Find an equivalent standard translation for FO formula Let $\psi(x):=\forall y\exists z\forall u(\neg R(x,y)\vee (R(y,z)\wedge \neg R(z,u)))$. We must find a equivalent form of $\psi(x)$ such that we can apply standard translation on it and find modal formula for it. I now have $st_{x}(\varphi):=\forall y(R(x,y)\rightarrow\neg(\forall x(R(y,x)\wedge P(x)\rightarrow \exists x(R(y,x)\wedge P(x))))$. This would equal $\varphi:=\square(\square p\wedge \neg\lozenge p)$. I feel that I am missing something...

$p \land \lnot q$ is $\lnot (p \to q)$.

Thus, we can rewrite $\varphi := (p∧¬p)$ as $¬(p \to p)$.