Particular cases leading to compatibility

In the context of an argumentation system with a finite relation ${\mathcal{R}}$ without cycles²⁹, the stable and the preferred semantics provide only one extension and the levels of uni-accepted, exi-accepted, cleanly-accepted coincide.

In this context, there are at least two particular cases leading to compatibility.

First case: It deals with the global valuation with tuples.

Theorem 1   Let ${\mathcal{G}}$ be the graph associated with ${{<}{\mathcal{A}},{\mathcal{R}}{>}}$, ${{<}{\mathcal{A}},{\mathcal{R}}{>}}$ being an argumentation system with a finite relation ${\mathcal{R}}$ without cycles and satisfying the following condition: $\exists$ $A \in {\mathcal{A}}$ such that

• $\forall X_i$, leaf of ${\mathcal{G}}$, $\exists$ only one path from $X_i$ to $A$, $X_i^1-\ldots-X_i^{l_i}-A$ with $X_i^1 = X_i$ and $l_i$ the length of this path (if $l_i$ is even, this path is a defence branch for $A$, else it is an attack branch),
• all the paths from $X_i$ to $A$ are root-dependent in $A$,
• $\forall A_i \in {\mathcal{A}}$, $\exists X_j$ a leaf of ${\mathcal{G}}$ such that $A_i$ belongs to a path from $X_j$ to $A$.

Let $v$ be a valuation with tuples. Let $S$ be a semantics $\in$ {preferred, stable}.

1. $\forall B \in {\mathcal{A}}$, $B \neq A$, $B$ (exi, uni, cleanly) accepted for $S$ iff $B$ well-defended for $v$.
2. If $A$ is (exi, uni, cleanly) accepted for $S$ then $A$ is well-defended for $v$ (the converse is false).
3. If $A$ is well-defended for $v$ and if all the branches leading to $A$ are defence branches for $A$ then $A$ is (exi, uni, cleanly) accepted for $S$.

Note that Theorem 1 is, in general, not satisfied by a local valuation. See the following counterexample for the valuation of [4]:

The graph satisfies the condition stated in Theorem 1. The set of well-defended arguments is $\{C_1, C_2, C_3\}$ (so, $A$ is not well-defended). Nevertheless, $\{C_1, C_2, C_3, A\}$ is the preferred extension.

Second case: This second case concerns the generic local valuation:

Theorem 2   Let ${{<}{\mathcal{A}},{\mathcal{R}}{>}}$ be an argumentation system with a finite relation ${\mathcal{R}}$ without cycles. Let $S$ be a semantics $\in$ {preferred, stable}. Let $v$ be a generic local valuation satisfying the following condition $(*)$:

$(\forall i = 1 \ldots n, g(x_i) \geq x_i) \Rightarrow (g(h(x_1, \ldots, x_n)) \geq h(x_1, \ldots, x_n))$ $(*)$

$\forall A \in {\mathcal{A}}$, $A$ (exi, uni, cleanly) accepted for $S$ iff $A$ well-defended for $v$.

This theorem is a direct consequence of the following lemma:

Lemma 1   Let ${{<}{\mathcal{A}},{\mathcal{R}}{>}}$ be an argumentation system with a finite relation ${\mathcal{R}}$ without cycles. Let $S$ be a semantics $\in$ {preferred, stable}. Let $v$ be a generic local valuation satisfying the condition $(*)$.

$(i)$

If $A$ is exi-accepted and $A$ has only one direct attacker $B$ then $A \succeq B$.

$(ii)$

If $B$ is not-accepted and $B$ has only one direct attacker $C$ then $C \succeq B$.

Remark: The condition $(*)$ stated in Theorem 2 is:

• false for the local valuation proposed by [4], as shown in the following graph:
  

  We know that $g(x) = \frac{1}{1 + x}$ and $h(x_1,\ldots,x_n) = \Sigma_{i=1}^nx_i$ (see Property 6). We get:

  • $\forall i = 1\ldots 3$, $x_i = v(B_i) = 0.5$,
  • $\forall i = 1\ldots 3$, $g(x_i) = 0.66$, so $g(x_i) \geq x_i$,
  • and nevertheless $g(h(x_1,x_2,x_3)) = v(A) = 0.4 \not\geq h(x_1,x_2,x_3) = 1.5$.

• false for the local valuations defined with $h$ such that $\exists n > 1$ with $h(x_1,\ldots, x_n) > \max(x_1,\ldots, x_n)$ (for all the functions $g$ strictly non-increasing): see the previous graph where $h(x_1,x_2,x_3) = 1.5$ and $\max(x_1,x_2,x_3) = 0.5$.

• true for the local valuations defined with $h = \max$ (for all the functions $g$): if $h = \max$ then $g(h(x_1,\ldots,x_n)) = g(\max(x_1,\ldots,x_n)) = g(x_j)$, $x_j$ being the maximum of the $x_i$; and, by assumption, $g(x_i) \geq x_i$, $\forall x_i$, so in particular for $x_j$; so, we get:
  $g(h(x_1,\ldots,x_n)) = g(x_j) \geq x_j = \max(x_1,\ldots,x_n) = h(x_1,\ldots,x_n)$.