How can I proof this:
Show that if $\sigma$ is a $\forall\exists-$sentence in the language $\mathcal{L} = \{<\}$ which is true for all finite linear orders then $\sigma$ is true in any lineal order.
I have tried to use the union of elemental chains but I am not sure if lineal orders is the crecent union of finite lineal orders. I also tried to use the transference principle from $\forall$ to $\exists$ and viceversa but I do not get to any place
Let $L$ be a linear order and let $\forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$ be an $\forall \exists$-sentence that is true in every finite linear order (so $\phi(\bar{x}, \bar{y})$ is quantifier-free). Now let $\bar{a}$ be an arbitrary tuple in $L$, matching the length of $\bar{x}$. We write $A$ for the substructure of $L$ with universe $\bar{a}$. So $A$ is a finite linear order. By assumption we have that $A \models \forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$, so in particular $A \models \exists \bar{y} \phi(\bar{a}, \bar{y})$. Since $\exists \bar{y} \phi(\bar{a}, \bar{y})$ is an $\exists$-formula (with parameters from $A$) it is preserved upwards by the substructure relation, so we conclude that $L \models \exists \bar{y} \phi(\bar{a}, \bar{y})$. As $\bar{a}$ was arbitrary we thus have $L \models \forall \bar{x} \exists \bar{y} \phi(\bar{x}, \bar{y})$, as required.