Theorem 1 Let D be a digraph of order n. There exists an injection D→ D′ which associates with D a fully irregular digraph D′ of order n+2⎡√n  ⎤ such that D is an induced subdigraph of D′ and such that deleting all arcs of D from D′ results in an oriented graph.

Proof. Let V = {v₁,…, v_n} be the vertex set of D. Let t = ⎡√n  ⎤-1.Consider two disjoint linearly ordered sets U and W which comprise altogether 2(t+1) new vertices respectively u_i and w_i, which are ordered by increasing subscripts i, i = 0,1, …,t. Let B be the bipartite digraph whose vertex set is U∪W and all arcs are of the form (w_j, u_i) for each i ∈ {0,1,…,t−j} where j = 0,1,…,t. Let D′ be a digraph of order n+2t+2 which includes disjoint digraphs D and B, all arcs both from V to u₀ and from w₀ to V, and possibly arcs (v,u_i) and/or (w_i,v) where v ∈ V and, moreover, the neighbours of any such v both in U and W make up precisely initial segments of U and W, respectively. Hence the outdegrees and indegrees of vertices from U and W, respectively, are all zero. Therefore the two obligatory arcs (v,u₀) and (w₀,v) for each vertex v of D enable us to identify D as the subdigraph of D′ induced by all vertices whose outdegrees and indegrees are all positive.

     For any vertex v of D the optional arcs (possibly none) from v to a segment of U can be chosen in t+1 ways. The same is the number of choices for remaining optional arcs from a segment of W to v. Thus the degree pair in D′ of each vertex v can be any of some (t+1)²   ( ≥ n) points in the plane integral lattice. Therefore distinct degree pairs in D′ for all n vertices of D can be designed and realized. The construction associates mutually distinct degree pairs with all remaining vertices, too. Therefore a required injection exists.