$□p$, $□q$ 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.
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1$\begingroup$ 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. $\endgroup$– Christian E. RamirezCommented Jan 27, 2023 at 10:07
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$\begingroup$ 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 $\vdash [](p\rightarrow q)$? $\endgroup$– Rodrigo Nicolau AlmeidaCommented Jan 29, 2023 at 13:23
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1 Answer
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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.
