# On Philosophy

## March 10, 2007

### Identity And Necessity

Filed under: Language,Logic — Peter @ 12:00 am

I was trying out the new latex rendering engine in wordpress with this post. As you can see below it kind of sucks (it puts way too much space around the equations and doesn’t adjust to the viewer’s font size settings), so I won’t be using it again. But try to bear with me for this post, since I have everything in the latex format already and don’t exactly relish trying to get it into html formatted math. If anything is unreadable because of these issues just leave a comment and I will do my best to clarify matters. Now, with that out of the way, on to the post.

[Edit: It turns out this was partly a problem with my theme, which I have changed. Things look much better now, so I think I will stick with doing math this way in the future, assuming no one has any objections.]

There statement $x=y \rightarrow (\varphi(x/c) \rightarrow \varphi(y/c))$ ( $\varphi(x/c)$ means that all free occurrences of c are replaced with x) is an axiom in first order logic, for all formulas $\varphi$ as long as x and y are free for c in $\varphi$. Now consider $\varphi = \Box (x=c)$. If we instantiate our axiom with that formula then it becomes $x=y \rightarrow (\Box (x=x) \rightarrow \Box (x=y))$, which seems to say, since $\Box (x=x)$ is true, that if x and y are identical then they are necessarily identical, or at least this is how Kripke reasons.

But the axiom that the argument begins with is not one from modal logic, it is from first order logic, so we can’t simply assume that it is consistent with modal logic. To explore that possibility let us describe modal logic in terms of first order logic. In first order logic we already have the concept of satisfaction. We say that a set $\Gamma$ satisfies a formula $\varphi$ when for all models $\mathcal{M}$ and assignments s that satisfy $\Gamma$ ( $\mathcal{M} \models \Gamma [s]$) then that model and assignment satisfy $\varphi$ ( $\mathcal{M} \models \varphi [s]$). Each model corresponds to a set of objects and defines which properties and relations hold for which objects. Thus a model is like a possible world. And so it seems natural to say that $\Box \varphi$ ( $\varphi$ is necessary) means that $\varphi$ is true under all models and assignments, which means that all models and assignments satisfy $\varphi$ ( $\mathcal{M} \models \varphi [s]$). Which means that the empty set satisfies $\varphi$ ( $\emptyset \models \varphi$). And that means that $\varphi$ is a logical truth. Now certainly this is sufficient when we are trying to determine whether a formula is a necessary truth. But it doesn’t quite capture what we mean when we specify that a formula is necessary. When we specify that a formula is necessary we are adding it to the axioms, so to speak, indicating that it is true in all models, even if it isn’t a logical truth.

Now we can recast our original substitution in these terms. We let $\varphi = \emptyset \models x=c$. Thus after substituting we obtain: $x=y \rightarrow (\emptyset \models x=x \rightarrow \emptyset \models x=y)$, Since $\emptyset \models x=x$ is in fact a logical truth we obtain $x=y \rightarrow \emptyset \models x=y$. Now, using the idea that a necessary truth is true under all possible models we can check this claim. And it is in fact false. So what went wrong? Well, as mentioned when the axiom was first introduced that x and y had to be free for the c that they were replacing. Usually this just means that $\forall x , \exists x , \forall y , \textrm{ and } \exists y$ don’t occur where they might be able to capture x or y when they replace c. I propose that $\Box$ actually captures all free variables, and thus that x and y are not free for c in $\Box (x=c)$. This is because $\Box (x=c)$ expands to $\emptyset \models x=c$ which expands to $\textrm{for all models } \mathcal{M} \textrm{ and assignments s } \mathcal{ M} \models x=c [s]$. An assignment effectively captures free variables by assigning them to members of the domain of $\mathcal{M}$. This is just as much as a capture as $\forall x(x \leq c)$ which if it was allowed to fill the place of $\varphi$ would result in the claim that $x = 0 \rightarrow (\forall x(x \leq x) \rightarrow \forall x(x \leq 0))$. Assuming that our domain is the positive integers including zero this would imply that $x = 0 \rightarrow \forall x(x \leq 0)$, which is clearly ridiculous.

The conclusion to draw from all of this is not that a posteriori necessary truths are impossible (formulas that are necessary but not logical truths). However it does show that modal logic and first order logic don’t mix well. Modal logic is defined as an extension to sentential logic, and thus trying to integrate it into first order logic is not the easiest task. At the very least this exercise shows that the modal operator should be considered to capture free variables, but other complications may arise as well.

Create a free website or blog at WordPress.com.