Proving $\Box(p \land q) \rightarrow (\Box p \land \Box q)$ in modal system $K$

Question

I need to prove $\Box(p \land q) \rightarrow (\Box p \land \Box q)$.

Currently, I know of a proof that utilizes the tautology $(p \land q) \rightarrow p$ as a first premise, from which we use the necessity axiom to induce $\Box((p \land q) \rightarrow p))$. We can then use the $K$ axiom and go from there.

However, my professor is adamant that if I am to use this tautology as a premise, I have to prove it's a tautology without truth tables, so I am looking to avoid using it.

My question is: Is there any other way of proving my original modal statement?

Answer 1

Seems that you are doing an axiomatic proof \begin{align} 1.& p\land q\to p\tag*{PL}\\ 2.& \square(p\land q\to p)\tag*{1, NEC}\\ 3.& \square(p\land q)\to \square p\tag*{2, K}\\ 4.& p\land q\to q\tag*{PL}\\ 5.& \square(p\land q\to q)\tag*{4, NEC}\\ 6.& \square(p\land q)\to \square q\tag*{5, K}\\ 7.& \square(p\land q)\to(\square p\land\square q)\tag*{3,6 PL} \end{align}