Abstract
This paper presents a novel abstract argumentation framework, called MultiAttack Argumentation Framework (MAAF), which supports different types of attacks. The introduction of types gives rise to a new family of nonstandard semantics which can support applications that classical approaches cannot, while also allowing classical semantics as a special case. The main novelty of the proposed semantics is the discrimination among two different roles that attacks play, namely an attack as a generator of conflicts, and an attack as a means to defend an argument. These two roles have traditionally been considered together in the argumentation literature. Allowing some attack types to serve one of those roles only, gives rise to the different semantics presented here.
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 1.
The proofs of all results appear in the Appendix.
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Acknowledgements
This project has received funding from the Hellenic Foundation for Research and Innovation (HFRI) and the General Secretariat for Research and Technology (GSRT), under grant agreement No 188.
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A Appendix
A Appendix
Proof of Proposition 1
For the first result, \(\mathcal {E}\cup \{a\}\) \(\mathbf {\theta }\)defends \(\mathcal {E}\cup \{a\}\), since, by our assumptions, \(\mathcal {E}\) \(\mathbf {\theta }\)defends \(\mathcal {E}\), and \(\mathcal {E}\) \(\mathbf {\theta }\)defends a. So it suffices to show that, \(\mathcal {E}\cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\).
Let us consider the case of firm semantics first. Suppose that \(\mathcal {E}\cup \{a\}\) is not \(\mathbf {fr}\)\(\mathbf {cf}\). Then, there exist \(a_1,a_2 \in \mathcal {E}\cup \{a\}\) such that \(a_1 \rightarrow a_2\). We consider four cases, all of which lead to a contradiction, thus proving the point:

1.
If \(a_1, a_2 \in \mathcal {E}\), then \(\mathcal {E}\) is not \(\mathbf {fr}\)\(\mathbf {cf}\), a contradiction.

2.
If \(a_1 \in \mathcal {E}, a_2 = a\), then, since \(\mathcal {E}\) \(\mathbf {fr}\)defends a, it follows that there exists some \(a_3 \in \mathcal {E}\) such that \(a_3 \rightarrow _{{\mathcal {T}_{0}}}a_1\), a contradiction by case #1.

3.
If \(a_1 = a, a_2 \in \mathcal {E}\), then, since \(\mathcal {E}\) is an \(\mathbf {fr}\)\(\mathbf {ad}\)extension, it follows that there exists \(a_3 \in \mathcal {E}\), such that \(a_3 \rightarrow _{{\mathcal {T}_{0}}}a\), i.e., \(a_3 \rightarrow a\), a contradiction by case #2.

4.
If \(a_1 = a_2 = a\), then, since \(\mathcal {E}\) \(\mathbf {fr}\)defends a, it follows that there exists some \(a_3 \in \mathcal {E}\) such that \(a_3 \rightarrow _{{\mathcal {T}_{0}}}a\), i.e., \(a_3 \rightarrow a\), a contradiction by case #2.
The case of restricted semantics is completely analogous and omitted.
For the second result, using the same reasoning we note that \(\mathcal {E}\cup \{a\}\) \(\mathbf {lo}\)defends \(\mathcal {E}\cup \{a\}\). Given that \(\mathcal {E}\cup \{a\}\) is \(\mathbf {lo}\)\(\mathbf {cf}\) by our assumptions, the result follows. \(\square \)
Proof of Proposition 2
By Proposition 1 when \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\), and \(\mathcal {E}\) \(\mathbf {\theta }\)defends a, then \(\mathcal {E}\cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\), for \(\mathbf {\theta }\in \{\mathbf {fr},\mathbf {re}\}\). The result then follows trivially. \(\square \)
Proof of Proposition 3
It suffices to show that when \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\), and \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}a\) whenever \(a \notin \mathcal {E}\), then \(\mathcal {E}\) is maximally \(\mathbf {\theta }\)\(\mathbf {cf}\). Indeed, suppose that \(\mathcal {E}'\) is \(\mathbf {\theta }\)\(\mathbf {cf}\) and \(\mathcal {E}' \supset \mathcal {E}\). Then, take some \(a \in \mathcal {E}' \setminus \mathcal {E}\). By our hypothesis, \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}a\), i.e., \(\mathcal {E}' \rightarrow _{{\mathcal {T}_{0}}}\mathcal {E}'\), a contradiction by our hypothesis that \(\mathcal {E}'\) is \(\mathbf {\theta }\)\(\mathbf {cf}\). \(\square \)
Proof of Proposition 4
Since \(\mathcal {A}= \mathcal {A}'\), take any \(a,b \in \mathcal {A}\), \(\mathcal {E}\subseteq \mathcal {A}\). Then, apparently:

a attacks b in \(\mathcal {F}'\) if and only if \(a \rightarrow _{{\mathcal {T}_{0}}}b\) in \(\mathcal {F}\)

\(\mathcal {E}\) defends a in \(\mathcal {F}'\) if and only if \(\mathcal {E}\) \(\mathbf {re}\)defends a in \(\mathcal {F}\)
Using the above two statements and Propositions 2, 3 (necessary for the case of \(\mathbf {co}\) and \(\mathbf {st}\)extensions respectively), it is easy to show the result. \(\square \)
Proof of Proposition 5
Since \(\mathcal {T}_{0} = \mathcal {T}\), we note that \(a\rightarrow _{{\mathcal {T}_{0}}}b\) if and only if \(a \rightarrow b\). The equivalence among #1, #2, #3 is then obvious by the respective definitions on \(\mathbf {\theta }\)extensions. Moreover, the equivalence among #2 and #4 is obvious from Proposition 4, whereas the equivalence among #4 and #5 follows from the fact that \(\mathcal {F}' = \mathcal {F}''\).
\(\square \)
Proof of Proposition 6
For \(\mathbf {re}\) semantics, all results follow from Proposition 4 and the corresponding results on the AAF (e.g., [8]), so let us consider the case of \(\mathbf {fr}\) and \(\mathbf {lo}\) semantics.
#1, #2 and #3 are obvious by the respective definitions.
For #4, let \(\mathbf {\theta }\in \{\mathbf {fr},\mathbf {lo}\}\), and take \(\mathcal {E}\) to be a \(\mathbf {\theta }\)\(\mathbf {pr}\)extension. Then it is \(\mathbf {\theta }\)\(\mathbf {ad}\). Suppose that it is not \(\mathbf {\theta }\)\(\mathbf {co}\). Then, there is some \(a \notin \mathcal {E}\), such that \(\mathcal {E}\) \(\mathbf {\theta }\)defends a and \(\mathcal {E}\cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\). But then, it is easy to see that \(\mathcal {E}\cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\), which is a contradiction by the definition of \(\mathbf {\theta }\)\(\mathbf {pr}\)extensions and the fact that \(\mathcal {E}\cup \{a\} \supset \mathcal {E}\).
For #5, let us consider the case of firm semantics first, and take \(\mathcal {E}\) to be an \(\mathbf {fr}\)\(\mathbf {st}\)extension. Then, it is \(\mathbf {fr}\)\(\mathbf {cf}\) (and maximally so). We will show that it is also \(\mathbf {fr}\)\(\mathbf {ad}\). Indeed, take some \(a,b \in \mathcal {A}\), such that \(a \in \mathcal {E}\) and \(b \rightarrow a\). Then \(b \notin \mathcal {E}\) (since \(\mathcal {E}\) is \(\mathbf {fr}\)\(\mathbf {cf}\)), thus \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}b\) (since \(\mathcal {E}\) is \(\mathbf {fr}\)\(\mathbf {st}\)), which implies that \(\mathcal {E}\) \(\mathbf {fr}\)defends a. Thus, \(\mathcal {E}\) is also \(\mathbf {fr}\)\(\mathbf {ad}\). It is also maximal, because \(\mathcal {E}\) is maximally \(\mathbf {fr}\)\(\mathbf {cf}\). Therefore, \(\mathcal {E}\) is an \(\mathbf {fr}\)\(\mathbf {pr}\)extension.
For the \(\mathbf {lo}\) case, take \(\mathcal {E}\) to be a \(\mathbf {lo}\)\(\mathbf {st}\)extension. Then, it is \(\mathbf {lo}\)\(\mathbf {cf}\) (and maximally so). We will show that it is also \(\mathbf {lo}\)\(\mathbf {ad}\). Indeed, take some \(a,b \in \mathcal {A}\), such that \(a \in \mathcal {E}\) and \(b \rightarrow _{{\mathcal {T}_{0}}}a\). Then \(b \notin \mathcal {E}\) (since \(\mathcal {E}\) is \(\mathbf {lo}\)\(\mathbf {cf}\)), thus \(\mathcal {E}\rightarrow b\) (since \(\mathcal {E}\) is \(\mathbf {lo}\)\(\mathbf {st}\)), which implies that \(\mathcal {E}\) \(\mathbf {lo}\)defends a. Thus, \(\mathcal {E}\) is also \(\mathbf {lo}\)\(\mathbf {ad}\). It is also maximal, because \(\mathcal {E}\) is maximally \(\mathbf {lo}\)\(\mathbf {cf}\). Therefore, \(\mathcal {E}\) is a \(\mathbf {lo}\)\(\mathbf {pr}\)extension. \(\square \)
Proof of Proposition 7
We will prove the claim constructively. First, we will describe a construction over \(\mathcal {F}\), and then we will show that this construction generates some \(\mathcal {E}\) with the above properties. The proof is broken down in steps, represented as claims proved individually below. The last claim (Claim 5) shows the result.
Construction. We assume a wellorder < over \(\mathcal {A}\) (its existence is guaranteed by the Axiom of Choice). For a given set \(E \subseteq \mathcal {A}\), we denote by \(\min _< E\) the minimal element of E according to <.
Moreover, for \(E \subseteq \mathcal {A}\), set : \(\mathbf {\theta }\)defends a, \(E \cup \{a\}\): \(\mathbf {\theta }\)\(\mathbf {cf}\) \(\}\), i.e., the arguments that are defended by E, and do not conflict with E.
We define the function: \(\phi : 2^{\mathcal {A}} \mapsto 2^{\mathcal {A}}\) as follows:
Finally, we define a function \(\mathcal {G}\) recursively on the ordinals as follows:
Claim 1
For two ordinals \(\beta ,\gamma \), if \(\beta < \gamma \), then \(\mathcal {G}(\beta ) \subseteq \mathcal {G}(\gamma )\).
Proof of Claim 1
We will use transfinite induction on \(\gamma \).
If \(\gamma = 0\), then the result holds trivially as there is no \(\beta \) for which \(\beta < \gamma \). Suppose that the result holds for all \(\gamma < \delta \); we will show that it holds for \(\gamma = \delta \).
If \(\delta \) is a successor ordinal, then there exists some \(\delta ^\) such that \(\delta = \delta ^ + 1\). Clearly, by the definition of \(\mathcal {G}\) and \(\phi \), \(\mathcal {G}(\delta ) \supseteq \mathcal {G}(\delta ^)\). Furthermore, by the inductive hypothesis, \(\mathcal {G}(\delta ^) \supseteq \mathcal {G}(\beta )\), which shows the result.
If \(\delta \) is a limit ordinal, then the result follows directly by the definition of \(\mathcal {G}\). \(\circ \)
Claim 2
For any ordinals \(\beta \), \(\mathcal {G}(\beta ) \supseteq \mathcal {E}_{*}\).
Proof of Claim 2
If \(\beta = 0\) the result follows by the definition of \(\mathcal {G}\). If \(\beta > 0\), the result follows by Claim 1. \(\circ \)
Claim 3
For any ordinal \(\beta \), \(\mathcal {G}(\beta )\) is \(\mathbf {\theta }\)\(\mathbf {ad}\).
Proof of Claim 3
We will use transfinite induction over \(\beta \). For \(\beta = 0\), the result follows by our assumption on \(\mathcal {E}_{*}\). Now suppose that it holds for all \(\beta <\gamma \). We will show that it holds for \(\beta = \gamma \).
If \(\gamma \) is a successor ordinal, then take \(\gamma ^\) such that \(\gamma = \gamma ^ + 1\). Then, by definition, \(\mathcal {G}(\gamma ) = \phi (\mathcal {G}(\gamma ^))\). By the inductive hypothesis \(\mathcal {G}(\gamma ^)\) is \(\mathbf {\theta }\)\(\mathbf {ad}\). Moreover, by the definition of \(\phi \), \(\phi (E)\) is \(\mathbf {\theta }\)\(\mathbf {ad}\) whenever E is \(\mathbf {\theta }\)\(\mathbf {ad}\), so \(\mathcal {G}(\gamma )\) is \(\mathbf {\theta }\)ad.
If \(\gamma \) is a limit ordinal, then suppose that \(\mathcal {G}(\gamma )\) is not \(\mathbf {\theta }\)\(\mathbf {cf}\). Then, there exist \(a_1, a_2 \in \mathcal {G}(\gamma )\) such that \(\{a_1,a_2\}\) is not \(\mathbf {\theta }\)\(\mathbf {cf}\), and, thus, there exist ordinals \(\delta _1, \delta _2\) such that \(\delta _1 < \gamma \), \(\delta _2 < \gamma \), \(a_1 \in \mathcal {G}(\delta _1)\), \(a_2 \in \mathcal {G}(\delta _2)\). If \(\delta _1 = \delta _2\) then \(\mathcal {G}(\delta _1)\) is not \(\mathbf {\theta }\)\(\mathbf {cf}\), a contradiction by the inductive hypothesis. If \(\delta _1 < \delta _2\) then \(\mathcal {G}(\delta _2) \supseteq \mathcal {G}(\delta _1)\) (by Claim 1), so \(a_1, a_2 \in \mathcal {G}(\delta _2)\), a contradiction by the inductive hypothesis. The case of \(\delta _2 < \delta _1\) is analogous. Thus, \(\mathcal {G}(\gamma )\) is \(\mathbf {\theta }\)\(\mathbf {cf}\).
Now consider some \(a \in \mathcal {G}(\gamma )\). Then, by the definition of \(\mathcal {G}\), there exists some \(\delta < \gamma \) such that \(a \in \mathcal {G}(\delta )\). Since \(\mathcal {G}(\delta )\) is \(\mathbf {\theta }\)\(\mathbf {ad}\) by the inductive hypothesis, it follows that \(\mathcal {G}(\delta )\) \(\mathbf {\theta }\)defends a, so, given that \(\mathcal {G}(\gamma ) \supseteq \mathcal {G}(\delta )\) (Claim 1), we conclude that \(\mathcal {G}(\gamma )\) \(\mathbf {\theta }\)defends a. Thus, \(\mathcal {G}(\gamma )\) is \(\mathbf {\theta }\)\(\mathbf {ad}\). \(\circ \)
Claim 4
There exists ordinal \(\beta \) such that \(\mathcal {G}(\beta ) = \mathcal {G}(\beta +1)\).
Proof of Claim 4
By Claim 1, we conclude that \(\mathcal {G}\) is an increasing function from the ordinals into \(2^{\mathcal {A}}\). It cannot be strictly increasing, as if it were we would have an injective function from the ordinals into a set, violating Hartogs’ lemma. Therefore the function must be eventually constant, so for some \(\beta \), \(\mathcal {G}(\beta ) = \mathcal {G}(\beta +1)\). \(\circ \)
Claim 5
There exists some \(\mathcal {E}\) such that \(\mathcal {E}\supseteq \mathcal {E}_{*}\), and the following hold:

1.
\(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {co}\).

2.
For any \(\mathcal {E}'\) such that \(\mathcal {E}_{*} \subseteq \mathcal {E}' \subset \mathcal {E}\), there exists \(a \in \mathcal {E}\setminus \mathcal {E}'\) which is \(\mathbf {\theta }\)defended by \(\mathcal {E}'\) and \(\mathcal {E}' \cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\).

3.
For any \(\mathcal {E}'\) such that \(\mathcal {E}_{*} \subseteq \mathcal {E}' \subset \mathcal {E}\), \(\mathcal {E}'\) is not \(\mathbf {\theta }\)\(\mathbf {co}\).
Proof of Claim 5
By Claim 4, there exists ordinal \(\beta \) such that \(\mathcal {G}(\beta ) = \mathcal {G}(\beta +1)\). Set \(\mathcal {E}= \mathcal {G}(\beta )\). By Claim 2, \(\mathcal {E}\supseteq \mathcal {E}_{*}\), so it is an adequate choice. We will show that \(\mathcal {E}\) satisfies the required properties.
For the first result, note that by Claim 3, \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\). Moreover, \(\mathcal {E}= \mathcal {G}(\beta ) = \mathcal {G}(\beta + 1) = \phi (\mathcal {G}(\beta )) = \phi (\mathcal {E})\), which implies that , which, in tandem with the fact that \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\) leads to the conclusion that \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {co}\).
For the second result, take some \(\mathcal {E}'\) such that \(\mathcal {E}_{*} \subseteq \mathcal {E}' \subset \mathcal {E}\).
Set \(S = \{\gamma \mid \mathcal {G}(\gamma ) \not \subseteq \mathcal {E}'\}\). We observe that \(\beta \in S\), so \(S \ne \emptyset \). Set \(\delta = \min _< S\). Obviously, \(\delta = \beta \) or \(\delta < \beta \).
If \(\delta = 0\), then \(\mathcal {G}(\delta ) = \mathcal {E}_{*} \subseteq \mathcal {E}'\), a contradiction.
If \(\delta \) is a successor ordinal, then take \(\delta ^\) such that \(\delta = \delta ^+1\). Thus, \(\mathcal {G}(\delta ) = \phi (\mathcal {G}(\delta ^))\). By construction, \(\mathcal {G}(\delta ^) \subseteq \mathcal {E}'\) and \(\mathcal {G}(\delta ) \not \subseteq \mathcal {E}'\), therefore \(\mathcal {G}(\delta ) = \mathcal {G}(\delta ^) \cup \{a\}\), for some a for which \(\mathcal {G}(\delta ^)\) \(\mathbf {\theta }\)defends a and \(\mathcal {G}(\delta ^) \cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\). If \(a \in \mathcal {E}'\), then \(\mathcal {G}(\delta ) \subseteq \mathcal {E}'\), a contradiction by the choice of \(\delta \), so \(a \notin \mathcal {E}'\). Moreover, \(a \in \mathcal {G}(\delta )\). If \(\delta = \beta \) then \(\mathcal {G}(\delta ) = \mathcal {E}\), so \(a \in \mathcal {E}\). If \(\delta < \beta \) then \(a \in \mathcal {G}(\delta ) \subseteq \mathcal {G}(\beta )\) (by Claim 1), so \(a \in \mathcal {E}\). We conclude that \(a \in \mathcal {E}\setminus \mathcal {E}'\). Thus, we have found some a with the required properties.
If \(\delta \) is a limit ordinal, then, by the definition of \(\delta \), \(\mathcal {G}(\delta ') \subseteq \mathcal {E}'\) for all \(\delta ' < \delta \). Therefore, \(\mathcal {G}(\delta ) = \bigcup _{\delta ' < \delta } \mathcal {G}(\delta ') \subseteq \mathcal {E}'\), a contradiction by the choice of \(\delta \).
The third result follows from the second: indeed, as there exists \(a \in \mathcal {E}\setminus \mathcal {E}'\) which is \(\mathbf {\theta }\)defended by \(\mathcal {E}'\) and \(\mathcal {E}' \cup \{a\}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\), it cannot be the case that \(\mathcal {E}'\) is \(\mathbf {\theta }\)\(\mathbf {co}\). \(\circ \) \(\square \)
Proof of Proposition 8
For the case where \(\mathbf {\theta }= \mathbf {re}\), the proof follows directly by Proposition 4 and the related results from the AAF literature. So suppose that \(\mathbf {\theta }\in \{\mathbf {fr},\mathbf {lo}\}\).
We first note that \(\emptyset \) is \(\mathbf {\theta }\)\(\mathbf {cf}\) and \(\mathbf {\theta }\)\(\mathbf {ad}\) w.r.t. \(\mathcal {T}_{0}\), so the claim is true for \(\mathbf {\sigma }\in \{\mathbf {cf}, \mathbf {ad}\}\).
Let us now turn our attention to the case where \(\mathbf {\sigma }= \mathbf {pr}\). Our proof follows the lines of the respective proof in [4]. Set (\(\mathcal {AD} \ne \emptyset \), as shown above). We will show that, any \(\subseteq \)chain \((\mathcal {E}_{i})_{i\in I}\) in \(\mathcal {AD}\) possesses an upper bound. Indeed, set \(\mathcal {E}= \bigcup \mathcal {E}_{i}\). Obviously \(\mathcal {E}\supseteq \mathcal {E}_{i}\), so it is an upper bound; it remains to show that \(\mathcal {E}\in \mathcal {AD}\), i.e., that \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\).
Now suppose that \(\mathcal {E}\) is not \(\mathbf {\theta }\)\(\mathbf {cf}\). Then there exist \(a_1, a_2 \in \mathcal {E}\) that attack each other (\(a \rightarrow b\) for \(\mathbf {\theta }= \mathbf {fr}\), \(a \rightarrow _{{\mathcal {T}_{0}}}b\) for \(\mathbf {\theta }= \mathbf {lo}\)). By the definition of \(\mathcal {E}\), there exist \(\mathcal {E}_{i}\), \(\mathcal {E}_{j}\) such that \(a_1 \in \mathcal {E}_{i}\), \(a_2 \in \mathcal {E}_{j}\) for some \(i,j \in I\). It is the case that \(\mathcal {E}_{i} \subseteq \mathcal {E}_{j}\) or \(\mathcal {E}_{i} \subseteq \mathcal {E}_{j}\), so suppose, without loss of generality, that \(\mathcal {E}_{i} \subseteq \mathcal {E}_{j}\). Then \(a_1,a_2 \in \mathcal {E}_{j}\), a contradiction, since \(\mathcal {E}_{j}\) is \(\mathbf {\theta }\)\(\mathbf {ad}\) (thus \(\mathbf {\theta }\)cf). Thus, \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {cf}\). It remains to show that \(\mathcal {E}\) defends all \(a \in \mathcal {E}\). Indeed, take some \(a \in \mathcal {E}\). Then, \(a \in \mathcal {E}_{i}\) for some \(i \in I\), and, thus \(\mathcal {E}_{i}\) \(\mathbf {\theta }\)defends a, which implies that \(\mathcal {E}\) \(\mathbf {\theta }\)defends a, since \(\mathcal {E}\supseteq \mathcal {E}_{i}\). Thus, any \(\subseteq \)chain \((\mathcal {E}_{i})_{i\in I}\) in \(\mathcal {AD}\) possesses an upper bound, which, by Zorn’s Lemma, implies that \(\mathcal {AD}\) has a maximal element, i.e., that there exists a \(\mathbf {\theta }\)\(\mathbf {pr}\) extension.
By proposition 6, this implies that there exists a \(\mathbf {\theta }\)\(\mathbf {co}\) extension as well.
For \(\mathbf {\theta }\)\(\mathbf {gr}\) extensions, note that \(\emptyset \) is \(\mathbf {\theta }\)\(\mathbf {ad}\), so applying Proposition 7 for \(\mathcal {E}_{*} = \emptyset \) we ensure the existence of some \(\mathcal {E}\) which is minimally \(\mathbf {\theta }\)\(\mathbf {co}\), i.e., \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {gr}\). \(\square \)
Proof of Proposition 9
Given that \(\emptyset \) is \(\mathbf {\theta }\)\(\mathbf {ad}\), we can apply Proposition 7 for \(\mathcal {E}_{*} = \emptyset \) to get some \(\mathcal {E}\) which is minimally \(\mathbf {\theta }\)\(\mathbf {co}\), i.e., \(\mathcal {E}\) is \(\mathbf {\theta }\)\(\mathbf {gr}\). Now suppose that there is a second \(\mathbf {\theta }\)\(\mathbf {gr}\) extension, say \(\mathcal {E}'\) (\(\mathcal {E}' \ne \mathcal {E}\)). Obviously, \(\mathcal {E}\not \subseteq \mathcal {E}'\) and \(\mathcal {E}' \not \subseteq \mathcal {E}\). Set \(\mathcal {E}_{0} = \mathcal {E}\cap \mathcal {E}'\). It follows that \(\emptyset \subseteq \mathcal {E}_{0} \subset \mathcal {E}\), so by Proposition 7 again there exists some \(a \in \mathcal {E}\setminus \mathcal {E}_{0}\) which is \(\mathbf {\theta }\)defended by \(\mathcal {E}_{0}\) and \(\mathcal {E}_{0} \cup \{a\}\) \(\mathbf {\theta }\)\(\mathbf {cf}\). Moreover, \(\mathcal {E}_{0} \subset \mathcal {E}'\), so a is \(\mathbf {\theta }\)defended by \(\mathcal {E}'\). Thus, \(\mathcal {E}'\) is \(\mathbf {\theta }\)\(\mathbf {gr}\), thus \(\mathbf {\theta }\)\(\mathbf {co}\), and also \(\mathcal {E}'\) \(\mathbf {\theta }\)defends a, so by Proposition 2, \(a \in \mathcal {E}'\), a contradiction by the choice of a. \(\square \)
Proof of Proposition 10
The first case is direct from Definition 3 and the definition of \({\mathcal {F}_{F}}\). The second case is direct using proof by contradiction and the fact that \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}\mathcal {E}\) implies \(\mathcal {E}\rightarrow \mathcal {E}\). The third is direct from Definition 3. \(\square \)
Proof of Proposition 11
The first case follows from the fact that \(b \rightarrow _{{\mathcal {T}_{0}}}c\) implies that \(b \rightarrow c\) for any \(b,c \in \mathcal {A}\). For the second and third cases, note that \(a \rightarrow b\) if and only if \((a,b) \in \mathcal {R}_{F}\), and that \(a \rightarrow _{{\mathcal {T}_{0}}}b\) implies that \(a \rightarrow b\). From these, and the definition of defense in AAFs and MAAFs, the results follow easily. \(\square \)
Proof of Proposition 12
The first four cases are direct from Propositions 10, 11. For the fifth case, note that, since \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {ad}\), it follows that for all \(a \in \mathcal {E}\), \(\mathcal {E}\) \(\mathbf {re}\)defends a for \(\tau _{0}\), and, thus, by Proposition 11, \(\mathcal {E}\) defends a in \({\mathcal {F}_{F}}\). Combining this with the fact that \(\mathcal {E}\) is \(\mathbf {cf}\) in \({\mathcal {F}_{F}}\), we get the result. \(\square \)
Proof of Proposition 13
For the first case, it suffices to show that, if \(\mathcal {E}\) \(\mathbf {fr}\)defends a w.r.t. \(\mathcal {T}_{0}\), then \(a \in \mathcal {E}\). Indeed, if \(\mathcal {E}\) \(\mathbf {fr}\)defends a, then, by Proposition 11, \(\mathcal {E}\) \(\mathbf {re}\)defends a, so, given that \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {co}\), it follows that \(a \in \mathcal {E}\). The proofs for the other cases are analogous. \(\square \)
Proof of Proposition 14
For the first case, we note that \(\emptyset \) is \(\mathbf {fr}\)ad, so applying Proposition 7 for \(\mathcal {E}_{*} = \emptyset \), we will get a \(\mathbf {fr}\)\(\mathbf {co}\) extension (say \(\mathcal {E}\)) that is minimal among \(\mathbf {fr}\)\(\mathbf {co}\) extensions, thus it is the (only) \(\mathbf {fr}\)\(\mathbf {gr}\) extension of \(\mathcal {F}\). By Proposition 7 again, we observe that, for any \(\mathcal {E}' \subset \mathcal {E}\), there exists some \(a \in \mathcal {E}\setminus \mathcal {E}'\) such that \(\mathcal {E}'\) \(\mathbf {fr}\)defends a, i.e., \(\mathcal {E}'\) \(\mathbf {re}\)defends a, i.e., \(\mathcal {E}'\) is not \(\mathbf {re}\)\(\mathbf {co}\). Thus, \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {gr}\).
The second case is totally analogous.
The third case uses a similar proof (and the same reasoning, except that the existence of a is guaranteed by the results in [8] (instead of Proposition 7). \(\square \)
Proof of Proposition 15
For the first case, suppose that \(\mathcal {E}\) is not \(\mathbf {fr}\)\(\mathbf {pr}\). Then, there exists some \(\mathcal {E}' \supset \mathcal {E}\) such that \(\mathcal {E}'\) is \(\mathbf {fr}\)\(\mathbf {pr}\). But then, \(\mathcal {E}'\) is \(\mathbf {fr}\)\(\mathbf {ad}\) so (by Proposition 12) \(\mathcal {E}'\) is \(\mathbf {re}\)\(\mathbf {ad}\), a contradiction by the fact that \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {pr}\). The other cases are analogous. \(\square \)
Proof of Proposition 16
For the first: observe that, by Proposition 10, \(\mathcal {E}\) is maximally \(\mathbf {fr}\)\(\mathbf {cf}\) if and only if \(\mathcal {E}\) is maximally \(\mathbf {re}\)\(\mathbf {cf}\). Then, the result is obvious by Definition 8.
For the second: we obtain by Proposition 10 that \(\mathcal {E}\) is \(\mathbf {cf}\) in \({\mathcal {F}_{F}}\). Also, since \(\mathcal {E}\) is \(\mathbf {fr}\)\(\mathbf {st}\), \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}a\) for all \(a \notin \mathcal {E}\), thus \(\mathcal {E}\rightarrow a\) in \({\mathcal {F}_{F}}\). We conclude that \(\mathcal {E}\) is \(\mathbf {st}\) in \({\mathcal {F}_{F}}\).
For the third: we observe that, by Proposition 10, \(\mathcal {E}\) is maximally \(\mathbf {re}\)\(\mathbf {cf}\) if and only if it is maximally \(\mathbf {lo}\)\(\mathbf {cf}\). Now take some \(a \notin \mathcal {E}\). If \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {st}\), then \(ext \rightarrow _{{\mathcal {T}_{0}}}a\), so \(\mathcal {E}\rightarrow a\), so \(\mathcal {E}\) is \(\mathbf {lo}\)\(\mathbf {st}\). If \(\mathcal {E}\) is \(\mathbf {lo}\)\(\mathbf {st}\), then \(\mathcal {E}\rightarrow a\), and suppose that it is not the case that \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}a\). Then, \(\mathcal {E}\cup \{a\} \supset \mathcal {E}\) and \(\mathbf {lo}\)\(\mathbf {cf}\), a contradiction.
For the fourth: since \(\mathcal {E}\) is \(\mathbf {re}\)\(\mathbf {st}\), we get that \(\mathcal {E}\rightarrow _{{\mathcal {T}_{0}}}a\) whenever \(a \notin \mathcal {E}\), thus \(\mathcal {E}\rightarrow a\) in \({\mathcal {F}_{F}}\) for all \(a \notin \mathcal {E}\), and \(\mathcal {E}\) is \(\mathbf {cf}\) in \({\mathcal {F}_{F}}\) by the hypothesis, so \(\mathcal {E}\) is \(\mathbf {st}\) in \({\mathcal {F}_{F}}\). \(\square \)
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Vassiliades, A., Flouris, G., Patkos, T., Bikakis, A., Bassiliades, N., Plexousakis, D. (2021). A Multi Attack Argumentation Framework. In: Baroni, P., Benzmüller, C., Wáng, Y.N. (eds) Logic and Argumentation. CLAR 2021. Lecture Notes in Computer Science(), vol 13040. Springer, Cham. https://doi.org/10.1007/9783030893910_23
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