Consider a directed acyclic graph $G(V,E)$ such that, if the path $v_0 v_1 \dots v_n$ exists, then no shortcuts, i.e. paths that skip some vertices in the original path, exist in the graph. More specifically a shortcut is of the form $v_{a_0} v_{a_1} \dots v_{a_k}$, where the sequence $\langle a_i \mid i \in [0..k] \rangle$ is a subsequence of $[0..n]$.
This question has been partially answered in this questions but it only considers paths of length $3$. The condition in this question is even stricter.
Some examples of legal and illegal graphs:
A --> B
| | This graph is okay. Although two paths from A to D exist,
v v neither is a shortcut of the other
C --> D
A --> B
| | illegal. The path A - C is a shortcut of A - B - C
| v
----> C
I am interested in the order of maximal number of edges in such graphs for proving an algorithmic bound. From the linked question, we know it to be strictly smaller than $|E| \lt \frac{1}{4} |V|^2$ but I fail to construct an example where $|E| = O(|V|^2)$ in general.
Also helpful would be to see if this property has an established name in literature since my crude attempts at searching the internet have failed to find such a name.
Equivalently - I believe, how many edges can there be in a directed graph such that more than one edge has to be flipped to introduce a cycle?
The previous question's bound of Turán's theorem still applies here, and the extremal example for Turán's theorem is the complete bipartite graph $K_{\lfloor n/2\rfloor ,\lceil n/2 \rceil}$.
If we orient the all edges of this graph consistently from one side of the bipartition to the other, then the graph we get does not contain any path of length more than $1$, and so it trivially contains no shortcuts.
And since any graph with more edges than this would contain a triangle (which means either a cycle or a shortcut, and both are forbidden), we know that this construction is the best possible.