I've come to the following problem:
Let n $\geq$ 1 and $\phi_1 \dots \phi_n$ are modal formulas:
Prove or disprove that the following are equivalent:
$\bullet$ At least one of the $\phi_1 \dots \phi_n$ is from the smallest normal modal logic (K)
$\bullet$ $\square \phi_1 \lor \square \phi_1 \dots \lor \square \phi_n $ is from the smallest normal modal logic (K).
I've come that the statement is true, but I can't seem to proove it formaly (in terms of modal logic). Any tips?
I agree that it's true. The forward direction is easy (if $\phi_i$ is a theorem of K, then so is $\square \phi_i$ by necessitation, which implies the disjunction).
For the backward direction it is useful to consider the contrapositive statement: suppose that for all $i$, $\phi_i$ is not a theorem of K, and we want to show that $\square \phi_1 \lor \cdots \lor \square \phi_n$ is not a theorem of K. Here it is useful to think in terms of Kripke structures. For each $i$ because $\phi_i$ is not a theorem, there must be a Kripke structure $\mathcal{S}_i$ in which world $w_i$ is a counterexample. Then we can use this to construct a counterexample to $\square \phi_1 \lor \cdots \lor \square \phi_n$: just combine all the Kripke structures and add a new world $w$ which has arrows to $w_i$ for each $i$. The statement $\square \phi_i$ is false at $w$ because $\phi_i$ is false at $w_i$ and $w_i$ is a possible world from $w$.
P.S.: It is important to understand what we are showing here: that if $\square \phi_1 \lor \cdots$ is a theorem of K, then $\phi_i$ is a theorem of K for some $i$. Taking the case $n = 1$, what we showed is that if $\square \phi_1$ is a theorem, then $\phi_i$ is a theorem. This is not the same as saying that $\square \phi_1 \to \phi_1$ for any $\phi_1$, which is not a valid reasoning principle in K. In other words, what we showed was only for theorems, not for an arbitrary choice of $\phi_1$.