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, Condition 2 can be proved by induction as follows: First note that, if E = ?, then a is e-supported by E iff a = ? iff a ? P iff a ? Sup(A E ) (note that there is no support ? ? S with s(?) = ?). Hence, we assume as induction hypothesis that the lemma statement holds for every set B ? E, First, note that 1 follows directly from the observation R A = R a and that 3 follows directly from 2
, R e and every c ? C is e-supported by E\{a}. Pick any c ? C. Since every c ? C is e-supported by E\{a}, from Lemma A.15, it follows that every c ? C is e-supported by E\{a, c} ? E. From induction hypothesis and Lemma A.1, this implies that every C ? Sup(A E \{a, c}) ? Sup(A E \{a}) which
, Then, there is some support ? ? S such that s(?) ? (E ? Sup(A E \{a})) and t(?) = a. If s(?) = {?}, then ? is e. Pick any c ? s(?), Assume now that a ? Sup(A E )
, by induction hypothesis, we get that c is e-supported by E\{c}. Hence, every c ? s(?) is e-supported by E\{c}, which
, Let E, B ? A be two sets of arguments such that B is self-supporting w.r.t. some EBAF. Then, B ? Sup(A ) ? with A
, Since B is a self-supporting set, it follows that b is e-supported by B and, from Proposition 4, that b ? Sup(A B ). Suppose, for the sake of contradiction, that B ? Def A (A E ) = ?. Then, B ? Def A (A E ) which implies that A B A . From Lemma A.1, this implies that b ? Sup(A B ) ? Sup(A )
, Let EBAF be some EBA framework and E ? A be a conflict-free, self-supporting set w.r.t. EBAF. An argument a ? A
, First note that, by definition, a ? A being a acceptable w.r.t. AF EBAF implies a ? Sup(A) which
, Let ? ? K be the attack name such that s(?) = C and t(?) = a. Since a is acceptable w.r.t. A E and AF EBAF , it follows that either ? ? UnAcc(A E ) or s(?) ? UnAcc(A) ?. Furthermore, since there is no attack targeting ? and ? ? K ? P, it immediately follows that ? UnAcc(A E ). Hence, we assume without loss of generality that s(?) ? UnAcc(A E ) ?. This implies the existence of some argument c ? s(?) s.t. either c ? Def A (A E ) or c Sup(A ) with A = Def A (A E ), K, S . Furthermore, the latter implies that there is some c ? s(?) ? Def A (A E ) (see Lemma A.16). Hence, in both cases, there is some b ? s(?) ? Def A (A E ), it holds that E e-attacks some b ? B. Since B e-attacks a, there is some
, First note that, from Definition 14, we have that (C, a) ? R a with C = s(?). Suppose, for the sake of contradiction, that s(?) ? UnAcc(A), For the if direction, we will show that s(?) ? UnAcc(A) ? for any attack ? ? K with t(?) = a