Consider an acyclic directed graph $G = <V, E>$. I am going to use $i \leadsto j$ in the case there is a directed path leading from $i$ to $j$. Further more I am going to define a spanning tree for a directed graph. From what I can tell there isn't exactly agreed upon terminology and if it seems that I should be using some other term, let me know.
In any case I will say a graph contains a directed spanning tree if there exists some vertex $r$ such that $i \in V$ implies $ i \leadsto r$. Stated informally, every vertex has a path to some common vertex. If I need to call it something, I will call $r$ the root of this directed spanning tree.
I am considering pairs of nodes $\{i,j\}$ and will say $\{i,j\}$ have property $A$ if for all vertices $k \neq j \neq i$ $$ \neg (i \leadsto j) $$ $$\neg(j \leadsto i)$$ $$i \leadsto k \implies \neg (j \leadsto k)$$ $$j \leadsto k \implies \neg(i \leadsto k)$$.
This is rather cumbersome property to work with. Simplifying it may help. Stated in words it says there is no path from $i$ to $j$ or vice versa, and if $i$ has a path to some other vertex $k$, then there is no path from $j$ to $k$ and vice versa.
Some sort of hint or solution to the following lemma would help me.
Lemma: Suppose G does not contain a directed spanning tree, then there exists nodes $i$ and $j$ where $\{i,j\}$ have property A.
I am able to show there exists $i$ and $j$ that satisfy the first two conditions of property $A$ but am unable to show there exists and $i$ and $j$ that also has the last two conditions. It seems intuitive whenever I draw and example, but I am struggling to pin down a rigorous argument.
Among all vertices of $G$, let $r$ be one that maximizes $I(r):=\{\,i\in V\mid i\leadsto r\,\}$.
Note that if $r\to v$ is any edge, then $I(v)\subseteq I(r)$ and hence by maximality $I(v)= I(r)$. And if $w$ is any vertex with $r\leadsto w$, then by repeating this argument (i.e., by induction on path length) $w\in I(r)$.
If $I(r)=V$, then we have a spanning tree. So assume $I(r)\ne V$ and $j\notin I(r)$. Assume $j\leadsto k$ and $r\leadsto k$ for some $k$. Then by the above $k\in I(r)$, hence $i\leadsto k\leadsto r$ and $j\in I(r)$, contradiction. In particular (wirh $k=r$ and $k=i$, respectively), $\neg(i\leadsto r)$ and $\neg(r\leadsto i)$. In other words, the pair $\{r,j\}$ have property $A$.