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D ?? ) satisfies the previous conditions. For every rule r ? (S ?? ? D ?? ) \ (S ? ? D ? ), there is at least an argument a = (d, x) s.t. r ? Strict(d) ? Def(d) ,
Let us show that E ? E ? . Let a = (d, x) ? E. Then, d is a derivation for x in NOption(E) ,
, Contradiction, since NOption(E) = NOption(E ? ). We show similarly that E ? ? E
, Th(E) and from the definition of functions Th and Arg it is obvious that E ? Arg(NOption(E)). Now let a = (d, x) ? Arg(NOption(E)). This means that a = (d, x) is constructed from NOption(E). So, x ? CN(NOption(E)) and Def(d) ? Defs(NOption(E))? ? CN(NOption(E)) and Def(d ? ) ? Defs(NOption(E)). But then
, Let H = (Arg(T ), R u ) be a system built over a theory T . Let O =
But since d and d ? are derivation schemas for x and x ? respectively in O we have: x ? CN(O) and Def(d ? ) ? D ? , so x ? D ? . Contradiction with the coherence of naive option O. E = Arg(O) is conflictfree. Now, suppose that E is not maximal, NOption(E ? ) = Th(E ? ) = O ? is a naive option of T . Let O ? =, vol.3 ,
Let Body(r) = {x 1 , . . . , x k } and Head(r) = y. Since for 1 ? i ? k, x i ? CN((F, S, D ? )), then there is an argument a i = (d i , x i ) ? E (1 ? i ? k) for each x i . Thus, we can construct an argument a for y using r as last rule, i.e., a = ( d, (y, r) , y) where Facts(d) = i Facts(d i ), Strict(d) = i Strict(d i ) and Def(d) = i Def, x, r) , x) is an argument outside E but since E is a stable extension, there is b ? E s.t. bR u a. So ,
,
, Let us show that E ? E ? . Let a = (d, x) ? E. Then, d is a derivation for x in SOption(E)
, Contradiction, since SOption(E) = SOption(E ? ). We show similarly that E ? ? E
, SOption(E) and from the definition of functions Th and Arg it is obvious that E ? Arg(SOption(E)). Now let a = (d, x) ? Arg(SOption(E)). a = (d, x) is constructed from SOption(E). So, Def(d) ? Defs(SOption(E))
, Since E is a stable extension then there is b = (d ? , x ? ) ? E such that bR u a. From b ? E we easily deduce that x ? ? CN(SOption(E)). But then, from bR u a, SOption(E) must be incoherent. Contradiction with the fact that SOption(E) is a stable option. Proof of Theorem 6, Let H =
We prove that E is conflict-free and ?b ? Arg(T ) \ E, ?a ? E s.t. aR u b. H = (Arg(T ), R u ) s.t. Ext p (H) = ?. Let us show that for all E ? Ext p (H) ,
Def(d) ? D ? and Strict(d) ? S ? . Thus, a = ( d, (x, r) , x) is an argument outside E. a does not attack any argument of E. Indeed, if we suppose the contrary then, since E is a preferred extension, there is b ? E s.t. bR u a. So, there is a sub-argument of a: a ? = ( d ? , (x ? , r ? ) , x ? ) with r ? ? D ? and b = (d ?? , r ? ). However since a ? ? E (because it uses only rules from S ? ?D ? ), this means that E is not conflict-free which contradicts the fact that E is a preferred extension. So E ? {a} is conflict free. Moreover, for every c ? Arg(T ) \ (E ? {a}), if cR u a then there is a sub-argument of a: a ? = ( d ? , (x ? , r ? ) , x ? ) with x ? ? D ? and c = (d ?? , x ? ). However since a ? ? E (because it uses only rules from S ? ? D ? ) and E is a preferred extension, we complete S ? by the remaining strict rules). Clearly, O is uniquely determined from E. We have that Concs(E) = CN(Th(E)). Let us show that: CN ,
Since E is a preferred extension, there is an argument c = (d ?? , x ?? ) ? E s.t. cR u b, i.e., there is a derivation d ?? for r ?? in (F, S, D ? ) s.t. d ?? ? Def(d ? ), then there is a minimal derivation d ? for r ? in ,
Def(d)) and x ? Def(d ? ), i.e. x ? D ?? . But, from the fourth condition of preferred options, there is r ?? ? Def(d) such that r ?? ? CN(O). So, there is an argument a ? ? E ? such that a ? ,
We show by a similar way as in the second point of Theorem 5 that: for all E, E ? ? Ext p (H) if POption(E) = POption ,
, Suppose that a / ? E. Since E is a preferred extension then we have two cases. The first case is that there is b = (d ? , x ? ) ? E such that bR u a. From b ? E we easily deduce that x ? ? CN(POption(E)). But then, from bR u a, POption(E) must be incoherent. Contradiction with the fact that POption(E) is a preferred option, Since Th(E) ? POption(E) and from the definition of functions Th and Arg it is obvious that E ? Arg(POption(E))
, Since Def(d) ? Defs(POption(E)) then x ? Defs(POption(E)). So, x is used in at least an argument c = (d ?? , x ?? ) of E i.e., x ? d ?? . Thus, c is attacked by b. But since E is a preferred extension, then it must contain an argument which attacks b. This contradict the hypothesis that E does not attack b. Proof of Theorem 8, ? E such that bR u a and E does not attack b. From bR u a we have x ? ? d
We prove that E is conflict-free, ?b ? Arg(T ) \ E, if ?a ? E s.t. bR u a then ?c ? E s.t. cR u b and E is a maximal subset of Arg(T ) satisfying the previous two conditions. Suppose that there are two argument a = (d, x) and b = (d ? , x ? ) in E s.t. aR u b, i.e., x ? Def(d ? ). But since d and d ? are derivation schemas for x and x ? respectively in O we have: x ? CN(O) and Def(d ? ) ? D ? ,
, )) and x ? Def(d ? ). From the fourth conditions of the definition of a preferred option, Now, let us show that: ?b ? Arg(T ) \ E, if ?a ? E s.t. bR u a then ?c ? E s.t. cR u b
E ? is an maximal conflict-free set of arguments that defends all its elements, is an argument c = (d ?? , r ?? ) with d ?? a minimal derivation of r ?? in O. Clearly, cR u b. Finally ,
) , x ? ) with r ? ? D ? and b = (d ?? , r ? ). However since a ? ? E (because it uses only rules from S ? ?D ? ), this means that E is not conflict-free which contradicts the fact that E is a complete extension, we complete S ? by the remaining strict rules). Clearly, O is uniquely determined from E. We have that Concs(E) = CN(Th(E)). Let us show that: CN ,
It follows that there is an arguments b = (d ? , x ? ) s.t. Defs(d ? ) ? D ?? and x ? ? D 1 , i.e., b R u a. In this case, there is an argument c = (d 1 , x 1 ) s.t. c R u b, i.e., there exists r ?? ? Defs(d ? ) hence r ?? ? D ?? s.t. x 1 = r ??, Since clearly x 1 ? CN ,
, Follows immediately from the bijection between complete options and complete extensions, pp.9-10