A formal framework for deliberated judgment

Abstract

While the philosophical literature has extensively studied how decisions relate to arguments, reasons and justifications, decision theory almost entirely ignores the latter notions. In this article, we elaborate a formal framework to introduce in decision theory the stance that decision-makers take towards arguments and counter-arguments. We start from a decision situation, where an individual requests decision support. We formally define, as a commendable basis for decision-aid, this individual’s deliberated judgment, a notion inspired by Rawls’ contributions to the philosophical literature, and embodying the requirement that the decision-maker should carefully examine arguments and counter-arguments. We explain how models of deliberated judgment can be validated empirically. We then identify conditions upon which the existence of a valid model can be taken for granted, and analyze how these conditions can be relaxed. We then explore the significance of our framework for the practice of decision analysis. Our framework opens avenues for future research involving both philosophy and decision theory, as well as empirical implementations.

A Proofs, and additional explanatory results

Our main goal in this section is to prove Theorem 2. We do this by first proving that if a set $S_\gamma$ is CAC, then it includes enough decisive arguments to settle the issue (we will call such a set $S\subseteq S^*$efficient). This requires a few intermediate lemmas. Efficiency will bring a number of consequences of interest to us, among which Theorem 2. As a second goal, we want to give some further results that help understand the relationship between the notions of clear-cut, validity and operational validity, existence of a CAC set of arguments, and efficiency.

Let us start with the formal definition of efficiency.

Definition 13

(Efficiency) Given a decision situation , given $S\subseteq S^*$, $S$ is efficient iff , and .

Recall that designates the arguments not trumped by any argument, thus, the decisive arguments, and hence, designates the arguments always trumped by some decisive argument in $S$.

In all this section, we assume we are given a decision situation and a subset of arguments $S_\gamma \subseteq S^*$ (except in Theorem 5).

Our strategy for proving that CAC implies efficiency, roughly speaking, involves excluding “undecided” situations from $S_\gamma$. For example, we want to show that it is impossible that an argument has no decisive argument trumping it in $S_\gamma$, but also fails to be defended in $S_\gamma$. We will do this by progressively promoting or degrading arguments, e.g., show that, in $S_\gamma$, if an argument is resistant (has no argument that decisively trumps it), then it must also be defended, and if it is defended, it must be replaceable by decisive arguments.

Define as the decisive arguments from $S_\gamma$.

Define an argument $s$ as finitely defended iff some finite set of arguments from $S_{\gamma \text {dec}}$ defends it, thus, iff $\exists S\subseteq S_{\gamma \text {dec}}$ such that , $S$ finite. Define $S_{\gamma \text {def}}$ as the arguments from $S_\gamma$ that are finitely defended.

Define $R_{\gamma \text {dec}}\subseteq S^*$ as the arguments that are replaceable by $S_{\gamma \text {dec}}$. Recall that $S$ replaces $s$ iff and .

Define as the resistant arguments from $S_\gamma$, namely, those not trumped by any argument from $S_{\gamma \text {dec}}$.

Define $E_{\gamma \text {res}}\subseteq S^*$ as the arguments that are essentially replaceable by $S_{\gamma \text {res}}$. Recall that $S$ essentially replaces $s$ iff and .

Similarly, $E_{\gamma \text {dec}}$ are the arguments essentially replaceable by $S_{\gamma \text {dec}}$.

Lemma 1

($S_{\gamma \text {def}}\subseteq R_{\gamma \text {dec}}$) If $S_\gamma$ is Closed under reinstatement and Answerable, then the arguments from $S_\gamma$ that are finitely defended are replaceable by decisive arguments from $S_\gamma$; formally: $S_{\gamma \text {def}}\subseteq R_{\gamma \text {dec}}$.

Proof

The strategy for this proof is the following. If $s\in S_{\gamma \text {def}}$, some finite set of arguments defends $s$. We wish to pick defenders one by one, replacing $s$ by applying Closed under reinstatement to $s$ and the chosen defender, obtaining an argument that fewer arguments trump, and then show that iterating the process yields a decisive argument replacing $s$.

We need the following intermediate result. Assume a set of arguments $S\subseteq S_{\gamma \text {dec}}$ is given, together with an argument $s_1 \in S$ and an argument $s^\text {r}_1 \in S_\gamma$ defended by $S$. Then, there exists an argument $s^\text {r}_2 \in S_\gamma$ replacing $s^\text {r}_1$ and defended by $S{\setminus } \{s_1\}$.

Indeed, from Answerability, because $s_1 \in S_{\gamma \text {dec}}$, . Also, as $s_1 \in S_{\gamma \text {dec}}$, we can assume that $s_1 \ne s^\text {r}_1$, otherwise $s^\text {r}_1 \in S_{\gamma \text {dec}}$ and the result is obtained by taking $s^\text {r}_2 = s^\text {r}_1$. And $s_1$ does not trump $s^\text {r}_1$, otherwise $s^\text {r}_1$ is trumped by a decisive argument and thus not defended. We can thus apply Closed under reinstatement to $(s_1, s^\text {r}_1)$. We obtain that for some $s^\text {r}_2 \in S_\gamma$, and $s^\text {r}_2$ replaces $s^\text {r}_1$. Thus, $S{\setminus } \{s_1\}$ defends $s^\text {r}_2$: any argument trumping $s^\text {r}_2$ already trumped $s^\text {r}_1$, hence, is trumped by $S$ (because that set defends $s^\text {r}_1$), and is not trumped by $s_1$. This proves our intermediate result.

Coming back to the main point, we know that a finite coalition $S\subseteq S_{\gamma \text {dec}}$ defends $s\in S_\gamma$. Define $s^\text {r}_1 = s$ and apply the intermediate result repetitively to obtain an argument $s^\text {r}_2 \in S_\gamma$ replacing $s$ and defended by $S$ minus one element, then $s^\text {r}_3 \in S_\gamma$ replacing $s^\text {r}_2$, thus, replacing $s$ (because replacement is transitive) and defended by $S$ minus two elements, and so on, until obtaining a replacer defended by $\emptyset$, thus, decisive. $\square$

Lemma 2

() If $S_\gamma$ is covering, then any argument is either essentially replaceable by $S_{\gamma \text {res}}$, or attacked by an argument from $S_{\gamma \text {res}}$; formally: .

Proof

We consider in turn three sets whose union yields $S^*$: ${\overline{S_\gamma }}$, and .

First, : from covering, if $s\notin S_\gamma$, $s$ is unnecessary, and by definition, $s$ is unnecessary iff $s\in E_{\gamma \text {res}}$ or .

Second, , because $S_{\gamma \text {dec}}\subseteq S_{\gamma \text {res}}$.

Third, , because $S_\gamma \cap \overline{\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})} = S_{\gamma \text {res}}$ by definition.

We have considered all three possible cases, and the conclusion obtains in all cases. $\square$

Lemma 3

($S_{\gamma \text {res}}\subseteq S_{\gamma \text {def}}$) If $S_\gamma$ is CAC, any argument in $S_\gamma$ that has no argument that decisively trumps it is finitely defended; formally: $S_{\gamma \text {res}}\subseteq S_{\gamma \text {def}}$.

Proof

Recall that the relation Q is defined in Bounded length (Condition 9) as $Q = (\mathbin {\vartriangleright _\exists }\cup (\mathbin {\vartriangleright _\exists }\circ \mathbin {\vartriangleright _\exists })) \cap (S_\gamma \times S_\gamma )$. Observe that, given any set $S\ne \emptyset$, Bounded Length forbids that $\forall s\in S: S\cap Q^{-1}(s) \ne \emptyset$. Otherwise, applying $Q^{-1}$ to an element of $S$ would always yield some element in $S$, and $Q^{-1}$ could then be applied any desired number of times starting from any $s\in S$, thereby building a chain as long as desired. Accordingly, for any set S, Bounded Length imposes that if $\forall s\in S: S\cap Q^{-1}(s) \ne \emptyset$, then $S= \emptyset$.

Define $S= S_{\gamma \text {res}}\cap {\overline{S_{\gamma \text {def}}}}$. We show that, given any $s\in S$, $S\cap Q^{-1}(s) \ne \emptyset$. This suffices to obtain $S= \emptyset$ and, therefore, our desired conclusion.

Pick any $s\in S$. Towards exhibiting an argument in $S\cap Q^{-1}(s)$, we want first to exhibit some argument $s'$ that is a) trumped by some argument $s^* \in S_{\gamma \text {res}}$, thus $s' \in \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$; b) not trumped by any argument in $S_{\gamma \text {dec}}$, thus $s' \notin \mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})$; c) equal to $s$ or trumping $s$. As a second step, from the existence of such an $s'$ we will then prove that $s^*$, the particular trumping argument in part a), belongs to $S$ (thanks to parts a) and b)), and belongs to $Q^{-1}(s)$ (thanks to part c)).

Our first step thus amounts to show that some $s'$ satisfies our three conditions above.

From $s\notin S_{\gamma \text {def}}$ and $s\in S_\gamma$, we know that $s$ is not finitely defended, and using the contrapositive of Bounded width, we obtain that $s$ is not infinitely defended either. Hence, by definition of defense, there exists some $s_1 \in \overline{\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})} \cap \mathbin {\vartriangleright _\exists ^{-1}}(s)$. And, applying [$S^*= E_{\gamma \text {res}}\cup \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$], either $s_1 \in E_{\gamma \text {res}}$, or $s_1 \in \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$.

If $s_1 \in E_{\gamma \text {res}}$, $s\in \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$. Besides, because $s\in S$, $s\in S_{\gamma \text {res}}$. Thus taking $s' = s$ satisfies our three conditions.

And if $s_1 \in \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$, because $s_1 \in \overline{\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})}$), taking $s' = s_1$ satisfies our three conditions.

For our second step, consider an argument $s^* \in S_{\gamma \text {res}}$ that trumps $s'$ (we know this is possible thanks to part a)). Thanks to part b), we know that $s'$ is not trumped by any argument in $S_{\gamma \text {dec}}$, and from [$S_{\gamma \text {def}}\subseteq R_{\gamma \text {dec}}$], we know that if $s'$ was trumped by an argument in $S_{\gamma \text {def}}$, it would be trumped by an argument in $S_{\gamma \text {dec}}$, thus, $s'$ is not trumped by any argument in $S_{\gamma \text {def}}$. Because $s^* \mathbin {\vartriangleright _\exists }s'$, we know that $s^* \notin S_{\gamma \text {def}}$. Thus, $s^* \in S$. Finally, $s^* \mathbin {\vartriangleright _\exists }s$ or $s^* \mathbin {\vartriangleright _\exists }s' \mathbin {\vartriangleright _\exists }s$ (thanks to part c)), thus, $s^* \in Q^{-1}(s)$. $\square$

Lemma 4

($S^*= E_{\gamma \text {dec}}\cup \mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})$) If $S_\gamma$ is CAC, any argument is either essentially replaceable by decisive arguments from $S_\gamma$, or attacked by a decisive argument from $S_\gamma$; formally: $S^*= E_{\gamma \text {dec}}\cup \mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})$.

Proof

This follows from [$S^*= E_{\gamma \text {res}}\cup \mathbin {\vartriangleright _\exists }(S_{\gamma \text {res}})$], [$S_{\gamma \text {res}}\subseteq S_{\gamma \text {def}}$] and [$S_{\gamma \text {def}}\subseteq R_{\gamma \text {dec}}$]. $\square$

Theorem 3

(CAC implies efficiency) If $S_\gamma$ is CAC, $S_\gamma$ is efficient.

Proof

We prove that .

This proves the point, as it shows that

• (i) , and

• (ii) , because .

That follows from the definitions of $E_{\gamma \text {dec}}$ and $T_i$.

The next subset relation holds because if some decisive argument supports t, that argument is not in $\mathbin {\vartriangleright _\forall }(S_{\gamma \text {dec}})$.

Finally, Answerability mandates that $\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}}) \subseteq \mathbin {\vartriangleright _\forall }(S_{\gamma \text {dec}})$, hence, $\overline{\mathbin {\vartriangleright _\forall }(S_{\gamma \text {dec}})} \subseteq \overline{\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})}$, and using [$S^*= E_{\gamma \text {dec}}\cup \mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})$], $\overline{\mathbin {\vartriangleright _\exists }(S_{\gamma \text {dec}})} \subseteq E_{\gamma \text {dec}}$. $\square$

Theorem 4

(Validity of $\eta )$ Assume $S_\gamma$ is efficient and $\eta$, a model of the decision situation, is $S_\gamma$-operationally valid. Then $T_i = T_\eta$.

Proof

Recall that a model is $S_\gamma$-operationally valid iff for all , $s\in S^*$, we have and $[\forall s_c \in S_\gamma : (s_c \mathbin {\ntriangleright _\exists }s) \vee (\exists s_{cc} \in S^*\;|\;s_{cc} \mathbin {\vartriangleright _{\eta }}s_c \wedge s_{cc} \mathbin {\vartriangleright _\exists }s_c)]$, and when t is not supported by $\eta$, .

Consider $t \in T_\eta$. By definition, some . From operational validity of $\eta$, we obtain that and $\forall s_c \mathbin {\vartriangleright _\forall }s: s_c \notin S_\gamma \cap \overline{\mathbin {\vartriangleright _\exists ^{-1}}(S^*)}$ (because $[s_c \mathbin {\vartriangleright _\forall }s\wedge s_c \in S_\gamma ] \Rightarrow s_c \in \mathbin {\vartriangleright _\exists ^{-1}}(S^*)$). Hence, $s\notin \mathbin {\vartriangleright _\forall }(S_\gamma \cap \overline{\mathbin {\vartriangleright _\exists ^{-1}}(S^*)})$, thus . Efficiency of $S_\gamma$ brings $t \in T_i$.

If $t \notin T_\eta$, from operational validity of $\eta$, no decisive argument in $S_\gamma$ may support t, equivalently, , and from efficiency, $t \notin T_i$. $\square$

We can now prove Theorem 2.

Proof of theorem 2

From [CAC implies efficiency], we obtain that $S_\gamma$ is efficient. It then follows from the efficiency of $S_\gamma$ that the decision situation is clear-cut and that a $S_\gamma$-operationally valid model exists. The last consequence is given by Theorem 4. $\square$

The following theorem may help clarify the relationship between efficiency, existence of CAC arguments, and the situation admitting a model as we conceive it.

Theorem 5

(CAC subset equivalent to efficiency) Given a decision situation and a subset of arguments $S\subseteq S^*$, there exists a set $S_\gamma \subseteq S$ that is CAC iff $S$ is efficient.

Proof

From [CAC implies efficiency], if some set $S_\gamma \subseteq S$ is CAC, then $S_\gamma$ is efficient, and because efficiency propagates to supersets, $S$ is efficient.

If $S$ is efficient (thus, the decision situation is clear-cut), then a CAC subset $S_\gamma$ exists: suffices to choose as members of $S_\gamma$ only the decisive arguments required to support the justifiable propositions and trump the supporters of untenable propositions. Observing that no arguments trump any argument in the resulting set (thus $s\mathbin {\vartriangleright _\exists }s_\gamma$ for no $s\in S^*, s_\gamma \in S_\gamma$), most of the conditions for $S_\gamma$ to be CAC are immediately seen to be satisfied. About arguments $s\in S^*{\setminus } S_\gamma$ being unnecessary, we only have to show that when , either $s$ is trumped by an argument from $S_\gamma$ that is not decisively trumped, or $s$ is essentially replaceable by arguments from $S_\gamma$. Indeed, by our construction of $S_\gamma$, if $s$ supports an accepted t, it is essentially replaceable, and otherwise, it is trumped by a decisive argument. $\square$