  We consider the complexity of deciding the winner of an election under the
Slater rule. In this setting we are given a tournament $T = (V, A)$, where the
vertices of V represent candidates and the direction of each arc indicates
which of the two endpoints is preferable for the majority of voters. The Slater
score of a vertex $v\in V$ is defined as the minimum number of arcs that need
to be reversed so that $T$ becomes acyclic and $v$ becomes the winner. We say
that $v$ is a Slater winner in $T$ if $v$ has minimum Slater score in $T$.
  Deciding if a vertex is a Slater winner in a tournament has long been known
to be NP-hard. However, the best known complexity upper bound for this problem
is the class $\Theta_2^p$, which corresponds to polynomial-time Turing machines
with parallel access to an NP oracle. In this paper we close this gap by
showing that the problem is $\Theta_2^p$-complete, and that this hardness
applies to instances constructible by aggregating the preferences of 7 voters.
