Is the distribution axiom in modal logic biconditional and does it hold for diamond as well as box?

Question

I'm a new learner of (normal) modal logic. I'm slowly teaching myself and have some questions related (broadly) to the distribution axiom: $\square (p \to q) \to (\square p \to \square q)$. I have no one else to ask, so I thought I'd turn here.

1. Does the distribution axiom also hold in the opposite direction? Can the '$\to$' be replaced with a biconditional?

2. Does a similar thing hold if we swap in diamonds for the boxes? That is, is either of the directions of the following any good: $\lozenge (p \to q) \leftrightarrow (\lozenge p \to \lozenge q)$?

I'm very sorry for the noob question. The book I'm using, Garson's Modal Logic for Philosophers, isn't particularly helpful. I might need to look elsewhere for a better book. I'm struggling to move from first order logic, which I'm fine with, to modal stuff. In particular, I'm struggling to know how to determine for myself whether a sentence in the language with modal operators is true or false and how to evaluate more complex (to me!) strings...

Answer 1

Welcome to MSE!

Evaluating modal formulas, and trying to understand their truth, is a fairly intricate topic. Let's start basic, and build our way up. In particular, let's start with your last (implicit) question, where you say you're struggling to see whether a sentence is true or false.

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A modal formula itself is neither true nor false. Just like the (classical) formula $A \lor B$ is neither true nor false. It's only after we specify a model (that is, an assignment of $v$ sending $A$ and $B$ to truth values) that we can figure out the truth of this formula in that model. For instance, if $v(A) = \top$ and $v(B) = \bot$, then $A \lor B$ should be true. We might write this as $v \models A \lor B$. However, if $v'(A) = v'(B) = \bot$, then $v' \not \models A \lor B$.

The "truth" of a formula is only defined relative to a choice of model!

Analogously, a naked modal formula $\square (A \lor B)$ is neither true nor false. We need to fix a model in order to interpret this formula, but in the modal setting our models are more complicated. There are various types of models that logicians consider, which correspond to different applications of modal logic in various settings. Perhaps the most fundamental, though, are the kripke models, which I'll describe briefly below:

To start, say we have a set of "possible worlds" $W$. Now our valuation $v$ takes a world $w$ and a proposition $A$, and spits out a truth value $v(w,A)$, which we think of as checking if $A$ is true in the world $w$. We also have a relation $R \subseteq W \times W$, and we say that $w_1$ "sees" $w_2$ whenever $(w_1, w_2) \in R$. This is where the modality really happens, since we'll say that "$w$ thinks $\square A$ is true" exactly when $w$ can only see worlds that think $A$ is true.

There's a formal definition of a relation $(W,R,v), w \models \varphi$, which you should think of as "In the model $(W,R,v)$, $\varphi$ is true in the world $w$" which is surely in your book, but can also be found on wikipedia.

To gain some experience computing the truth of a formula at a world, you might want to play around with the Modal Logic Playground, which lets you build a kripke model, type in a formula, and ask if it's true or false at a given world. This is a really great way to test your own understanding. Take a formula, see if you can figure out whether it's true or false in some world, then ask the computer if you're right.

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Now, for your stated questions. You ask whether the distribution axiom can be reversed. That is, does every world of every model think that $(\square p \to \square q) \to \square (p \to q)$?

I'll tell you this is false, and indeed I'll tell you the model which falisfies it too:

• $W = \{w_0, w_1\}$
• $R = \{(w_0, w_0), (w_0, w_1)\}$
• $v(w_0, p) = \top$
• $v(w_0,q) = v(w_1,p) = v(w_1,q) = \bot$

Can you show that $w_0 \not \models (\square p \to \square q) \to \square(p \to q)$?

Next, you ask about distributivity for $\lozenge$. It turns out that both of these implications are invalid! Though this time, I'll leave it as a (fun?) exercise to try and find models which invalidate both of them. If you get stuck, you can check this website, which (after you type in a formula) will tell you "yes, that formula is valid" and will give you a derivation in System K, or will tell you "no, that formula is not valid" and will give you a model which invalidates it.

Lastly, if you're interested in a better book, I've heard good things about the Open Introduction to Modal Logic, which is freely available here.

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I hope this helps ^_^

Answer 2

One way is to use the Kripke semantics for modal logics. The "accessible worlds" interpretation.

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Assume $(\square p\to\square q)$ holds in every world of some frame. Let this frame contain a world that can access two distinct worlds, such that in one $p$ holds and $q$ does not, while in the other $q$ holds while $p$ does not. Thus in our world, $\square p\to\square q$ holds as required, because neither $\square p$ nor $\square q$ hold. However, $\square (p\to q)$ fails to hold, since $p\to q$ fails to hold in one of the accessible worlds (the one where $p$ holds but $q$ does not).

So we can construct a frame where $(\square p\to\square q)$ does not entail $\square( p\to q)$.

Therefore $(\square p\to\square q)\to\square (p\to q)$ is not a theorem .