If the union of two theories is not consistent there exists a sentence $P$ for which one of them implies $P$ and the other one implies $\lnot P$.
Can someone please check if this is true or not?
I used to prove this by contradiction. I assumed for every sentence $P$, if $T_1$ implies $P$, then $T_2$ satisfies $P$ or $T_2$ doesn't imply $P$ but never $T_2$ implies $\lnot P$. So i think there exists a model of $T_2$ as a substructure of a model of $T_1$( i'm not sure if this is true... I just say it intuitively ), so I take that model of $T_1$ as a model of $T_1 \cup T_2$. Since the union doesn't have a model so there exists a sentence $P$ such that $T_1$ implies $P$ and $T_2$ implies the $\lnot P$.
If the language of both theories is a first-order language, so that the compactness theorem applies, then the statement is true.
So, suppose $T_1 \cup T_2$ is inconsistent. Then, by compactness, there is a finite set of axioms from $T_1 \cup T_2$ which is inconsistent. Let this finite set of axioms be $\phi_1, \ldots, \phi_m \in T_1$ and $\psi_1, \ldots, \psi_n \in T_2$. Then if the contradiction entailed by $\phi_i, \psi_j$ is $Q \wedge \lnot Q$, then we get both $T_1 \vdash (\psi_1 \wedge \cdots \wedge \psi_n \rightarrow Q)$ and $T_1 \vdash (\psi_1 \wedge \cdots \wedge \psi_n \rightarrow \lnot Q)$. It follows that $T_1 \vdash \lnot (\psi_1 \wedge \cdots \wedge \psi_n)$, whereas of course $T_2 \vdash \psi_1 \wedge \cdots \wedge \psi_n$.