TITLE: Overspill in models of arithmetic
QUESTION [5 upvotes]: 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 $\varphi$, ...).

REPLY [4 votes]: You asked about positive instances, so here is a natural one.
Theorem. If $M\models\text{PA}$ is nonstandard, and $S< R$ are cuts in $M$ with $R\prec M$, then the double overspill principle holds. 
Proof. Assume $S<R\prec M$, and suppose $\varphi(s,r)$ holds in $M$ for all $s\in S$ and $r\in R$. Since by elementarity $R$ sees that $\forall r\varphi(s,r)$ holds for each particular $s\in S$, and since $S$ cannot be definable in $R$, as it satisfies $\text{PA}$, it must be that there is some $s_0\in R-S$ with $\varphi(s_0,r)$ for all $r\in R$. But now, $M$ sees all those facts, and since $M$ cannot define the cut determined by $R$, it must be that $\varphi(s_0,r_0)$ for some $r_0> R$. So we've achieved the double overspill principle. QED 
We don't really need $\varphi(s,r)$ for all $s\in S$ and $r\in R$, but rather only for unboundedly many such $s$ and $r$, and the principle is perhaps more interesting and useful in that form. 
Indeed, we achieved a stronger principle: if $S<R\prec M$ and $\varphi(s,r)$ for unboundedly many $s\in S, r\in R$, then there is $s_0\in R-S$ and $r_0>R$ with $\varphi(s_0,r_0)$.