# Consistency of a sentence in $S5$

True or false: if a modal sentence $$\phi$$ is consistent in K, then it is consistent in S5.

This is equivalent to the contrapositive: if $$\phi$$ is not consistent in S5, then it's not consistent in K. This seems to be false because if we can prove $$\neg \phi$$ in S5, it is not necessarily the case that we could prove it in K.

Here is a counterexample. Let $$\phi=\top\land\square\bot$$. This sentence is consistent in K because it is satisfiable in K. But its negation $$\neg(\square \bot\land \top)$$, which is logically equivalent to $$\square \bot\to\bot$$ is provable in S5.

In particular this proves the stronger statement that a modal sentence consistent in K need not be consistent in KT (an in particular need not be consistent in S5).

Is this counterexample and argument correct?

Your example is right, but you could have just chosen $$\square \bot.$$ It is satisfied if and only if the node is a dead end (no successors). These nodes can exist in general, but can't exist in a reflexive frame since every node is its own successor, so certainly can't exist in an S5 frame.
In general, the example in these sort of situations is the negation of an axiom particular to the stronger system. For instance, here we could use $$\lnot(\lozenge A\to \square\lozenge A).$$ Then we would find a (necessarily non-S5) frame where this holds at some node. Then we would argue that the sentence is false at every node of any S5-frame (and this has probably already been shown at some point).
• I found my sentence this way: we need a sentence $\phi$ such that $\neg \phi$ is provable in S5 but not in K. One of the axioms that can be used in S5 is $\square \psi\to \psi$. So I decided to find a sentence whose negation is provable only by using this axiom. This reduced a problem to finding a sentence $\phi$ whose negation is of the form $\square \psi\to\psi$. Such a sentence is $\square \psi\land \neg \psi$. Now $\psi$ can be only $\bot$ or $\top$, but if it's $\top$ then this sentence is not consistent in K. This gives $\square\bot\land \top$. – user643175 Mar 23 at 21:19
• Regarding your first paragraph: but the fact that $\square \bot$ isn't satisfiable on a reflexive model isn't sufficient to prove that $\square \bot$ is inconsistent in S5, right? I have to derive a proof of $\neg \square \bot$ in S5 anyway to prove that. – user643175 Mar 23 at 21:24