I'm currently working on exercises of the book "Modal Logic" by A.Chagrov and M.Zakharyaschev (for pleasure, not homework).
One exercise asks to prove this version of Gabbay rule (exercise $3.10$): A frame $F$ validates the rule $(\Box p \rightarrow p) \vee \psi$ $/$ $\psi$, with $p$ not appearing in $\psi$, if and only if $F$ is irreflexive.
(Note that: "$p$ not appearing in $\psi$").
I have the $\leftarrow$ part, but I'm having a hard time proving the other implication. I would really like to understand this excercise, and I know that the excercises in this book are sometimes hard. But I think this has to be fairly easy.
Until now I tried proving it by reductio ad absurdum: I have one reflexive node $x$, and I try to find a formula $\phi$ that $x$ doesn't validate, but that every other node does validate. If I can prove such formula exists then I'm done, because I plug it in the rule.
I hope I was clear, thanks in advance!