two finite models of complete orderings are elementary equivalent let M , N be finite simple orderings. prove that M and N satisfy exactly the same universal sentences.
I know for two finite models if they are elementary equivalent they would be isomorphic, I don't know if I can use this fact to prove this problem or not.
 A: EDIT: replacing "finite" with "infinite," the statement is true, and follows from the more general fact

If $\mathcal{M}$ is a relational structure and $\varphi$ is a universal sentence true of $\mathcal{M}$, then every finite substructure $\mathcal{A}\subseteq\mathcal{M}$ has $\mathcal{A}\models\varphi$

and the particular feature of linear orders that

Any two infinite linear orders have the same finite suborders.

(In the context of finite non-relational languages, replace "finite" with "finitely generated" in the first fact above; the point is that "finite" = "finitely generated" when there are only relation symbols in the language.)

The statement is extremely false as stated. Let $L_n$ be the unique up to isomorphism linear order with $n$ elements. Then "$\forall x, y(x=y)$" is true in $L_1$ but not $L_2$.
And this just keeps going: e.g. "$\forall x_1, ..., x_{17}(\bigvee_{i<j<18}x_i=x_j)$"  is true in $L_{16}$ but not $L_{17}$.
Note that these are even positive universal formulas, and that this has nothing to do with the order structure.
