Consider the monadic second order logic over the natural numbers with $<$ as a predicate, i.e. the second order logic over $(\mathbb N, 1, <)$, where we can quantify over sets and individual variables, denote it by $MSO[<]$.
Also consider the first order theory over the natural numbers with addition, i.e. $(\mathbb N, 1, +)$ where we can just quantify over individual elements (the so called Presburger arithmetic), call it $FO[+]$.
Of course in $FO[+]$ we can express $<$ by a formula, i.e. $x < y$ is equivalent to$$ \exists z ( x + z = y ).$$
But what is the relation of $MSO[<]$ to $FO[+]$? Both logics are decidable, but are their
- any formulaes in $FO[+]$ that could not be expressed in $MSO[<]$, and
- similar any formulas in $MSO[<]$ that could not be expressed in $FO[+]$?
I guess even such a simple formula as $\varphi(x,y,z) = (x + y = z)$ could not be described in monadic second order $MSO[<]$, but how to prove that? Also for some $MSO[<]$ formulas, like\begin{align*} \varphi(x_1) & = \exists X ( 1 \in X \land \forall u,v ( \neg \exists z ( u < z \land z < v ) \to ( u \in X \leftrightarrow v \notin X ) \land x_1 \in X )\end{align*}we have the equivalent $FO[+]$ formula$$ \psi(x_1) = \exists x ( x + x + 1 = x_1 ) \lor x_1 = 1.$$
