Rxx \leftrightarrow Ryy))$ is true for $x=a$, it is not true for $x=b$, since we cannot pick $y=a$ for $x=a$.
OK, so we need a third object $c$, so that we have $Rbc$ and $\neg Rcb$, and since we need $\neg (Rbb \leftrightarrow Rcc)$ and we don't have $Rbb$, we do need $Rcc$. But that means that for $x=c$ we can neither pick $y=a$ nor $y=b$ to make the $\exists y ((Rxy \land \neg Ryx) \land \neg (Rxx \leftrightarrow Ryy))$ true, so we need a fourth object $d$!
OK, so we have $Rcd$ and $\neg Rdc$, and since we have $Rcc$ we need $\neg Rdd$. OK, can we make $\exists y ((Rxy \land \neg Ryx) \land \neg (Rxx \leftrightarrow Ryy))$ true for $x=d$ by picking $y=a$? Yes, as long as we make we make sure that $Rda$ and $\neg Rad$.
To sum up, then, the counterexample is one that has $4$ objects in the domain, call them $a$, $b$, $c$, $d$, and where we have $Raa$, $Rcc$, $Rab$, $Rbc$, $Rcd$, and $Rda$, and no other pairs in relation $R$