Maybe I am missing something, but found potential counterexample to a conjecture of Nash-Williams.
According to HAMILTONIAN DEGREE SEQUENCES IN DIGRAPHS
The outdegree and indegree sequences of digraph $G$ are $d_1^+ \le \cdots \le d_n^+$ and $d_1^- \le \cdots \le d_n^-$. Note that the terms $d_i^+$ and $d_i^-$ do not necessarily corresponds to the degree of the same vertex of $G$.
Conjecture 1 (Nash-Williams). Suppose that $G$ is a strongly connected digraph on $n \ge 3$ vertices such that for all $i < n/2$
(i) $d_i^+ \ge i + 1$ or $d_{n-i}^- \ge n - i$,
(ii) $d_i^- \ge i + 1$ or $d_{n-i}^+ \ge n - i$,
Then $G$ contains a Hamilton cycle.
The potential counterexample is $G$ on $6$ vertices with edges:
[(0, 3), (0, 5), (1, 4), (1, 5), (2, 3), (2, 4), (3, 0), (3, 2), (3, 4), (3, 5), (4, 0), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)]
$G$ is strongly connected and by inspection the degree sequences satisfy the hypotheses for $i \in [1,2]$ (both degree sequences are $[2, 2, 2, 4, 4, 4]$).
Nonhamitlonicity was shown using exhaustive search, sage 5.6 and Max Alekseyev's hamiltonian cycle counting pari program.
Is this really a counterexample the the conjecture of Nash-Williams?
Drawing of $G$:
