In a directed series-parallel graph, is every pair of non-terminal nodes connected by at most one path?

Question

The title pretty much says it all. A directed graph $G = (V, E)$ is two-terminal series-parallel, with terminals $s_G$ and $t_G$, if it can be produced by a sequence of the following operations:

1. Create a new graph, consisting of a single directed edge directed from $s_G$ to $t_G$.
2. Given two two-terminal series parallel graphs $H$ and $K$ with terminals $s_H$, $t_H$, $s_K$, $t_K$, form a new graph $G = P_c(H, K)$ by identifying $s_G = s_H = s_K$ and $t_G = t_H = t_K$. This is known as the parallel composition of $H$ and $K$.
3. Given two two-terminal series parallel graphs $H$ and $K$ with terminals $s_H$, $t_H$, $s_K$, $t_K$, form a new graph $G = S_c(H, K)$ by identifying $s_G = s_H$, $t_H = s_K$, and $t_G = t_K$. This is known as the series composition of $H$ and $K$.

From what I have noticed in several examples, it seems to be the case that if $v_1, v_2\in V$ are not terminal nodes, then they are connected by at most one path. However, I can't seem to prove it definitively. Is there a theorem which already states this? From my search of papers online I cannot seem to find one which states this.

Answer 1

No, this is not true.

You've probably already found examples with multiple paths between the terminal nodes: a parallel composition of graphs $a \to b \to d$ (with terminals $a,d$) and $a \to c \to d$ (with terminals $a,d$) does the job.

This can be turned into an example in which there are still two paths from $a$ to $d$, but they are no longer terminal nodes. Just take a series composition of three graphs:

1. $s \to a$ (with terminals $s,a$)
2. The graph above (with terminals $a,d$)
3. $d \to t$ (with terminals $d,t$)