How to prove that if $\mathfrak{K}_2$ is $\Delta$-elementary, then $\mathfrak{K}_1$ is elementary?

Question

I have some problems with this exercise in Chapter VI.3 of Mathematical Logic by Ebbinghaus.

3.9 Exercise. Let $\mathfrak{K}$ and $\mathfrak{K}_1$ be classes of $S$-structures such that $\mathfrak{K}_1 \subseteq \mathfrak{K}$. Let $\mathfrak{K}_2$ be the class of $S$-structures which are in $\mathfrak{K}$ but not in $\mathfrak{K}_1$, that is $\mathfrak{K}_2 = \mathfrak{K} \setminus \mathfrak{K}_1$. Furthermore, let $\mathfrak{K}$ be elementary and $\mathfrak{K}_1$ be $\Delta$-elementary. Show:

(a) $\mathfrak{K}_1$ is elementary $\quad$ iff $\quad$ $\mathfrak{K}_2$ is $\Delta$-elementary $\quad$ iff $\quad$ $\mathfrak{K}_2$ is elementary.

In particular I don't know how to prove the opposite direction of the first line ($\Leftarrow$), that if $\mathfrak{K}_2$ is $\Delta$-elementary, then $\mathfrak{K}_1$ is elementary. I don't know either if the exercise in question is wrong, I've tried to find a counter example, but without success, if anyone can help me, I'd appreciate it.

Answer 1

Say $\mathfrak K_1= \operatorname{Mod}(\Phi_1),$ $\mathfrak K_2= \operatorname{Mod}(\Phi_2),$ and $\mathfrak K = \operatorname{Mod}(\{\varphi\})$. We know $\Phi_1\cup \Phi_2$ is inconsistent, so there is a finite subset $\Phi_2^0\subseteq \Phi_2$ such that $\Phi_1\cup \Phi_2^0$ is inconsistent. Let $$\psi = \left(\bigvee_{\sigma\in \Phi_2^0}\lnot\sigma\right)\land\varphi.$$ Then $\mathfrak K_1= \operatorname{Mod}(\{\psi\}):$

• If $M\models \psi,$ then $M\models \varphi,$ so $M\in \mathfrak K,$ and $M\models \lnot\sigma$ for some $\sigma\in \Phi_2^0\subseteq \Phi_2,$ so $M\notin \mathfrak K_2$ and hence $M\in \mathfrak K_1.$
• If $M\in \mathfrak K_1,$ then $M\in \mathfrak K,$ so $M\models \varphi.$ Also $M\models \Phi_1,$ so $M\not\models \Phi_2^0,$ so there is a $\sigma\in \Phi_2^0$ such that $M\models \lnot \sigma.$ Thus $M\models \psi.$