If $\operatorname{Mod}(T_1 \cup T_2) = \emptyset$ then for some $\sigma$, $T_1 \vDash \sigma$ and $T_2 \vDash \neg \sigma$

Question

Problem description: if $T_1$ and $T_2$ are theories such that $\operatorname{Mod}(T_1 \cup T_2) = \emptyset$, then there is a $\sigma$ such that $T_1 \vDash \sigma$ and $T_2 \vDash \neg \sigma$.

I don’t understand how to approach this. Is it even true if I’m not given that $T_1$ and $T_2$ are consistent?

Hints only please; no full solutions. Thanks!

Answer 1

If $T_1 \cup T_2$ is inconsistent, then $T_1 \cup T_2 \vdash \bot$, where $\bot$ is some falsehood such as $\exists x\,(x\neq x)$. Let $S_i$ be all the sentences of $T_i$ that occur in the proof of $\bot$, $i = 1,2$. Then $S_1 \cup S_2\vdash \bot$.

Now let $\sigma_i = \bigwedge_{\varphi\in S_i} \varphi, i = 1,2$. Then $$ \vdash (\sigma_1 \land \sigma_2) \to \bot, $$ i.e. $$ \vdash \sigma_1 \to \neg\, \sigma_2. $$ Note that $T_i \vdash \sigma_i, i = 1,2$.

Answer 2

Theorem (Compactness). A theory $T$ is consistent iff every finite $S\subset T$ is consistent.Theorem (Henkin). A theory is consistent iff it has a model. So if $T_1\cup T_2$ has no model, then it is inconsistent.

Suppose $T_1$ and $T_2$ are consistent. By Compactness, we have $\neg$ Con $(T_1\cup U)$ for some finite $U\subset T_2,$ and since Con $(T_1)$ we have $U\neq \phi.$ Take $u \in U$ and let $W=U\backslash \{u\}.$ Let $\sigma$ be $[(\land_{w\in W})\implies \neg u].$

I will leave the cases $\neg$ Con$(T_1)$ and $\neg$ Con$(T_2)$ to someone else.