## Beth’s Definability Theorem

We’ll set this up in the most general way and later see how Hellman’s account in terms of $\phi$ and $\psi$ applies.

Letting $\textup{L}$ be a language with $p$ a $k$-ary relation symbol not appearing in the relation-symbol set of $\textup{L}$, $\textup{L}^{+}$ is the language that extends $\textup{L}$ and which includes $p$. $\textup{T}^{+}$ is a theory in $\textup{L}^{+}$.

We say that $p$ is explicitly definable if, and only if, there is an $\textup{L}$ formula $\theta$ with free variables $x_{0}\dots x_{k-1}$ such that $\textup{T}^{+} \models p x_{0}\dots x_{k-1} \leftrightarrow \theta$.  And we say that $p$ is implicitly definable if, and only if, for any $\textup{L}$- structure $\mathfrak{A}$ and any $\textup{P}, \textup{Q} \subseteq \ \vert \mathfrak{A}\vert^{k}$ (i.e., the universe of the structure), if both $(\mathfrak{A}, \textup{P})$, and $(\mathfrak{A}, \textup{Q})$ are models of $\textup{T}^{+}$, then $\textup{P} = \textup{Q}$.

Beth’s definability theorem says that if $\textup{L}$ a language with $p$ a $k$-ary relation symbol not appearing in the relation-symbol set of $\textup{L}$, $\textup{L}^{+}$ is the language that extends $\textup{L}$ and which includes $p$ and $\textup{T}^{+}$ is a theory in $\textup{L}^{+}$, then $p$ is explicitly definable if, and only if, $p$ is implicitly definable.

To prove that explicit definability implies implicit definability is straightforward. Assume $\textup{T}^{+} \models p x_{0}\dots x_{k-1} \leftrightarrow \theta$.  Show for $\textup{P}, \textup{Q} \subseteq \ \vert \mathfrak{A}\vert^{k}$, if $(\mathfrak{A}, \textup{P}), (\mathfrak{A}, \textup{Q}) \in \textup{M} od (\textup{T}^{+}) \rightarrow \textup{P} = \textup{Q}$.  Since we’re assuming $\textup{T}^{+} \models p x_{0}\dots x_{k-1} \leftrightarrow \theta$, $\textup{P}$ is definable over $\mathfrak{A}$ (i.e., for $k$-ary $p$, $\textup{P} = \phi^{\mathfrak{A}}$, for $\phi$ with $k$ free variables.  Let $\phi = \theta$.  So $\textup{P} = \theta$;  but $\textup{Q} \subseteq \ \vert \mathfrak{A}\vert^{k}$ and $(\mathfrak{A},\textup{Q})$ models $\textup{T}^{+}$. So $\textup{P} = \theta^{\mathfrak{A}}$ $\square$.

To prove that implicit definability implies explicit definability takes just a bit longer, but it’s simple; using the Compactness Theorem  and the Craig Interpolation Theorem.

Next time I’ll turn to how Hellman deals with the problems posed by this theorem.