as we are in case 2.
```
v1 -> ...v[i-1] -> v[i] -> ... vn
^
v
```

If we have $v[i-1] \rightarrow v$ we are done because we get to insert $v$
into the path as $v[i-1] \rightarrow v \rightarrow v[i]$. Because $v[i]$ is the
smallest index that $v$ beats, we must that $v[i-1]$ beats $v$ --- otherwise $i$
is no longer the smallest index!
#### § Chicken tournament

Either a chicken $u$ pecks a chicken $v$ then $u \rightarrow v$ or the other
direction, $v \rightarrow u$. We say that $u$ *virtually* pecks $v$ if there's
a patch of pecking for $u$ to peck $v$.
The chicken king is the chicken that virtually pecks all other chickens.

We can have multiple king chickens. We want to find at least one chicken king.
We may want to show that the vertex with the most number of outgoing edges
is going to be a king.
#### § Theorem: chicken with highest out degree is the king

Proof by contradiction: assume $u$ has highest out degree and is not the king.
So there is some vertex $v$ such that $\not u \rightarrow v$. Hence we have
that $v \rightarrow u$. In the other case, we have that $\not u \rightarrow w \xrightarrow{\star} v$.