Prove $M\Vdash\Box(\Box p\rightarrow p)\rightarrow\Box p$ for a model $M=\langle W,R,V\rangle$ in which $R$ is transitive and converse well-founded.
I proceed by reductio. So suppose $M\nVdash\Box(\Box p\rightarrow p)\rightarrow\Box p$.
Then, $M,w\nVdash\Box(\Box p\rightarrow p)\rightarrow\Box p$ for some $w\in W$.
But if $M,w\nVdash\Box(\Box p\rightarrow p)\rightarrow\Box p$, then $M,w\Vdash\Box(\Box p\rightarrow p)$ and $M,w\nVdash\Box p$.
And if $M,w\nVdash\Box p$, then $M,x\nVdash p$ for some world $x\in W$ such that $Rwx$.
But since $Rwx$ and $M,w\nVdash\Box(\Box p\rightarrow p)\rightarrow\Box p$, $M,x\Vdash\Box p\rightarrow p$ also.
And since $M, x\nVdash p $, it must be that $M,x\nVdash\Box p $
But if $M,x\nVdash \Box p $, then $M,y\nVdash p $ for some world $y\in W$ such that $Rxy$.
But since $R$ is transitive since $Rwx$ and $Rxy$, $Rwy$ also.
And so, since $M,w\Vdash\Box(\Box p\rightarrow p)$, $M,y\Vdash\Box p\rightarrow p$ also.
And since $M,y\nVdash p$, $M,y\Vdash\Box p\rightarrow p$ only if $M,y\nVdash\Box p$.
So, $M,z\nVdash p$ for some world $z\in W$ such that $Ryz$
But since $R$ is transitive and $Rwy$ and $Ryz$, $Rwz$ also and so on ad infinitum...
My question is this: just as a well-founded relation needs to have a minimal element, does a converse well-founded relation need to have a maximal element?
And so in this model I was constructing where we just have an infinite sequence of worlds at which $\lnot p$ is true there would be no maximal element and so it would contradict our assumptions that $R$ is both transitive and converse well-founded and $M\nVdash\Box(\Box p\rightarrow p)\rightarrow\Box p$?
Thanks for any and all help!