TITLE: How many disproofs are needed to complete an implication graph?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: We may assume that equivalent statements have been identified, so that the relation of implication is antisymmetric as well as being reflexive and transitive, i.e., it's a partial order. Thus the problem may be stated as follows:

Let $V$ be a finite set partially ordered by a relation $R.$ What is the smallest set $D\subseteq(V\times V)\setminus R$ such that every transitive relation $T\subseteq V\times V,$ which contains $R$ and is disjoint from $D,$ is equal to $R?$

Clearly, any such set $D$ must contain any pair $(a,b)\in(V\times V)\setminus R$ such that $R\cup\{(a,b)\}$ is a transitive relation; otherwise $T=R\cup\{(a,b)\}$ would be a transitive relation containing $R$ and disjoint from $D$ and unequal to $R.$ Let $D_0$ be the set of all such pairs:

$$D_0=\{(a,b)\in(V\times V)\setminus R:R\cup\{(a,b)\}\text{ is transitive}\}.$$

I claim that $D_0$ is the desired set. Let $T\subseteq V\times V$ be a transitive relation such that $R\subseteq T$ and $T\cap D_0=\emptyset;$ I have to show that $T=R.$ Assume for a contradiction that $T\setminus R\ne\emptyset.$ Choose a minimal (with respect to the partial order $R$) element $a\in V$ such that $(a,y)\in T\setminus R$ for some $y,$ and then choose a maximul element $b\in V$ such that $(a,b)\in T\setminus R.$ It is easy to see that $R\cup\{(a,b)\}$ is transitive, whence $(a,b)\in D_0,$ contradiction our assumption that $T\cap D_0=\emptyset.$
In other words, writing $x\le y$ for $(x,y)\in R,$ we can describe $D_0$ as the set of ordered pairs $(a,b)\in(V\times V)$ satisfying the conditions:

$$a\not\le b$$
  $$\forall x\in V\ (x\lt a\implies x\le b);$$
  $$\forall x\in V\ (b\lt x\implies a\le x.$$

For $a\in V$ let $F(a)$ be the set of all maximal elements of $\{x:x\lt a\},$ and let $G(a)$ be the set of all minimal elements of $\{x:b\lt x\}.$ If we wish to compute $D_0,$ we might start by computing the sets $F(a)$ and $G(a)$ for each $a\in V;$ then, for each ordered pair $a\not\le b,$ we just have to check the inequalities $x\le b$ for $x\in F(a),$ and $a\le x$ for $x\in G(b).$
Example. If $V=\{v_0,v_1,v_2,v_3\}$ and $R=\{(v_0,v_0),(v_0,v_1),(v_1,v_1),(v_2,v_2),(v_2,v_3),(v_3,v_3)\},$ then $D_0=\{(v_0,v_3),(v_1,v_0),(v_2,v_1),(v_3,v_2)\}.$