I was reading a book about modal logic and in the book I read:
Irreflexivity ($\neg Rxx\ \forall x$) is not modally definable They prove it with the frames $A$ and $B$, where $A = (\{a\}, \{(a, a)\})$ and $B = (\{b1, b2\}, \{(b1, b2),(b2, b1)\})$, by giving two frames that are bisimilar and where irreflexivity holds in $B$ but not in $A$.
My question is how can frame A and B bisimilar?
Denote the relation of $A$ and $B$ as $R$ and $R'$ respectively. We can define a bisimulation $Z \subseteq A \times B$ as $Z = \{(a,b_1),(a,b_2)\}$. It is straightforward to verify that this is indeed a bisimulation by checking the forth and back condition. If $(s,s') \in Z$, then
(forth). For all $t \in A$ with $Rst$, $\exists t' \in B$ with $R's't'$ and $Ztt'$. We only have $a \in A$ and $Raa$, so if $s' = b_1$ we can take $t' = b_2$ and vice versa.
(Back). For all $t' \in B$ with $Rs't'$, $\exists t \in A$ with $Rst$ and $Ztt'$. Again, we could only have $Rb_1b_2$ or $Rb_2b_1$, in both cases we can take $t = a \in A$.
Since two frames that are bisimilar satisfy the same formulas, but one is irreflexive and the other is not, it follows that irreflexivity is not modally definable.
