
## Appendix: Technical Proofs

Proof of Lemma 1. Let C = (x1, z1, . . . , xk, zk, t) be a certifying path for x1 in F. Take a specific audience ≤ such that η(x1) < · · · < η(xk) < η(t) and all other values in V are smaller than η(x1). We claim that the unique preferred extension P = GE(F≤) of F≤ includes {x1, . . . , xk, t} and excludes {z1**, . . . , z**k}, which means that x1 is subjectively accepted in F. It follows from C5 that t is not attacked by any other argument in F≤ and hence t ∈ P (see also Section 2 for a description of an algorithm to find the unique preferred extension of an acyclic abstract argumentation system). From C4 it follows that zk ∈/ P. Furthermore, if there exists an argument z 6= t, η(t) = η(z) then either (**t, z**) ∈ A≤ or z does not attack an argument in {x1, . . . , xk, t}. In the first case z /∈ P
and does not influence the membership in P for any other arguments in X. In the second case z ∈ P but it does not attack any argument in {x1, . . . , xk, t}. In both cases it follows that xk ∈ P. Using C3 it follows that zk−1 ∈/ P and since we already know that zk ∈/ P it follows that xk−1 ∈ P. A repeated application of the above arguments establishes the claim, and hence x1 ∈ P follows.

Conversely, suppose that there exists a specific audience ≤ such that x1 is contained in the unique preferred extension P = GE(F≤) of F≤. We will now construct a certifying path C for x1 in F. Clearly, if there is no z1 ∈ X \ {x1} with η(z1) = η(x1) and (z1, x1) ∈ A, then (x1) is a certifying path for x1 in F. Hence, it remains to consider the case where such a z1 exists. Since x1 ∈ P it follows that z1 ∈/ P. The sequence (x1, z1)
clearly satisfies properties C1–C3. We now show that we can always extend such a sequence until we have found a certifying path for x1 in F. Hence, let S = (x1, z1, . . . , xl, zl) be such a sequence satisfying conditions C1–C3, and in addition assume S satisfies the following two conditions:
S1 It holds that η(x1) < · · · < η(xl).

S2 For every 1 ≤ i ≤ l we have xi ∈ P and zi ∈/ P.

Clearly, the sequence (x1, z1) satisfies S1 and S2, hence we can include these conditions in our induction hypothesis. It remains to show how to extend S to a certifying path. Let Y := { y ∈ P | (y, zl) ∈ A ∧ η(y) > η(xl) =
η(zl) }. Then Y 6= ∅ because zl ∈/ P by condition S2 and the assumption that P is a preferred extension.

For each y ∈ Y let Cy = (x1, z1, . . . , xl, zl, y). If there is an argument y ∈ Y such that Cy is a certifying path for x1 in F we are done. Hence assume there is no such y ∈ Y .

We choose xl+1 ∈ Y arbitrarily. Note that Cxl+1 satisfies the condition C4; (xl+1, zl) ∈ A (as xl+1 ∈ Y )
and (xl+1, xi) ∈/ A for 1 ≤ i ≤ l (as xl+1, xi ∈ P and P is conflict-free). Since we assume that Cxl+1 is not a certifying path, Cxl+1 must violate C5.

It follows that there exists some argument zl+1 with η(zl+1) = η(xl+1) such that (xl+1, zl+1) ∈/ A and
(zl+1, xi) ∈ A for some 1 ≤ i ≤ l + 1. We conclude that S
′ = (x1, z1, . . . , xl, zl, xl+1, zl+1) satisfies conditions C1–C3 and S1–S2. Hence, we are indeed able to extend S and will eventually obtain a certifying path for x1 in F.

Proof of Lemma 2. Assume that x1 is objectively accepted in F. Suppose there is a p ∈ X that attacks x1 and η(p) = η(x1). If we take a specific audience ≤ where η(x1) is the greatest element, then x1 is not in the unique preferred extension of F≤, a contradiction to the assumption that x1 is objectively accepted. Hence η(p) 6= η(x1) for all arguments p ∈ X that attack x1. Next suppose there is an argument p ∈ X that attacks x1 and is subjectively accepted in F − η(x1). Let ≤ be a specific audience such that p is in the unique preferred extension of (F − η(x1))≤. We extend ≤ to a total ordering of V ensuring η(x1) ≤ η(p). Clearly x1 is not in the unique preferred extension of F≤, again a contradiction. Hence indeed for all p ∈ X that attack x1 we have η(p) 6= η(x1) and p is not subjectively accepted in F − η(x1)
We establish the reverse direction by proving its counter positive. Assume that x1 is not objectively accepted in F. We show that there exists some p ∈ X that attacks x1 and where either η(p) = η(x1) or p is subjectively accepted in F − η(x1). Let ≤ be a specific audience of F such that x1 is not in the unique preferred extension P = GE(F≤) of F≤. In view of the labeling procedure for finding P as sketched in Section 2, it follows that there exists some p ∈ P that attacks x1 with η(x1) ≤ η(p). If η(x1) = η(p) then we are done. On the other hand, if η(p) 6= η(x1), then p is in the unique preferred extension of (F − η(x1))≤, and so p is subjectively accepted in F − η(x1).