Constructing a Kripke model where $p \rightarrow \Box \Diamond q$ is false.

discrete mathematicskripke-modelslogicmodal-logicpropositional-calculus

I have constructed the following Kripke model for this problem:

enter image description here

My idea is the following: Implication is false iff we have $ \top \implies \bot$.

For world $0$, we have that $p$ is true. Now we need to evaluate $\Box \Diamond q$. For world $0$, $\Box \Diamond q$ is true iff $\Diamond q$ is true in every world which is reachable from world $0$ (that would be world $1$). Now, in world $1$, $\Diamond q$ is true if $q$ is true in at least one world reachable from world $1$, and the only such world is $2$. Now, since $q$ is false in world $2$, we can say that $\Diamond q$ is false in $1$, and so $\Box \Diamond q$ is false in $0$. We have our implication $\top \implies \bot$.

My questions are:

  1. Obviously, the value of $q$ in world $1$ is $\neg q$. However, we don't really need that value for anything because we only use world $1$ as an "intermediary" world of sorts, right?

  2. Can different worlds in the same Kripke model have the same truth values for the same variables? I.e., could I have added a world $3$ with $p,q$ as values for $p,q$?

Best Answer

  1. You're exactly right. For the purposes of evaluating the truth of $\square \lozenge q$ in world $0$, it doesn't matter what world $1$ thinks about $p$ or $q$ (although it would matter if there were an edge from $1$ to itself. Do you see why?)

  2. It's completely fine for different worlds to have the same truth values. For instance, you can have something like this:

a kripke model

where not only do $0$ and $3$ agree on all the primitive propositions ($p$ and $q$) they actually agreee on all modal formulas! (This is a nice exercise)


I hope this helps ^_^

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