# $\phi$ satisfiable implies $\square \phi$ satisfiable?

Below $$\phi$$ stands for a modal sentence.

The question is to decide whether it is true that 1) if $$\phi$$ is satisfiable, then $$\square \phi$$ and $$\diamond \phi$$ are satisfiable, and 2) if $$\phi$$ is satisfiable on a transitive model, then the same is true of $$\square \phi$$.

For 1): I think both are true. Suppose $$\phi$$ is satisfiable in some world $$x$$ of some model $$M$$. Consider the model $$M'$$ obtained from $$M$$ by adding an extra world $$y$$ and the relation $$y\to x$$. Then $$y\models \square\phi\land \diamond \phi$$. Is this a correct argument?

Could I prove these claims using the fact that consistent=satisfiable? For example, I tried to prove that if $$\diamond\phi$$ is not consistent, then $$\phi$$ is not consistent. The former tells us that $$\square \neg \phi$$ is provable in K. But I don't see a way to deduce that $$\neg \phi$$ is provable unless we are working in KT, which is not the case.

For 2): The above argument doesn't work because $$M'$$ can fail to be transitive. But I couldn't come up with a counterexample.