Valid attacks in argumentation frameworks with recursive attacks

Abstract

The purpose of this work is to study a generalisation of Dung’s abstract argumentation frameworks that allows representing recursive attacks, that is, a class of attacks whose targets are other attacks. We do this by developing a theory of argumentation where the classic role of attacks in defeating arguments is replaced by a subset of them, which is “extension-dependent” and which, intuitively, represents a set of “valid attacks” with respect to the extension. The studied theory displays a conservative generalisation of Dung’s semantics (complete, preferred, stable and grounded) and also of its principles (conflict-freeness, acceptability and admissibility). Furthermore, despite its conceptual differences, we are also able to show that our theory agrees with the AFRA interpretation of recursive attacks for the complete, preferred, stable and grounded semantics and with a recent flattening method.

Appendices

Appendix

A Proofs

2.1 A.1 Proofs of Section 3

Lemma 4

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}}$ be a conflict-free structure. Then, it follows that ${Acc}({\mathfrak {A}}) \cap {Def}({\mathfrak {A}}) = \varnothing$ and $\mathit {Acc}{{\mathfrak {A}}} \cap {Inh}({\mathfrak {A}}) = \varnothing$

Proof

Assume that $a \in ({Acc}({\mathfrak {A}}) \cap {Def}({\mathfrak {A}}))$. Then, there is α ∈Γ with s(α) ∈ S and t(α) = a. Since $a \in {Acc}({\mathfrak {A}})$, it follows that either $s(\alpha ) \in {Def}({\mathfrak {A}})$ or $\alpha \in {Inh}({\mathfrak {A}})$ holds. Both situations are impossible since ${\mathfrak {A}}$ is conflict-free, meaning that $S \cap \mathit {Def}({\mathfrak {A}}) = \varnothing$ and ${\varGamma } \cap {Inh}({\mathfrak {A}}) = \varnothing$.

The same reasoning holds for $\beta \in ({Acc}({\mathfrak {A}}) \!\cap \! {Inh}({\mathfrak {A}}))$ replacing a by β. □

Lemma 5

Let RAF = 〈A, K, s, t〉 be some framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some an admissible structure. Then, any acceptable argument $a \in ({Acc}({\mathfrak {A}}) \cap \mathbf {A})$ satisfies that ${\mathfrak {A}}^{\prime } = \langle {S \cup {\{a\}},{\varGamma }}\rangle$ is conflict-free.

Proof

Let $S^{\prime } = (S \cup {\{a\}})$ and suppose, for the sake of contradiction, that ${\mathfrak {A}}^{\prime }$ is not conflict-free, that is, that either $(S^{\prime } \cap {Def}({\mathfrak {A}}^{\prime })) \neq \varnothing$ or $({\varGamma } \cap {Inh}({\mathfrak {A}}^{\prime })) \neq \varnothing$ holds.

• a) In the first case, there is $a^{\prime } \in (S^{\prime } \cap {Def}({\mathfrak {A}}^{\prime }))$. So there is α ∈Γ such that $\mathbf {t}(\alpha ) = a^{\prime }$ and $\mathbf {s}(\alpha ) \in S^{\prime }$. Either $a^{\prime } = a$ or $a^{\prime } \in S$.

  Assume first that $a^{\prime } = a$. Since $a \in {Acc}({\mathfrak {A}})$, it follows that either $\alpha \in {Inh}({\mathfrak {A}})$ or $s(\alpha ) \in {Def}({\mathfrak {A}})$ holds. However, we know that ${\mathfrak {A}}$ is conflict-free, so it is impossible that $\alpha \in ({\varGamma } \cap {Inh}({\mathfrak {A}}))$. So, it must be the case that $s(\alpha ) \in {Def}({\mathfrak {A}})$ holds and, thus, that s(α)∉S (also because ${\mathfrak {A}}$ is conflict-free). Hence, s(α) = a. Due to Lemma 4, it is impossible to have $a \in ({Acc}({\mathfrak {A}}) \cap {Def}({\mathfrak {A}}))$ and, thus, $s(\alpha ) \in {Def}({\mathfrak {A}})$ plus s(α) = a imply that $a \notin {Acc}({\mathfrak {A}})$. This is a contradiction with the fact $a \in {Acc}({\mathfrak {A}})$.

  Assume now that $a^{\prime }\neq a$ and, thus, that $a^{\prime } \in S$. Since $(S \cap {Def}({\mathfrak {A}})) = \varnothing$, we have that $a^{\prime }\not \in {Def}({\mathfrak {A}})$. Hence, s(α)∉S and s(α) = a hold. Since ${\mathfrak {A}}$ is admissible and $a^{\prime } \in S$, it follows that $a^{\prime } \in {Acc}({\mathfrak {A}})$. Furthermore, since $\mathbf {t}(\alpha ) = a^{\prime }$, it also follows that either $\alpha \in {Inh}({\mathfrak {A}})$ or $s(\alpha ) \in {Def}({\mathfrak {A}})$. The former is in contradiction with the fact that ${\mathfrak {A}}$ is admissible (and thus conflict-free). Furthermore, from Lemma 4 and the fact that s(α) = a, the latter implies that $a \notin {Acc}({\mathfrak {A}})$ which is a contradiction, too.

• b) If $({\varGamma } \cap {Inh}({\mathfrak {A}}^{\prime })) \neq \varnothing$, then there is some attack β ∈Γ with $\beta \in {Inh}({\mathfrak {A}}^{\prime })$ and thus, there is also some α ∈Γ such that t(α) = β and $\mathbf {s}(\alpha ) \in S^{\prime }$. Since ${\mathfrak {A}}$ is conflict-free, $\beta \not \in {Inh}({\mathfrak {A}})$ which implies that s(α)∉S and thus, s(α) = a holds. Since ${\mathfrak {A}}$ is admissible and β ∈Γ, it follows that $\beta \in {Acc}({\mathfrak {A}})$. Furthermore, since t(α) = β, it must be that either $\alpha \in {Inh}({\mathfrak {A}})$ or $s(\alpha ) \in {Def}({\mathfrak {A}})$ holds. The former is in contradiction with the fact that α ∈Γ and the latter implies that $a \notin {Acc}({\mathfrak {A}})$ which is in contradiction with the hypothesis.

Consequently, ${\mathfrak {A}}^{\prime }$ is conflict-free. □

Lemma 6

Let RAF = 〈A, K, s, t〉 be some framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some admissible structure. Then, any attack $\alpha \in ({Acc}({\mathfrak {A}}) \cap \mathbf {K})$ satisfies that ${\mathfrak {A}}^{\prime } = {\langle S,{\varGamma } \cup {\{\alpha \}} \rangle }$ is conflict-free.

Proof

Let ${{\varGamma }}^{\prime } = ({\varGamma } \cup {\{\alpha \}})$ and suppose, for the sake of contradiction, that ${\mathfrak {A}}^{\prime }$ is not conflict free, that is, either $(S \cap {Def}({\mathfrak {A}}^{\prime })) \neq \varnothing$ or $({{\varGamma }}^{\prime } \cap {Inh}({\mathfrak {A}}^{\prime })) \neq \varnothing$.

• 1. In the first case, there is a ∈ S with $a \in {Def}({\mathfrak {A}}^{\prime })$. So there is $\beta \in {{\varGamma }}^{\prime }$ such that t(β) = a and s(β) ∈ S. Since ${\mathfrak {A}}$ is conflict-free, $a \not \in \mathit {Def}{{\mathfrak {A}}}$ and thus, β∉Γ and β = α follow. Then, $\beta \in \mathit {Acc}{{\mathfrak {A}}}$. Furthermore, since ${\mathfrak {A}}$ is admissible and a ∈ S, we have that $a \in \mathit {Acc}{{\mathfrak {A}}}$ and thus, that either $\beta \in \mathit {Inh}{{\mathfrak {A}}}$ or $s(\beta ) \in \mathit {Def}{{\mathfrak {A}}}$ holds. From Lemma 4, the former is in contradiction with the fact that $\beta \in \mathit {Acc}{{\mathfrak {A}}}$ and, since ${\mathfrak {A}}$ is conflict-free, the latter is in contradiction with the fact that s(β) ∈ S.

• 2. If $({{\varGamma }}^{\prime } \cap {Inh}({\mathfrak {A}}^{\prime })) \neq \varnothing$, there is some attack $\alpha ^{\prime } \in {{\varGamma }}^{\prime }$ with $\alpha ^{\prime } \in {Inh}({\mathfrak {A}}^{\prime })$. So there is $\beta \in {{\varGamma }}^{\prime }$ such that $\mathbf {t}(\beta ) = \alpha ^{\prime }$ and s(β) ∈ S. Furthermore, either $\alpha ^{\prime } = \alpha$ or $\alpha ^{\prime } \in {\varGamma }$.

  Assume first that $\alpha ^{\prime } = \alpha$. Since $\alpha \in {Acc}({\mathfrak {A}})$, it follows that either $\beta \in {Inh}({\mathfrak {A}})$ or $s(\beta ) \in {Def}({\mathfrak {A}})$. However, we know that ${\mathfrak {A}}$ is conflict-free, so it is impossible that $s(\beta ) \in (S \cap {Def}({\mathfrak {A}}))$. So we must have $\beta \in \mathit {Inh}{{\mathfrak {A}}}$ and thus, that β∉Γ (also because ${\mathfrak {A}}$ is conflict-free). Hence β = α. Due to Lemma 4, it is impossible to have $\beta \in ({Acc}({\mathfrak {A}}) \cap {Inh}({\mathfrak {A}}))$ and thus, $\alpha \not \in {Acc}({\mathfrak {A}})$. This is in contradiction with the hypothesis on α.

  Assume now that $\alpha ^{\prime }\neq \alpha$ and thus that $\alpha ^{\prime } \in {\varGamma }$. Since ${\mathfrak {A}}$ is conflict-free, it follows that $({\varGamma } \cap {Inh}({\mathfrak {A}})) = \varnothing$ and thus, that $\alpha ^{\prime }\notin {Inh}({\mathfrak {A}})$. So β∉Γ, that is, β = α and, thus, that $\beta \in \mathit {Acc}{{\mathfrak {A}}}$. Furthermore, since ${\mathfrak {A}}$ is admissible and $\alpha ^{\prime } \in {\varGamma }$, we have that $\alpha ^{\prime } \in \mathit {Acc}{{\mathfrak {A}}}$. As $\mathbf {t}(\beta ) = \alpha ^{\prime }$, either $\beta \in \mathit {Inh}{{\mathfrak {A}}}$ or $s(\beta ) \in {Def}({\mathfrak {A}})$. From Lemma 4, the former is in contradiction with the fact that $\beta \in {Acc}({\mathfrak {A}})$ and, since ${\mathfrak {A}}$ is conflict-free, the latter is in contradiction with the fact that s(β) ∈ S.

Consequently, ${\mathfrak {A}}^{\prime }$ is conflict-free. □

Lemma 7

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}} = {\langle S,{\varGamma } \rangle }$ be some an admissible structure. Then, any element $x \in {Acc}({\mathfrak {A}})$ satisfies that ${\mathfrak {A}}^{\prime } = {\mathfrak {A}} \cup \{{x}\}$ is conflict-free.

Proof

If x ∈A, the result follows directly from Lemma 5. Otherwise, x ∈K, and the result follows from Lemma 6. □

Lemma 8

Any conflict-free structure HCode $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ satisfies: ${Acc}({\mathfrak {A}}) \subseteq (\overline {{Def}({\mathfrak {A}})} \cup \overline {{Inh}({\mathfrak {A}})})$.

Proof

It follows directly from Lemma 4 and the definitions of $\overline {{Def}({\mathfrak {A}})}$ and $\overline {{Inh}({\mathfrak {A}})}$. □

Lemma 9

Let RAF = 〈A, K, s, t〉 be some non-recursive framework and ${\mathfrak {A}}_{0} \sqsubseteq {\mathfrak {A}}_{1} \sqsubseteq \dotsc$ be some sequence of conflict-free structures such that ${\mathfrak {A}}_{i} = {\langle S_{i},{\varGamma }_{i} \rangle }$. Let us also define ${\mathfrak {A}} = {\langle \bigcup _{0\leq i} S_{i}, \bigcup _{0\leq i} {\varGamma }_{i} \rangle }$. Then, ${\mathfrak {A}}$ is conflict-free.

Proof

Suppose, for the sake of contradiction, that ${\mathfrak {A}}$ is not conflict-free. Then, either $(S \cap {Def}({\mathfrak {A}})) \neq \varnothing$ or $({\varGamma } \cap {Inh}({\mathfrak {A}})) \neq \varnothing$ (with $S=\bigcup _{0\leq i} S_{i}$ and ${\varGamma }=\bigcup _{0\leq i} {\varGamma }_{i}$). Pick any argument $x \in (S \cap {Def}({\mathfrak {A}}))$ (resp. attack $x \in {\varGamma } \cap {Inh}({\mathfrak {A}}))$). Then, $x \in {Def}({\mathfrak {A}})$ (resp. $x \in {Inh}({\mathfrak {A}})$) implies that there is α ∈Γ such that t(α) = x and s(α) ∈ S. Hence, there is 0 ≤ i such that α ∈Γ_i and 0 ≤ j such that s(α) ∈ S_j. Let $k = \max \limits {\{i,j\}}$. Then, α ∈Γ_k and s(α) ∈ S_k which means that $x \in {Def}({\mathfrak {A}}_{k})$ (resp. $x \in {Inh}({\mathfrak {A}}_{k})$). Moreover, there is 0 ≤ l such that x ∈ S_l (resp. x ∈Γ_l). Let $m = \max \limits {\{k,l\}}$. Then, x ∈ S_m (resp. x ∈Γ_m), and from Observation 1, we have that ${Def}({\mathfrak {A}}_{k}) \subseteq {Def}({\mathfrak {A}}_{m})$ (resp. ${Inh}({\mathfrak {A}}_{k}) \subseteq {Inh}({\mathfrak {A}}_{m})$). That is in contradiction with the fact that ${\mathfrak {A}}_{m}$ is conflict-free.

Hence, ${\mathfrak {A}}$ must be conflict-free. □

2.2 A.2 Proofs of Section 4

Lemma 10

Let RAF = 〈A, K, s, t〉 be a framework and ${\mathfrak {A}}$ be a structure. Then, $x \in ({Def}({\mathfrak {A}}) \cup {Inh}({\mathfrak {A}}))$ implies that there is some $\alpha \in \texttt {Afra}({\mathfrak {A}})$ such that α defeats x.

Proof

Since $x \in ({Def}({\mathfrak {A}}) \cup {Inh}({\mathfrak {A}}))$, there is α ∈Γ s.t. t(α) = x and s(α) ∈ S. Note that α ∈Γ and s(α) ∈Γ imply that $\alpha \in \texttt {Afra}({\mathfrak {A}})$ and that t(α) = x implies that α defeats x. □

Lemma 11

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}}$ be some structure. Then, every $a \in ({Acc}({\mathfrak {A}}) \cap \mathbf {A})$ is AFRA-acceptable w.r.t. $\texttt {Afra}{{\mathfrak {A}}}$.

Proof

Pick any attack α ∈K such that α defeats a. Then, t(α) = a and, since $a \in {Acc}({\mathfrak {A}})$ it follows that either $\alpha \in {Inh}({\mathfrak {A}})$ or $\mathbf {s}(\alpha ) \in {Def}({\mathfrak {A}})$.

If $\alpha \in {Inh}({\mathfrak {A}})$, then there is β ∈Γ such that s(β) ∈ S and t(β) = α. Note that β ∈Γ plus s(β) ∈ S imply $\beta \in \texttt {Afra}({\mathfrak {A}})$ and that t(β) = α implies that β defeats α. Hence, the fact that a is AFRA-acceptable w.r.t. ${\mathcal {E}}$ follows.

Otherwise, $\mathbf {s}(\alpha ) \in {Def}({\mathfrak {A}})$, and, there is β ∈Γ such that s(β) ∈ S and t(β) = s(α). As above, β ∈Γ plus s(β) ∈ S imply $\beta \in \texttt {Afra}({\mathfrak {A}})$, and t(β) = s(α) implies that β defeats α. Hence, the fact that a is AFRA-acceptable w.r.t. ${\mathcal {E}}$ follows.

In consequence, it holds that a is AFRA-acceptable w.r.t. ${\mathcal {E}}$. □

Lemma 12

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}} = {\langle S,{\varGamma } \rangle }$ be some structure. Then, every $\alpha \in ({Acc}({\mathfrak {A}}) \cap \mathbf {K})$ that satisfies $\mathbf {s}(\alpha ) \in {Acc}({\mathfrak {A}})$, is also AFRA-acceptable w.r.t. $\texttt {Afra}({\mathfrak {A}})$.

Proof

Pick any attack β ∈K such that β defeats α. Then, either t(β) = α or t(β) = s(α).

If the latter, then Lemma 11 plus $\mathbf {s}(\alpha ) \in {Acc}({\mathfrak {A}})$ imply that s(α) is AFRA-acceptable w.r.t. $\texttt {Afra}({\mathfrak {A}})$ and, thus, that there is some $\gamma \in \texttt {Afra}{{\mathfrak {A}}}$ that defeats β.

If the former, $\alpha \in {Acc}({\mathfrak {A}})$ implies that either $\beta \in {Inh}({\mathfrak {A}})$ or $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}})$. Assume $\beta \in {Inh}({\mathfrak {A}})$. Then, there is γ ∈Γ such that s(γ) ∈ S and t(γ) = β. Note that γ ∈Γ plus s(γ) ∈ S imply $\gamma \in \texttt {Afra}({\mathfrak {A}})$ and that t(γ) = β implies that γ defeats β.

Otherwise, $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}})$ and there is γ ∈Γ such that s(γ) ∈ S and t(γ) = s(β). As above, γ ∈Γ plus s(γ) ∈ S imply $\gamma \in \texttt {Afra}({\mathfrak {A}})$ and t(γ) = s(β) implies that γ defeats β.

Hence, for any attack β ∈K that defeats α, there is some attack $\gamma \in \texttt {Afra}({\mathfrak {A}})$ that defeats β. That is, the fact that α is AFRA-acceptable w.r.t. $\texttt {Afra}({\mathfrak {A}})$ follows. □

Lemma 13

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}}$ be some structure. Then, $\alpha \in ({Acc}({\mathfrak {A}}) \cap \mathbf {K})$ and s(α)∉S imply $\alpha \in {\varGamma }_{\texttt {Afra}{{\mathfrak {A}}}}$.

Proof

Let ${\mathfrak {A}}^{\prime } = {\langle S,{\varGamma }_{\texttt {Afra}({\mathfrak {A}})} \rangle }$ and let us show that $\alpha \in {Acc}({\mathfrak {A}}^{\prime })$. Pick any β ∈K such that t(β) = α. Since $\alpha \in {Acc}({\mathfrak {A}})$, it follows that either $\beta \in {Inh}({\mathfrak {A}})$ or $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}})$. If the former, there is γ ∈Γ such that t(γ) = β and s(γ) ∈ S. Note that γ ∈Γ plus s(γ) ∈ S imply that $\gamma \in \texttt {Afra}({\mathfrak {A}})$ and, thus, that $\gamma \in {\varGamma }_{\texttt {Afra}{{\mathfrak {A}}}}$ and that $\beta \in {Inh}({\mathfrak {A}}^{\prime })$. Similarly, $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}})$ implies that there is γ ∈Γ such that t(γ) = s(β) and s(γ) ∈ S and, thus, $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}}^{\prime })$. Hence, any β ∈K with t(β) = α satisfies either $\beta \in {Inh}({\mathfrak {A}}^{\prime })$ or $\mathbf {s}(\beta ) \in {Def}({\mathfrak {A}}^{\prime })$. That is, $\alpha \in {Acc}({\mathfrak {A}}^{\prime })$. Hence, to show that $\alpha \in {\varGamma }_{\texttt {Afra}({\mathfrak {A}})}$ it is enough to prove that $\mathbf {s}(\alpha ) \notin \texttt {Afra}({\mathfrak {A}})$ which directly follows from the fact that s(α)∉S. □

Lemma 14

Let RAF = 〈A, K, s, t〉 be an framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some admissible structure. Then, $S = S_{\texttt {Afra}({\mathfrak {A}})}$ and ${\varGamma } \subseteq {\varGamma }_{\texttt {Afra}({\mathfrak {A}})}$ hold.

Proof

Note that, by definition, it follows that $S_{\texttt {Afra}({\mathfrak {A}})} = (\texttt {Afra}({\mathfrak {A}}) \cap \mathbf {A}) = S$.

Then, to show that ${\varGamma } \subseteq {\varGamma }_{\texttt {Afra}({\mathfrak {A}})}$ holds, pick any attack α ∈Γ. If s(α) ∈ S, then $\alpha \in (\texttt {Afra}({\mathfrak {A}}) \cap \mathbf {K})$ and thus, $\alpha \in {\varGamma }_{\texttt {Afra}({\mathfrak {A}})}$. Otherwise, s(α)∉S and $\alpha \in ({Acc}({\mathfrak {A}}) \cap \mathbf {K})$ as $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ is admissible. So, from Lemma 13, it follows that $\alpha \in {\varGamma }_{\texttt {Afra}({\mathfrak {A}})}$. □

Lemma 15

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}}$ be a complete structure. Then, it follows that ${Acc}({\mathfrak {A}}) \subseteq (S_{\texttt {Afra}({\mathfrak {A}})} \cup {\varGamma }_{\texttt {Afra}{{\mathfrak {A}}}})$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$. Note that, since ${\mathfrak {A}}$ is complete, it follows that ${Acc}({\mathfrak {A}}) \subseteq (S \cup {\varGamma })$ and, thus, the result follows directly from Lemma 14. □

Lemma 16

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be a complete structure and x be some AFRA-acceptable element w.r.t. $\texttt {Afra}({\mathfrak {A}})$. Then, x ∈ (S ∪Γ) holds.

Proof

From Lemma 2, the hypothesis implies that $x \in {Acc}({\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})})$. Note that, since ${\mathfrak {A}}$ is complete, Proposition 10 implies that ${\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})} = {\mathfrak {A}}$ and, thus, that $x \in {Acc}({\mathfrak {A}})$ and that x ∈ (S ∪Γ) (recall that ${\mathfrak {A}}$ is complete). □

Lemma 17

Let RAF = 〈A, K, s, t〉 be some framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be a complete structure. Then, $\texttt {Afra}({\mathfrak {A}})$ is AFRA-complete.

Proof

Since ${\mathfrak {A}}$ is a complete structure, it is admissible and, in addition, it satisfies $(S \cup {\varGamma }) = {Acc}({\mathfrak {A}})$. From Proposition 6 the former implies that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-admissible. Hence, to show that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-complete, it is enough to prove that every acceptable element x w.r.t. $\texttt {Afra}({\mathfrak {A}})$ belongs to $\texttt {Afra}({\mathfrak {A}})$.

Pick any AFRA-acceptable element x ∈ (A ∪K) w.r.t. $\texttt {Afra}({\mathfrak {A}})$. From Lemma 16, this implies that x ∈ (S ∪Γ). Note that, by construction, we have that $S \subseteq \texttt {Afra}({\mathfrak {A}})$. Furthermore, if x ∈Γ, then Lemma 1 in [4] plus the fact that x is AFRA-acceptable w.r.t. $\texttt {Afra}({\mathfrak {A}})$, imply that s(x) is AFRA-acceptable w.r.t. $\texttt {Afra}({\mathfrak {A}})$ and, from Lemma 16 again, this implies that s(x) ∈ S. By definition, x ∈Γ plus s(x) ∈ S imply $x \in \texttt {Afra}({\mathfrak {A}})$ and, thus, that ${\varGamma } \subseteq \texttt {Afra}({\mathfrak {A}})$. Therefore, we have that every AFRA-acceptable element x w.r.t. $\texttt {Afra}({\mathfrak {A}})$ belongs to $\texttt {Afra}{{\mathfrak {A}}}$ and, thus, that $\texttt {Afra}({\mathfrak {A}})$ is an AFRA-complete extension. □

Lemma 18

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a closed set. Then, it follows that $\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) = {{\mathcal {E}}}$.

Proof

Note that, by definition, it follows that

$$ (\texttt{Afra}({\mathfrak{A}}_{{\mathcal{E}}}) \cap \mathbf{A}) \ = \ S_{{\mathcal{E}}} \ = \ ({{\mathcal{E}}} \cap \mathbf{A}) $$

It remains to be shown that $(\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K}) = ({{\mathcal {E}}} \cap \mathbf {K})$. By definition, it follows that ${\varGamma }_{{\mathcal {E}}} \supseteq ({{\mathcal {E}}} \cap \mathbf {K})$. Therefore, $\alpha \in {\mathcal {E}}$ implies that $\alpha \in {\varGamma }_{{\mathcal {E}}}$ and, since ${\mathcal {E}}$ is closed, this implies that $\mathbf {s}(\alpha ) \in ({{\mathcal {E}}} \cap \mathbf {A}) = S_{{\mathcal {E}}}$ and, thus, that $\alpha \in (\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K})$. In its turn, this implies $(\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K}) \supseteq ({{\mathcal {E}}} \cap \mathbf {K})$.

On the other hand, every $\alpha \in (\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K})$ satisfies $\alpha \in {\varGamma }_{{\mathcal {E}}}$ and $\mathbf {s}(\alpha ) \in S_{{\mathcal {E}}} \subseteq {{\mathcal {E}}}$. Together, these two facts imply $\alpha \in ({{\mathcal {E}}} \cap \mathbf {K})$. Hence, $(\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K}) = ({\mathcal {E}} \cap \mathbf {K})$ holds and, thus, $\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) = {{\mathcal {E}}}$ follows. □

Lemma 19

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathcal {E}}$ be some AFRA-complete extension. Then, ${\mathcal {E}}$ is closed.

Proof

Pick any attack $\alpha \in ({{\mathcal {E}}} \cap \mathbf {K})$. Then, since ${\mathcal {E}}$ is AFRA-complete, this implies that α is AFRA-acceptable w.r.t. ${\mathcal {E}}$ and, from Lemma 1 in [4], this implies that s(α) is also AFRA-acceptable w.r.t. ${\mathcal {E}}$. This plus the fact that ${\mathcal {E}}$ is AFRA-complete imply that $\mathbf {s}(\alpha ) \in {\mathcal {E}}$. □

Lemma 20

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a AFRA-complete extension. Then, ${\mathfrak {A}}_{{\mathcal {E}}}$ is a complete structure.

Proof

By definition, every AFRA-complete extension is also AFRA-admissible. Furthermore, from Lemma 19, every AFRA-complete extension is also closed, thus, Proposition 7 implies that ${\mathfrak {A}}_{{\mathcal {E}}}$ is an admissible structure. Then, to show that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a complete structure, it is enough to prove the following inclusion: ${Acc}({\mathfrak {A}}_{{\mathcal {E}}}) \subseteq (S_{{\mathcal {E}}} \cup {\varGamma }_{{\mathcal {E}}})$. Let us recall that, from Lemma 15, it follows that

$$ {Acc}({\mathfrak{A}}_{{\mathcal{E}}}) \ \ \subseteq \ \ (S_{\texttt{Afra}({\mathfrak{A}}_{{\mathcal{E}}})} \cup {\varGamma}_{\texttt{Afra}({\mathfrak{A}}_{{\mathcal{E}}})}) $$

and that, from Lemma 18, it follows $\texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) = {{\mathcal {E}}}$. As a result, ${Acc}({\mathfrak {A}}_{{\mathcal {E}}}) \subseteq (S_{{\mathcal {E}}} \cup {\varGamma }_{{\mathcal {E}}})$ holds and, thus, ${\mathfrak {A}}_{{\mathcal {E}}}$ is complete. □

Lemma 21

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq {{\mathcal {E}}^{\prime }} \subseteq (\mathbf {A} \cup \mathbf {K})$ be two AFRA-complete extensions. Then, ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubseteq {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$.

Proof

First, note that

$$ S_{{\mathcal{E}}} \ = \ ({{\mathcal{E}}} \cap \mathbf{A}) \ \subseteq \ ({{\mathcal{E}}^{\prime}} \cap \mathbf{A}) \ = \ S_{{\mathcal{E}}^{\prime}} $$

Let $S = S_{{\mathcal {E}}}$, $S^{\prime } = S_{{\mathcal {E}}^{\prime }}$, ${\varGamma } = ({{{\mathcal {E}}} \cap \mathbf {K}})$ and ${{\varGamma }}^{\prime } = ({{{\mathcal {E}}^{\prime }} \cap \mathbf {K}})$. Let also $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{\prime } = \langle {S^{\prime },{{\varGamma }}^{\prime }}\rangle$. Then,

$$ \begin{array}{@{}rcl@{}} {\varGamma}_{{\mathcal{E}}} &=& ({{\mathcal{E}}} \cap \mathbf{K}) \cup \left\{ \alpha \in {Acc}({\mathfrak{A}}) \big| \mathbf{s}(\alpha) \notin S\right\}\\ {\varGamma}_{{\mathcal{E}}^{\prime}} &=& ({{\mathcal{E}}^{\prime}} \cap \mathbf{K}) \cup \left\{ \alpha \in {Acc}({\mathfrak{A}}^{\prime}) \big| \mathbf{s}(\alpha) \notin S^{\prime} \right\} \end{array} $$

Hence, to show ${\varGamma }_{{\mathcal {E}}} \subseteq {\varGamma }_{{\mathcal {E}}^{\prime }}$, it is enough to prove

$$ {\{\ \alpha \!\in\! {Acc}({\mathfrak{A}}) \ \big|\ \mathbf{s}(\alpha) \!\notin\! S \ \}} \subseteq {{\mathcal{E}}^{\prime}} \cup {\{\ \alpha \!\in\! {Acc}({\mathfrak{A}}^{\prime}) \ \big|\ \mathbf{s}(\alpha) \!\notin\! S^{\prime}\ \}} $$

Furthermore, from Observation 2 and the fact that ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime }$, it follows that ${Acc}({\mathfrak {A}}) \subseteq {Acc}({\mathfrak {A}}^{\prime })$ and, thus, it is enough to show that every $\alpha \in {Acc}({\mathfrak {A}})$ satisfies that: s(α)∉S implies that either $\mathbf {s}(\alpha ) \notin S^{\prime }$ or $\alpha \in {{\mathcal {E}}^{\prime }}$.

Suppose, for the sake of contradiction, that there is some $\alpha \in ({Acc}({\mathfrak {A}})\setminus {{\mathcal {E}}^{\prime }})$ that satisfies s(α)∉S and $\mathbf {s}(\alpha ) \in S^{\prime }$. Since by hypothesis ${\mathcal {E}}^{\prime }$ is AFRA-complete, $\alpha \notin {{\mathcal {E}}^{\prime }}$ implies that α is not AFRA-acceptable w.r.t. ${\mathcal {E}}^{\prime }$ and, thus, there is some β ∈K that defeats α and is not defeated by any $\gamma \in {\mathcal {E}}^{\prime }$. That is, either t(β) = α or t(β) = s(α). If the latter, then β also defeats s(α). But, then $\mathbf {s}(\alpha ) \in S^{\prime }$ implies $\mathbf {s}(\alpha ) \in {{\mathcal {E}}^{\prime }}$ which, in its turn, implies that s(α) is AFRA-acceptable w.r.t. ${\mathcal {E}}^{\prime }$ and, thus, that β is defeated by some $\gamma \in {\mathcal {E}}^{\prime }$ which is a contradiction with the above. Hence, it must be that s(β) = α and, thus, that $\alpha \notin {Acc}({\mathfrak {A}})$ which is a contradiction with the assumption. Hence, ${\varGamma } \subseteq {{\varGamma }}^{\prime }$ holds. □

Lemma 22

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subset {{\mathcal {E}}^{\prime }} \subseteq (\mathbf {A} \cup \mathbf {K})$ be two AFRA-complete extensions. Then, ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubset {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$.

Proof

First note that from Lemma 22, we have ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubseteq {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$. Suppose, for the sake of contradiciton that ${\mathfrak {A}}_{{\mathcal {E}}} = {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$. Then, $S \subset S^{\prime }$ implies ${\mathfrak {A}} \sqsubset {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$, which is a contradiction with the assumption. Hence, it must be that $S = S^{\prime }$ holds. Furthermore, since ${\mathcal {E}} \subset {{\mathcal {E}}^{\prime }}$, there is some element $x \in ({{\mathcal {E}}^{\prime }} \setminus {\mathcal {E}})$ and, since $S = S^{\prime }$, it follows that x ∈K. From Lemma 19 and the fact that ${\mathcal {E}}^{\prime }$ is AFRA-complete, it follows that ${\mathcal {E}}^{\prime }$ is closed and, thus, $x \in {{\mathcal {E}}^{\prime }}$ implies that $\mathbf {s}(x) \in {{\mathcal {E}}^{\prime }}$. This implies s(x) ∈ S and, since $S = S^{\prime }$, that s(x) ∈ S and $\mathbf {s}(x) \in {\mathcal {E}}$. This plus $x \notin {\mathcal {E}}$ imply that x∉Γ and, thus, that ${\varGamma } \subset {{\varGamma }}^{\prime }$ and ${\mathfrak {A}} \sqsubset {\mathfrak {A}}^{\prime }$ hold. □

Lemma 23

Let RAF = 〈A, K, s, t〉 be some framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be a preferred structure. Then, $\texttt {Afra}({\mathfrak {A}})$ is AFRA-preferred.

Proof

Since ${\mathfrak {A}}$ is a preferred structure, it is admissible and, in addition, there is no admissible structure ${\mathfrak {A}}$ such that ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime }$. From Proposition 6, the former implies that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-admissible. Hence, to show that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-preferred, it is enough to prove that there does not exist any AFRA-admissible extension ${\mathcal {E}}$ such that $\texttt {Afra}({\mathfrak {A}}) \subset {{\mathcal {E}}}$.

Suppose, for the sake of contradiction, that there exists any AFRA-admissible extension ${\mathcal {E}}$ such that $\texttt {Afra}({\mathfrak {A}}) \subset {{\mathcal {E}}}$. Since ${\mathcal {E}}$ is AFRA-admissible, from Theorem 2 in [4], there is some AFRA-preferred extension $\mathcal {E}^{\prime }$ such that $\texttt {Afra}{{\mathfrak {A}}} \subset {\mathcal {E}} \subseteq {\mathcal {E}^{\prime }}$. Furthermore, from Lemma 4 in [4], it follows that $\mathcal {E}^{\prime }$ is also AFRA-complete and, thus, from Lemma 20, that ${\mathfrak {A}}_{{\mathcal {E}^{\prime }}}$ is a complete structure. Furthermore, since ${\mathfrak {A}}$ is a preferred structure, from Theorem 2, it follows that ${\mathfrak {A}}$ is also complete and thus, from Lemma 17, that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-complete. From Lemma 22 and the fact that both $\texttt {Afra}({\mathfrak {A}})$ and ${\mathcal {E}}^{\prime }$ are complete, $\texttt {Afra}({\mathfrak {A}}) \subset {{\mathcal {E}}}$ implies that ${\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})} \sqsubset {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$ Moreover, since ${\mathfrak {A}}$ is complete, from Proposition 10, it follows that ${\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})} = {\mathfrak {A}}$ and, thus, that ${\mathfrak {A}} \sqsubset {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$. This is a contradiction with the assumption that ${\mathfrak {A}}$ is a preferred structure. Hence, $\texttt {Afra}({\mathfrak {A}})$ is an AFRA-preferred set. □

Lemma 24

Let RAF = 〈A, K, s, t〉 be some framework and $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be a stable structure. Then, $\texttt {Afra}({\mathfrak {A}})$ is AFRA-stable.

Proof

Since ${\mathfrak {A}}$ is a stable structure, it is conflict-free and, in addition, it satisfies $S = \overline {{Def}({\mathfrak {A}})}$ and ${\varGamma } = \overline {\mathit {Inh}{{\mathfrak {A}}}}$. From Proposition 4, the former implies that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-conflict-free. Hence, to show that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-stable, it is enough to prove that, for every $x \in ((\mathbf {A} \cup \mathbf {K}) \setminus \texttt {Afra}({\mathfrak {A}}))$, there is $\alpha \in \texttt {Afra}({\mathfrak {A}})$ such that α defeats x.

First, note that $x \in ((\mathbf {A} \cup \mathbf {K}) \setminus \texttt {Afra}({\mathfrak {A}}))$ implies that either x∉(S ∪Γ) or x ∈Γ but s(x)∉S. Since ${\mathfrak {A}}$ is stable, the former implies that $x \in ({Def}({\mathfrak {A}}) \cup {Inh}({\mathfrak {A}}))$ and, from Lemma 10, this implies that there is some $\alpha \in \texttt {Afra}({\mathfrak {A}})$ such that α defeats x. On the other hand, the latter implies that s(x)∉S and, thus, that $\mathbf {s}(x) \in {Def}({\mathfrak {A}})$. From Lemma 10, this implies that there is some $\alpha \in \texttt {Afra}({\mathfrak {A}})$ such that α defeats s(x) and, thus, that defeats x. □

Proof of Proposition 8

Conditions i), iii), and iv) follow directly from Lemmas 17, 23 and 24, respectively. For ii), note that ${\mathfrak {A}}$ being grounded implies that it is also complete and, thus, that $\texttt {Afra}({\mathfrak {A}})$ is AFRA-complete. Suppose, for the sake of contradiction, that $\texttt {Afra}({\mathfrak {A}})$ is not AFRA-grounded and, thus, that there is some set $\mathcal {E}$ which is AFRA-complete and satisfies ${\mathcal {E}} \subset \texttt {Afra}({\mathfrak {A}})$. Then, from Lemmas 20 and 22, it respectively follows that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a complete structure and that ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubset {\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})}$. Note that, from Proposition 10, we have that ${\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})} = {\mathfrak {A}}$ and, thus, we get ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubset {\mathfrak {A}}$ which is a contradiction with the fact that ${\mathfrak {A}}$ is a grounded structure. Hence, $\texttt {Afra}({\mathfrak {A}})$ is AFRA-grounded. □

Lemma 25

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a closed set. Then, it follows that $x \in (\overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})})$ implies that there is no $\alpha \in {\mathcal {E}}$ such that α directly defeats x.

Proof

Suppose, for the sake of contradiction, that there is $\alpha \in {\mathcal {E}}$ such that α directly defeats x. If α directly defeats x, then t(α) = x. Furthermore, since ${\mathcal {E}}$ is closed, $\alpha \in {\mathcal {E}}$ implies that $\mathbf {s}(\alpha ) \in {\mathcal {E}}$ and, thus, that $\alpha \in {\varGamma }_{{\mathcal {E}}}$ and that $\mathbf {s}(\alpha ) \in S_{{\mathcal {E}}}$. This implies that $x \in ({Def}({\mathfrak {A}}_{{\mathcal {E}}}) \cup {Inh}({\mathfrak {A}}_{{\mathcal {E}}}))$ which is a contradiction with the assumption. □

Lemma 26

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a closed set. Then, $a \in \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$ implies that there is no $\alpha \in {\mathcal {E}}$ such that α defeats a.

Proof

$a \in \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$ implies that a ∈A and, thus, α defeats a only if α directly defeats a. Then, the result follows directly from Lemma 25. □

Lemma 27

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a closed set. Then, $\alpha \in \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$ and $\mathbf {s}(\alpha ) \in \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$ imply that there is no $\beta \in {\mathcal {E}}$ such that β defeats α.

Proof

From Lemma 25, it follows that there is is no $\beta \in {\mathcal {E}}$ such that β directly defeats α. Suppose, for the sake of contradiction, that there is $\beta \in {\mathcal {E}}$ such that β indirectly defeats α. This implies that t(β) = s(α). Furthermore, since ${\mathcal {E}}$ is closed, it follows that $\beta \in {\mathcal {E}}$ implies that $\mathbf {s}(\beta ) \in {\mathcal {E}}$ and, thus, that $\beta \in {\varGamma }_{{\mathcal {E}}}$ and that $\mathbf {s}(\beta ) \in S_{{\mathcal {E}}}$. This implies that $\mathbf {s}(\alpha ) \in {Def}({\mathfrak {A}}_{{\mathcal {E}}})$ which is a contradiction with the assumption. □

Lemma 28

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a closed AFRA-conflict-free set. Then, $x \in (\overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})})$ implies that there is no $\alpha \in {\mathcal {E}}$ such that α defeats x.

Proof

Pick any $x \in (\overline {{Def}({\mathfrak {A}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})})$. If x ∈A, from Lemma 26, it follows that there is no $\alpha \in {\mathcal {E}}$ such that α defeats x. Otherwise, x ∈K and $x \in \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$. Since ${\mathcal {E}}$ is closed, it follows that $\mathbf {s}(x) \in {\mathcal {E}}$ and $\mathbf {s}(x) \in S_{{\mathcal {E}}}$. Furthermore, since ${\mathcal {E}}$ is conflict-free, Proposition 5 implies that ${\mathfrak {A}}_{{\mathcal {E}}}$ is conflict-free. Then, $\mathbf {s}(x) \in S_{{\mathcal {E}}}$ implies $\mathbf {s}(x) \in \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$. From Lemma 27, this plus $x \in \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$ imply that there is no $\alpha \in {\mathcal {E}}$ such that α defeats x. □

Lemma 29

Let RAF = 〈A, K, s, t〉 be some framework. Then, every AFRA-stable extension is closed.

Proof

Note that every AFRA-stable extension is also AFRA-complete (Lemmas 4 and 5 in [4]) and, thus, Lemma 19 implies that every AFRA-stable extension is also closed. □

Lemma 30

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a AFRA-stable extension. Then, it follows that $(\overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}) \subseteq {\mathcal {E}}$.

Proof

By definition every AFRA-stable extension is AFRA-conflict-free. Furthermore, from Lemma 29, every AFRA-stable extension is closed. Then, Lemma 28 implies that, for every $x \in (\overline {{Def}({\mathfrak {A}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})})$, there is no $\alpha \in {\mathcal {E}}$ such that α defeats x. Then, since ${\mathcal {E}}$ is AFRA-stable, this implies that $x \in {{\mathcal {E}}}$ and, consequently, that $(\overline {{Def}({\mathfrak {A}})} \cup \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}) \subseteq {\mathcal {E}}$ holds. □

Lemma 31

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a AFRA-stable extension. Then, ${\mathfrak {A}}_{{\mathcal {E}}}$ is a stable structure.

Proof

Since by definition every AFRA-stable extension is AFRA-conflict-free, it follows that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a conflict-free structure (Proposition 5). Then, to show that ${\mathfrak {A}}_{{\mathcal {E}}}$ is stable, it is enough to prove $S_{{\mathcal {E}}} = \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$ and ${\varGamma }_{{\mathcal {E}}} = \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$. Note that, since ${\mathfrak {A}}_{{\mathcal {E}}}$ is conflict-free, it follows that $S \subseteq \overline {\mathit {Def}{{\mathfrak {A}}_{\mathcal {E}}}}$ and ${\varGamma } \subseteq \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$ hold. Furthermore, from Lemma 30, it follows that

$$ (\overline{{Def}({\mathfrak{A}}_{{\mathcal{E}}})} \cup \overline{{Inh}({\mathfrak{A}}_{{\mathcal{E}}})}) \ \ \subseteq \ \ {{\mathcal{E}}} \ \ \subseteq \ \ (S_{{\mathcal{E}}} \cup {\varGamma}_{{\mathcal{E}}}) $$

and, thus, that $S = \overline {{Def}({\mathfrak {A}}_{{\mathcal {E}}})}$ and ${\varGamma } = \overline {{Inh}({\mathfrak {A}}_{{\mathcal {E}}})}$ hold. Consequently, ${\mathfrak {A}}_{{\mathcal {E}}}$ is a stable structure. □

Lemma 32

Let RAF = 〈A, K, s, t〉 be some framework. Every AFRA-preferred extension is closed.

Proof

Note that every AFRA-preferred extension is also AFRA-complete (Lemma 4 in [4]) and, thus, Lemma 19 implies that every AFRA-preferred extension is also closed. □

Lemma 33

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime }$ be two structures. Then, $\texttt {Afra}({\mathfrak {A}}) \subseteq \texttt {Afra}({\mathfrak {A}}^{\prime })$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$. Then,

$$ (\texttt{Afra}({\mathfrak{A}}) \cap \mathbf{A}) \ = \ S \ \subseteq \ S^{\prime} \ = \ (\texttt{Afra}({\mathfrak{A}}^{\prime}) \cap \mathbf{A}) $$

Furthermore,

$$ \begin{array}{@{}rcl@{}} (\texttt{Afra}({\mathfrak{A}}) \cap \mathbf{K}) &=& \left\{ \alpha \in {\varGamma} \big| \mathbf{s}(\alpha) \in S \right\}\\ (\texttt{Afra}({\mathfrak{A}}) \cap \mathbf{K}) &=& \left\{ \alpha \in {{\varGamma}}^{\prime} \big| \mathbf{s}(\alpha) \in S^{\prime} \right\} \end{array} $$

and, thus, $(\texttt {Afra}({\mathfrak {A}}) \cap \mathbf {K}) \subseteq (\texttt {Afra}({\mathfrak {A}}) \cap \mathbf {K})$. These two facts together imply $\texttt {Afra}({\mathfrak {A}}) \subseteq \texttt {Afra}({\mathfrak {A}}^{\prime })$. □

Lemma 34

Let RAF = 〈A, K, s, t〉 be some framework and ${\mathfrak {A}} \sqsubset {\mathfrak {A}}^{\prime }$ be two complete structures. Then, $\texttt {Afra}({\mathfrak {A}}) \subset \texttt {Afra}({\mathfrak {A}}^{\prime })$.

Proof

First note that, from Lemma 34, we have that $\texttt {Afra}({\mathfrak {A}}) \subset \texttt {Afra}({\mathfrak {A}}^{\prime })$. Suppose, for the sake of contradiction that $\texttt {Afra}({\mathfrak {A}}) = \texttt {Afra}({\mathfrak {A}}^{\prime })$. Then, it is obvious that ${\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}})} = {\mathfrak {A}}_{\texttt {Afra}({\mathfrak {A}}^{\prime })}$. However, from Proposition 10, this implies that ${\mathfrak {A}} = {\mathfrak {A}}^{\prime }$ which is a contradiction with the assumption. Hence, $\texttt {Afra}({\mathfrak {A}}) \subset \texttt {Afra}({\mathfrak {A}}^{\prime })$ holds. □

Lemma 35

Let RAF = 〈A, K, s, t〉 be some framework and ${{\mathcal {E}}} \subseteq (\mathbf {A} \cup \mathbf {K})$ be a AFRA-preferred extension. Then, ${\mathfrak {A}}_{{\mathcal {E}}}$ is a preferred structure.

Proof

By definition every AFRA-preferred extension is AFRA-admissible. Furthermore, from Lemma 32, it follows that AFRA-preferred extensions are closed. Then, from Proposition 7, it follows that ${\mathfrak {A}}_{{\mathcal {E}}}$ is an admissible structure. Suppose, for the sake of contradiction, that ${\mathfrak {A}}_{{\mathcal {E}}}$ is not preferred and, thus, that there is some admissible structure ${\mathfrak {A}}^{\prime }$ such that ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubset {\mathfrak {A}}^{\prime }$. Then, from Proposition 1, there is some preferred structure ${\mathfrak {A}}^{\prime \prime }$ such that ${\mathfrak {A}}_{{\mathcal {E}}} \sqsubset {\mathfrak {A}}^{\prime } \sqsubseteq {\mathfrak {A}}^{\prime \prime }$ and, from Theorem 2, we have that ${\mathfrak {A}}^{\prime \prime }$ is complete. Then, from Proposition 7, it follows that $\texttt {Afra}({\mathfrak {A}}^{\prime \prime })$ is admissible and, from Lemma 34 and Proposition 10, that ${\mathcal {E}} = \texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}}) \subset \texttt {Afra}({\mathfrak {A}}^{\prime \prime })$. This is a contradiction with the fact that ${\mathcal {E}}$ is AFRA-preferred. Hence, ${\mathfrak {A}}_{{\mathcal {E}}}$ must be a preferred structure. □

Proof of Proposition 9

Conditions i), iii), and iv) directly follow from Lemmas 20, 31 and 35, respectively. For ii) note that ${\mathcal {E}}$ being AFRA-grounded implies that it is also AFRA-complete and, thus, that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a complete structure. Suppose, for the sake of contradiction, that ${\mathfrak {A}}_{{\mathcal {E}}}$ is not AFRA-grounded and, thus, that there is some complete structure ${\mathfrak {A}}$ which satisfies ${\mathfrak {A}} \sqsubset {\mathfrak {A}}_{{\mathcal {E}}}$. Then, from Proposition 8 and Lemma 34, it respectively follows that $\texttt {Afra}({\mathfrak {A}})$ is a AFRA-complete and that $\texttt {Afra}({\mathfrak {A}}) \subset \texttt {Afra}({\mathfrak {A}}_{{\mathcal {E}}})$. Note that, from Proposition 10, we have that $\texttt {Afra}{{\mathfrak {A}}_{{\mathcal {E}}}} = {\mathcal {E}}$ and, thus, we get $\texttt {Afra}({\mathfrak {A}}) \subset {\mathcal {E}}$ which is a contradiction with the fact that ${\mathcal {E}}$ is a AFRA-grounded. Hence, ${\mathfrak {A}}_{{\mathcal {E}}}$ must be a grounded structure. □

Proof of Proposition 11

Let a ∈A be some argument. Then, from Theorem 3 and Proposition 12 in [4], it follows that some structure $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ is preferred (resp. stable) w.r.t. RAF

iff $\texttt {Afra}({\mathfrak {A}})$ is a preferred (resp. stable) extension w.r.t. RAF

iff $\texttt {Afra}({\mathfrak {A}})$ is a preferred (resp. stable) extension w.r.t. RAF_AF with RAF_AF is the corresponding Dung framework of RAF as given by Def. 19 in [4].

Hence, a is credulous accepted w.r.t. RAF and the preferred (resp. stable) semantics

iff there is some preferred (resp. stable) structure $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ of RAF such that a ∈ S

iff there is some preferred (resp. stable) extension ${\mathcal {E}} \subseteq (\mathbf {A} \cup \mathbf {K})$ of RAF such that $a \in {{\mathcal {E}}}$

iff there is some preferred (resp. stable) extension ${\mathcal {E}} \subseteq (\mathbf {A} \cup \mathbf {K})$ of RAF_AF such that $a \in {{\mathcal {E}}}$.

iff a is credulous accepted w.r.t. RAF_AF and the preferred (resp. stable) semantics.

Then, since credulous acceptance for Dung’s frameworks w.r.t. the preferred and the stable semantics is NP-complete [17] and RAF_AF can be computed in polynomial time, it follows that credulous acceptance for RAFs is in NP. Hardness, follows from the fact that every Dung’s framework is also a RAF and that, from Theorem 5, the preferred (resp. stable) semantics for RAFs are conservative generalisations.

Analogously, since sceptical acceptance for Dung’s frameworks w.r.t. the preferred (resp. stable) semantics is coNP-complete (resp. ${\mathrm {\Pi }^{\mathrm {P}}_{2}}$-complete) [17], it follows that sceptical acceptance for RAFs w.r.t. the preferred (resp. stable) semantics is coNP-complete (resp. ${\mathrm {\Pi }}_{2}^{\mathrm {P}}$-complete).

Finally, for the complete semantics, note that from Theorem 2, every preferred structure is also a complete structure and, thus, if an argument is credulous accepted w.r.t. the preferred semantics, it is also credulous accepted w.r.t. the complete semantics. Furthermore, every complete ${\mathfrak {A}}$ structure is admissible and, from Proposition 1, this implies that there is a preferred structure ${\mathfrak {A}}^{\prime }$ such that ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime }$. This implies that, if an argument is credulous accepted w.r.t. the complete semantics, it is also credulous accepted w.r.t. the preferred semantics. □

2.3 A.3 Proofs of Section 5

Lemma 36

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathcal {E}}$ be an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$.

• 1. If ${\mathcal {E}} \cap \mathbf {K}$ contains α, then ${\mathcal {E}}$ contains s(α).

• 2. If ${\mathcal {E}}$ contains N_s(α)α then ${\mathcal {E}} \cap \mathbf {K}$ attacks s(α).

• 3. Moreover, if ${\mathcal {E}}$ is complete the equivalence holds: ${\mathcal {E}}$ contains N_s(α)α if and only if ${\mathcal {E}} \cap \mathbf {K}$ attacks s(α).

Proof

Let N_s(α)α ∈N. Due to the definition of $\mathbf {R}^{\prime }$ (see Definition 13), the only attack to N_s(α)α is the attack (s(α),N_s(α)α).

• 1. As ${\mathcal {E}}$ is admissible, α is acceptable w.r.t. ${\mathcal {E}}$. As s(α) is the only defender of α against N_s(α)α, s(α) must belong to $\mathcal {E}$.

• 2. If ${\mathcal {E}}$ is admissible and contains N_s(α)α, N_s(α)α is acceptable w.r.t. ${\mathcal {E}}$ so ${\mathcal {E}}$ must attack s(α). Due to the definition of $\mathbf {R}^{\prime }$ again (see Definition 13), this attack comes from $\mathcal {E} \cap \mathbf {K}$.

• 3. If ${\mathcal {E}}$ is complete and ${\mathcal {E}} \cap \mathbf {K}$ attacks s(α) then N_s(α)α is acceptable w.r.t. ${\mathcal {E}}$. So N_s(α)α must belong to ${\mathcal {E}}$.

□

Lemma 37

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathfrak {A}}= \langle S,{\Gamma }\rangle$ be a structure. Let ${\mathfrak {A}}^{\prime }$ be the structure $\langle S, \texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {K}\rangle$. It holds that $Acc({\mathfrak {A}})$ = $Acc({\mathfrak {A}}^{\prime })$.

Proof

By definition $\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}= \{\alpha \in {\Gamma }$ s.t. s(α) ∈ S}). Let x ∈A ∪K. By definition, $x \in Acc({\mathfrak {A}})$ if and only if for each attack β ∈K such that t(β) = x, there exists γ ∈Γ with s(γ) ∈ S and t(γ) ∈{β, s(β)}. Obviously, γ ∈Γ with s(γ) ∈ S is equivalent to $\gamma \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}$ with s(γ) ∈ S. So $x \in Acc({\mathfrak {A}})$ if and only if $x \in Acc({\mathfrak {A}}^{\prime })$. □

Lemma 38

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathfrak {A}}= \langle S,{\Gamma }\rangle$ be a structure, $\texttt {Maf}({\mathfrak {A}})$ be the associated MAF-extension, and ${\mathfrak {A}}_{\texttt {Maf}({\mathfrak {A}})} = \langle S_{\texttt {Maf}({\mathfrak {A}})}, {\Gamma }_{\texttt {Maf}({\mathfrak {A}})}\rangle$ be the structure associated with the extension $\texttt {Maf}({\mathfrak {A}})$. The following assertions hold:

• 1. $S_{\texttt {Maf}({\mathfrak {A}})} = S$

• 2. If ${\mathfrak {A}}$ is an admissible structure, then ${\Gamma } \subseteq {\Gamma }_{\texttt {Maf}({\mathfrak {A}})}$

• 3. If $(Acc({\mathfrak {A}}) \cap \mathbf {K}) \subseteq {\Gamma }$, then ${\Gamma }_{\texttt {Maf}({\mathfrak {A}})} \subseteq {\Gamma }$

• 4. If ${\mathfrak {A}}$ is a complete structure, then ${\Gamma } = {\Gamma }_{\texttt {Maf}({\mathfrak {A}})}$

Proof

• 1. By definition of ${\mathfrak {A}}_{\texttt {Maf}({\mathfrak {A}})}$ we have $S_{\texttt {Maf}({\mathfrak {A}})} = \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}$. By definition of $\texttt {Maf}({\mathfrak {A}})$, we have $\texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {A} = S$.

• 2. By definition, ${\Gamma }_{\texttt {Maf}({\mathfrak {A}})} = (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}) \cup \{\alpha \not \in (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K})$ s.t. $\alpha \in Acc((\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}), (\texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {K}))$ and $\mathbf {s}(\alpha ) \not \in (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}) \}$. Note that by definition, $(\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}) = S$ and $(\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}) = \{\gamma \in {\Gamma }$ with s(γ) ∈ S}. And from Lemma 37, we have $Acc((\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}), (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K})) = Acc({\mathfrak {A}})$.

  Let α ∈Γ. If s(α) ∈ S, $\alpha \in (\texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {K})$ so $\alpha \in {\Gamma }_{\texttt {Maf}({\mathfrak {A}})}$. If s(α)∉S, as ${\mathfrak {A}}$ is admissible, we have $\alpha \in Acc({\mathfrak {A}})$, so α belongs to the second part of ${\Gamma }_{\texttt {Maf}{{\mathfrak {A}}}}$.

• 3. Assume that $(Acc({\mathfrak {A}}) \cap \mathbf {K}) \subseteq {\Gamma }$. Let $\alpha \in {\Gamma }_{\texttt {Maf}({\mathfrak {A}})}$. If $\alpha \in (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}) \subseteq {\Gamma }$, we have α ∈Γ. If $\alpha \not \in (\texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {K})$, by definition of ${\Gamma }_{\texttt {Maf}{{\mathfrak {A}}}}$, we have $\alpha \in Acc({\mathfrak {A}})$. Due to the assumption, it follows that α ∈Γ.

• 4. The proof follows directly from the above items of this lemma.

□

Lemma 39

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathcal {E}}$ be an extension of MAF. Let ${\mathfrak {A}}_{{\mathcal {E}}}$ be its associated structure and $\texttt {Maf}{{\mathfrak {A}}_{\mathcal {E}}}$ be the MAF-extension associated with ${\mathfrak {A}}_{{\mathcal {E}}}$. The following assertions hold:

• 1. $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {A}$ = ${\mathcal {E}} \cap \mathbf {A}$

• 2. $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K} \subseteq {\mathcal {E}} \cap \mathbf {K}$

• 3. If ${\mathcal {E}}$ is admissible, then $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K} = {\mathcal {E}} \cap \mathbf {K}$

• 4. If ${\mathcal {E}}$ is admissible, then ${\mathcal {E}} \cap \mathbf {N} \subseteq \texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N}$

• 5. If ${\mathcal {E}}$ is complete then $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N}$ = ${\mathcal {E}} \cap \mathbf {N}$

Proof

By definition, ${\mathfrak {A}}_{{\mathcal {E}}} = \langle S_{{\mathcal {E}}},{\Gamma }_{{\mathcal {E}}}\rangle$ with $S_{{\mathcal {E}}} = {\mathcal {E}}_{a} = {\mathcal {E}} \cap \mathbf {A}$, ${\mathcal {E}}_{k} = {\mathcal {E}} \cap \mathbf {K}$, and ${\Gamma }_{{\mathcal {E}}} = {\mathcal {E}}_{k} \cup \{\alpha \not \in {\mathcal {E}}_{k}$ s.t. $\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ and $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a} \}$.

Then, $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}})$ is such that $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {A} = S_{\mathcal {E}}$, $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K} = \{\alpha \in {\Gamma }_{{\mathcal {E}}}$ such that $\mathbf {s}(\alpha ) \in S_{{\mathcal {E}}}\}$, and $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N} = \{N_{\mathbf {s}(\alpha )\alpha }$ s.t. $\mathbf {s}(\alpha ) \not \in S_{{\mathcal {E}}}$ and $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}_{{\mathcal {E}}}) \}$.

• 1. Obviously, $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {A}$ = ${\mathcal {E}} \cap \mathbf {A}$.

• 2. By definition of ${\Gamma }_{{\mathcal {E}}}$, if $\alpha \in {\Gamma }_{{\mathcal {E}}}$ with $\mathbf {s}(\alpha ) \in S_{{\mathcal {E}}}= {\mathcal {E}}_{a}$ then $\alpha \in {\mathcal {E}}_{k}$. So, $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K} \subseteq \mathcal {E}_{k}$.

• 3. Assume that ${\mathcal {E}}$ is admissible. Let $\alpha \in {\mathcal {E}}_{k}$. From Lemma 36, we have $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$. As ${\mathcal {E}}_{k} \subseteq {\Gamma }_{{\mathcal {E}}}$ it follows that $\alpha \in {\Gamma }_{\mathcal {E}}$ and $\mathbf {s}(\alpha ) \in \mathcal {E}_{a}$. So $\alpha \in \texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K}$. From the above item we conclude that $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {K} = {\mathcal {E}}_{k}$.

• 4. Assume that ${\mathcal {E}}$ is admissible. Let $N_{\mathbf {s}(\alpha )\alpha } \in {\mathcal {E}} \cap \mathbf {N}$. From Lemma 36, we know that ${\mathcal {E}}_{k}$ attacks s(α). So there exists $\beta \in {\mathcal {E}}_{k}$ that attacks s(α). From Lemma 36 again, we have $\mathbf {s}(\beta ) \in {\mathcal {E}}$. So, s(α) is attacked by $\beta \in {\mathcal {E}}_{k} \subseteq {\Gamma }_{{\mathcal {E}}}$ with $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$. That means that $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}_{{\mathcal {E}}})$. Moreover, as ${\mathcal {E}}$ is conflict-free, $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$. It follows that $N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N}$.

• 5. Assume that ${\mathcal {E}}$ is complete. Let $N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N}$. By definition, $\mathbf {s}(\alpha ) \not \in S_{{\mathcal {E}}} = {\mathcal {E}}_{a}$ and $\mathbf {s}(\alpha ) \in Def(S_{\mathcal {E}}, {\Gamma }_{\mathcal {E}})$. So there exists $\beta \in {\Gamma }_{{\mathcal {E}}}$ with $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$ such that β attacks s(α). By definition of ${\Gamma }_{{\mathcal {E}}}$ it follows that $\beta \in {\mathcal {E}}_{k}$. Then from Lemma 36, we conclude that ${\mathcal {E}}$ contains N_s(α)α. Then from the above item of this lemma, as E is also admissible, we conclude that $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) \cap \mathbf {N}$ = ${\mathcal {E}} \cap \mathbf {N}$.

□

Lemma 40

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathcal {E}}$ and ${\mathcal {E}}^{\prime }$ be two complete extensions of MAF such that ${\mathcal {E}} \subseteq {\mathcal {E}}^{\prime }$ then ${\mathfrak {A}}_{{\mathcal {E}}} \subseteq {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$.

Proof

By definition, ${\mathfrak {A}}_{{\mathcal {E}}} = \langle S_{{\mathcal {E}}},{\Gamma }_{{\mathcal {E}}}\rangle$ with $S_{{\mathcal {E}}} = {\mathcal {E}}_{a} = {\mathcal {E}} \cap \mathbf {A}$, ${\mathcal {E}}_{k} = {\mathcal {E}} \cap \mathbf {K}$, and ${\Gamma }_{{\mathcal {E}}} = {\mathcal {E}}_{k} \cup \{\alpha \not \in {\mathcal {E}}_{k}$ s.t. $\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ and $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a} \}$.

First, we have ${\mathcal {E}}_{a} \subseteq {\mathcal {E}}^{\prime }_{a}$ and ${\mathcal {E}}_{k} \subseteq {\mathcal {E}}^{\prime }_{k}$. Then, due to Lemma 36, as ${\mathcal {E}}$ (resp. ${\mathcal {E}}^{\prime }$) is admissible, $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$ (resp. $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}^{\prime }_{a}$) implies that $\alpha \not \in {\mathcal {E}}_{k}$ (resp. $\alpha \not \in {\mathcal {E}}^{\prime }_{k}$). So it is enough to prove that: $\{\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ such that $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a} \}$ $\subseteq$ ${\mathcal {E}}^{\prime }_{k} \cup \{\alpha \in Acc({\mathcal {E}}^{\prime }_{a}, {\mathcal {E}}^{\prime }_{k})$ such that $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}^{\prime }_{a} \}$.

Furthermore, we have $Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k}) \subseteq Acc({\mathcal {E}}^{\prime }_{a}, {\mathcal {E}}^{\prime }_{k})$ so it is enough to show that: For each $\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ such that $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$, either $\alpha \in {\mathcal {E}}^{\prime }_{k}$ or $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}^{\prime }_{a}$.

Assume that the contrary holds. So there is $\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ such that $\alpha \not \in {\mathcal {E}}^{\prime }_{k}$, $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$ and $\mathbf {s}(\alpha ) \in {\mathcal {E}}^{\prime }_{a}$. As ${\mathcal {E}}^{\prime }$ is a complete extension of MAF, α is not acceptable w.r.t. ${\mathcal {E}}^{\prime }$. Moreover, as $\mathbf {s}(\alpha ) \in {\mathcal {E}}^{\prime }_{a}$, α is defended by ${\mathcal {E}}^{\prime }$ against its attacker N_s(α)α. So there must exist another attacker of α, say β, such that ${\mathcal {E}}^{\prime }$ does not attack β. That implies that $N_{\mathbf {s}(\beta )\beta } \not \in {\mathcal {E}}^{\prime }$, and from Lemma 36 that ${\mathcal {E}}^{\prime }$ does not attack s(β). So ${\mathcal {E}}^{\prime }$ attacks neither β, nor s(β); this fact will be denoted by (*).

Moreover, as $\alpha \in Acc({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$, either $\beta \in Inh({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ (Case 1), or $\mathbf {s}(\beta ) \in Def({\mathcal {E}}_{a}, {\mathcal {E}}_{k})$ (Case 2). It follows that there is $\gamma \in {\mathcal {E}}_{k} \subseteq {\mathcal {E}}^{\prime }_{k}$ with $\mathbf {s}(\gamma ) \in {\mathcal {E}}_{a}$ such that t(γ) = β in Case 1 (resp. t(γ) = s(β) in Case 2). So ${\mathcal {E}}^{\prime }_{k}$ attacks β in Case 1 (resp. s(β) in Case 2), which is in contradiction with the fact (*). □

Lemma 41

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let ${\mathfrak {A}}$ and ${\mathfrak {A}}^{\prime }$ be two conflict-free structures of RAF such that ${\mathfrak {A}} \subseteq {\mathfrak {A}}^{\prime }$ then $\texttt {Maf}({\mathfrak {A}}) \subseteq \texttt {Maf}({\mathfrak {A}}^{\prime })$.

Proof

By definition, ${\mathfrak {A}}$ being the structure 〈S,Γ〉, $\texttt {Maf}({\mathfrak {A}}) = S \cup \{\alpha \in {\Gamma }$ s.t. s(α) ∈ S}∪{N_s(α)α s.t. s(α)∉S and $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}) \}$.

First, we have $S \subseteq S^{\prime }$ and {α ∈Γ s.t. s(α) ∈ S} $\subseteq$ $\{\alpha \in {\Gamma }^{\prime }$ s.t. $\mathbf {s}(\alpha ) \in S^{\prime }\}$. So it is enough to prove that $(\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {N}) \subseteq (\texttt {Maf}({\mathfrak {A}}^{\prime }) \cap \mathbf {N})$.

Let $x = N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {N}$. As ${\mathfrak {A}} \subseteq {\mathfrak {A}}^{\prime }$, we have $Def({\mathfrak {A}}) \subseteq Def({\mathfrak {A}}^{\prime })$. So $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}^{\prime })$. Hence, as ${\mathfrak {A}}^{\prime }$ is a conflict-free structure, it is impossible to have $\mathbf {s}(\alpha ) \in S^{\prime }$. So $x = N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}({\mathfrak {A}}^{\prime }) \cap \mathbf {N}$. □

Proposition 1

Let RAF = 〈A, K, s, t〉 and its associated $\mathbf {MAF} = {\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$.

• 1. Let ${\mathfrak {A}}$ be an admissible structure of RAF. Then, $\texttt {Maf}({\mathfrak {A}})$ is an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$.

• 2. Let ${{\mathcal {E}}}$ be an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. ${\mathfrak {A}}^{\prime }_{{\mathcal {E}}}$ and ${\mathfrak {A}}_{{\mathcal {E}}}$ are admissible structures of RAF.

Proof

• 1. Let ${\mathfrak {A}}=\langle S,{\Gamma }\rangle$ be an admissible structure of RAF.

  From Proposition 13, $\texttt {Maf}({\mathfrak {A}})$ is conflict-free in MAF. It remains to prove that $\forall x \in \texttt {Maf}{{\mathfrak {A}}}$, x is acceptable w.r.t. $\texttt {Maf}{{\mathfrak {A}}}$. Three cases must be considered for x:

  • (a) Let $x \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {A}$. So x ∈ S. As ${\mathfrak {A}}$ is admissible, x is acceptable w.r.t. ${\mathfrak {A}}$. From Proposition 14, it follows that x is acceptable w.r.t. $\texttt {Maf}{{\mathfrak {A}}}$ in MAF.

  • (b) Let $x \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}$. As ${\mathfrak {A}}$ is admissible, and x ∈Γ, x is acceptable w.r.t. ${\mathfrak {A}}$. Moreover from the definition of $\texttt {Maf}{{\mathfrak {A}}}$ we have s(x) ∈ S. So Proposition 14 applies and we conclude that x is acceptable w.r.t. $\texttt {Maf}{{\mathfrak {A}}}$ in MAF.

  • (c) Let $x \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {N}$. So x has the form N_s(α)α with s(α)∉S and $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}})$.

    The only possible attack to x is from s(α). As $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}})$, ∃β ∈Γ s.t. s(β) ∈ S and t(β) = s(α). From the definition of $\texttt {Maf}({\mathfrak {A}})$, it follows that $\beta \in \texttt {Maf}({\mathfrak {A}})$ and then that $\texttt {Maf}({\mathfrak {A}})$ attacks s(α). So N_s(α)α is acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$ in MAF.

  Hence we have proved that $\texttt {Maf}({\mathfrak {A}})$ is an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$.

• 2. Let ${\mathcal {E}}$ be an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$.

  • (a) From Proposition 13, ${\mathfrak {A}}^{\prime }_{{\mathcal {E}}}$ is a conflict-free structure in RAF.

    It remains to prove that $\forall x \in ({\mathcal {E}}_{a} \cup {\mathcal {E}}_{k})$, x is acceptable w.r.t. ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$. If x is unattacked in RAF, then it is obviously acceptable w.r.t. ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$. Otherwise two cases must be considered for x:

    • (i) Let $x \in {\mathcal {E}}_{a}$. Assume that x is attacked by α ∈K. We have to prove that either $\alpha \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ or $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}^{\prime }_{\mathcal {E}})$.

      The attack α is encoded in the MAF with the three following attacks in $\mathbf {R}^{\prime }$: (s(α),N_s(α)α), (N_s(α)α,α) and (α, x).

      As $\mathcal {E}$ is assumed to be admissible, $\mathcal {E}$ attacks α. So either $N_{\mathbf {s}(\alpha )\alpha } \in {\mathcal {E}}$ (Case 1), or ${\mathcal {E}}_{k}$ attacks α (Case 2). Moreover, due to Lemma 36, Case 1 also implies that ${\mathcal {E}}_{k}$ attacks s(α).

      So there exists $\beta \in {\mathcal {E}}_{k}$ s.t. β attacks s(α) in Case 1 (resp. β attacks α in Case 2). As ${\mathcal {E}}$ is admissible, ${\mathcal {E}}$ defends β against N_s(β)β (Observation 4). So $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$. That fact with $\beta \in {\mathcal {E}}_{k}$ prove that $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ (resp. $\alpha \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$).

    • (ii) Let $x \in {\mathcal {E}}_{k}$. Assume that x is attacked by α ∈K. We have to prove that either $\alpha \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ or $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}^{\prime }_{\mathcal {E}})$. We can do exactly the same reasoning as for the first case ($x \in \mathcal {E}_{a}$).

      So we have proved that the structure ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$ is admissible.

  • (b) From Proposition 13, ${\mathfrak {A}}_{{\mathcal {E}}}$ is a conflict-free structure in RAF.

    It remains to prove that $\forall x \in ({\mathcal {E}}_{a} \cup {\Gamma }_{{\mathcal {E}}})$, x is acceptable w.r.t. ${\mathfrak {A}}_{\mathcal {E}}$. We recall that ${\Gamma }_{\mathcal {E}} = \mathcal {E}_{k} \cup \{\alpha \not \in \mathcal {E}_{k}$ s.t. $\alpha \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}}) \}$.

    From the first part of the proof, ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$ is admissible. So $\forall x \in (\mathcal {E}_{a} \cup \mathcal {E}_{k})$, x is acceptable w.r.t. ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$ and then w.r.t. ${\mathfrak {A}}_{{\mathcal {E}}}$. It remains to consider $x \in {\Gamma }_{{\mathcal {E}}} \setminus {\mathcal {E}}_{k}$. In that case, due to the definition of ${\Gamma }_{{\mathcal {E}}}$, $x \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}})$ so x is acceptable w.r.t. ${\mathfrak {A}}_{{\mathcal {E}}}$.

  So we have proved that the structure ${\mathfrak {A}}_{{\mathcal {E}}}$ is admissible.

□

Proof of Proposition 12

• 1. Let ${\mathfrak {A}}$ be a structure of RAF.

  • (a) (σ = complete) Let ${\mathfrak {A}}=\langle S,{\Gamma }\rangle$ be a complete structure of RAF. Let us recall that $\texttt {Maf}({\mathfrak {A}}) = S \cup \{\alpha \in {\Gamma }$ s.t. s(α) ∈ S}∪{N_s(α)α s.t. s(α)∉S and $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}}) \}$.

    By definition, ${\mathfrak {A}}$ is an admissible structure and satisfies $Acc({\mathfrak {A}}) \subseteq S \cup {\Gamma }$. From Proposition 20, we have that $\texttt {Maf}({\mathfrak {A}})$ is an admissible extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. So it remains to prove that $\forall x \in \mathbf {A}^{\prime }$, if x is acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$, then $x \in \texttt {Maf}({\mathfrak {A}})$.

    Three cases must be considered for x:

    • (i) Let x ∈A being acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$. Assume that $x \not \in \texttt {Maf}({\mathfrak {A}})$. Then x∉S and so $x \not \in Acc({\mathfrak {A}})$, due to the assumption on ${\mathfrak {A}}$. $x \not \in Acc({\mathfrak {A}})$ means that there is β ∈K with t(β) = x and such that $\beta \not \in Inh({\mathfrak {A}})$ and $\mathbf {s}(\beta ) \not \in Def({\mathfrak {A}})$; this fact will be denoted by (*). So we have the attack (β, x) in MAF.

      As x is acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$, we know that $\texttt {Maf}({\mathfrak {A}})$ attacks β in MAF. So, either $N_{\mathbf {s}(\beta )\beta } \in \texttt {Maf}({\mathfrak {A}})$ (Case 1), or there exists $\gamma \in (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K})$ that attacks β (Case 2).

      In Case 1, by definition of $\texttt {Maf}{{\mathfrak {A}}}$, $\mathbf {s}(\beta ) \in Def({\mathfrak {A}})$, which is in contradiction with the fact (*). In Case 2, by definition of $\texttt {Maf}({\mathfrak {A}})$, we have γ ∈Γ and s(γ) ∈ S. So $\beta \in Inh({\mathfrak {A}})$, which is in contradiction with the fact (*).

      So we have proved that x must belong to $\texttt {Maf}({\mathfrak {A}})$.

    • (ii) Let α ∈K being acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$. From Observation 5, it follows that s(α) ∈ S. Now, assume that $\alpha \not \in \texttt {Maf}({\mathfrak {A}})$. By definition of $\texttt {Maf}({\mathfrak {A}})$, it follows that α∉Γ and so $\alpha \not \in Acc({\mathfrak {A}})$, due to the assumption on ${\mathfrak {A}}$.

      The rest of the proof is analogous to the proof of the first item. $\alpha \not \in Acc({\mathfrak {A}})$ means that there is β ∈K with t(β) = α and such that $\beta \not \in Inh({\mathfrak {A}})$ and $\mathbf {s}(\beta ) \not \in Def({\mathfrak {A}})$; this fact will be denoted by (*). So we have the attack (β, α) in MAF.

      As α is acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$, we know that $\texttt {Maf}({\mathfrak {A}})$ attacks β in MAF. So, either $N_{\mathbf {s}(\beta )\beta } \in \texttt {Maf}({\mathfrak {A}})$ (Case 1), or there exists $\gamma \in (\texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K})$ that attacks β (Case 2).

      In Case 1, by definition of $\texttt {Maf}{{\mathfrak {A}}}$, $\mathbf {s}(\beta ) \in Def({\mathfrak {A}})$, which is in contradiction with the fact (*). In Case 2, by definition of $\texttt {Maf}({\mathfrak {A}})$, we have γ ∈Γ and s(γ) ∈ S. So $\beta \in Inh({\mathfrak {A}})$, which is in contradiction with

      the fact (*).

      So we have proved that α must belong to $\texttt {Maf}({\mathfrak {A}})$.

    • (iii) Let x ∈N being acceptable w.r.t. $\texttt {Maf}({\mathfrak {A}})$. x has the form N_s(α)α. As s(α) is the only attacker of x, we know that $\texttt {Maf}({\mathfrak {A}})$ attacks s(α) in MAF. So there exists $\beta \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {K}$ with $(\beta , \mathbf {s}(\alpha )) \in \mathbf {R}^{\prime }$. By definition of $\texttt {Maf}({\mathfrak {A}})$, we have β ∈Γ and s(β) ∈ S. Hence $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}})$. As ${\mathfrak {A}}$ is conflict-free, it implies that s(α)∉S. So we have $N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}({\mathfrak {A}}) \cap \mathbf {N}$.

  • (b) (σ = stable) Let ${\mathfrak {A}}=\langle S,{\Gamma }\rangle$ be a stable structure of RAF. By definition, ${\mathfrak {A}}$ is a conflict-free structure that satisfies: $\mathbf {A} \setminus S \subseteq Def({\mathfrak {A}})$ and $\mathbf {K} \setminus {\Gamma } \subseteq Inh({\mathfrak {A}})$. First, from Proposition 13, we have that $\texttt {Maf}({\mathfrak {A}})$ is conflict-free in ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Then, we have to prove that $\forall x \in \mathbf {A}^{\prime } \setminus \texttt {Maf}({\mathfrak {A}})$, x is attacked by $\texttt {Maf}({\mathfrak {A}})$.

    Three cases must be considered for x:

    • (i) Let $x \in \mathbf {A} \setminus \texttt {Maf}({\mathfrak {A}})$. Then x ∈A ∖ S. By assumption on ${\mathfrak {A}}$, it follows that $x \in Def({\mathfrak {A}})$ and from Observation 5it follows that $\texttt {Maf}{{\mathfrak {A}}}$ attacks x.

    • (ii) Let $\alpha \in \mathbf {K} \setminus \texttt {Maf}({\mathfrak {A}})$. Then either α∉Γ (Case 1), or α ∈Γ and s(α)∉S (Case 2).

      In Case 1, as ${\mathfrak {A}}$ is a stable structure, $\alpha \in Inh({\mathfrak {A}})$ and from Observation 5 it follows that $\texttt {Maf}{{\mathfrak {A}}}$ attacks α. In Case 2, as ${\mathfrak {A}}$ is a stable structure, $\mathbf {s}(\alpha ) \in Def({\mathfrak {A}})$. Moreover, s(α)∉S, so $N_{\mathbf {s}(\alpha )\alpha } \in \texttt {Maf}{{\mathfrak {A}}} \cap \mathbf {N}$. As $(N_{\mathbf {s}(\alpha )\alpha }, \alpha ) \in \mathbf {R}^{\prime }$, we conclude that $\texttt {Maf}{{\mathfrak {A}}}$ attacks α.

    • (iii) Let $x \in \mathbf {N} \setminus \texttt {Maf}({\mathfrak {A}})$. x has the form N_s(α)α with s(α) ∈ S or $\mathbf {s}(\alpha ) \not \in Def({\mathfrak {A}})$. Note that, as ${\mathfrak {A}}$ is stable, if $\mathbf {s}(\alpha ) \not \in Def({\mathfrak {A}})$ then s(α) ∈ S. Then, if s(α) ∈ S, as $(\mathbf {s}(\alpha ),N_{\mathbf {s}(\alpha )\alpha }) \in \mathbf {R}^{\prime }$ and $S \subseteq \texttt {Maf}({\mathfrak {A}})$, we conclude that $\texttt {Maf}({\mathfrak {A}})$ attacks x.

    So we have proved that $\texttt {Maf}({\mathfrak {A}})$ is a stable extension of MAF.

  • (c) (σ = preferred) Let ${\mathfrak {A}}$ be a preferred structure. By definition, ${\mathfrak {A}}$ is a $\subseteq$-maximal admissible structure. Moreover ${\mathfrak {A}}$ is a complete structure.

    So, from Proposition 12 (item 1 (a)), $\texttt {Maf}({\mathfrak {A}})$ is a complete extension of MAF.

    Assume that $\texttt {Maf}({\mathfrak {A}})$ is not a preferred extension of MAF. Then there exists ${\mathcal {E}}^{\prime }$ an admissible extension of MAF that strictly contains $\texttt {Maf}({\mathfrak {A}})$. It can be assumed that ${\mathcal {E}}^{\prime }$ is a $\subseteq$-maximal admissible extension of MAF. So ${\mathcal {E}}^{\prime }$ is preferred and thus complete.

    From Lemma 40, it follows that ${\mathfrak {A}}_{\texttt {Maf}({\mathfrak {A}})} \subseteq {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$. From Proposition 15, we have ${\mathfrak {A}}_{\texttt {Maf}({\mathfrak {A}})} = {\mathfrak {A}}$. So, ${\mathfrak {A}} \subseteq {\mathfrak {A}}_{{\mathcal {E}}^{\prime }}$. As ${\mathfrak {A}}$ is preferred, it follows that ${\mathfrak {A}} = {\mathfrak {A}}_{\mathcal {E}^{\prime }}$.

    From Proposition 15 again, $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}^{\prime }}) = {\mathcal {E}}^{\prime }$ so $\texttt {Maf}({\mathfrak {A}}) = {\mathcal {E}}^{\prime }$. That is in contradiction with the assumption that ${\mathcal {E}}^{\prime }$ strictly contains ${\mathcal {E}}_{{\mathfrak {A}}}$. Hence, we have proved that $\texttt {Maf}({\mathfrak {A}})$ is a preferred extension of MAF.

  • (d) (σ = grounded) Let ${\mathfrak {A}}=\langle S,{\Gamma }\rangle$ be the grounded structure of RAF. By definition, ${\mathfrak {A}}$ is the $\subseteq$-minimal complete structure. From Proposition 12 (item 1 (a)) $\texttt {Maf}{{\mathfrak {A}}}$ is a complete extension of MAF. Assume that there is $\mathcal {E}^{\prime }$ a complete extension that is strictly included in $\texttt {Maf}{{\mathfrak {A}}}$. From Lemma 40, we have ${\mathfrak {A}}_{\mathcal {E}^{\prime }} \subseteq {\mathfrak {A}}_{\texttt {Maf}{{\mathfrak {A}}}}$. As ${\mathfrak {A}}_{\texttt {Maf}{{\mathfrak {A}}}}$ = ${\mathfrak {A}}$, due to Proposition 15, we have ${\mathfrak {A}}_{\mathcal {E}^{\prime }} \subseteq {\mathfrak {A}}$. Proposition 12 (item 2 (a)), ${\mathfrak {A}}_{\mathcal {E}^{\prime }}$ is a complete structure, so by assumption on ${\mathfrak {A}}$ it follows that ${\mathfrak {A}}_{\mathcal {E}^{\prime }} ={\mathfrak {A}}$. Hence, $\texttt {Maf}{{\mathfrak {A}}_{\mathcal {E}^{\prime }}}$ = $\texttt {Maf}{{\mathfrak {A}}}$, and from Proposition 15 again, $\mathcal {E}^{\prime } = \texttt {Maf}({\mathfrak {A}})$. That is in contradiction with the fact that ${\mathcal {E}}^{\prime }$ is strictly included in $\texttt {Maf}({\mathfrak {A}})$. So we have proved that $\texttt {Maf}({\mathfrak {A}})$ is a $\subseteq$-minimal complete extension of MAF, or in other words the grounded extension of MAF.

• 2. Let ${{\mathcal {E}}}$ be a set of $\mathbf {A}^{\prime }$.

  • (a) (σ = complete) Let ${\mathcal {E}}$ be a complete extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. Let us recall that ${\mathfrak {A}}_{{\mathcal {E}}} = \langle {\mathcal {E}}_{a},{\Gamma }_{{\mathcal {E}}}\rangle$, where ${\Gamma }_{{\mathcal {E}}} = {\mathcal {E}}_{k} \cup \{\alpha \not \in {\mathcal {E}}_{k}$ s.t. $\alpha \in Acc({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ and $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a} \}$, and ${\mathfrak {A}}^{\prime }_{\mathcal {E}} = \langle {\mathcal {E}}_{a},{\mathcal {E}}_{k}\rangle$.

    By definition, ${\mathcal {E}}$ is an admissible extension of MAF and $\forall x \in \mathbf {A}^{\prime }$, if x is acceptable w.r.t. ${\mathcal {E}}$, then $x \in {\mathcal {E}}$.

    From Proposition 20, we have that ${\mathfrak {A}}_{{\mathcal {E}}}$ is an admissible structure. So it remains to prove that $Acc({\mathfrak {A}}_{{\mathcal {E}}}) \subseteq {\mathcal {E}}_{a} \cup {\Gamma }_{{\mathcal {E}}}$. Two cases must be considered:

    • (i) Let $a \in \mathbf {A} \cap Acc({\mathfrak {A}}_{{\mathcal {E}}})$. Assume that $a \not \in {\mathcal {E}}_{a}$. As ${\mathcal {E}}$ is a complete extension of MAF, a is not acceptable w.r.t. ${\mathcal {E}}$. So there exists an attack (β, a) in $\mathbf {R}^{\prime }$ such that $\mathcal {E}$ does not attack β. That implies that $N_{\mathbf {s}(\beta )\beta } \not \in \mathcal {E}$, and from Lemma 36 that $\mathcal {E}$ does not attack s(β). So $\mathcal {E}$ attacks neither β, nor s(β); this fact will be denoted by (*).

      Moreover, as $a \in Acc({\mathfrak {A}}_{\mathcal {E}})$, either $\beta \in Inh({\mathfrak {A}}_{\mathcal {E}})$ (Case 1), or $\mathbf {s}(\beta ) \in Def({\mathfrak {A}}_{\mathcal {E}})$ (Case 2). It follows that there is $\gamma \in {\Gamma }_{\mathcal {E}}$ with $\mathbf {s}(\gamma ) \in \mathcal {E}_{a}$ such that t(γ) = β in Case 1 (resp. t(γ) = s(β) in Case 2). These conditions on γ and the definition of ${\Gamma }_{{\mathcal {E}}}$ imply that γ must belong to ${\mathcal {E}}_{k}$. So we have that ${\mathcal {E}}$ attacks β in Case 1 (resp. s(β) in Case 2). Hence we obtain a contradiction with the fact (*) and consequently we have proved that $a \in {\mathcal {E}}_{a}$.

    • (ii) Let $\alpha \in \mathbf {K} \cap Acc({\mathfrak {A}}_{{\mathcal {E}}})$. Assume that $\alpha \not \in {\Gamma }_{{\mathcal {E}}}$. It follows that $\alpha \not \in {\mathcal {E}}_{k}$ and either $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$ or $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}})$. Let us successively consider the two cases.

      • (A) Assume that $\alpha \not \in {\mathcal {E}}_{k}$ and $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$. As ${\mathcal {E}}$ is a complete extension of MAF, α is not acceptable w.r.t. ${\mathcal {E}}$. As $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$, α is defended by ${\mathcal {E}}$ against its attacker N_s(α)α. So there must exist another attacker of α, say β, such that ${\mathcal {E}}$ does not attack β. That implies that $N_{\mathbf {s}(\beta )\beta } \not \in {\mathcal {E}}$, and from Lemma 36 that ${\mathcal {E}}$ does not attack s(β). So ${\mathcal {E}}$ attacks neither β, nor s(β); this fact will be denoted by (*).

        Moreover, as $\alpha \in Acc({\mathfrak {A}}_{{\mathcal {E}}})$, either $\beta \in Inh({\mathfrak {A}}_{{\mathcal {E}}})$ (Case 1), or $\mathbf {s}(\beta ) \in Def({\mathfrak {A}}_{{\mathcal {E}}})$ (Case 2). It follows that there is $\gamma \in {\Gamma }_{{\mathcal {E}}}$ with $\mathbf {s}(\gamma ) \in {\mathcal {E}}_{a}$ such that t(γ) = β in Case 1 (resp. t(γ) = s(β) in Case 2). These conditions on γ and the definition of ${\Gamma }_{{\mathcal {E}}}$ imply that γ must belong to ${\mathcal {E}}_{k}$. So we have that ${\mathcal {E}}$ attacks β in Case 1 (resp. s(β) in Case 2). Hence we obtain a contradiction with the fact (*).

      • (B) It remains to consider the case when $\alpha \not \in {\mathcal {E}}_{k}$, $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$ and $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. Let us recall that ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$ is the structure $\langle \mathcal {E}_{a},\mathcal {E}_{k}\rangle$. $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}})$ implies that α is attacked in RAF by β ∈K such that $\beta \not \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ and $\mathbf {s}(\beta ) \not \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$; this fact will be denoted by (**).

        Moreover, as $\alpha \in Acc({\mathfrak {A}}_{{\mathcal {E}}})$, either $\beta \in Inh({\mathfrak {A}}_{{\mathcal {E}}})$ (Case 1), or $\mathbf {s}(\beta ) \in Def({\mathfrak {A}}_{{\mathcal {E}}})$ (Case 2). It follows that there is $\gamma \in {\Gamma }_{\mathcal {E}}$ with $\mathbf {s}(\gamma ) \in {\mathcal {E}}_{a}$ such that t(γ) = β in Case 1 (resp. t(γ) = s(β) in Case 2). These conditions on γ and the definition of ${\Gamma }_{{\mathcal {E}}}$ imply that γ must belong to ${\mathcal {E}}_{k}$. So we have that $\beta \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ in Case 1 (resp. $\mathbf {s}(\beta ) \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ in Case 2). Hence we obtain a contradiction with with the fact (**).

      So we have proved that α must belong to ${\Gamma }_{{\mathcal {E}}}$.

  • (b) (σ = stable) Let ${\mathcal {E}}$ be a stable extension of ${\langle \mathbf {A}^{\prime }, \mathbf {R}^{\prime }\rangle }$. First, from Proposition 13, we have that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a conflict-free structure in RAF. Then, we have to prove that $\mathbf {A} \setminus \mathcal {E}_{a} \subseteq Def({\mathfrak {A}}_{\mathcal {E}})$ and $\mathbf {K} \setminus {\Gamma }_{\mathcal {E}} \subseteq Inh({\mathfrak {A}}_{\mathcal {E}})$.

    • (i) Let $x \in \mathbf {A} \setminus {\mathcal {E}}_{a}$. As ${\mathcal {E}}$ is stable, ${\mathcal {E}}$ attacks x. As x ∈A all the attackers of x belong to K. So there is $\alpha \in {\mathcal {E}}_{k}$ that attacks x. Note that $\alpha \in {\Gamma }_{\mathcal {E}}$ by definition of ${\Gamma }_{\mathcal {E}}$.

      Moreover, as $\mathcal {E}$ is a stable extension of MAF, $\mathcal {E}$ is admissible. As α is attacked by N_s(α)α, $\mathcal {E}$ contains the only attacker of N_s(α)α, that is s(α). So $\mathbf {s}(\alpha )\in {\mathcal {E}}_{a}$. By definition, $\alpha \in {\mathcal {E}}_{r}$ and $\mathbf {s}(\alpha )\in {\mathcal {E}}_{a}$ imply that $x \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. As ${\Gamma }_{{\mathcal {E}}}$ contains ${\mathcal {E}}_{k}$ it follows that we also have $x \in Def({\mathfrak {A}}_{{\mathcal {E}}})$.

    • (ii) Let $\alpha \in \mathbf {K} \setminus {\Gamma }_{{\mathcal {E}}}$. It follows that $\alpha \not \in {\mathcal {E}}_{k}$ and either $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$ or $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}})$. Let us successively consider the two cases.

      • (A) Assume that $\alpha \not \in {\mathcal {E}}_{k}$ and $\mathbf {s}(\alpha ) \in {\mathcal {E}}_{a}$. As ${\mathcal {E}}$ is stable, ${\mathcal {E}}$ attacks α. As ${\mathcal {E}}$ is conflict-free and contains s(α), it follows that $N_{\mathbf {s}(\alpha )\alpha } \not \in {\mathcal {E}}$. So there exists $\beta \in {\mathcal {E}}_{k}$ that attacks α.

        Moreover, as ${\mathcal {E}}$ is a stable extension of MAF, ${\mathcal {E}}$ is admissible. As β is attacked by N_s(β)β, ${\mathcal {E}}$ contains the only attacker of N_s(β)β, that is s(β). So $\mathbf {s}(\beta )\in {\mathcal {E}}_{a}$. By definition, $\beta \in {\mathcal {E}}_{k}$ and $\mathbf {s}(\beta )\in {\mathcal {E}}_{a}$ imply that $\alpha \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. As ${\Gamma }_{{\mathcal {E}}}$ contains ${\mathcal {E}}_{k}$ it follows that we also have $\alpha \in Inh({\mathfrak {A}}_{{\mathcal {E}}})$.

      • (B) It remains to consider the case when $\alpha \not \in {\mathcal {E}}_{k}$, $\mathbf {s}(\alpha ) \not \in {\mathcal {E}}_{a}$ and $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. Let us recall that ${\mathfrak {A}}^{\prime }_{\mathcal {E}}$ is the structure $\langle \mathcal {E}_{a},\mathcal {E}_{k}\rangle$. $\alpha \not \in Acc({\mathfrak {A}}^{\prime }_{\mathcal {E}})$ implies that α is attacked in RAF by β ∈K such that $\beta \not \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$ and $\mathbf {s}(\beta ) \not \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$; this fact will be denoted by (*).

        If $\mathbf {s}(\beta ) \not \in {\mathcal {E}}_{a}$, from the first part of this proof, it follows that $\mathbf {s}(\beta ) \in Def({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. That is in contradiction with the fact (*). So we have $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$.

        If $\beta \not \in {\Gamma }_{{\mathcal {E}}}$, as $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$, from the first item of the second part of this proof, it follows that $\beta \in Inh({\mathfrak {A}}^{\prime }_{{\mathcal {E}}})$. That is in contradiction with the fact (*). So we have $\beta \in {\Gamma }_{{\mathcal {E}}}$.

        By definition, $\mathbf {s}(\beta ) \in {\mathcal {E}}_{a}$, $\beta \in {\Gamma }_{\mathcal {E}}$ and β attacks α imply that $\alpha \in Inh({\mathfrak {A}}_{{\mathcal {E}}})$.

      In both cases, we have proved that $\alpha \in Inh({\mathfrak {A}}_{{\mathcal {E}}})$.

  • (c) (σ = preferred) Let ${\mathcal {E}}$ be a preferred extension of MAF. By definition, ${\mathcal {E}}$ is a $\subseteq$-maximal admissible extension. Moreover ${\mathcal {E}}$ is a complete extension. So, from Proposition 12 (item 2 (a)), ${\mathfrak {A}}_{\mathcal {E}}$ is a complete structure of RAF.

    Assume that ${\mathfrak {A}}_{\mathcal {E}}$ is not a preferred structure of RAF. Then there exists ${\mathfrak {A}}^{\prime }$ an admissible structure that strictly contains ${\mathfrak {A}}_{\mathcal {E}}$. It can be assumed that ${\mathfrak {A}}^{\prime }$ is a $\subseteq$-maximal admissible structure of RAF. So ${\mathfrak {A}}^{\prime }$ is preferred and thus complete.

    From Lemma 41, it follows that $\texttt {Maf}{{\mathfrak {A}}_{\mathcal {E}}} \subseteq \texttt {Maf}{{\mathfrak {A}}^{\prime }}$. From Proposition 15, we have $\texttt {Maf}{{\mathfrak {A}}_{\mathcal {E}}} = \mathcal {E}$. So, ${\mathcal {E}} \subseteq \texttt {Maf}({\mathfrak {A}}^{\prime })$. As ${\mathcal {E}}$ is preferred, it follows that ${\mathcal {E}} = \texttt {Maf}({\mathfrak {A}}^{\prime })$.

    From Proposition 15 again, ${\mathfrak {A}}_{\texttt {Maf}{{\mathfrak {A}}^{\prime }}} = {\mathfrak {A}}^{\prime }$ so ${\mathfrak {A}}_{{\mathcal {E}}} = {\mathfrak {A}}^{\prime }$.

    That is in contradiction with the assumption that ${\mathfrak {A}}^{\prime }$ strictly contains ${\mathfrak {A}}_{\mathcal {E}}$. Hence, we have proved that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a preferred structure.

  • (d) (σ = grounded) Let ${\mathcal {E}}$ be the grounded extension of MAF. By definition, ${\mathcal {E}}$ is the $\subseteq$-minimal complete extension. From Proposition 12 (item 2 (a)), ${\mathfrak {A}}_{\mathcal {E}}$ is a complete structure of RAF. Assume that there is ${\mathfrak {A}}^{\prime }$ a complete structure that is strictly included in ${\mathfrak {A}}_{{\mathcal {E}}}$. From Lemma 41, we have $\texttt {Maf}({\mathfrak {A}}^{\prime }) \subseteq \texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}})$ As $\texttt {Maf}({\mathfrak {A}}_{{\mathcal {E}}}) = {\mathcal {E}}$, due to Proposition 15, we have $\texttt {Maf}({\mathfrak {A}}^{\prime }) \subseteq {\mathcal {E}}$.

    From Proposition 12 (item 1 (a)), $\texttt {Maf}({\mathfrak {A}}^{\prime })$ is a complete extension, so by assumption on ${\mathcal {E}}$ it follows that $\texttt {Maf}({\mathfrak {A}}^{\prime }) = {\mathcal {E}}$. Hence, ${\mathfrak {A}}_{\texttt {Maf}({\mathfrak {A}}^{\prime })} = {\mathfrak {A}}_{{\mathcal {E}}}$, and from Proposition 15 again, ${\mathfrak {A}}^{\prime } = {\mathfrak {A}}_{{\mathcal {E}}}$. That is in contradiction with the fact that ${\mathfrak {A}}^{\prime }$ is strictly included in ${\mathfrak {A}}_{{\mathcal {E}}}$. So we have proved that ${\mathfrak {A}}_{{\mathcal {E}}}$ is a $\subseteq$-minimal complete structure of RAF, or in other words is the grounded extension of RAF.

□

Proof of Proposition 15

• 1. The proof follows directly from Lemma 38.

• 2. The proof follows directly from Lemma 39.

□

2.4 A.4 Proofs of Section 7

Note

By $\downarrow \!\!\alpha = \{\ \beta \in \mathbf {K} \ \left |\ \beta \preceq \alpha \right .\}$ we denote the down set generated by α. Furthermore, for some argumentation framework RAF and structure ${\mathfrak {A}}$, by ${Def}{\mathbf {RAF}}{{\mathfrak {A}}}$ and ${Inh}{\mathbf {RAF}}{{\mathfrak {A}}}$ we respectively denote the defeated arguments and inhibited attacks w.r.t. RAF and ${\mathfrak {A}}$. This allows us to relate defeated arguments (resp. inhibited attacks) w.r.t. different argumentation frameworks.

Lemma 42

Let RAF = 〈A, K, s, t〉 be some framework and α, β ∈K be be two attacks and x ∈ (A ∪K) be some argument or attack. Then, α≠β, t(β) = x and x∉ ↓ α imply β∉ ↓ α.

Proof

Suppose, for the sake of contradiction, that β ∈↓ α. Then, since α≠β, it follows that there is some chain δ₀,δ₁,δ₂,δ_n such that t(δ_i) = δ_i+ 1 and δ₀ = β and δ_n = α. But δ₀ = β plus t(β) = x imply that δ₁ = x and, thus, that x ∈↓ α. This is a contradiction with the assumption. Consequently, it must be that β∉ ↓ α. □

Lemma 43

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some structure and α ∈K be some attack. Then, ${Def}(\mathbf {RAF},{\mathfrak {A}}) \supseteq {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick any $a \in {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. Then, there is some β ∈Γ^−α such that t(β) = a and s(β) ∈ S. Furthermore, β ∈Γ^−α plus ${{\varGamma }}^{-\alpha } \subseteq {\varGamma }$ imply β ∈Γ which, in its turn, implies that $a \in {Def}(\mathbf {RAF},{\mathfrak {A}})$. □

Lemma 44

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some structure and α ∈K be some attack. Then, ${Inh}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha } \supseteq {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick any $\beta \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. Then, there is some γ ∈Γ^−α such that t(γ) = β and s(γ) ∈ S. Furthermore, γ ∈Γ^−α plus ${{\varGamma }}^{-\alpha } \subseteq {\varGamma }$ imply γ ∈Γ which, in its turn, implies that $\beta \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$. Furthermore, $\beta \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ implies β ≰ α. Hence, $\beta \in {Inh}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }$. □

Lemma 45

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some conflict-free structure w.r.t. RAF and α ∈K be some attack. Then, ${\mathfrak {A}}^{-\alpha }$ is conflict-free w.r.t. RAF^−α.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick a ∈ S. Then, since ${\mathfrak {A}}$ is conflict-free, it follows that $a \notin {Def}(\mathbf {RAF},{\mathfrak {A}})$ and, from Lemma 43, that $a \notin {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. Similarly, β ∈Γ^−α implies $\beta \notin {Inh}(\mathbf {RAF},{\mathfrak {A}})$. From Lemma 44, this implies $\beta \notin {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. Hence, ${\mathfrak {A}}^{-\alpha }$ is conflict-free w.r.t. RAF^−α. □

Lemma 46

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some structure and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack.

Then, it follows that ${Acc}{\mathbf {RAF}}{{\mathfrak {A}}}^{-\alpha } \supseteq {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick any ${x} \in {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ and γ ∈K such that t(γ) = x. Then, ${x} \in {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ implies x ∈ (A ∪K^−α) and, thus, that x∉ ↓ α. From Lemma 42, this plus t(γ) = x imply that either γ = α or γ∉ ↓ α. On the one hand, by hypothesis we have that $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ and, thus, the former implies $\gamma \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$. On the other hand, the latter implies γ ∈K^−α and, thus, ${x} \in {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ implies that $\gamma \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ or $\mathbf {s}(\gamma ) \in {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. From Lemmas 43 and 44, this implies that $\gamma \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ or $\mathbf {s}(\gamma ) \in {Def}(\mathbf {RAF},{\mathfrak {A}})$. Hence, ${x} \in {Acc}(\mathbf {RAF},{\mathfrak {A}})$. Finally, we have that x∉ ↓ α implies ${x} \in {Acc}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }$. □

Lemma 47

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some structure and $\alpha \in ({Inh}(\mathbf {RAF},{\mathfrak {A}})\setminus {\varGamma })$ be some inhibited attack w.r.t. ${\mathfrak {A}}$. Then, it follows that ${Def}(\mathbf {RAF},{\mathfrak {A}}) = {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. From Lemma 43, it follows that ${Def}(\mathbf {RAF},{\mathfrak {A}}) \supseteq {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. Pick now any argument $a \in {Def}(\mathbf {RAF},{\mathfrak {A}})$. Then, there is some β ∈Γ such that t(β) = a and s(β) ∈ S. Furthermore, since α∉Γ, it follows that β≠α. Moreover, every γ ∈ ↓ α satisfies either γ = α or t(γ) ∈K and, thus, that β∉ ↓ α. This implies that β ∈Γ^−α and, thus, that $a \in {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$□

Lemma 48

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some structure and $\alpha \in ({Inh}(\mathbf {RAF},{\mathfrak {A}})\setminus {\varGamma })$ be some inhibited attack w.r.t. ${\mathfrak {A}}$. Then, it follows that ${Inh}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha } = {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

First note that, from Lemma 44, it follows that

$$ {Inh}(\mathbf{RAF},{\mathfrak{A}})^{-\alpha} \supseteq {Inh}(\mathbf{RAF}^{-\alpha},{\mathfrak{A}}^{-\alpha}) $$

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick now any $\beta \in {Inh}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }$.

Then, there is some γ ∈Γ such that t(γ) = β and s(γ) ∈ S. Furthermore, since α∉Γ, it follows that γ≠α. Moreover, $\beta \in {Inh}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }$ implies that β ≰ α. Then, since γ ≺ β, that γ ≰ α. Hence, it follows that γ ∈Γ^−α and $\beta \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. □

Lemma 49

Let RAF = 〈A, K, s, t〉 be some framework, ${\mathfrak {A}}$ be some structure and $\alpha \in ({Inh}(\mathbf {RAF},{\mathfrak {A}})\setminus {\varGamma })$ be some inhibited attack w.r.t. ${\mathfrak {A}}$.

Then, it follows that $({Acc}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }) = {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$.

Proof

First note that, from Lemma 46, it follows that

$$ {Acc}(\mathbf{RAF},{\mathfrak{A}})^{-\alpha} \supseteq {Acc}(\mathbf{RAF}^{-\alpha},{\mathfrak{A}}^{-\alpha}) $$

Let $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ and ${\mathfrak {A}}^{-\alpha } = {\langle S,{{\varGamma }}^{-\alpha } \rangle }$. Pick any $\beta \in {Acc}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }$ and $\gamma \in \mathbf {K}^{-\alpha } \subseteq \mathbf {K}$ such that t(γ) = β. Since $\beta \in {Acc}(\mathbf {RAF},{\mathfrak {A}})$, it follows that either $\gamma \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ or $\mathbf {s}(\gamma ) \in {Def}(\mathbf {RAF},{\mathfrak {A}})$. Furthermore, γ ∈K^−α implies γ ≰ α and, from Lemmas 47 and 48, this implies that either $\gamma \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{\prime })$ or $\mathbf {s}(\gamma ) \in {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{\prime })$. Hence, $\beta \in {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{\prime })$ and, thus, ${Acc}(\mathbf {RAF},{\mathfrak {A}})^{-\alpha }) = {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. □

Lemma 50

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some admissible structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, it follows that ${\mathfrak {A}}^{-\alpha }$ is admissible w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is an admissible structure w.r.t. RAF, it is conflict-free and, from Lemma 45, this implies that ${\mathfrak {A}}^{-\alpha }$ is conflict-free w.r.t. RAF^−α. Furthermore, since ${\mathfrak {A}}$ is admissible, it follows that $(S \cup {\varGamma }) \subseteq {Acc}(\mathbf {RAF},{\mathfrak {A}})$. From Lemma 49, this implies $(S \cup {{\varGamma }}^{-\alpha }) \subseteq {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ and, thus, that ${\mathfrak {A}}^{-\alpha }$ is admissible w.r.t. RAF^−α. □

Lemma 51

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some complete structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, it follows that ${\mathfrak {A}}^{-\alpha }$ is complete w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is a complete structure w.r.t. RAF, it follows that it is admissible and, from Lemma 50, this implies that ${\mathfrak {A}}^{-\alpha }$ is admissible w.r.t. RAF^−α. Furthermore, since ${\mathfrak {A}}$ is complete, it follows that $(S \cup {\varGamma }) \supseteq {Acc}(\mathbf {RAF},{\mathfrak {A}})$. From Lemma 49, this implies $(S \cup {{\varGamma }}^{-\alpha }) \supseteq {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ and, thus, that ${\mathfrak {A}}^{-\alpha }$ is complete w.r.t. RAF^−α. □

Lemma 52

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some conflict-free structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Let ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$ some conflict-free structure w.r.t. RAF^−α such that ${{\varGamma }}^{-\alpha } \subseteq {{\varGamma }}^{\prime }$. Then, the structure ${\mathfrak {A}}^{\prime \prime } = {\langle S^{\prime }, {\varGamma } \cup {{\varGamma }}^{\prime } \rangle }$ is conflict-free w.r.t. RAF.

Proof

Let ${{\varGamma }}^{\prime \prime } = {\varGamma } \cup {{\varGamma }}^{\prime }$. Pick first $a \in S^{\prime }$ and any $\beta \in {{\varGamma }}^{\prime \prime }$ such that t(β) = a. Then, since t(β) = a, it follows that β ≰ α and, thus, that β ∈K^−α. Hence, β ∈Γ implies $\beta \in {{\varGamma }}^{-\alpha } \subseteq {{\varGamma }}^{\prime }$. Then, since ${\mathfrak {A}}^{\prime }$ is conflict-free, it follows that $\mathbf {s}(\beta ) \notin S^{\prime }$. This implies that every $\beta \in {{\varGamma }}^{\prime \prime }$ with t(β) = a satisfies $\mathbf {s}(\beta ) \notin S^{\prime }$ and, thus, that $a \notin {Def}(\mathbf {RAF},{\mathfrak {A}}^{\prime \prime })$.

Pick now $\gamma ,\beta \in {{\varGamma }}^{\prime \prime }$ such that t(β) = γ. Suppose, for the sake of contradiction, that $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$ and $\beta \in ({{\varGamma }}^{\prime \prime }\setminus {{\varGamma }}^{\prime })$. Then, $\gamma \notin ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$ implies $\gamma \in {{\varGamma }}^{\prime } \subseteq \mathbf {K}^{-\alpha }$ and, thus, it follows that γ ≰ α and β ≰ α. On the other hand, $\beta \in ({{\varGamma }}^{\prime \prime }\setminus {{\varGamma }}^{\prime })$ implies β ∈Γ which, since β ≰ α, implies $\beta \in {{\varGamma }}^{-\alpha } \subseteq {{\varGamma }}^{\prime }$. This is a contradiction with the assumption. Similarly, suppose that $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {{\varGamma }}^{\prime })$ and $\beta \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$. Then, $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {{\varGamma }}^{\prime })$ implies and γ ∈Γ. Furthermore, since ${{\varGamma }}^{-\alpha } \subseteq {{\varGamma }}^{\prime }$, it follows that $\gamma \notin {{\varGamma }}^{\prime }$ implies γ∉Γ^−α and, this plus γ ∈Γ, imply γ ≼ α. Since t(β) = γ, the latter implies that β ≼ α holds and, thus, that $\beta \notin {{\varGamma }}^{\prime }$. This is a contradiction with the assumption that $\beta \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$. Hence, either γ, β ∈Γ or $\gamma ,\beta \in {{\varGamma }}^{\prime }$ must hold. In both cases, the fact that ${\mathfrak {A}}$ and ${\mathfrak {A}}^{\prime }$ are conflict-free imply s(β)∉S. This implies that every $\beta \in {{\varGamma }}^{\prime \prime }$ with t(β) = γ satisfies s(β)∉S and, thus, that $\gamma \notin {Inh}(\mathbf {RAF},{\mathfrak {A}}^{\prime \prime })$. Consequently, ${\mathfrak {A}}^{\prime \prime }$ is conflict-free. □

Lemma 53

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some naive structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, ${\mathfrak {A}}^{-\alpha }$ is naive w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is a naive structure w.r.t. RAF, it follows that it is conflict-free and, from Proposition 18, this implies that ${\mathfrak {A}}^{-\alpha }$ is conflict-free w.r.t. RAF^−α. Suppose, for the sake of contradiction, that there exists some structure ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$ which is conflict-free w.r.t. RAF^−α and that satisfies ${\mathfrak {A}}^{-\alpha } \sqsubset {\mathfrak {A}}^{\prime }$. Then, we may assume without loss of generality that ${\mathfrak {A}}^{\prime }$ is also naive w.r.t. RAF^−α.

Let ${{\varGamma }}^{\prime \prime } = ({\varGamma } \cup {{\varGamma }}^{\prime })$ and let ${\mathfrak {A}}^{\prime \prime } = {\langle S^{\prime },{{\varGamma }}^{\prime \prime } \rangle }$ be some structure. Then, from Lemma 52, it follows that ${\mathfrak {A}}^{\prime \prime }$ is conflict-free. Furthermore, note that by construction, ${\mathfrak {A}},{\mathfrak {A}}^{\prime } \sqsubseteq {\mathfrak {A}}^{\prime \prime }$ holds and, thus, the fact that ${\mathfrak {A}}$ is a naive structure implies ${\mathfrak {A}} \sqsupseteq {\mathfrak {A}}^{\prime \prime }$. On the other hand, ${\mathfrak {A}}^{-\alpha } \sqsubset {\mathfrak {A}}^{\prime }$ implies that there is some element $x \in ((S^{\prime } \cup {{\varGamma }}^{\prime }) \setminus (S^{-\alpha } \cup {{\varGamma }}^{-\alpha })$. Moreover, $x \in (S^{\prime } \cup {{\varGamma }}^{\prime })$ implies that x ≰ α and that $x \in (S^{\prime } \cup {{\varGamma }}^{\prime \prime })$. Since ${\mathfrak {A}} \sqsupseteq {\mathfrak {A}}^{\prime \prime }$, the latter implies that x ∈ (S ∪Γ) which, together with x ≰ α, implies that x ∈ (S^−α ∪Γ^−α). This is a contradiction and, consequently, we have that ${\mathfrak {A}}^{-\alpha }$ must be naive. □

Lemma 54

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some grounded structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, ${\mathfrak {A}}^{-\alpha }$ is grounded w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is a grounded structure w.r.t. RAF, it follows that it is complete and, from Lemma 51, this implies that ${\mathfrak {A}}^{-\alpha }$ is complete w.r.t. RAF^−α. Suppose, for the sake of contradiction, that there is some structure ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$ which is complete w.r.t. RAF^−α and that satisfies ${\mathfrak {A}}^{-\alpha } \sqsupset {\mathfrak {A}}^{\prime }$. Then, we also have ${\mathfrak {A}} \sqsupset {\mathfrak {A}}^{\prime }$. So ${\mathfrak {A}}^{\prime }$ must not be complete w.r.t. RAF. Obviously, since ${\mathfrak {A}}$ is conflict-free and ${\mathfrak {A}} \sqsupseteq {\mathfrak {A}}^{\prime }$, we have that ${\mathfrak {A}}^{\prime }$ is also conflict-free. Suppose, for the sake of contradiction, that ${\mathfrak {A}}^{\prime }$ is not admissible w.r.t. RAF. Then, there is $x \in (S^{\prime } \cup {{\varGamma }}^{\prime })$ and β ∈K such that t(β) = x and there is no $\gamma \in {\varGamma } \supseteq {{\varGamma }}^{\prime }$ such that t(γ) ∈{β, s(β)} and $\mathbf {s}(\gamma ) \in S \supseteq S^{\prime }$. Since ${\mathfrak {A}}$ is admissible w.r.t. RAF^−α, it must be that β∉K^−α and, thus, that β ≼ α. However, this is a contradiction with the fact that ${\mathfrak {A}}^{-\alpha } \sqsupset {\mathfrak {A}}^{\prime }$. Hence, ${\mathfrak {A}}^{\prime }$ must be admissible. Suppose now that ${\mathfrak {A}}^{\prime }$ is not complete w.r.t. RAF. Then, there is some $x \in {Acc}(\mathbf {RAF},{\mathfrak {A}}^{\prime })$ such that $x \notin (S^{\prime } \cup {{\varGamma }}^{\prime })$. Hence, for every β ∈K such that t(β) = x and there is $\gamma _{\beta } \in {{\varGamma }}^{\prime }$ such that t(γ_β) ∈{β, s(β)} and $\mathbf {s}(\gamma _{\beta }) \in S^{\prime }$. On the other hand, since ${\mathfrak {A}}^{\prime }$ is complete w.r.t. RAF^−α, this means that $x \in {Acc}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{\prime })$ and, thus, that γ_β ≼ α. This is a contradiction with the fact that $\gamma _{\beta } \in {{\varGamma }}^{\prime } \subseteq {{\varGamma }}^{-\alpha }$. Hence, ${\mathfrak {A}}^{\prime }$ is complete w.r.t. RAF which is a contradiction with the fact that ${\mathfrak {A}}$ is grounded. Thus, ${\mathfrak {A}}^{-\alpha }$ must be grounded. □

Lemma 55

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some admissible structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Let ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$ some admissible structure w.r.t. RAF^−α such that ${\mathfrak {A}}^{-\alpha } \sqsubseteq {\mathfrak {A}}^{\prime }$. Then, the structure ${\mathfrak {A}}^{\prime \prime } = {\langle S^{\prime }, {\varGamma } \cup {{\varGamma }}^{\prime } \rangle }$ is admissible w.r.t. RAF.

Proof

From Lemma 52, it follows that ${\mathfrak {A}}^{\prime \prime }$ is conflict-free. Furthermore, since ${\mathfrak {A}}$ is admissible, it follows that $(S \cup {\varGamma }) \subseteq {Acc}(\mathbf {RAF},{\mathfrak {A}})$. Note also that ${\mathfrak {A}}^{-\alpha } \sqsubseteq {\mathfrak {A}}^{\prime }$ and α ∈K implies that ${\varGamma } = {{\varGamma }}^{-\alpha } \subseteq {{\varGamma }}^{\prime }$ and, thus, that ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime \prime }$. Then, from Observation 2, it follows that $(S \cup {\varGamma }) \subseteq {Acc}(\mathbf {RAF},{\mathfrak {A}}^{\prime \prime })$. Pick now any attack $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$ and any attack β ∈K such that t(β) = γ. Since $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$, it follows that $\gamma \in {{\varGamma }}^{\prime }\subseteq \mathbf {K}^{-\alpha }$ and, thus, that γ ≰ α and β ≰ α. Hence, β ∈K implies β ∈K^−α. Furthermore, since ${\mathfrak {A}}^{\prime }$ is admissible w.r.t. RAF^−α, this implies that either $\mathbf {s}(\beta ) \in {Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$ or $\beta \in {Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })$. From Lemmas 47 and 48 respectively, this implies that either

$$ \mathbf{s}(\beta) \in {Def}(\mathbf{RAF},{\mathfrak{A}})^{-\alpha} \subseteq {Def}(\mathbf{RAF},{\mathfrak{A}}) $$

or

$$ \beta \in {Inh}(\mathbf{RAF},{\mathfrak{A}})^{-\alpha} \subseteq {Inh}(\mathbf{RAF},{\mathfrak{A}}) $$

holds.

Furthermore, since ${\mathfrak {A}} \sqsubseteq {\mathfrak {A}}^{\prime \prime }$, Observation 1 implies either $\mathbf {s}(\beta ) \in {Def}(\mathbf {RAF},{\mathfrak {A}}^{\prime \prime })$ or $\beta \in {Inh}{\mathbf {RAF}}{{\mathfrak {A}}^{\prime \prime }}$. Hence, every $\gamma \in ({{\varGamma }}^{\prime \prime }\setminus {\varGamma })$ satisfies that $\gamma \in {Acc}{\mathbf {RAF}}{{\mathfrak {A}}^{\prime \prime }}$ and, thus, $(S \cup {{\varGamma }}^{\prime \prime }) \subseteq {Acc}{\mathbf {RAF}}{{\mathfrak {A}}^{\prime \prime }}$. This means that $\mathfrak {A}^{\prime \prime }$ is admissible. □

Lemma 56

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some preferred structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, ${\mathfrak {A}}^{-\alpha }$ is preferred w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is a preferred structure w.r.t. RAF, it follows that it is admissible and, from Lemma 50, this implies that ${\mathfrak {A}}^{-\alpha }$ is admissible w.r.t. RAF^−α. Suppose, for the sake of contradiction, that there is some structure ${\mathfrak {A}}^{\prime } = {\langle S^{\prime },{{\varGamma }}^{\prime } \rangle }$ which is admissible w.r.t. RAF^−α and that satisfies ${\mathfrak {A}}^{-\alpha } \sqsubset {\mathfrak {A}}^{\prime }$. Then, from Proposition 1, we may assume without loss of generality that ${\mathfrak {A}}^{\prime }$ is also preferred w.r.t. RAF^−α.

Let ${{\varGamma }}^{\prime \prime } = ({\varGamma } \cup {{\varGamma }}^{\prime })$ and let ${\mathfrak {A}}^{\prime \prime } = {\langle S^{\prime },{{\varGamma }}^{\prime \prime } \rangle }$ be some structure. Then, from Lemma 55, it follows that ${\mathfrak {A}}^{\prime \prime }$ is admissible. Furthermore, note that by construction, ${\mathfrak {A}},{\mathfrak {A}}^{\prime } \sqsubseteq {\mathfrak {A}}^{\prime \prime }$ holds and, thus, the fact that ${\mathfrak {A}}$ is a preferred structure implies ${\mathfrak {A}} \sqsupseteq {\mathfrak {A}}^{\prime \prime }$. On the other hand, ${\mathfrak {A}}^{-\alpha } \sqsubset {\mathfrak {A}}^{\prime }$ implies that there is some element $x \in ((S^{\prime } \cup {{\varGamma }}^{\prime }) \setminus (S^{-\alpha } \cup {{\varGamma }}^{-\alpha })$. Moreover, $x \in (S^{\prime } \cup {{\varGamma }}^{\prime })$ implies that x ≰ α and that $x \in (S^{\prime } \cup {{\varGamma }}^{\prime \prime })$. Since ${\mathfrak {A}} \sqsupseteq {\mathfrak {A}}^{\prime \prime }$, the latter implies that x ∈ 7(S ∪Γ) which, together with x ≰ α, implies that x ∈ (S^−α ∪Γ^−α). This is a contradiction and, consequently, we have that ${\mathfrak {A}}^{-\alpha }$ must be preferred. □

Lemma 57

Let RAF = 〈A, K, s, t〉 be some framework, $\mathfrak {A} = \langle {S,{\varGamma }}\rangle$ be some stable structure w.r.t. RAF and $\alpha \in {Inh}(\mathbf {RAF},{\mathfrak {A}})$ be some inhibited attack. Then, it follows that ${\mathfrak {A}}^{-\alpha }$ is stable w.r.t. RAF^−α.

Proof

Since ${\mathfrak {A}}$ is a stable structure w.r.t. RAF, it follows that it is conflict-free and, from Lemma 45, this implies that ${\mathfrak {A}}^{-\alpha }$ is conflict-free w.r.t. RAF^−α. Furthermore, since ${\mathfrak {A}}$ is stable, it follows that $S = \overline {{Def}{\mathbf {RAF}}{{\mathfrak {A}}}}$ and ${\varGamma } = \overline {{Inh}(\mathbf {RAF},{\mathfrak {A}})}$. From Lemmas 47 and 48 respectively, this implies that $S = \overline {{Def}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })}$ and ${{\varGamma }}^{-\alpha } = \overline {{Inh}(\mathbf {RAF}^{-\alpha },{\mathfrak {A}}^{-\alpha })}$. This implies that ${\mathfrak {A}}^{-\alpha }$ is stable w.r.t. RAF^−α. □

Proof of Proposition 19

The fact that ${\mathfrak {A}}^{-\alpha }$ is complete, naive, grounded, preferred or stable respectively follows from Lemmas 51, 53, 54, 56 and 57. □