Let $P$ be a partial order on a set $X,$ $P^\prime=(X\times X)\setminus P,$ and $A\subset P^\prime$ be the converse of $P$ minus $\{(x,x):x\in X\}.$ By transitivity of $P,$ it is obvious that
$$(a,b)\in P'\implies\left\{\begin{array}{lr} (a,x)\in P^\prime\mbox{ for all }x\mbox{ such that }(x,b)\in P\\ (x,b)\in P^\prime\mbox{ for all }x\mbox{ such that }(a,x)\in P. \end{array}\right.$$
For each $(a,b)\in P'\setminus A,$ let $V(a,b)$ be the subset of $P^\prime$ generated by this implication and for each $Y\subset P^\prime\setminus A,$ let $V(Y)=\bigcup\{V(a,b):(a,b)\in Y\}.$ Define $$V=\hskip-10pt\bigcap_{\scriptstyle Y\ \subset\ P^\prime\ \setminus\ A\atop\scriptstyle V(Y)\ \cup\ A\ =\ P^\prime}\hskip-14ptY.$$
For example, suppose $X=\{1,\dots,7\}$ and $P$ is represented by the following Hasse diagram.
It is easy to verify that $|P|=17$ and $|A|=10,$ hence $|P^\prime\setminus A|=22.$ Since
$$\begin{array}{l} V(1,7)=\{1,\dots,5\}\times\{6,7\}\\ V(6,5)=\{6,7\}\times\{1,\dots,5\}\\ V(3,4)=\{(3,4)\}\\ V(4,3)=\{(4,3)\},\\ \end{array}$$
it follows that $V=\{(1,7),(6,5),(3,4),(4,3)\}.$
Does $V$ have a name? Assuming $X$ is finite, is there an easy algorithm to compute it from the adjacency matrix of $P?$