Transitive feedback arc set (TFAS): NP-complete?
To add a little context, here's a construction for a graph that doesn't have a transitive feedback arc set. For this construction, I'll use the following gadget graph:



This tournament has the following properties (which I checked using a program, I didn't prove it formally):


if (2,7) is not in a given TFAS, then (1,3) is
if (5,1) is in a given TFAS, then so is (3,6)
if (7,3) is in a given TFAS, then (5,1) is not


or slightly abusing predicate logic notation:


$\neg (2,7) \rightarrow (1,3)$
$(5,1) \rightarrow (3,6)$
$(7,3) \rightarrow \neg (5,1)$


You'll notice that for each implication, the two edges are pairwise disjoint, so the following construction works:



I hope you can make out the idea of the graph: using the implication properties of the tournament above, we can construct a graph in which each transitive feedback arc set both includes and doesn't include the edge $A$, i.e. a contradiction, which means the graph doesn't have a transitive feedback arc set. Any completion of that graph cannot have one either since the same contradiction will remain in any completion. I left out a large number of vertices, all of which can be derived from substituting the tournament above for the implications.