I have 4 possible properties of a kind of mathematical object, and I want to prove their implications. I've proved $P_1 \rightarrow P_2$, $P_2 \rightarrow P_3$, and thus also $P_1 \rightarrow P_3$.
Now, I think no other implications are possible between them, and in fact can disprove all other implications. What is the least number of implications I need to disprove to disprove all of them?
Is there a general algorithm for such question? Given a directed graph of implications between properties, and asking how many implications must be disproved in order to show that no other arrows exist in the graph? Well, yes, if the algorithm simply checks all possibilities exhaustively. But is there a more efficient one?
In response to bof, Here I only ask about disproving the arrows of the form $P_n \rightarrow P_m$, not of the more general form $P_n \wedge P_m \rightarrow P_k$, though that could make a nice extension to the problem.
After some looking, I think a formal way to state the question is: given a directed graph $G$ with vertices $V$, denote its transitive closure to be $G_t$. Find a minimal (smallest in number) set of "forbidden" directed edges, such that $G_t$ is the unique transitive directed graph with vertices $V$, that contains $G$, and does not contain any of the forbidden directed edges.