Assume that $M$ is a non-standard model of complete arithmetic, i.e. of the theory $Th(\mathbb{N})$. Suppose that $R$ and $S$ are proper cuts of $M$. (With a cut, I mean a subset of the universe of $M$ which is closed under successor and which is downward closed.) Let $\varphi(x,y)$ be a formula in the language of arithmetic, possibly with parameters from $M$. My question is whether the following variant of the overspill principle holds: 


If for all $s \in S$ and all $r \in R$, it holds that $M\vDash \varphi(s,r)$, then there is an element $s\_0 > S$ and an element $r\_0 > R$ with $M \vDash \varphi(s\_0,r\_0)$.


It will also be helpful for me to learn about some "interesting" special cases where the above statement may hold (e.g. restrictions on the cuts $R$ and $S$, restrictions on the formula $\varphi$, ...).



Overspill in models of arithmetic
No, this "double overspill" can fail; here's an example. Fix a non-standard element $q\in M$. Let $S$ be the cut consisting of just the standard numbers, and let $R$ consist of those elements $r\in M$ that are infinitely far below $q$ (i.e., $r+n<q$ for all standard $n$). Then $R$ is also a cut, and all elements $s\in S$ and $r\in R$ satisfy $s+r<q$. But if $s\_0>S$ (i.e., $s\_0$ is infinite) and $r\_0>R$ (i.e., for some standard $n\_0$, $r\_0+n\_0\geq q$), then $s\_0+r\_0>n\_0+r\_0\geq q$. 


