Suppose we have a logic $\mathbf{L}$ containing all the propositional tautologies and the formula $\square(p\wedge q)\leftrightarrow(\square p \wedge\square q)$, and it is closed by Modus Ponens, Substitution and the following rule (R): if $\varphi\leftrightarrow\psi\in\mathbf{L}$ then $\square\varphi\leftrightarrow\square\psi\in\mathbf{L}$. Prove that $\square(p\to q)\to(\square p\to\square q)\in\mathbf{L}$.
I'm really stuck with this exercise. I'm used to propositional calculus, where Modus Ponens and Deduction Thereom are the main tools. Any help is appreciated.
Since $\mathbf{L}$ contains all propositionnal tautologies, you have $$ \Big(p \wedge (p \to q) \Big) \longleftrightarrow \Big(p \wedge q \Big) \in \mathbf{L}$$
whence $$ \Big(\operatorname{\square} p \wedge \operatorname{\square} (p\to q) \Big) \longleftrightarrow \Big(\operatorname{\square} p \wedge \operatorname{\square} q \Big) \in \mathbf{L}$$
Since $\mathbf{L}$ is closed under Modus Ponens, we get the weakening : $$ \Big(\operatorname{\square} p \wedge \operatorname{\square} (p\to q) \Big) \longrightarrow \operatorname{\square} q $$ which is logically equivalent to the Distribution Axiom indeed : $$\Big(\operatorname{\square} p \wedge \operatorname{\square} (p\to q) \Big) \longrightarrow \operatorname{\square} q \equiv \operatorname{\square} (p\to q) \longrightarrow (\operatorname{\square} p \to \operatorname{\square} q)$$ Hence the conclusion since $\mathbf{L}$ is closed under substitutions.