# Prove that the following argument is invalid

$$□p$$, $$□q$$ therefore, $$□(p→q)$$

according to the K system of modal logic, the argument is invalid. I tried proving it using a truth tree, but all the branches unfortunately close, I don't know how to proceed.

• Your math is not displaying right, so it might be a good idea to format it using MathJax. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Commented Jan 27, 2023 at 10:07
• I am not sure whether the display is showing correctly (as mentioned above), but I think the argument you mention might be valid. Could you confirm whether it is $\vdash []p$ and $\vdash []q$ then $\vdash [](p\rightarrow q)$? Commented Jan 29, 2023 at 13:23

Your claim that $$\Box(p \rightarrow q)$$ is not a logical consequence of $$\{\Box p, \Box q\}$$ in $$K$$ is ambiguous. It can be read as saying that $$\Box(p \rightarrow q)$$ is a not local consequence of $$\{\Box p, \Box q\}$$ and as saying that the formula is not a global consequence of the formula set. For present purposes, local consequence can be defined as preservation of satisfaction at every point from every $$K$$-model; global consequence means preservation of validity on every $$K$$-frame.
However, your claim is wrong under both of its interpretations. To see that the formula $$\Box(p \rightarrow q)$$ is a local consequence, note that it is a local consequence of $$\Box q$$ and that local consequence is monotonic. Analogous reasoning shows that the formula is a global consequence.