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Intuitively, it is a tautology. Imagine two possible worlds $m0$ and $m1$, such that $m1$ is accessible from $m0$, i.e., we have the folowing scheme of possible worlds: $m0 \rightarrow m1$. Whatever is the truth value of $q$ in $m1$, $\lozenge(q \rightarrow q)$ is true in $m0$. It follows that $p\rightarrow \lozenge(q\rightarrow q)$ is true in $m0$, wathever is the truth value of $p$ in $m0$ and $m1$. It is a tautology. But I've avaliated this expression by using MOLTAP, and the result was NOT VALID. Is MOLTAP wrong?

Walter r
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    It's only provable in $K$ if it is valid in all Kripke frames, you could have a Kripke frame where there is a world from which no other world is accessible, and such that $p$ is true at that world, and then your formula wouldn't be true at that world in that Kripke frame. So it's not valid in all Kripke frames and so for that reason not provable in $K$. – Rupert Dec 08 '16 at 14:34
  • Good Afternoon, Rupert. Let's prove it formally: $\lnot( p \rightarrow \lozenge(q \rightarrow q))$ equals $p \land$, $\square(q \land \lnot q)$. Suppose $m0 \rightarrow m1$. So I can write that in $m1$ $q \land \lnot q$ holds. But it is a contradition. Shouldn't it prove that it is a tautology? – Walter r Dec 08 '16 at 15:58
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    You've shown that the negation of your formula cannot hold in that particular Kripke frame. But the negation of the formula might hold in some other Kripke frame of the kind I described. Formulas which are provable in $K$ are valid in all Kripke frames. – Rupert Dec 08 '16 at 16:06
  • Ok. So, is it possible to prove formally that this formula is not valid in K anyway? – Walter r Dec 08 '16 at 16:17
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    To show that the formula is not provable in $K$ it is sufficient to exhibit a Kripke frame and a world in that Kripke frame at which the formula is not true. I showed you how to do this. – Rupert Dec 09 '16 at 16:59
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    @Walter: Even in your frame the formula is not valid at $m1$. Take $p$ to be $T$ at $m1$, then $\Diamond(q\rightarrow q)$ is $\Diamond T$ which is $F$ at $m1$, so here the formula evaluates to $T\rightarrow F$ i.e. $F$. – JuneA Dec 09 '16 at 17:21

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Yet another way to see that $p\rightarrow\Diamond(q\rightarrow q)$ is not a tautology in $K$ is as follows: if it is, then $T\rightarrow\Diamond(q\rightarrow q)$ is a theorem, then so is $\Diamond(q\rightarrow q)$ i.e. $\Diamond T$, where $T$ stands for $true$. But $\Diamond T$ is not a theorem in $K$. It is only a theorem in modal logic $D$, which corresponds to serial frames. In fact $\Diamond T$ is equivalent to the $D$ axiom $\Box p \rightarrow \Diamond p$, see answer to 3-rd question below this one.

JuneA
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