If $T= T_1 \cup T_2$ is inconsistent, then can we conclude that $T_1 \models \neg \phi$ for some $\phi\in T_2$

Question

I have seen this reasoning elsewhere and I am not sure wether it is correct.

I think the reasoning behind it should be as follows: Suppose that T is inconsistent. Then $T_2$ is inconsistent with $T_1$ and so there is $\phi \in T_2$ such that $T_1 \nvDash \phi$.

But I can't find a proof of why it is the case that $T_2$ is inconsistent with $T_1$, as I only know that they both should be inconsistent by compactness. Also, we need to know that $T_1$ is complete to deduce to know that $T_1 \models \neg \phi$.

What is missing in this argument?

Answer 1

Not necessarily. Let $T_1=\emptyset$ and $T_2=\{\phi,\lnot\phi\}$ where neither $\phi$ nor $\lnot \phi$ is valid. ${}{}{}{}$

Answer 2

The best you can do is get:

$$T_1\models \lnot(P_1\land \cdots \land P_n)$$

For some $P_1,\dots,P_n\in T_2.$

This is true because if there is a proof of $\phi\land\lnot\phi$ in $T_1\cup T_2,$ let $P_1,\dots,P_n$ be the statements from $T_2$ used in the proof.

Then $$T_1\models (P_1\land\cdots \land P_n)\implies (\phi\land\lnot \phi)$$

From which you get, by proof by contradiction:

$$T_1\models \lnot(P_1\land\cdots \land P_n)$$

There is no particular way to decide which of the $P_i$ is the “cause” of the contradiction. It could possibly be true that $T_1\cup T_2\setminus \{P_i\}$ is consistent for each $i.$ So which $P_i$ is false is not decidable in $T_1.$

Answer 3

Suppose that $T=T_1 \cup T_2$ is inconsistent, and $T, T_1, T_2 \neq \emptyset$ then by compactness, it is not the case that every finite subset of T is consistent.

As $T_1, T_2 \subseteq T$, w.l.o.g suppose that there is M such that $M \models T_1$, then $M \nvDash T_2$ so there is some $\phi \in T_2$ such that $T_1 \nvDash \phi$.

If additionally $T_1$ is complete: $T_1 \models \neg \phi$.