We say that a partially ordered set $(P,<)$ is self-additive if the two natural embeddings of $P$ in $P\oplus P$ (the linear sum of $P$ and itself) are elementary.

We have the following.

Suppose $(L,<)$ is a self-additive linear order. Then for any linear order $X$ and any nonempty $Y\subseteq X$, the natural embedding of $\sum_{x\in Y} L$ in $\sum_{x\in X} L$ is elementary.

My question is:

Does this hold if we replace $L$ with a partial order (but keep $X$ linear)?

This is Theorem 13.93 in Rosenstein's book on linear orderings. The key ingredient of the proof is the Corollary 13.39, which says that if $L$ is a linear order and $K\subseteq L$ is convex, then for any $U\subseteq L^n$ definable with parameters in $L\setminus K$, the set $U\cap K^n$ is definable in $(K,<)$ without parameters.

The proof of 13.39 is quite hard to understand without digging through the notation introduced before, but it seems to be related to binarity of linear orders, which is (I believe) not true for arbitrary posets. Thus, I suspect that the analogue of 13.39 is not true (well, maybe not the naive analogue). Still, the analogue of 13.93 seems plausible.

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    $\begingroup$ It looks like Corollary 13.39 does indeed fail for partial orders, if we use the same definition for "convex" as we do for linear orders. Let $X$ be a countable set of atoms, and put the following partial order on $X \cup \mathcal{P}(X)$: $x < Z$ iff $x \in Z$ (so distinct elements all from $X$ or all from $P(X)$ are incomparable). Then $X$ is a convex subset of $X \cup \mathcal{P}(X)$ (for a trivial reason). Every subset of $X$ is definable if you are allowed to use parameters from outside $X$, but within $X$ all definable sets are finite and cofinite (even if you are allowed parameters). $\endgroup$ – James Sep 19 '18 at 1:25
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    $\begingroup$ @James: You're right. This is a very nice example. I guess to obtain a version of 13.39 that could work (and be helpful in proving the analogue of 13.93), you should replace "convex" with the property that for every $x_1,x_2\in K$ and $y\in L\setminus K$, $x_1<y$ iff $x_2<y$ and $y<x_1$ iff $y<x_2$. I'm not sure if it has a name. It coincides with convexity for linear orders. $\endgroup$ – tomasz Sep 19 '18 at 1:36

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