In The logic of provability, by G. Boolos, we are asked to ponder about this statement:
If this statement is consistent, then you will have an exam tomorrow, but you cannot deduce from this statement that you will have an exam tomorrow.
I tried formalizing it the following way:
Let $p$ stand for the whole statement, and $q$ is going to be true iff I have an exam tomorrow.
Then the paragraph is stating that: $$ p\leftrightarrow \diamond (p\rightarrow q)\wedge \neg\square(p\rightarrow q) $$
Solving the fixed point using any of the usual methods will give us that: $$ GL\vdash \boxdot [p\leftrightarrow A(p,q)]\implies \boxdot[p\leftrightarrow \bot] $$ Where $A(p,q)$ stands for the right hand side of the first formula.
Therefore, we can conclude by neccesitation and modus ponens that if $p$ and $q$ are suitable chosen statements of $PA$ satisfying the first formula then: $$ PA\vdash \neg p $$ Therefore we conclude that the statement is false under every interpretation of $PA$, and therefore it gives no information about $q$.
Is my reasoning correct? Is my conclusion pertinent?