Proof that an property is definable if and only if its axiomatiazable and its complement its axiomatizable

Question

For a homework of first-order logic I need to prove that a property, lets call it P, is definable if and only if P is axiomatizable and the complement of P is axiomatizable. I have no idea of how to solve this, if you can give me a hint I will appreciate it.

Thanks.

Answer 1

Hint: Let $T_1$ and $T_2$ axiomatize $P$ and its complement, respectively. Can you think of a sentence $\psi$ such that $T_1 \models \psi$ and $T_2 \models \neg \psi$, provided that $T_1 \cup T_2$ is inconsistent?

Solution:

By compactness theorem, there exist finitely many $\phi^i_1,\dots,\phi^i_{n_i} \in T_i$ $(i = 1,2)$ such that $\{\phi^i_k : 1 \le k \le n_i, i = 1,2\}$ is inconsistent. This implies $\models \bigwedge _ {1\le k \le n_1} \phi^1_k \rightarrow \bigvee_ {1 \le k \le n_2} \neg \phi^2_k$. Let $\psi$ denote the consequent of the conditional. Then $T_1 \models \psi$ and $T_2 \models \neg \psi$. Let me show $\psi$ defines $P$. Clearly any structure $\mathfrak M \in P$ satisfies $\psi$, since $\mathfrak M \models T_1$. Conversely, suppose $\mathfrak M \models \psi$. Then $\mathfrak M \not \models \neg \psi$, hence $\mathfrak M \not \models T_2$. Since $T_2$ axiomatizes the complement of $P$, we have $\mathfrak M \in P$.