A formal characterization of the outcomes of rule-based argumentation systems

Abstract

Rule-based argumentation systems are developed for reasoning about defeasible information. As a major feature, their logical language distinguishes between strict rules (encoding strict information) and defeasible rules (describing general behavior with exceptional cases). They build arguments by chaining such rules, define attacks between them, use a semantics for evaluating the arguments and finally identify the plausible conclusions that follow from the rules. Focusing on the family of inconsistency-based attack relations, this paper presents the first study of the outcomes of such systems under various acceptability semantics, namely naive, stable, semi-stable, preferred, grounded and ideal. It starts by extending the existing list of rationality postulates that any rule-based system should satisfy. Then, it defines the key notion of option of a theory (a theory being a set of facts, a set of strict rules and a set of defeasible rules). For each of the cited semantics, it characterizes the extensions of a rule-based system that satisfies all the postulates in terms of options of the theory under which the system is built. It also fully characterizes the set of plausible conclusions of the system. The results show that designing a rule-based argumentation system requires great care.

Appendix: Proofs

Proof of Property 2

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ be a theory and $x \in {\mathcal {L}}$. Let $d = \langle (x_1,r_1), \ldots , (x_n,r_n)\rangle$ be a derivation schema for x from ${\mathcal {T}}$.

$(\longrightarrow )$ Let us assume that there exist $x_i$ and $x_j$ such that $x_i = x_j$ but $i \ne j$. Clearly, we can further assume $i < j$ without loss of generality. For each $(x_k,r_k)$ in d where $k > j$, $x_j \in \mathtt {Body}(r_k)$ is trivially equivalent to $x_i \in \mathtt {Body}(r_k)$ hence $\mathtt {Body}(r_k) \subseteq \{x_1,\ldots ,x_{j-1},x_{j+1},\ldots ,x_{k-1}\}$. Therefore, $\langle (x_1,r_1), \ldots , (x_{j-1},r_{j-1})$, $(x_{j+1},r_{j+1}), \ldots , (x_n,r_n)\rangle$ is also a derivation schema, but it is a proper subsequence of d, a contradiction arises. Now, let us assume that d fails to be focused. There exists $i \in \{1,\ldots ,n-1\}$ such that $x_i \not \in \mathtt {Body}(r_j)$ for every $j > i$. Consequently, $\langle (x_1,r_1),\ldots ,(x_{i-1},r_{i-1}),(x_{i+1},r_{i+1}),\ldots ,(x_n,r_n)\rangle$ is also a derivation schema for x in ${\mathcal {T}}$, contradicting the minimality of d.

$(\longleftarrow )$ Let us assume that d fails to be minimal although d is focussed and the literals $x_1,\ldots ,x_n$ are pairwise distinct. As d is not minimal, there exists a proper subsequence $d'$ of d which is a derivation schema for x in ${\mathcal {T}}$. Let us write $\langle (x_{k+1},r_{k+1}), \ldots , (x_n,r_n)\rangle$ for the largest common final subsequence of d and $d'$. Now, k exists (and $k > 0$) because $d'$ is a proper subsequence of d. As $d'$ is a derivation schema for $x_n$ and $d'$ is a subsequence of d and $x_1,\ldots ,x_n$ are pairwise distinct, $k < n$ ensues. Since d is focussed, $x_k \in \mathtt {Body}(r_j)$ for some $j > k$. So, $(x_j,r_j)$ is in $\langle (x_{k+1},r_{k+1}),\ldots ,(x_n,r_n)\rangle$. As $d'$ is a derivation schema, $\mathtt {Body}(r_j) \subseteq \{x_1,\ldots ,x_{k-1}\}$ (remember, $d'$ is a subsequence of $\langle (x_1,r_1), \ldots , (x_{k-1}$, $r_{k-1})$, $(x_{k+1}$, $r_{k+1}), \ldots , (x_n,r_n)\rangle$). Hence, $x_k \in \{x_1,\ldots ,x_{k-1}\}$. That is, $x_1,\ldots ,x_n$ are not pairwise distinct. $\square$

Proof of Property 3

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ be a theory.

• The inclusions $\mathtt {CN}({\mathcal {T}}) \subseteq {\mathcal {F}}\cup \{\mathtt {Head}(r) \mid r \in {\mathcal {S}}\cup {\mathcal {D}}\} \subseteq {\mathcal {L}}$ follow trivially from Definition 6.

• If ${\mathcal {T}}$ is finite, then ${\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}}$ are finite. Thus, the set ${\mathcal {F}}\cup \{\mathtt {Head}(r) \mid r \in {\mathcal {S}}\cup {\mathcal {D}}\}$ is finite. From the first item, $\mathtt {CN}({\mathcal {T}})$ is finite.

• For any $x \in {\mathcal {F}}$, the sequence $\langle (x,\sigma )\rangle$ is a derivation schema for x from ${\mathcal {T}}$. Thus, $x \in \mathtt {CN}({\mathcal {T}})$ and this proves the inclusion ${\mathcal {F}}\subseteq \mathtt {CN}({\mathcal {T}})$.

• $\top \in \mathtt {CN}({\mathcal {T}})$ since $\top \in {\mathcal {F}}$ and ${\mathcal {F}}\subseteq \mathtt {CN}({\mathcal {T}})$.

• Assume that ${\mathcal {F}}= \{\top \}$ and $\not \exists r \in {\mathcal {D}}$ such that $\mathtt {Body}(r) = \{\top \}$. Thus, since the body of any other rule in ${\mathcal {T}}$ is assumed to be non-empty, no rule in ${\mathcal {S}}\cup {\mathcal {D}}$ can be applied, hence $\mathtt {CN}({\mathcal {T}}) = \{\top \}$. Conversely, if $\mathtt {CN}({\mathcal {T}}) = \{\top \}$, then ${\mathcal {F}}= \{\top \}$ (since ${\mathcal {F}}\subseteq \mathtt {CN}({\mathcal {T}})$) and $\not \exists r \in {\mathcal {D}}$ such that $\mathtt {Body}(r) = \{\top \}$ (since each such rule is applicable when it exists).

• Let $d = \langle (x_1,r_1),\ldots ,(x_n,r_n)\rangle$ be a derivation schema for $x \in {\mathcal {L}}$ from ${\mathcal {T}}$. From Definition 6, for each $x_i$ ($i = 1, \ldots , n$), there exists a derivation schema from ${\mathcal {T}}$ for $x_i$. Thus, $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {T}})$. $\square$

Proof of Property 4

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ and ${\mathcal {T}}' = ({\mathcal {F}}', {\mathcal {S}}', {\mathcal {D}}')$ be two theories such that ${\mathcal {T}}\sqsubseteq {\mathcal {T}}'$. Let $x \in \mathtt {CN}({\mathcal {T}})$. So, there exists a derivation schema $d = \langle (x_1,r_1), \ldots , (x_n,r_n)\rangle$ for x from ${\mathcal {T}}$. Since ${\mathcal {T}}\sqsubseteq {\mathcal {T}}'$, $\mathtt {Facts}(d) \subseteq {\mathcal {F}}'$ and $\mathtt {Strict}(d) \subseteq {\mathcal {S}}'$ and $\mathtt {Def}(d) \subseteq {\mathcal {D}}'$. Therefore, d is also a derivation schema for x from ${\mathcal {T}}'$. $\square$

Proof of Property 5

The two properties follow trivially from the definition of option. $\square$

Proof of Property 6

The inclusion $\mathtt {POpt}({\mathcal {T}})\subseteq \mathtt {Opt}({\mathcal {T}})$ follows trivially from Definitions 9 and 10.

Assume that $\mathtt {CN}({\mathcal {T}})$ is consistent. From Property 5, $\mathtt {Opt}({\mathcal {T}}) = \{{\mathcal {T}}\}$. Since $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubseteq {\mathcal {T}}$, $\mathtt {POpt}({\mathcal {T}})= \{{\mathcal {T}}\}$. Assume now that $\mathtt {Opt}({\mathcal {T}}) \subseteq \mathtt {POpt}({\mathcal {T}})$. Since $\mathtt {Opt}({\mathcal {T}}) \ne \emptyset$, for all ${\mathcal {O}}\in \mathtt {Opt}({\mathcal {T}})$ it holds that ${\mathcal {O}}\in \mathtt {POpt}({\mathcal {T}})$. Thus, for all ${\mathcal {O}}\in \mathtt {Opt}({\mathcal {T}})$, $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubseteq {\mathcal {O}}$. It follows that $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubseteq \mathtt {Free}({\mathcal {T}})$.² Assume that $\mathtt {CN}({\mathcal {T}})$ is inconsistent. Then, there exists a minimal conflict$^2$$C = (X, Y, Z) \sqsubseteq {\mathcal {T}}$. Since $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubseteq \mathtt {Free}({\mathcal {T}})$, $X = Y = \emptyset$. But, by assumption, the body of every defeasible rule is not empty. Thus, $\mathtt {CN}(C) = \emptyset$. This contradicts the fact that $\mathtt {CN}(C)$ is inconsistent. $\square$

Proof of Property 7

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ be a theory.

Assume that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ is consistent. Thus, there exists a preferred option ${\mathcal {O}}$ such that either (i) for all $r \in {\mathcal {D}}$, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \{r\}))$ is inconsistent meaning that ${\mathcal {O}}= \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ or (ii) there exists $r \in {\mathcal {D}}$ such that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \{r\}))$ is consistent thus $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubset {\mathcal {O}}$. In both cases, $\mathtt {POpt}({\mathcal {T}})\ne \emptyset$. Assume now that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ is inconsistent. Since ${\mathcal {F}}$ and ${\mathcal {S}}$ should be part of any preferred option and the set of consequences of a preferred option should be consistent, then $\mathtt {POpt}({\mathcal {T}})= \emptyset$.

Let $r \in {\mathcal {D}}$ and assume that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \{r\}))$ is consistent. From Definition 10, $({\mathcal {F}}, {\mathcal {S}}, \{r\})$ is either a preferred option (iff for all $r' \in {\mathcal {D}}$ such that $r \ne r'$, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \{r, r'\}))$ is inconsistent). Or, there exists a preferred option ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, \{r\} \cup {\mathcal {D}}')$ where ${\mathcal {D}}' \subseteq {\mathcal {D}}{\setminus } \{r\}$. $\square$

Proof of Property 8

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ be a theory and $\mathtt {Free}({\mathcal {T}})= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}}')$. From the definition of $\mathtt {Free}({\mathcal {T}})$, $\mathtt {Free}({\mathcal {T}})\sqsubseteq {\mathcal {O}}$ for all ${\mathcal {O}}\in \mathtt {POpt}({\mathcal {T}})$. From Property 4, $\mathtt {CN}(\mathtt {Free}({\mathcal {T}})) \subseteq \mathtt {CN}({\mathcal {O}})$. Since $\mathtt {CN}({\mathcal {O}})$ is consistent, then so is for $\mathtt {CN}(\mathtt {Free}({\mathcal {T}}))$. $\square$

Proof of Property 9

Let (d, x) be a sub-argument of $(d',x')$. Let $x_i \in \mathtt {Seq}(d)$. There are two possibilities:

• $x_i \in \mathtt {Facts}(d)$, thus $x_i \in \mathtt {Facts}(d')$ since $\mathtt {Facts}(d) \subseteq \mathtt {Facts}(d')$. So, $x_i \in \mathtt {Seq}(d')$.

• $x_i = \mathtt {Head}(r)$ with $r \in \mathtt {Strict}(d) \cup \mathtt {Def}(d)$; thus, $r \in \mathtt {Strict}(d') \cup \mathtt {Def}(d')$ since $\mathtt {Strict}(d) \subseteq \mathtt {Strict}(d')$ and $\mathtt {Def}(d) \subseteq \mathtt {Def}(d')$. So, $x_i \in \mathtt {Seq}(d')$.

$\square$

Proof of Property 10

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ be a theory. Since $\top \in {\mathcal {F}}$ by Definition 4, $(\langle \top , \sigma \rangle , \top ) \in {\mathtt {Arg}}({\mathcal {T}})$ and thus ${\mathtt {Arg}}({\mathcal {T}}) \ne \emptyset$. $\square$

Proof of Property 11

Let ${\mathcal {H}}= (\mathtt {Arg}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system over a theory ${\mathcal {T}}$ and $\mathtt {Ext}({\mathcal {H}})$ its set of extensions under any extension-based semantics. Assume that $\mathtt {Ext}({\mathcal {H}})\ne \emptyset$.

Let $x \in \mathtt {Output}({\mathcal {H}})$. Thus, for all ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$, $\exists a \in {\mathcal {E}}$ such that $\mathtt {Conc}(a) = x$. It follows that $x \in \mathtt {Concs}({\mathcal {E}}_i)$, $\forall {\mathcal {E}}_i \in \mathtt {Ext}({\mathcal {H}})$ and hence $x \in \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i)$.

Assume now that $x \in \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i)$. Thus, $\forall {\mathcal {E}}_i$, $\exists a_i \in {\mathcal {E}}_i$ such that $\mathtt {Conc}(a_i) = x$. Consequently, $x \in \mathtt {Output}({\mathcal {H}})$. $\square$

Proof of Property 12

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$. Let $x \in \mathtt {Output}({\mathcal {H}})$. From Definition 17, $\exists (d,x) \in {\mathtt {Arg}}({\mathcal {T}})$. From Definition 12, d is a derivation for x from ${\mathcal {T}}$. Thus, $x \in \mathtt {CN}({\mathcal {T}})$. $\square$

Proof of Property 16

Let ${\mathcal {R}}\in \mathfrak {R}_p$ and let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be a rule-based argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$. Since ${\mathcal {H}}$ satisfies the five postulates, thus for all ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$,

• $\mathtt {Concs}({\mathcal {E}})$ is consistent

• $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Concs}({\mathcal {E}}), {\mathcal {S}}, \emptyset )$

• For all $a \in {\mathcal {E}}$, $\mathtt {Sub}(a) \subseteq {\mathcal {E}}$

• for all $(d,x) \in {\mathtt {Arg}}({\mathcal {T}})$, if $\mathtt {Seq}(d) \subseteq \mathtt {Concs}({\mathcal {E}})$, then $(d,x) \in {\mathcal {E}}$.

From Property 1, $\mathtt {Ext}_s({\mathcal {H}})\subseteq \mathtt {Ext}_p({\mathcal {H}})$, then for all ${\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$, ${\mathcal {E}}$ satisfies the above four properties. Thus, ${\mathcal {H}}$ satisfies consistency, exhaustiveness and closure under both sub-arguments and strict rules. Let us now show that it also satisfies strict precedence under stable semantics. From Property 11, $\mathtt {Output}({\mathcal {H}})= \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}_p({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i)$. Thus, for all ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$, $\mathtt {Output}({\mathcal {H}})\subseteq \mathtt {Concs}({\mathcal {E}})$. Since ${\mathcal {H}}$ satisfies strict precedence under preferred semantics, $\mathtt {CN}({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \subseteq \mathtt {Concs}({\mathcal {E}})$. Thus, the property is satisfied by every stable extension. $\square$

Proof of Proposition 1

Let ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$. Let ${\mathcal {O}}, {\mathcal {O}}' \in \mathtt {Opt}({\mathcal {T}})$ be such that $\mathtt {CN}({\mathcal {O}}) = \mathtt {CN}({\mathcal {O}}')$. Let ${\mathcal {O}}= (X, Y, Z)$ and ${\mathcal {O}}' = (X', Y', Z')$. For all $x \in X$, $x \in \mathtt {CN}({\mathcal {O}})$ and thus $x \in X'$. The same holds for all $x' \in X'$. Thus, $X = X'$.

Let $r \in Y \cup Z$. There are two cases: (i) $\mathtt {Body}(r) \not \subseteq \mathtt {CN}({\mathcal {O}})$. Consequently, $\mathtt {Body}(r) \not \subseteq \mathtt {CN}({\mathcal {O}}')$. Thus, $\mathtt {CN}({\mathcal {O}}' \oplus r)$ is consistent. So, $r \in Y' \cup Z'$ (by definition of an option).

(ii) $\mathtt {Body}(r) \subseteq \mathtt {CN}({\mathcal {O}})$. Consequently, $\mathtt {Body}(r) \subseteq \mathtt {CN}({\mathcal {O}}')$. Thus, $\mathtt {CN}({\mathcal {O}}' \oplus r)$ is consistent. So, $r \in Y' \cup Z'$ (by definition of an option). $\square$

Proof of Proposition 2

If ${\mathcal {T}}$ is finite, then $\mathtt {CN}({\mathcal {T}})$ is finite (apply Property 3). Consequently, ${\mathtt {Arg}}({\mathcal {T}})$ is finite. $\square$

Proof of Proposition 3

Let ${\mathcal {T}}$ and ${\mathcal {T}}'$ be two theories such that ${\mathcal {T}}\sqsubseteq {\mathcal {T}}'$. Let (d, x) be an argument defined from ${\mathcal {T}}$. All items in Definition 12 are independent from ${\mathcal {T}}$ except for d being a derivation schema for x from ${\mathcal {T}}$. Hence, for (d, x) to be an argument defined from ${\mathcal {T}}'$, it is enough that d be a derivation schema for x from ${\mathcal {T}}'$. Now, this is equivalent to $x \in \mathtt {CN}({\mathcal {T}}')$. By Property 4, the latter follows from $x \in \mathtt {CN}({\mathcal {T}})$ (which is itself proved from the fact that d is a derivation schema for x from ${\mathcal {T}}$). Thus, (d, x) is an argument defined from ${\mathcal {T}}'$. $\square$

Proof of Proposition 4

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}})$, ${\mathcal {R}})$ be an argumentation system such that ${\mathcal {R}}$ is conflict-dependent. Let $a = (d,x) \in {\mathtt {Arg}}({\mathcal {T}})$ be such that $(a,a) \in {\mathcal {R}}$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d)$ is inconsistent. This is impossible since a is an argument (thus $\mathtt {Seq}(d)$ should be consistent). $\square$

Proof of Proposition 5

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that $\mathtt {CN}({\mathcal {T}})$ is consistent and ${\mathcal {R}}$ is conflict-dependent. Assume that ${\mathtt {Arg}}({\mathcal {T}})$ is not conflict-free. Thus, there exist $(d,x), (d',x') \in {\mathtt {Arg}}({\mathcal {T}})$ such that $(d,x) {\mathcal {R}}(d',x')$. Consequently, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. Besides, from Property 3, $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {T}})$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {T}})$. Thus, $\mathtt {CN}({\mathcal {T}})$ is inconsistent. Contradiction. $\square$

Proof of Proposition 6

Let ${\mathcal {H}}$ be an argumentation system which satisfies consistency and closure under sub-arguments. From Proposition 12, $\forall {\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$$\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Since ${\mathcal {H}}$ satisfies consistency, $\forall {\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$$\mathtt {Concs}({\mathcal {E}})$ is consistent. Thus, so is for $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. $\square$

Proof of Proposition 7

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}})$, ${\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that $\emptyset \in \mathtt {Ext}({\mathcal {H}})$. Thus, $\mathtt {Output}({\mathcal {H}})= \emptyset$. Assume that ${\mathcal {H}}$ satisfies strict precedence, then $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$$\subseteq$$\mathtt {Output}({\mathcal {H}})$. Since $\top \in {\mathcal {F}}$ and from Property 3, it holds that ${\mathcal {F}}\subseteq \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$, then $\top \in \mathtt {Output}({\mathcal {H}})$. $\square$

Proof of Proposition 8

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$. Assume that ${\mathcal {H}}$ satisfies consistency and strict precedence. From Property 13, it holds that $\mathtt {Output}({\mathcal {H}})$ is consistent. From strict precedence, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$$\subseteq$$\mathtt {Output}({\mathcal {H}})$. Thus, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ is consistent. $\square$

Proof of Proposition 9

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system such that $\mathtt {Ext}({\mathcal {H}})\ne \emptyset$ (under an extension-based semantics). Assume that ${\mathcal {H}}$ is closed under sub-arguments and satisfies exhaustiveness. Let ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$. From the monotonicity of ${\mathtt {Arg}}$, it holds that ${\mathcal {E}}\subseteq {\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}))$. Let $(d,x) \in {\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}))$. From Proposition 12, $\mathtt {Seq}(d) \subseteq \mathtt {Concs}({\mathcal {E}})$. From the exhaustiveness postulate, $(d,x) \in {\mathcal {E}}$. $\square$

Proof of Proposition 10

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system such that $\mathtt {Ext}({\mathcal {H}})\ne \emptyset$ (under an extension-based semantics). Assume that ${\mathcal {H}}$ satisfies exhaustiveness and strict precedence. Since ${\mathcal {H}}$ satisfies strict precedence, $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset )) \subseteq \mathtt {Output}({\mathcal {H}})$. From Property 11, for all ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$, $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset )) \subseteq \mathtt {Concs}({\mathcal {E}})$. Let $(d,x) \in {\mathtt {Arg}}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$. Thus, $\mathtt {Seq}(d) \subseteq \mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$. From exhaustiveness, $(d,x) \in {\mathcal {E}}$. $\square$

Proof of Proposition 11

In order to prove the compatibility of the postulates, it is sufficient to give an example of a system which satisfies all the five postulates. This system is ASPIC as defined in [17]. Proposition 1 in [17] shows that the system is closed under sub-arguments under any Dung’s semantics. Proposition 8 in [17] shows that the system is closed under strict rules under complete semantics, thus under stable semantics. Property 2 in [17] shows that the system satisfies consistency under any Dung’s semantics. From Proposition 13, the system satisfies exhaustiveness. Let us now show that the system satisfies strict precedence. This follows from the definition of attack relation (Definition 16 in [17]) according to which a strict argument cannot be attacked. Thus, it belongs to any stable extension. $\square$

Proof of Proposition 12

Let ${\mathcal {H}}$ be an argumentation system such that $\mathtt {Ext}({\mathcal {H}})\ne \emptyset$ where $\mathtt {Ext}({\mathcal {H}})$ is its set of extensions under an extension-based semantics. Assume that ${\mathcal {H}}$ is closed under sub-arguments and let ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$.

• Let $x \in \mathtt {Concs}({\mathcal {E}})$. Thus, $\exists (d, x) \in {\mathcal {E}}$ where d is a derivation for x from $(\mathtt {Facts}(d)$, $\mathtt {Strict}(d)$, $\mathtt {Def}(d))$. From Property 3, $x \in \mathtt {Facts}(d)$ (thus $x \in X$), or $x = \mathtt {Head}(r)$ where $r \in \mathtt {Strict}(d)$ (thus $r \in Y$) or $x \in \mathtt {Def}(d)$ (thus $x \in Z$).

  Assume now that $x \in X$. Thus, $\exists (d,y) \in {\mathcal {E}}$ such that $x \in \mathtt {Facts}(d)$. Besides, $(\langle (x,\sigma )\rangle , x)$ is a sub-argument of (d, y). Since ${\mathcal {H}}$ is closed under sub-arguments, $(\langle (x,\sigma )\rangle , x) \in {\mathcal {E}}$, and thus, $x \in \mathtt {Concs}({\mathcal {E}})$.

  Let $r \in Y \cup Z$. Thus, $\exists (d,x) \in {\mathcal {E}}$ such that $r \in \mathtt {Strict}(d) \cup \mathtt {Def}(d)$. Let $d = \langle (x_1,r_1), \ldots , (x_i,r), (x_{i+1},r_{i+1}) \ldots , (x_n = x,r_n)\rangle$ with $x_i = \mathtt {Head}(r)$. Thus, there exists a sub-sequence $d'$ of d which is a derivation for $x_i$. This derivation is minimal (for set inclusion since (d, x) is an argument). Thus, $(d', x_i)$ is an argument. Moreover, it is a sub-argument of (d, x). Since ${\mathcal {H}}$ is closed under sub-arguments, $(d',x_i) \in {\mathcal {E}}$. Consequently, $x_i \in \mathtt {Concs}({\mathcal {E}})$. Thus, $\mathtt {Concs}({\mathcal {E}}) = X \cup \{\mathtt {Head}(r) \mid r \in Y \cup Z\}$.

• From the definitions of the two functions $\mathtt {Concs}$ and $\mathtt {Th}$, it follows that $\mathtt {Concs}({\mathcal {E}})$$\subseteq$$\mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. From Property 3, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) \subseteq X \cup \{\mathtt {Head}(r) \mid r \in Y \cup Z\}$. From above, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) \subseteq \mathtt {Concs}({\mathcal {E}})$.

• Assume now that $a = (d, x) \in {\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}))$. For all $x_i \in \mathtt {Seq}(d)$, $x_i \in \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {Concs}({\mathcal {E}})$. Then, $x_i \in \mathtt {Concs}({\mathcal {E}})$. Thus, $\mathtt {Seq}(d) \subseteq \mathtt {Concs}({\mathcal {E}})$.

$\square$

Proof of Proposition 13

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments. Assume that ${\mathcal {H}}$ violates exhaustiveness. Thus, there exists ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ and there exists $a = (d,x) \in {\mathtt {Arg}}({\mathcal {T}})$ such that $\mathtt {Seq}(d) \subseteq \mathtt {Concs}({\mathcal {E}})$ but $(d,x) \notin {\mathcal {E}}$. So, $\exists b = (d',x') \in {\mathcal {E}}$ such that $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. Thus, $\exists y \in \mathtt {Seq}(d)$ such that $\lnot y \in d'$. But, $y, \lnot y \in \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {Concs}({\mathcal {E}})$. Thus, $y, \lnot y \in \mathtt {Concs}({\mathcal {E}})$. This contradicts the fact that ${\mathcal {H}}$ satisfies consistency. $\square$

Proof of Proposition 14

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments. From Proposition 13, ${\mathcal {H}}$ satisfies exhaustiveness. From Proposition 9, it follows that for all ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$, ${\mathcal {E}}= {\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}))$. $\square$

Proof of Proposition 15

The proof is similar to that of Propositions 13 and 14. $\square$

Proof of Theorem 1

Let ${\mathcal {T}}$ be a theory and $d = \langle (x_1,r_1),\ldots ,(x_n,r_n)\rangle$ a consistent sequence from ${\mathcal {T}}$.

$(\longrightarrow )$ Assume that (d, x) is an argument from ${\mathcal {T}}$. $d = \langle (x_1,r_1),\ldots ,(x_n,r_n)\rangle$ yields $x = x_n$ (Definition 12). Assume that d is not focused. Let $d^*$ be obtained from d by deleting all repeated pairs. Since d is not focused, $d^*$ is not minimal. Therefore, there exists $(x_k,r_k)$ in $d^*$ (hence in d) such that depriving $d^*$ from $(x_k,r_k)$ still gives a derivation of x from ${\mathcal {T}}$. Since $d^*$ contains no repeated pair, for every $(x_i,r_i)$ in $d^*$, if $i \ne k$ then either $x_i \ne x_k$ or $r_i \ne r_k$. For $r_k \ne \sigma$, the former implies the latter hence $r_i \ne r_k$ whenever $i \ne k$. Thus, depriving $d^*$ from $(x_k,r_k)$ gives a derivation $d'$ of x from ${\mathcal {T}}$ such that $\mathtt {Facts}(d') \subseteq \mathtt {Facts}(d)$ and $\mathtt {Strict}(d') \subseteq \mathtt {Strict}(d)$ and $\mathtt {Def}(d') \subseteq \mathtt {Def}(d)$, with one of the latter two inclusions being strict. That is, there exists ${\mathcal {T}}' \sqsubset (\mathtt {Facts}(d), \mathtt {Strict}(d), \mathtt {Def}(d))$ such that $x \in \mathtt {CN}({\mathcal {T}}')$, thereby contradicting Definition 12. The remaining case is $r_k = \sigma$. Since $d^*$ contains no repeated pair, no $(x_i,r_i)$ in $d^*$ is $(x_k,\sigma )$ except for $i = k$ and it follows that $d^*$ deprived from $(x_k,r_k)$ gives a derivation $d'$ of x from ${\mathcal {T}}$ such that $\mathtt {Facts}(d') \subset \mathtt {Facts}(d)$ while $\mathtt {Strict}(d') \subseteq \mathtt {Strict}(d)$ and $\mathtt {Def}(d') \subseteq \mathtt {Def}(d)$. As above, a contradiction arises.

$(\longleftarrow )$ Assume that d is a focused derivation schema from ${\mathcal {T}}$ such that $x_n = x$. By the definitions, $x \in {\mathcal {L}}$ and d is a derivation schema for x from ${\mathcal {T}}$. Due to the hypothesis in the statement of the theorem, $\mathtt {Seq}(d)$ is consistent. Assume that there exists ${\mathcal {T}}' = ({\mathcal {F}}', {\mathcal {S}}', {\mathcal {D}}') \sqsubset (\mathtt {Facts}(d),\mathtt {Strict}(d),\mathtt {Def}(d))$ such that $x \in \mathtt {CN}({\mathcal {T}}')$. That is, there exists a derivation $d' = \langle (x'_1,r'_1),\ldots ,(x'_m,r'_m)\rangle$ for some $m < n$ such that $\mathtt {Facts}(d') = {\mathcal {F}}'$ and $\mathtt {Strict}(d') = {\mathcal {S}}'$ and $\mathtt {Def}(d') = {\mathcal {D}}'$. Let $d^*$ be a minimal derivation schema for x from ${\mathcal {T}}$ obtained from d by deleting all repeated pairs. Accordingly, $\mathtt {Facts}(d^*) = \mathtt {Facts}(d)$ and $\mathtt {Strict}(d^*) = \mathtt {Strict}(d)$ and $\mathtt {Def}(d^*) = \mathtt {Def}(d)$. Since $\mathtt {Strict}(d') \subseteq \mathtt {Strict}(d)$ and $\mathtt {Def}(d') \subseteq \mathtt {Def}(d)$, if $(x'_i,r'_i)$ is in $d'$ with $r'_i \ne \sigma$ then there exists k such that $(x_k,r_k)$ is in $d^*$ where $r_k = r'_i$ (also, $x_k = x'_i$ because $x'_i = \mathtt {Head}(r'_i) = \mathtt {Head}(r_k) = x_k$). Similarly, since $\mathtt {Facts}(d') \subseteq \mathtt {Facts}(d)$, if $(x'_i,r'_i)$ is in $d'$ with $r'_i = \sigma$ then there exists k such that $(x_k,r_k)$ is in $d^*$ where $x'_i = x_k$ and $r_k = r'_i = \sigma$. That is, $d'$ is a proper subsequence of a reordering of $d^*$, thereby contradicting the minimality of $d^*$.

Indeed, let us show that no initial proper fragment of a reordering $d^{*}_\iota$ of $d^*$ is a minimal derivation of x. Assume a reordering $d^{*}_\iota = \langle (x^*_{\iota 1},r^*_{\iota 1}),\ldots ,(x^*_{\iota p},r^*_{\iota p})\rangle$ of $d^* = \langle (x^*_1,r^*_1),\ldots ,(x^*_p,r^*_p)\rangle$ such that $d_{\iota } = \langle (x^*_{\iota 1},r^*_{\iota 1}),\ldots ,(x^*_{\iota q},r^*_{\iota q})\rangle$ is a minimal derivation of x from ${\mathcal {T}}$ for some $q < p$. Let j be the greatest index from $1 \ldots p$ such that $x^*_j$ is in $\mathtt {Seq}(d^*)$ but not in $\mathtt {Seq}(d_{\iota })$ (clearly, $j < p$). Since $d^*$ is minimal, there must exist $h > j$ such that $x^*_j \in \mathtt {Body}(r^*_h)$ (otherwise $d^*$ deprived of $(x^*_j,r^*_j)$ would give a proper subsequence also being a derivation of x). By Property 2, $x^*_h \ne x^*_i$ for $i \ne h$. Hence, $\mathtt {Head}(r^*_i) \ne x^*_h$ for $i \ne h$. Together with $x^*_j \not \in \mathtt {Seq}(d_{\iota })$ and $x^*_j \in \mathtt {Body}(r^*_h)$, this entails $x^*_h \not \in \mathtt {Seq}(d_{\iota })$. Therefore, j is not the greatest index such that $x^*_j$ is in $\mathtt {Seq}(d^*)$ but not in $\mathtt {Seq}(d_{\iota })$. $\square$

Lemma 1

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$ is consistent and ${\mathcal {R}}$ is conflict-dependent and privileges strict arguments. For all $a \in {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$, $b \in {\mathtt {Arg}}({\mathcal {T}})$, if $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$, then $\exists a' \in \mathtt {Sub}(a)$ such that $a'$ is strict and $a' {\mathcal {R}}b$.

Proof

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$ is consistent. Assume that ${\mathcal {R}}$ is conflict-dependent and privileges strict arguments.

Assume that $\exists a = (d,x) \in {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ and $\exists b = (d',x') \in {\mathtt {Arg}}({\mathcal {T}})$ such that $b {\mathcal {R}}a$ or $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent and thus $\mathtt {CN}(\mathtt {Th}(\{a\})) \cup \mathtt {CN}(\mathtt {Th}(\{b\}))$ is inconsistent (since from Property 3, $\mathtt {Seq}(d) \subseteq \mathtt {CN}(\mathtt {Th}(\{a\}))$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}(\{b\}))$).

Let $\mathtt {Th}(\{a\}) = (X,Y,Z)$ and $\mathtt {Th}(\{b\}) = (X',Y',Z')$. Let us show that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z'))$ is inconsistent. Assume that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z'))$ is consistent. Thus, there exists a preferred option ${\mathcal {O}}\in \mathtt {POpt}({\mathcal {T}})$ such that $({\mathcal {F}}, {\mathcal {S}}, Z') \sqsubseteq {\mathcal {O}}$. Since $(X,Y,Z) \sqsubseteq \mathtt {Free}({\mathcal {T}})$ and $\mathtt {Free}({\mathcal {T}})\sqsubseteq {\mathcal {O}}$, $({\mathcal {F}}, {\mathcal {S}}, Z\cup Z') \sqsubseteq {\mathcal {O}}$. From Property 4, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z\cup Z')) \subseteq \mathtt {CN}({\mathcal {O}})$. Thus, $\mathtt {CN}({\mathcal {O}})$ is inconsistent. This contradicts the fact that ${\mathcal {O}}$ is an option.

Let $Z^*$ be the smallest (for set inclusion) subset of $Z'$ such that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z^*))$ is inconsistent. Thus, for all $r \in Z^*$, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z^*{\setminus }\{r\}))$ is consistent. It follows that for all $r \in Z^*$, there exists a preferred option ${\mathcal {O}}\in \mathtt {POpt}({\mathcal {T}})$ such that $({\mathcal {F}}, {\mathcal {S}}, Z \cup Z^*{\setminus }\{r\}) \sqsubseteq {\mathcal {O}}$ by Property 8.

Assume that for every strict $a'' \in \mathtt {Sub}(a)$, $\mathtt {Seq}(d'') \cup \mathtt {Seq}(d')$ is consistent. However, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent (since $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$ while ${\mathcal {R}}$ is conflict-dependent). Hence, $\mathtt {Head}(\mathtt {Def}(d)) \cup \mathtt {Head}(\mathtt {Def}(d'))$ is inconsistent, say $y \in \mathtt {Head}(\mathtt {Def}(d))$ and $\lnot y \in \mathtt {Head}(\mathtt {Def}(d'))$. Should no such y be in ${\mathcal {F}}\cup \mathtt {Head}(S)$, then there would be a preferred option $O=(F,S,D_O)$ with $\lnot y \in \mathtt {Head}(D_O)$. A contradiction arises, because $a=(d,x)$ being in ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ means that $\mathtt {Def}(d)$ is a subset of $D_O$ for every preferred option $O=(F,S,D_O)$.

That is, there exists $a'' = (d'',x'') \in \mathtt {Sub}(a)$ such that $a''$ is strict and $\mathtt {Seq}(d'') \cup \mathtt {Seq}(d')$ is inconsistent. Since ${\mathcal {R}}$ privileges strict arguments, $a'' {\mathcal {R}}b$. $\square$

Proof of Theorem 2

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$ is consistent. Assume that ${\mathcal {R}}$ is conflict-dependent and privileges strict arguments.

We show first that ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ is conflict-free. From Property 8, $\mathtt {CN}(\mathtt {Free}({\mathcal {T}}))$ is consistent. Since ${\mathcal {R}}$ is conflict-dependent, from Proposition 5, ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ is conflict-free.

Let us now show that ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ defends its elements. Assume that $\exists a = (d,x) \in {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ and $\exists b = (d',x') \in {\mathtt {Arg}}({\mathcal {T}})$ such that $b {\mathcal {R}}a$. From Lemma 1, there exists $a' = (d'',x'') \in \mathtt {Sub}(a)$ such that $a'$ is strict, hence $a \in {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$, and $a' {\mathcal {R}}b$. $\square$

Proof of Theorem 3

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}})$, ${\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that $\mathtt {Ext}({\mathcal {H}})\ne \emptyset$. Assume that ${\mathcal {H}}$ satisfies strict precedence and closure under both strict rules and sub-arguments. Let ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$. Since $X \subseteq {\mathcal {F}}$ and $Y \subseteq {\mathcal {S}}$, $(X, Y, Z) \sqsubseteq ({\mathcal {F}}, {\mathcal {S}}, Z)$. From Property 4, $\mathtt {CN}((X, Y, Z)) \subseteq \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z))$.

Let us now show that $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z)) \subseteq \mathtt {CN}((X, Y, Z))$. Since ${\mathcal {H}}$ satisfies strict precedence, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq \mathtt {Output}({\mathcal {H}})$. From Property 11, $\mathtt {Output}({\mathcal {H}})\subseteq \mathtt {Concs}({\mathcal {E}})$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ by Proposition 12. Hence, $\mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq \mathtt {CN}((X, Y, Z))$. Furthermore, from Property 3, ${\mathcal {F}}\subseteq \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$. Thus, $X = {\mathcal {F}}$, i.e., $\mathtt {Th}({\mathcal {E}}) = ({\mathcal {F}}, Y, Z)$. Let $x \in \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z))$. Then, there exists a derivation schema

$$\begin{aligned} d = \langle (x_1,r_1), \ldots , (x_n, r_n)\rangle \end{aligned}$$

for x from $({\mathcal {F}}, {\mathcal {S}}, Z)$. There are two cases:

• For any $i = 1, \ldots , n$, $r_i \in \{\sigma \} \cup Y \cup Z$. Hence, d is also a derivation schema for x from $({\mathcal {F}}, Y, Z)$. Thus, $x \in \mathtt {CN}(({\mathcal {F}}, Y, Z))$.

• There exists $1 < i \le n$ such that $r_i \in {\mathcal {S}}{\setminus } Y$ (note that the two theories $({\mathcal {F}}, Y, Z)$ and $({\mathcal {F}}, {\mathcal {S}}, Z)$ differ only on ${\mathcal {S}}{\setminus } Y$). Let i be the first step where an element of ${\mathcal {S}}{\setminus } Y$ is used in the derivation d. Note also that $i > 1$ since strict rules have non-empty bodies. Thus, for any $j < i$, $r_j \in \{\sigma \} \cup Y \cup Z$ and $\langle (x_1,r_1), \ldots , (x_j, r_j)\rangle$ is a derivation schema of $x_j$ from $({\mathcal {F}}, Y, Z)$. Thus, $x_j \in \mathtt {CN}(({\mathcal {F}}, Y, Z))$. Furthermore, $\mathtt {Body}(r_i) \subseteq \{x_1, \ldots , x_{i-1}\}$, so $\mathtt {Body}(r_i) \subseteq \mathtt {CN}(({\mathcal {F}}, Y, Z))$. Since $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$, $\mathtt {Body}(r_i) \subseteq \mathtt {Concs}({\mathcal {E}})$. Since ${\mathcal {H}}$ is closed under strict rules, $\mathtt {Head}(r_i) \in \mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(({\mathcal {F}}, Y, Z))$, i.e., $x_i \in \mathtt {CN}(({\mathcal {F}}, Y, Z))$. We repeat the same reasoning for showing that each $x_i \in \mathtt {CN}(({\mathcal {F}}, Y, Z))$ and conclude that $x \in \mathtt {CN}(({\mathcal {F}}, Y, Z))$.

$\square$

Proof of Theorem 4

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {H}}$ satisfies strict precedence, and closure under both strict rules and sub-arguments.

Let ${\mathcal {E}}\in \mathtt {Ext}({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$. From Theorem 3 and Proposition 12, it holds that

$$\begin{aligned} \mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z)) = \mathtt {Concs}({\mathcal {E}}). \end{aligned}$$

(*)

Let ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, Z \cup Z')$ where $Z' = \{r \ | \ r \in {\mathcal {D}}{\setminus } Z \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))\}$. Since $({\mathcal {F}}, {\mathcal {S}}, Z) \sqsubseteq {\mathcal {O}}$, then from Property 4 and (*), $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}})$. Let us now show that $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {Concs}({\mathcal {E}})$. Let $x \in \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z \cup Z'))$. Then, there exists a derivation schema

$$\begin{aligned} d = \langle (x_1,r_1), \ldots , (x_n, r_n)\rangle \end{aligned}$$

for x from $({\mathcal {F}}, {\mathcal {S}}, Z \cup Z')$. There are two cases:

• For any $i = 1, \ldots , n$, $r_i \in \{\sigma \} \cup {\mathcal {S}}\cup Z$. Hence, d is also a derivation schema for x from $({\mathcal {F}}, {\mathcal {S}}, Z)$. Thus, $x \in \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z))$, and from (1), $x \in \mathtt {Concs}({\mathcal {E}})$.

• Assume that there exists $1 < i \le n$ such that $r_i \in Z'$ (note that the two theories $({\mathcal {F}}, {\mathcal {S}}, Z)$ and $({\mathcal {F}}, {\mathcal {S}}, Z \cup Z')$ differ only on $Z'$). Let i be the first step where an element of $Z'$ is used in the derivation d. Since the bodies of defeasible rules are not empty, $i > 1$. It follows that for any $j < i$, $r_i \in \{\sigma \} \cup {\mathcal {S}}\cup Z$, thus $\langle (x_1,r_1), \ldots , (x_j, r_j)\rangle$ is a derivation schema of $x_j$ from $({\mathcal {F}}, {\mathcal {S}}, Z)$. Thus, $x_j \in \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z))$. Furthermore, by Definition 6, $\mathtt {Body}(r_i) \subseteq \{x_1, \ldots , x_{i-1}\}$. Then, $\mathtt {Body}(r_i) \subseteq \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, Z))$. This contradicts the fact that $r_i \in Z'$ and thus such $r_i$ does not exist.

$\square$

Proof of Theorem 5

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent. Assume that for all ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$, ${\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq {\mathcal {E}}$. Assume now that $\exists a \in {\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ and $\exists b \in {\mathtt {Arg}}({\mathcal {T}})$ such that $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $(b,b) \notin {\mathcal {R}}$ (cf. Proposition 4). Thus, $\exists {\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ such that $b \in {\mathcal {E}}$. Consequently, $a, b \in {\mathcal {E}}$, this contradicts the fact that ${\mathcal {E}}$ is conflict-free (since it is a naive extension).

Assume now that for all $a \in {\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$, $\not \exists b \in {\mathtt {Arg}}({\mathcal {T}})$ such that $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. This means that arguments of ${\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ are not attacked. Thus, they belong to every naive extension. $\square$

Proof of Theorem 6

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent. Assume that $\exists a, b \in {\mathtt {Arg}}({\mathcal {T}})$ such that $a \in {\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ and $\mathtt {Conc}(a) = \lnot \mathtt {Conc}(b)$. Thus, $(b,b) \notin {\mathcal {R}}$ and $\exists {\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ such that $b \in {\mathcal {E}}$. If ${\mathcal {H}}$ satisfies strict precedence, then $\mathtt {Conc}(a) \in \mathtt {Concs}({\mathcal {E}})$ meaning that $\mathtt {Concs}({\mathcal {E}})$ is inconsistent. Thus, ${\mathcal {H}}$ violates consistency. If ${\mathcal {H}}$ satisfies consistency, then $\mathtt {Conc}(a) \notin \mathtt {Concs}({\mathcal {E}})$ meaning that ${\mathcal {H}}$ violates strict precedence. $\square$

Proof of Theorem 7

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments. Let ${\mathcal {E}}, {\mathcal {E}}' \in \mathtt {Ext}_n({\mathcal {H}})$ such that $\mathtt {Concs}({\mathcal {E}}') \subseteq \mathtt {Concs}({\mathcal {E}})$.

Assume that $\exists a = (d,x) \in {\mathcal {E}}{\setminus }{\mathcal {E}}'$. Thus, $\exists b = (d',x') \in {\mathcal {E}}'$ such that $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. But ${\mathcal {H}}$ is closed under sub-arguments. Thus, Proposition 12 gives $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ and $\mathtt {Concs}({\mathcal {E}}') = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}'))$. Besides, $\mathtt {Seq}(d) \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}'))$ using Propositions 12 and 14. Since $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}')) \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Thus, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ is inconsistent. This contradicts the fact that ${\mathcal {H}}$ satisfies consistency.

The same reasoning holds for $a = (d',x') \in {\mathcal {E}}'{\setminus }{\mathcal {E}}$. $\square$

Proof of Theorem 8

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments. Let ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$.

From Proposition 6, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ is consistent. Thus, $\exists {\mathcal {O}}\in \mathtt {Opt}({\mathcal {T}})$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$. From Property 4, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) \subseteq \mathtt {CN}({\mathcal {O}})$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {Concs}({\mathcal {E}})$ by Proposition 12. Thus, $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}})$. Assume now that $\exists x \in \mathtt {CN}({\mathcal {O}}){\setminus }\mathtt {Concs}({\mathcal {E}})$. Then, there exists a minimal derivation d for x from ${\mathcal {O}}$. From Property 3, $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {O}})$. Since $\mathtt {CN}({\mathcal {O}})$ is consistent, (d, x) is an argument. In addition $(d,x) \notin {\mathcal {E}}$. Then, $\exists (d',x') \in {\mathcal {E}}$ such that $(d,x) {\mathcal {R}}(d',x')$ or $(d',x') {\mathcal {R}}(d,x)$. Since ${\mathcal {R}}$ is conflict-dependent, then $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. But, $\mathtt {Seq}(d') \subseteq \mathtt {Concs}({\mathcal {E}})$. So, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {O}})$. This contradicts the fact that ${\mathcal {O}}$ is an option. So, $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {Concs}({\mathcal {E}})$.

Since both $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}})$ and $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {Concs}({\mathcal {E}})$ have now been proved, the required $\mathtt {CN}({\mathcal {O}}) = \mathtt {Concs}({\mathcal {E}})$ follows.

Let us now show that ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. Assume that $\exists {\mathcal {O}}' \in \mathtt {Opt}({\mathcal {T}})$ such that $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. Thus, $\exists x \in \mathtt {CN}({\mathcal {O}}')$ and $x \notin \mathtt {CN}({\mathcal {O}})$. Thus, there exists a minimal derivation d for x from ${\mathcal {O}}'$. Since $\mathtt {CN}({\mathcal {O}}')$ is consistent and $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {O}}')$ (from Property 3), (d, x) is an argument. In addition $(d,x) \notin {\mathcal {E}}$ (since $x \notin \mathtt {CN}({\mathcal {O}})$). Then, $\exists (d',x') \in {\mathcal {E}}$ such that $(d,x) {\mathcal {R}}(d',x')$ or $(d',x') {\mathcal {R}}(d,x)$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. But, $\mathtt {Seq}(d') \subseteq \mathtt {Concs}({\mathcal {E}})$. So, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {O}}')$. This contradicts the fact that ${\mathcal {O}}'$ is an option.

From Proposition 1, it follows that for all ${\mathcal {O}}, {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$, if $\mathtt {CN}({\mathcal {O}}) = \mathtt {CN}({\mathcal {O}}') = \mathtt {Concs}({\mathcal {E}})$, then ${\mathcal {O}}= {\mathcal {O}}'$. $\square$

Proof of Theorem 9

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments.

• Let ${\mathcal {E}}, {\mathcal {E}}' \in \mathtt {Ext}_n({\mathcal {H}})$. From Theorem 8, $\exists {\mathcal {O}}, {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$ and $\mathtt {Concs}({\mathcal {E}}') = \mathtt {CN}({\mathcal {O}}')$. If ${\mathcal {O}}= {\mathcal {O}}'$, then $\mathtt {Concs}({\mathcal {E}}) = \mathtt {Concs}({\mathcal {E}}')$. From Theorem 7, ${\mathcal {E}}= {\mathcal {E}}'$.

• Let ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ and ${\mathcal {O}}= \mathtt {Option}({\mathcal {E}})$. Thus, $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$ and $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$. From Proposition 3, $\mathtt {Arg}(\mathtt {Th}({\mathcal {E}})) \subseteq \mathtt {Arg}({\mathcal {O}})$. From Proposition 14, $\mathtt {Arg}(\mathtt {Th}({\mathcal {E}})) = {\mathcal {E}}$. Thus, ${\mathcal {E}}\subseteq \mathtt {Arg}({\mathcal {O}})$. Assume now that $\exists a = (d,x) \in \mathtt {Arg}({\mathcal {O}})$ and $a \notin {\mathcal {E}}$. Thus, $\exists b = (d',x') \in {\mathcal {E}}$ and $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. Besides, $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {O}})$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {CN}({\mathcal {O}})$. Thus, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {O}})$. This contradicts the fact that ${\mathcal {O}}$ is an option.

$\square$

Proof of Theorem 10

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments.

• Let ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. Thus, $\mathtt {CN}({\mathcal {O}})$ is consistent. From Proposition 5, since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Arg}({\mathcal {O}})$ is conflict-free. Assume now that $\mathtt {Arg}({\mathcal {O}}) \notin \mathtt {Ext}_n({\mathcal {H}})$. Thus, $\exists a = (d,x) \in \mathtt {Arg}({\mathcal {T}})$ such that $a \notin \mathtt {Arg}({\mathcal {O}})$ and $\mathtt {Arg}({\mathcal {O}}) \cup \{a\}$ is conflict-free. Consequently, $\exists {\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ such that $\mathtt {Arg}({\mathcal {O}}) \cup \{a\} \subseteq {\mathcal {E}}$. It follows that $\mathtt {Concs}(\mathtt {Arg}({\mathcal {O}}) \cup \{a\}) \subseteq \mathtt {Concs}({\mathcal {E}})$. Since $\mathtt {CN}({\mathcal {O}})$ is consistent, $\mathtt {Concs}(\mathtt {Arg}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}})$. Thus, $\mathtt {CN}({\mathcal {O}}) \cup \{x\} \subseteq \mathtt {Concs}({\mathcal {E}})$. From Theorem 8, $\exists {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}}')$. Then, $\mathtt {CN}({\mathcal {O}}) \cup \{x\} \subseteq \mathtt {CN}({\mathcal {O}}')$. This contradicts the fact that ${\mathcal {O}}$ is a maximal option.

• Let ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. By definition of $\mathtt {Th}$, $\mathtt {Th}(\mathtt {Arg}({\mathcal {O}})) \sqsubseteq {\mathcal {O}}$. From Property 4, $\mathtt {CN}(\mathtt {Th}(\mathtt {Arg}({\mathcal {O}}))) \subseteq \mathtt {CN}({\mathcal {O}})$. Besides, from first item, $\mathtt {Arg}({\mathcal {O}}) \in \mathtt {Ext}_n({\mathcal {H}})$. From Theorem 8, $\exists {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ such that $\mathtt {Th}(\mathtt {Arg}({\mathcal {O}})) \sqsubseteq {\mathcal {O}}'$ and $\mathtt {Concs}(\mathtt {Arg}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}}')$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}(\mathtt {Arg}({\mathcal {O}}))) = \mathtt {Concs}(\mathtt {Arg}({\mathcal {O}}))$. Consequently, $\mathtt {CN}({\mathcal {O}}') \subseteq \mathtt {CN}({\mathcal {O}})$. From Proposition 1, ${\mathcal {O}}= {\mathcal {O}}'$.

• Let ${\mathcal {O}}, {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. Assume that $\mathtt {Arg}({\mathcal {O}}) = \mathtt {Arg}({\mathcal {O}}')$.

  It follows that $\mathtt {Option}(\mathtt {Arg}({\mathcal {O}})) = \mathtt {Option}(\mathtt {Arg}({\mathcal {O}}'))$. From item 2, it follows that ${\mathcal {O}}= {\mathcal {O}}'$.

$\square$

Proof of Theorem 11

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency, strict precedence and closure under both strict rules and sub-arguments.

Let us show that $\mathtt {Max}(\mathtt {Opt}({\mathcal {T}})) \subseteq \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. Let ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. From (item 1) of Theorem 10, ${\mathtt {Arg}}({\mathcal {O}}) \in \mathtt {Ext}_n({\mathcal {H}})$. From Theorem 8, $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}(\mathtt {Option}({\mathtt {Arg}}({\mathcal {O}})))$. From Corollary 1, $\mathtt {Option}({\mathtt {Arg}}({\mathcal {O}})) = {\mathcal {O}}$. Hence, $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}})$. From Theorem 4, there exists ${\mathcal {O}}' = ({\mathcal {F}}, {\mathcal {S}}, Z)$ such that

$$\begin{aligned} Z = \left( \bigcup _{(d,x) \in {\mathtt {Arg}}({\mathcal {O}})} \mathtt {Def}(d) \right) \cup \left\{ r \ | \ r \in {\mathcal {D}} \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}\left( \mathtt {Th}\left( {\mathtt {Arg}}({\mathcal {O}})\right) \right) \right\} \end{aligned}$$

and $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}}')$. Since $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}})$ and using Proposition 12, $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}(\mathtt {Th}({\mathtt {Arg}}({\mathcal {O}})))$, then $\mathtt {CN}(\mathtt {Th}({\mathtt {Arg}}({\mathcal {O}}))) = \mathtt {CN}({\mathcal {O}})$ and we get

$$\begin{aligned} Z = \left( \bigcup _{(d,x) \in {\mathtt {Arg}}({\mathcal {O}})} \mathtt {Def}(d) \right) \cup \left\{ r \ | \ r \in {\mathcal {D}} \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}({\mathcal {O}})\right\} \end{aligned}$$

and $\mathtt {CN}({\mathcal {O}}) = \mathtt {CN}({\mathcal {O}}')$. From Proposition 1, it follows that ${\mathcal {O}}= {\mathcal {O}}'$. Furthermore, ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ and is maximal (for set inclusion) up to consistency and contains the strict part of ${\mathcal {T}}$, then ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$.

Let us now show that $\mathtt {Max}(\mathtt {POpt}({\mathcal {T}})) \subseteq \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. Let ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. By definition of preferred option, $\mathtt {CN}({\mathcal {O}})$ is consistent. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Arg}({\mathcal {O}})$ is conflict-free by Proposition 5. Assume now that $\mathtt {Arg}({\mathcal {O}}) \notin \mathtt {Ext}_n({\mathcal {H}})$. Thus, $\exists a \in \mathtt {Arg}({\mathcal {T}}){\setminus }\mathtt {Arg}({\mathcal {O}})$ such that ${\mathtt {Arg}}({\mathcal {O}}) \cup \{a\}$ is conflict-free. Consequently, $\exists {\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$ such that ${\mathtt {Arg}}({\mathcal {O}}) \cup \{a\} \subseteq {\mathcal {E}}$. Thus, $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}}) \cup \{a\}) \subseteq \mathtt {Concs}({\mathcal {E}})$. Since $\mathtt {CN}({\mathcal {O}})$ is consistent, $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}})$. Thus, $\mathtt {CN}({\mathcal {O}}) \cup \{\mathtt {Conc}(a)\} \subseteq \mathtt {Concs}({\mathcal {E}})$. From Theorem 8, $\exists {\mathcal {O}}' \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}}')$. Then, $\mathtt {CN}({\mathcal {O}}) \cup \{\mathtt {Conc}(a)\} \subseteq \mathtt {CN}({\mathcal {O}}')$. This means that ${\mathcal {O}}' \in \mathtt {POpt}({\mathcal {T}})$ (since it contains all consequences of the strict part of ${\mathcal {T}}$). This contradicts the fact that ${\mathcal {O}}$ is a maximal preferred option. Consequently, ${\mathtt {Arg}}({\mathcal {O}}) \in \mathtt {Ext}_n({\mathcal {H}})$. From Theorem 8, ${\mathcal {O}}\in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$. $\square$

Proof of Theorem 12

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments. From Property 11, $\mathtt {Output}({\mathcal {H}}) = \bigcap \limits _{{\mathcal {E}}_i \in \mathtt {Ext}_n({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i)$. From Theorem 8, for all ${\mathcal {E}}_i \in \mathtt {Ext}_n({\mathcal {H}})$, there exists a unique ${\mathcal {O}}_i \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}_i) = \mathtt {CN}({\mathcal {O}}_i)$. Also, Corollary 1 guarantees that $\mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))$ does not have any additional elements that do not have a mapping in $\mathtt {Ext}_n({\mathcal {H}})$. Thus,

$$\begin{aligned} \mathtt {Output}({\mathcal {H}})= \bigcap \limits _{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))} \mathtt {CN}({\mathcal {O}}_i). \end{aligned}$$

$\square$

Lemma 2

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. For any ${\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$, it holds that $({\mathcal {F}}, {\mathcal {S}}, Z) \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ whenever

$$\begin{aligned} Z = \left( \bigcup _{(d,x) \in {\mathcal {E}}} \mathtt {Def}(d)\right) \cup \left\{ r \mid r \in {\mathcal {D}} \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))\right\} . \end{aligned}$$

Proof

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. Let ${\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$. Let ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, Z \cup Z')$ where $Z' = \{r \ | \ r \in {\mathcal {D}}{\setminus } Z \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))\}$. Clearly, $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$. From Theorem 4, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$. Since ${\mathcal {H}}$ satisfies consistency, then $\mathtt {Concs}({\mathcal {E}})$ is consistent, and thus, $\mathtt {CN}({\mathcal {O}})$ is consistent as well. Since $({\mathcal {F}}, {\mathcal {S}}, \emptyset ) \sqsubseteq {\mathcal {O}}$ and $\mathtt {CN}({\mathcal {O}})$ is consistent, from Property 7, $\exists {\mathcal {O}}' \in \mathtt {POpt}({\mathcal {T}})$ such that ${\mathcal {O}}\sqsubseteq {\mathcal {O}}'$. From Proposition 3, ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}})) \subseteq {\mathtt {Arg}}({\mathcal {O}}) \subseteq {\mathtt {Arg}}({\mathcal {O}}')$. From Proposition 9, ${\mathcal {E}}= {\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}))$. Hence, ${\mathcal {E}}\subseteq {\mathtt {Arg}}({\mathcal {O}}) \subseteq {\mathtt {Arg}}({\mathcal {O}}')$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {CN}({\mathcal {O}})$ and $\mathtt {CN}({\mathcal {O}}')$ are consistent, then from Proposition 5${\mathtt {Arg}}({\mathcal {O}})$ and ${\mathtt {Arg}}({\mathcal {O}}')$ are both conflict-free. From Property 1, ${\mathcal {E}}\in \mathtt {Ext}_n({\mathcal {H}})$. Then, ${\mathcal {E}}$ is maximal (for set inclusion) among conflict-free sets. Thus, ${\mathcal {E}}= {\mathtt {Arg}}({\mathcal {O}}) = {\mathtt {Arg}}({\mathcal {O}}')$. From consistency of $\mathtt {CN}({\mathcal {O}})$ and $\mathtt {CN}({\mathcal {O}}')$, it follows that $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}})) = \mathtt {CN}({\mathcal {O}})$ and $\mathtt {Concs}({\mathtt {Arg}}({\mathcal {O}}')) = \mathtt {CN}({\mathcal {O}}')$. Then, $\mathtt {CN}({\mathcal {O}}) = \mathtt {CN}({\mathcal {O}}')$ and ${\mathcal {O}}\in \mathtt {POpt}({\mathcal {T}})$.

Let us now show that ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. Assume that $\exists {\mathcal {O}}' \in \mathtt {POpt}({\mathcal {T}})$ such that $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. Since $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$, $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. Let $x \in \mathtt {CN}({\mathcal {O}}')$ and $x \notin \mathtt {Concs}({\mathcal {E}})$. Since $\mathtt {CN}({\mathcal {O}}')$ is consistent, there exists an argument $(d,x) \in {\mathtt {Arg}}({\mathcal {O}}')$, i.e., d is a derivation of x from ${\mathcal {O}}'$. Clearly, $(d,x) \notin {\mathcal {E}}$. Thus, $\exists (d',x') \in {\mathcal {E}}$ such that $(d', x') {\mathcal {R}}(d, x)$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. But, $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ and $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {O}}')$. Proposition 13 gives $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {Concs}({\mathcal {E}})$. Then, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) \subseteq \mathtt {CN}({\mathcal {O}}')$. Finally, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {O}}')$. This contradicts the fact that $\mathtt {CN}({\mathcal {O}}')$ is consistent. $\square$

Proof of Theorem 13

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency, exhaustiveness, strict precedence and closure under both strict rules and sub-arguments. Assume that $\mathtt {Ext}_s({\mathcal {H}})\ne \emptyset$. Let ${\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$. From Lemma 2, the option ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, Z') \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ with $Z' = Z \cup \{r \mid r \in {\mathcal {D}}{\setminus } Z \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))\}$. Since $X \subseteq {\mathcal {F}}$, $Y \subseteq {\mathcal {S}}$ and $Z \subseteq Z'$, $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$. From Theorem 4, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$.

Let us show that ${\mathcal {E}}$ has a unique corresponding preferred maximal option. Assume that $\exists {\mathcal {O}}_1, {\mathcal {O}}_2 \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}_1$, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}}_1)$, $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}_2$ and $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}}_2)$. Obviously, $\mathtt {CN}({\mathcal {O}}_1) = \mathtt {CN}({\mathcal {O}}_2)$. However, ${\mathcal {O}}_1, {\mathcal {O}}_2 \in \mathtt {POpt}({\mathcal {T}})$ according to Property 6 hence Proposition 1 gives ${\mathcal {O}}_1 = {\mathcal {O}}_2$.

Let us now show that ${\mathcal {E}}= {\mathtt {Arg}}({\mathcal {O}})$. Since $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$, from Proposition 3, ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}})) \subseteq {\mathtt {Arg}}({\mathcal {O}})$. From Proposition 15, ${\mathcal {E}}\subseteq {\mathtt {Arg}}({\mathcal {O}})$. Assume now that $\exists a = (d,x) \in \mathtt {Arg}({\mathcal {O}})$ such that $a \notin {\mathcal {E}}$. Thus, $\exists b = (d',x') \in {\mathcal {E}}$ and $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. Besides, $\mathtt {Seq}(d) \subseteq \mathtt {CN}({\mathcal {O}})$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Since ${\mathcal {H}}$ is closed under sub-arguments, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}})) = \mathtt {CN}({\mathcal {O}})$ by Proposition 12. Thus, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}({\mathcal {O}})$. This contradicts the fact that ${\mathcal {O}}$ is an option. $\square$

Proof of Theorem 14

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies consistency, strict precedence and closure under both strict rules and sub-arguments. Assume that $\mathtt {Ext}_s({\mathcal {H}})\ne \emptyset$. Let ${\mathcal {E}}, {\mathcal {E}}' \in \mathtt {Ext}_s({\mathcal {H}})$. From Theorem 13, $\exists {\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$ and $\exists {\mathcal {O}}' \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}') = \mathtt {CN}({\mathcal {O}}')$. If ${\mathcal {O}}= {\mathcal {O}}'$, then $\mathtt {Concs}({\mathcal {E}}) = \mathtt {Concs}({\mathcal {E}}')$. Assume that $\exists a = (d,x) \in {\mathcal {E}}{\setminus }{\mathcal {E}}'$. Thus, $\exists b = (d',x') \in {\mathcal {E}}'$ such that $b {\mathcal {R}}a$. Since ${\mathcal {R}}$ is conflict-dependent, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d')$ is inconsistent. But ${\mathcal {H}}$ is closed under sub-arguments. Thus, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ and $\mathtt {Concs}({\mathcal {E}}') = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}'))$. Besides, $\mathtt {Seq}(d) \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ and $\mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}'))$. Since $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}')) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$, $\mathtt {Seq}(d) \cup \mathtt {Seq}(d') \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$. Thus, $\mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ is inconsistent. This contradicts the fact that ${\mathcal {H}}$ satisfies consistency. The same reasoning holds for $a = (d',x') \in {\mathcal {E}}'{\setminus }{\mathcal {E}}$. $\square$

Proof of Theorem 15

Any argumentation system ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ that satisfies strict precedence should have ${\mathcal {F}}$ as plausible conclusions, i.e., ${\mathcal {F}}\subseteq \mathtt {CN}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq \mathtt {Output}({\mathcal {H}})$. Since $\top \in {\mathcal {F}}$, then $\mathtt {Output}({\mathcal {H}})\ne \emptyset$. However, since ${\mathcal {R}}\in \mathfrak {R}_{s_1}$, $\mathtt {Ext}_s({\mathcal {H}})= \emptyset$. Thus, $\mathtt {Output}({\mathcal {H}})= \emptyset$. $\square$

Proof of Theorem 16

Let ${\mathcal {H}}= (\mathtt {Arg}({\mathcal {T}})$, ${\mathcal {R}})$ be an argumentation system over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}\in \mathfrak {R}_{s_2}$. Let ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. Since $|\mathtt {Ext}_s({\mathcal {H}})| = |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$, from Theorems 13 and 14, $\exists {\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$ such that ${\mathcal {E}}= {\mathtt {Arg}}({\mathcal {O}})$, hence ${\mathtt {Arg}}({\mathcal {O}}) \in \mathtt {Ext}_s({\mathcal {H}})$. $\square$

Proof of Theorem 17

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}\in \mathfrak {R}_{s2}$. From Corollary 5, $\mathtt {Ext}_s({\mathcal {H}})= \mathtt {Ext}_{ss}({\mathcal {H}})$. Assume that $\exists {\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}}){\setminus }\mathtt {Ext}_s({\mathcal {H}})$. From Theorem 22, there exists ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$. Since $|\mathtt {Ext}_s({\mathcal {H}})| = |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$, from Theorem 16, ${\mathtt {Arg}}({\mathcal {O}}) \in \mathtt {Ext}_s({\mathcal {H}})$. From Theorems 13 and 14, ${\mathcal {O}}= \mathtt {Option}({\mathtt {Arg}}({\mathcal {O}}))$. From Theorem 24, ${\mathcal {E}}= {\mathtt {Arg}}({\mathcal {O}})$. $\square$

Proof of Theorem 18

Let ${\mathcal {H}}= (\mathtt {Arg}({\mathcal {T}})$, ${\mathcal {R}})$ be an argumentation system such that ${\mathcal {R}}\in \mathfrak {R}_{s_2}$. If ${\mathcal {H}}$ satisfies all the postulates under naive semantics, then from Corollary 1 and Theorem 11, there is a bijection between $\mathtt {Ext}_n({\mathcal {H}})$ and $\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. From Theorems 13 and 16, $|\mathtt {Ext}_s({\mathcal {H}})| = |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$. Since every stable extension is a naive one, $\mathtt {Ext}_n({\mathcal {H}})= \mathtt {Ext}_s({\mathcal {H}})$. $\square$

Proof of Theorem 19

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}\in \mathfrak {R}_{s_2}$. From Property 11,

$$\begin{aligned} \mathtt {Output}({\mathcal {H}})= \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}_s({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i). \end{aligned}$$

From Theorems 13 and 14, for all ${\mathcal {E}}_i \in \mathtt {Ext}_s({\mathcal {H}})$, there exists a unique ${\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}_i) = \mathtt {CN}({\mathcal {O}}_i)$. Thus,

$$\begin{aligned} \mathtt {Output}({\mathcal {H}})= \bigcap _{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))} \mathtt {CN}({\mathcal {O}}_i). \end{aligned}$$

$\square$

Proof of Theorem 20

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}\in \mathfrak {R}_{s_3}$. From Property 11,

$$\begin{aligned} \mathtt {Output}({\mathcal {H}})= \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}_s({\mathcal {H}})} \mathtt {Concs}({\mathcal {E}}_i). \end{aligned}$$

From Theorem 13, for all ${\mathcal {E}}_i \in \mathtt {Ext}_s({\mathcal {H}})$, there exists a unique ${\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Concs}({\mathcal {E}}_i) = \mathtt {CN}({\mathcal {O}}_i)$. Since ${\mathcal {R}}\in \mathfrak {R}_{s_3}$, $|\mathtt {Ext}_s({\mathcal {H}})| < |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$. Thus,

$$\begin{aligned} \mathtt {Output}({\mathcal {H}})= \bigcap _{{\mathcal {O}}_i \in {\mathcal {X}}} \mathtt {CN}({\mathcal {O}}_i) \end{aligned}$$

with ${\mathcal {X}}= \{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}})) \mid {\mathcal {E}}_i = {\mathtt {Arg}}({\mathcal {O}}_i) \in \mathtt {Ext}_s({\mathcal {H}})\}$. $\square$

Proof of Theorem 21

Let ${\mathcal {H}}$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {H}}$ satisfies the strict precedence postulate, i.e., ${\mathcal {F}}\subseteq \mathtt {Output}({\mathcal {H}})$. Since $\top \in {\mathcal {F}}$, $\mathtt {Output}({\mathcal {H}})\ne \emptyset$. Hence, $\mathtt {Ext}_p({\mathcal {H}})\ne \{\emptyset \}$. $\square$

Proof of Theorem 22

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}),{\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. Let ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$ and $\mathtt {Th}({\mathcal {E}}) = (X, Y, Z)$. From Theorem 4, $\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}({\mathcal {O}})$ where ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, Z \cup Z')$ and $Z' \subseteq {\mathcal {D}}{\setminus } Z$. Clearly $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$. From consistency, $\mathtt {Concs}({\mathcal {E}})$ is consistent. Then, $\mathtt {CN}({\mathcal {O}})$ is consistent as well. Then, there exists ${\mathcal {O}}' \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that ${\mathcal {O}}\sqsubseteq {\mathcal {O}}'$. Therefore, $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. Thus, $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. $\square$

Proof of Theorem 23

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates under preferred semantics. Since $\mathtt {Ext}_s({\mathcal {H}})\subseteq \mathtt {Ext}_p({\mathcal {H}})$, ${\mathcal {H}}$ satisfies the postulates under stable semantics. Consequently, from Theorem 15, $\mathtt {Ext}_s({\mathcal {H}})\ne \emptyset$. Let ${\mathcal {E}}\in \mathtt {Ext}_s({\mathcal {H}})$. From Theorem 13, $\exists {\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that ${\mathcal {E}}= {\mathtt {Arg}}({\mathcal {O}})$. $\square$

Proof of Theorem 24

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies exhaustiveness and closure under sub-arguments. Let ${\mathcal {E}}, {\mathcal {E}}' \in \mathtt {Ext}_p({\mathcal {H}})$ and ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$ and $\mathtt {Th}({\mathcal {E}}') \sqsubseteq {\mathcal {O}}$. We show that ${\mathcal {E}}\cup {\mathcal {E}}'$ is a preferred extension (which contradicts the fact that ${\mathcal {E}}$ and ${\mathcal {E}}'$ are preferred extensions).

From Proposition 3, ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}})) \subseteq {\mathtt {Arg}}({\mathcal {O}})$ and ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}')) \subseteq {\mathtt {Arg}}({\mathcal {O}})$. Since ${\mathcal {H}}$ satisfies exhaustiveness and closure under sub-arguments, from Proposition 9, ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}})) = {\mathcal {E}}$ and ${\mathtt {Arg}}(\mathtt {Th}({\mathcal {E}}')) = {\mathcal {E}}'$. Thus, ${\mathcal {E}}\cup {\mathcal {E}}' \subseteq {\mathtt {Arg}}({\mathcal {O}})$. Since $\mathtt {CN}({\mathcal {O}})$ is consistent and ${\mathcal {R}}$ is conflict-dependent, from Proposition 5${\mathtt {Arg}}({\mathcal {O}})$ is conflict-free. Consequently, ${\mathcal {E}}\cup {\mathcal {E}}'$ is also conflict-free. Moreover, ${\mathcal {E}}\cup {\mathcal {E}}'$ defends its elements since ${\mathcal {E}}$ and ${\mathcal {E}}'$ are preferred extensions. Thus, ${\mathcal {E}}\cup {\mathcal {E}}'$ is an admissible set. Due to ${\mathcal {E}}$ and ${\mathcal {E}}'$ being preferred extensions, it follows that ${\mathcal {E}}\cup {\mathcal {E}}' = {\mathcal {E}}= {\mathcal {E}}'$. $\square$

Proof of Theorem 25

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$ such that ${\mathcal {R}}$ is conflict-dependent and privileges strict arguments, and ${\mathcal {H}}$ satisfies consistency, exhaustiveness, strict precedence and closure under sub-arguments. From consistency and strict precedence, it follows by Proposition 8 that $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$ is consistent.

The conclusion of the theorem, i.e., ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}})) \subseteq \bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}_p({\mathcal {H}})} {\mathcal {E}}_i$, is trivial in the case that ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ is empty. Consider $a \in {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$. Let ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$.

Let us show that ${\mathcal {E}}\cup \{a\}$ is conflict-free. Assume that $\exists b = (d_2,x_2) \in {\mathcal {E}}$ such that $a {\mathcal {R}}b$ or $b {\mathcal {R}}a$. From Lemma 1, there exists $a' \in \mathtt {Sub}(a)$ such that $a' = (d'_1,x'_1) \in {\mathtt {Arg}}(({\mathcal {F}},{\mathcal {S}},\emptyset ))$ and $a' {\mathcal {R}}b$. Then, $\mathtt {Seq}(d'_1) \cup \mathtt {Seq}(d_2)$ is inconsistent. Since ${\mathcal {H}}$ satisfies strict precedence and exhaustiveness, ${\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq {\mathcal {E}}$ by Proposition 10, so $a' \in {\mathcal {E}}$. Consequently, $\mathtt {Seq}(d'_1) \cup \mathtt {Seq}(d_2) \subseteq \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ by Proposition 12. Since ${\mathcal {H}}$ satisfies consistency and closure under sub-arguments, by Proposition 6$\mathtt {Concs}({\mathcal {E}}) = \mathtt {CN}(\mathtt {Th}({\mathcal {E}}))$ is consistent. Contradiction.

Let us show that ${\mathcal {E}}$ defends a. Consider $b \in {\mathtt {Arg}}({\mathcal {T}})$ such that $b {\mathcal {R}}a$. From Lemma 1, there exists $a' \in \mathtt {Sub}(a)$ such that $a' \in {\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset ))$ and $a' {\mathcal {R}}b$. Since ${\mathcal {H}}$ satisfies strict precedence and exhaustiveness, ${\mathtt {Arg}}(({\mathcal {F}}, {\mathcal {S}}, \emptyset )) \subseteq {\mathcal {E}}$, thus $a' \in {\mathcal {E}}$.

Summing up, ${\mathcal {E}}\cup \{a\}$ is an admissible set. However, ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$ means that ${\mathcal {E}}$ is a maximal admissible set; hence, ${\mathcal {E}}\cup \{a\} \subseteq {\mathcal {E}}$. Therefore, $a \in {\mathcal {E}}$. $\square$

Proof of Theorem 26

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be a system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. From Theorem 22, for all ${\mathcal {E}}\in \mathtt {Ext}_p({\mathcal {H}})$, $\exists {\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$ and $\mathtt {Concs}({\mathcal {E}}) \subseteq \mathtt {CN}({\mathcal {O}})$. From Theorem 24, there cannot exist two maximal preferred extensions ${\mathcal {E}}$ and ${\mathcal {E}}'$ such that $\mathtt {Th}({\mathcal {E}}) \sqsubseteq {\mathcal {O}}$ and $\mathtt {Th}({\mathcal {E}}') \sqsubseteq {\mathcal {O}}$ for some ${\mathcal {O}}\in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))$. Thus, every maximal preferred option is captured by at most one preferred extension. Then, $|\mathtt {Ext}_p({\mathcal {H}})| \le |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$. $\square$

Proof of Theorem 27

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}= ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}})$. Assume that ${\mathcal {R}}\in \mathfrak {R}_{p_2}$. The following equalities hold by Theorems 16 and 17: $\mathtt {Ext}_p({\mathcal {H}})= \mathtt {Ext}_s({\mathcal {H}})= \mathtt {Ext}_{ss}({\mathcal {H}})= \{{\mathtt {Arg}}({\mathcal {O}}_i) \ | \ {\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))\}.$ Let us now show the equality

$$\begin{aligned} \bigcap \limits _{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))} {\mathtt {Arg}}({\mathcal {O}}_i) = {\mathtt {Arg}}(\bigcap \limits _{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))} {\mathcal {O}}_i). \end{aligned}$$

Let $\mathtt {Max}(\mathtt {POpt}({\mathcal {T}})) = \{{\mathcal {O}}_1 = ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}}_1), \ldots , {\mathcal {O}}_n = ({\mathcal {F}}, {\mathcal {S}}, {\mathcal {D}}_n)\}$. Assume that $(d,x) \in \bigcap \limits _{i = 1}^{n} {\mathtt {Arg}}({\mathcal {O}}_i)$. For any $i = 1, \ldots , n$, $(d,x) \in {\mathtt {Arg}}({\mathcal {O}}_i)$ and thus

$$\begin{aligned} (\mathtt {Facts}(d), \mathtt {Strict}(d), \mathtt {Def}(d)) \sqsubseteq {\mathcal {O}}_i. \end{aligned}$$

This means that $\mathtt {Def}(d) \subseteq \bigcap \nolimits _{i = 1}^{n} {\mathcal {D}}_i$. Consequently, d is also a derivation schema from $({\mathcal {F}}, {\mathcal {S}}, \bigcap \nolimits _{i = 1}^{n} {\mathcal {D}}_i) = \bigcap \nolimits _{i = 1}^{n} {\mathcal {O}}_i$. Finally, $(d,x) \in {\mathtt {Arg}}(\bigcap \nolimits _{i = 1}^{n} {\mathcal {O}}_i)$.

Assume now that $(d,x) \in {\mathtt {Arg}}(\bigcap \nolimits _{i = 1}^{n} {\mathcal {O}}_i)$. Then, d is a derivation schema from $({\mathcal {F}}, {\mathcal {S}}, \bigcap \nolimits _{i = 1}^{n} {\mathcal {D}}_i)$. Hence, $\mathtt {Def}(d) \subseteq \bigcap \nolimits _{i = 1}^{n} {\mathcal {D}}_i$. Hence, for any $i = 1, \ldots , n$, $\mathtt {Def}(d) \subseteq {\mathcal {D}}_i$. Thus, d is a derivation schema from each theory ${\mathcal {O}}_i$ and (d, x) is an argument in each ${\mathtt {Arg}}({\mathcal {O}}_i)$.

From above, it follows that

$$\begin{aligned} \bigcap _{{\mathcal {O}}_i \in \mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))} {\mathtt {Arg}}({\mathcal {O}}_i) = {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}})). \end{aligned}$$

$\square$

Proof of Theorem 28

Let ${\mathcal {H}}$ be an argumentation system which satisfies the five postulates. From strict precedence and the fact that $\mathtt {Output}({\mathcal {H}})= \mathtt {Concs}(\mathtt {IE}({\mathcal {H}}))$, it holds that $\mathtt {CN}(({\mathcal {F}},{\mathcal {S}},\emptyset )) \subseteq \mathtt {Concs}(\mathtt {IE}({\mathcal {H}}))$. From Theorem 4, $\mathtt {Concs}(\mathtt {IE}({\mathcal {H}})) = \mathtt {CN}({\mathcal {O}})$ such that ${\mathcal {O}}= ({\mathcal {F}}, {\mathcal {S}}, Z)$ where

$$\begin{aligned} Z = \left( \bigcup _{(d,x) \in \mathtt {IE}({\mathcal {H}})} \mathtt {Def}(d)\right) \cup \left\{ r \mid r \in {\mathcal {D}} \text{ and } \mathtt {Body}(r) \not \subseteq \mathtt {CN}(\mathtt {Th}(\mathtt {IE}({\mathcal {H}})))\right\} . \end{aligned}$$

It holds that $\mathtt {Th}(\mathtt {IE}({\mathcal {H}})) \sqsubseteq {\mathcal {O}}$. From consistency postulate, it follows that $\mathtt {CN}({\mathcal {O}})$ is consistent (since $\mathtt {Concs}(\mathtt {IE}({\mathcal {H}}))$ is consistent). Thus, there exists ${\mathcal {O}}' \in \mathtt {POpt}({\mathcal {T}})$ such that ${\mathcal {O}}\sqsubseteq {\mathcal {O}}'$. From Property 4, $\mathtt {CN}({\mathcal {O}}) \subseteq \mathtt {CN}({\mathcal {O}}')$. Consequently, $\mathtt {Concs}(\mathtt {IE}({\mathcal {H}}))$$\subseteq$$\mathtt {CN}({\mathcal {O}}')$. $\square$

Proof of Theorem 29

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$. Assume that ${\mathcal {R}}\in \mathfrak {R}_{p_2}$ and privileges strict arguments. From Theorem 27, $\bigcap _{{\mathcal {E}}_i \in \mathtt {Ext}_p({\mathcal {H}})} {\mathcal {E}}_i = {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$. From Theorem 2, ${\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$ is an admissible extension of ${\mathcal {H}}$. Thus, $\mathtt {IE}({\mathcal {H}})= {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$. $\square$

Proof of Theorem 30

The proof is similar to that of Theorem 28. $\square$

Proof of Corollary 1

It follows directly from Theorems 8 and 10. $\square$

Proof of Corollary 2

It follows directly from Theorem 11 and Corollary 1. $\square$

Proof of Corollary 3

It follows from Corollary 1. $\square$

Proof of Corollary 4

It follows from Corollary 3, i.e., the equality $|\mathtt {Ext}_n({\mathcal {H}})| = |\mathtt {Max}(\mathtt {Opt}({\mathcal {T}}))|$ and the fact that if a theory ${\mathcal {T}}$ is finite, then it has a finite number of options, thus of maximal options. $\square$

Proof of Corollary 5

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be such that ${\mathcal {R}}\in \mathfrak {R}_{s_2} \cup \mathfrak {R}_{s_3}$. From Theorem 15, $\mathtt {Ext}_s({\mathcal {H}})\ne \emptyset$. From Property 1, $\mathtt {Ext}_s({\mathcal {H}})= \mathtt {Ext}_{ss}({\mathcal {H}})$. $\square$

Proof of Corollary 6

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. From Theorem 15, $\mathtt {Ext}_s({\mathcal {H}})\ne \emptyset$. From Theorem 13, $|\mathtt {Ext}_s({\mathcal {H}})| \le |\mathtt {Max}(\mathtt {POpt}({\mathcal {T}}))|$. $\square$

Proof of Corollary 7

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system built over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}$ is conflict-dependent and ${\mathcal {H}}$ satisfies the five postulates. If ${\mathcal {T}}$ is finite, then ${\mathcal {T}}$ has a finite number of maximal preferred options. From Corollary 6, ${\mathcal {H}}$ has a finite number of stable extensions. $\square$

Proof of Corollary 8

It follows immediately from Theorem 26. $\square$

Proof of Corollary 9

It follows immediately from Theorems 17, 13 and 16. $\square$

Proof of Corollary 10

Let ${\mathcal {H}}= ({\mathtt {Arg}}({\mathcal {T}}), {\mathcal {R}})$ be an argumentation system over a theory ${\mathcal {T}}$ such that ${\mathcal {R}}\in \mathfrak {R}_{p_2}$ and privileges strict arguments. From Theorem 29, $\mathtt {IE}({\mathcal {H}})= {\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}}))$. Then, $\mathtt {Output}({\mathcal {H}})= \mathtt {Concs}(\mathtt {IE}({\mathcal {H}})) = \mathtt {Concs}({\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}})))$. Since $\mathtt {CN}(\mathtt {Free}({\mathcal {T}}))$ is consistent, $\mathtt {CN}(\mathtt {Free}({\mathcal {T}})) = \mathtt {Concs}({\mathtt {Arg}}(\mathtt {Free}({\mathcal {T}})))$. $\square$