Reasoning with dimensions and magnitudes

Abstract

This paper shows how two models of precedential constraint can be broadened to include legal information represented through dimensions. I begin by describing a standard representation of legal cases based on boolean factors alone, and then reviewing two models of constraint developed within this standard setting. The first is the “result model”, supporting only a fortiori reasoning. The second is the “reason model”, supporting a richer notion of constraint, since it allows the reasons behind a court’s decisions to be taken into account. I then show how the initial representation can be modified to incorporate dimensional information and how the result and reason models can be adapted to this new dimensional setting. As it turns out, these two models of constraint, which are distinct in the standard setting, coincide once they are transposed to the new dimensional setting, yielding exactly the same patterns of constraint. I therefore explore two ways of refining the reason model of constraint so that, even in the dimensional setting, it can still be separated from the result model.

Appendices

Appendices

1.1 Interpreting standard information

The body of this paper explores the notion of precedential constraint in two general settings, standard and dimensional. In this first appendix, we show how information from the standard setting can be interpreted in the dimensional setting.

The interpretation described here is based on an idea introduced by  Bench-Capon (1999), who represents the overall set of factors through two partial orders, each containing as elements subsets of the factors favoring one side or the other, $\pi$ or $\delta$; in an anticipation of the reason model, Bench-Capon then sees precedent cases as establishing further ordering relations between elements of these separate partial orders.⁹ Here, based on Bench-Capon’s representation, we postulate the two dimensions $d_\pi$ and $d_\delta$, each taking as values subsets of the standard factors favoring the appropriate side. More exactly, where s is a side, the possible values of the dimension $d_s$ are the subsets of $F^{s}$, the set of standard factors favoring that side, with these values ordered through the subset relation: if X and Y are subsets of $F^{s}$, then

$$\begin{aligned} X \le ^s Y \, {\mathrm{if}}\, {\mathrm{and}}\, {\mathrm{only}}\, {\mathrm{if}}\, X \subseteq Y. \end{aligned}$$

The idea, of course, is that subsets of $F^{s}$ containing more standard factors favoring s are values along the dimension $d_s$ that favor the side s more strongly.

Against this background, we now define a dimensionalization function $\mathcal{D}$, mapping items from the standard setting into their dimensional counterparts, in five steps. First, where X is a standard fact situation, we take

$$\begin{aligned} \mathcal{D}(X) = \{ \langle d_\pi , X^\pi \rangle , \langle d_\delta , X^\delta \rangle \}, \end{aligned}$$

where, as we recall, $X^s = X \cap F^{s}$. $\mathcal{D}(X)$ is thus the dimensional fact situation that assigns to each dimension $d_s$ the set $X^s$ containing those standard factors from X that favor the side s.

There is a slight wrinkle when it comes to interpreting standard reasons in the dimensional setting, since standard reasons are objects of the same type as standard fact situations—sets of standard factors—and so would likewise be mapped by $\mathcal{D}$ into dimensional fact situations, rather than magnitude reasons. As our second step, we therefore introduce an auxiliary function $\mathcal{D}'$ mapping standard reasons into their dimensional counterparts in such a way that, where W is a standard reason favoring the side s, its dimensionalization is

$$\begin{aligned} \mathcal{D}'(W) = \{ M_{d_s,W}^{s}\}. \end{aligned}$$

On the basis of this definition, we can note that, the standard fact situation X satisfies the standard reason W just in case the dimensionalization $\mathcal{D}(X)$ of this situation satisfies $\mathcal{D}'(W)$. This claim can be verified by observing that holds, by Definition 11, just in case (1) $W \le ^s \mathcal{D}(X)(d_s)$, which is equivalent, since $\mathcal{D}(X)(d_s)$ is $X^s$, to (2) $W \le ^s X^s$, which is equivalent by the current ordering relation on dimension values to (3) $W \subseteq X^s$, which is equivalent to by Definitions 1 and 2.

As the third step, where r is a standard rule supporting the outcome s, its dimensionalization $\mathcal{D}(r)$ is defined as a magnitude rule of the form

$$\begin{aligned} \mathcal{D}'({ Premise}(r)) \rightarrow s, \end{aligned}$$

supporting the same outcome as the original, and taking as its premise the dimensionalization of the standard reason that forms the premise of the original rule. Fourth, where $c = \langle X, r, s \rangle$ is a standard case, its dimensionalization

$$\begin{aligned} \mathcal{D}(c) = \langle \mathcal{D}(X), \mathcal{D}(r), s \rangle \end{aligned}$$

is the dimensional case containing the dimensionalization of the fact situation and rule from the original case, and the same outcome. Fifth, and finally, where $\Gamma$ is a standard case base, its dimensionalization is

$$\begin{aligned} \mathcal{D}(\Gamma ) = \{ \mathcal{D}(c) : c \in \Gamma \} \end{aligned}$$

containing dimensionalizations of each case belonging to the original.

We can see these definitions at work by calculating the dimensionalization of the case base $\Gamma _{1} = \{ c_{1} \}$, considered earlier, containing the single case $c_{1} = \langle X_{1}, r_{1}, s_{1} \rangle$, where $X_{1} = \{f^{\pi }_1, f^{\pi }_2, f^{\delta }_1, f^{\delta }_2 \}$, where $r_{1}$ is $\{ f^{\pi }_1 \} \rightarrow \pi$, and where $s_{1}$ is $\pi$. Descending through the steps in our definition, we have $\mathcal{D}(\Gamma _{1}) = \{ \mathcal{D}(c_{1}) \}$, with $\mathcal{D}(c_{1}) = \langle \mathcal{D}(X_{1}), \mathcal{D}(r_{1}), s_{1} \rangle$, where $\mathcal{D}(X_{1}) = \{ \langle d_\pi , \{ f^{\pi }_1, f^{\pi }_2 \} \rangle , \langle d_\delta , \{ f^{\delta }_1, f^{\delta }_2 \} \rangle \}$ and where $\mathcal{D}(r_{1}) = \{ M_{d_\pi ,\;\{f^{\pi }_1\}}^{\pi }\} \rightarrow \pi$.

The question now arises: to what extent are the constraint relations defined in the standard setting preserved under the mapping described here from standard to dimensional information? Or more exactly: given a standard case base $\Gamma$ and fact situation X, is it the case that a decision for the side s is required in the situation X just in case, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, a decision for s is also required in the dimensional fact situation $\mathcal{D}(X)$? Since we are working with two models of constraint, result and reason, we need to ask the question separately for each model.

Beginning with the result model, it turns out that, here, the notion of constraint defined in the standard setting carries over without change to the dimensional setting.

Observation 4

Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the result model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, the result model of constraint requires a decision for s in the dimensional fact situation $\mathcal{D}(X)$.

Things are different when we turn to the reason model: here, the concept of constraint defined in the standard setting fails to survive interpretation into the dimensional setting. We can see this by reconsidering our earlier example, introduced in Sect. 2 to illustrate the reason model, in which a court faces the standard fact situation $X_{3} = \{ f^{\pi }_1, f^{\delta }_1 \}$ against the background of the standard case base $\Gamma _{1} = \{ c_{1} \}$. There, we noted that the reason model of constraint requires the court to decide this situation for the plaintiff, since a decision for the defendant would prioritize $\{ f^{\delta }_1 \}$ over $\{ f^{\pi }_1 \}$, but the opposite priority ordering is already supported by the background case base. If the example is interpreted in the dimensional setting, however—that is, supposing the court faces the dimensional fact situation $\mathcal{D}(X_{3}) = \{ \langle d_\pi , \{ f^{\pi }_1 \} \rangle , \langle d_\delta , \{ f^{\delta }_1 \} \rangle \}$ against the background of the dimensional case base $\mathcal{D}(\Gamma _{1})$—then the reason model no longer requires a decision for the plaintiff. This fact follows at once from Observation 3, according to which, in the dimensional setting, the reason model requires a decision for a particular side only if the result model also requires a decision for that side. But the result model does not require a decision for the plaintiff in this case, since we do not have $\mathcal{D}(X_{1}) \le ^\pi \mathcal{D}(X_{3})$. Why not? Because there is at least one dimension, namely $d_\pi$, whose value in the fact situation $\mathcal{D}(X_{3})$ does not favor the plaintiff as strongly as its value in $\mathcal{D}(X_{1})$; more precisely, ${\mathcal{D}(X_{1})}(d_\pi )$—that is, the value assigned to $d_\pi$ in the situation $\mathcal{D}(X_{1})$—is $\{ f^{\pi }_1, f^{\pi }_2 \}$ while ${\mathcal{D}(X_{3})}(d_\pi )$ is $\{ f^{\pi }_1 \}$, and since $\{ f^{\pi }_1, f^{\pi }_2 \} \subseteq \{ f^{\pi }_1 \}$ fails, our ordering on dimension values entails that we do not have ${\mathcal{D}(X_{1})}(d_\pi ) \preceq ^\pi {\mathcal{D}(X_{3})}(d_\pi )$.

If the reason model no longer requires a decision for the plaintiff in the dimensional situation $\mathcal{D}(X_{3})$, against the background of the dimensional case base $\mathcal{D}(\Gamma _{1})$, there must be some rule on the basis of which this situation can consistently be decided for the defendant. What is this rule? Well, as we saw in our discussion of Example 5, any respect in which a situation is weaker for a side than a situation from a background case already settled for that side can consistently be used as the basis of a rule supporting a decision for the opposite side. In the current example, the situation $\mathcal{D}(X_{3})$ is weaker for the plaintiff along the dimension $d_\pi$ than the situation $\mathcal{D}(X_{1})$ from the background case, by taking the value $\{ f^{\pi }_1 \}$ rather than $\{ f^{\pi }_1, f^{\pi }_2 \}$, and so can be decided for the defendant on that basis. More precisely, $\mathcal{D}(X_{3})$ can be decided for the defendant on the basis of the factor $M_{d_\pi ,\{f^{\pi }_1\}}^{\delta }$—according to which the value of this situation along the dimension $d_\pi$ favors the defendant at least as strongly as $\{ f^{\pi }_1 \}$. The resulting decision would be represented by the dimensional case $c_{11} = \langle X_{12}, r_{11}, s_{11} \rangle$, where $X_{11} = \mathcal{D}(X_{3})$, where $r_{11} = \{ M_{d_\pi ,\;\{f^{\pi }_1\}}^{\delta }\} \rightarrow \delta$, and where $s_{11}$ is $\delta$. The reader can verify that the expanded case base $\Gamma _{1} \cup \{ c_{11} \}$ is consistent, so that this decision is allowed by the reason model.

It is now worth asking: Once standard information is interpreted into the dimensional setting, can the reason model be modified so that it allows, in the dimensional setting, a pattern of constraint that aligns with that of the standard reason model? In fact, it can—each of the two refinements of the reason model described earlier leads to a pattern of constraint in the dimensional setting matching that from the standard setting.

According to the first refinement of the reason model, set out in Definition 13, a court is required to base its decision in a new situation on a rule that is not only consistent with the existing case base but also separated from any existing rule that applies to the new situation and supports the opposite outcome. This refinement, it turns out, leads to a reason model of constraint in the dimensional setting matching that from the standard setting.

Observation 5

Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the reason model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, a decision for s in $\mathcal{D}(X)$ is also required by the reason model with the constraint on rule selection subject to the first refinement.

And the general point can be illustrated with our previous example, where, as we can see, the requirement of separation blocks appeal to the new rule $r_{11} = \{M_{d_\pi ,\;\{f^{\pi }_1\}}^{\delta }\} \rightarrow \delta$ since it is not separated from the existing $\mathcal{D}(r_{1}) = \{ M_{d_\pi ,\;\{f^{\pi }_1\}}^{\pi }\} \rightarrow \pi$—both of these rule address the dimension $d_\pi$.

According to the second refinement of the reason model, set out in Definition 14, a court is required to base its decision on a rule that is not only consistent with the background case base but also, as we said, acceptable—where the notion of acceptability is schematic and can be interpreted in various ways. Earlier, we explored an interpretation of the acceptable rules as those based on salient values. In the present context, where standard information is interpreted in the dimensional setting, let us now suppose that the acceptable rules are the dimensionalizations of standard rules—that is, let us suppose that a dimensional rule r is acceptable if there is some standard rule $r'$ such that r is $\mathcal{D}(r')$. If we think of the acceptable rules in this way, then the second refinement also leads to a reason model of constraint in the dimensional setting matching that from the standard setting.

Observation 6

Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the reason model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, a decision for s in $\mathcal{D}(X)$ is also required by the reason model with the constraint on rule selection subject to the second refinement.

And again the general point can be illustrated with our previous example, which depends on the new rule $r_{11} = \{M_{d_\pi ,\;\{f^{\pi }_1\}}^{\delta }\} \rightarrow \delta$. Although this new rule—suggesting that we should decide for the defendant because the set of factors favoring the plaintiff is not any stronger than $\{ f^{\pi }_1 \}$—is a perfectly legitimate rule in the dimensional setting, it is not the dimensionalization of any standard rule, since, in the standard setting, a set of factors favoring the plaintiff, no matter how weak, can support only the plaintiff, not the defendant.

1.2 Observations and proofs

Observation 1 Let $\Gamma$ be a case base with $<_\Gamma$ its derived priority relation. Then $\Gamma$ is inconsistent if and only if there are cases $c = \langle X, r, s \rangle$ and $c' = \langle Y, r', \overline{s}\rangle$ belonging to $\Gamma$ such that ${ Premise}(r') <_c { Premise}(r)$ and ${ Premise}(r) <_{c'} { Premise}(r')$.

Proof

Suppose $\Gamma$ is inconsistent. Then there are cases $c = \langle X, r, s \rangle$ and $c' = \langle Y, r', \overline{s}\rangle$ belonging to $\Gamma$ such that $A <_c B$ and $B <_{c'} A$ for some reasons A and B. Since $A <_c B$, we have (1) and (2) . Since $B <_{c'} A$, we have (3) and (4) . From (1) and (4) we have , and of course , so that ${ Premise}(r') <_c { Premise}(r)$. From (2) and (3) we have , and of course , so that ${ Premise}(r) <_{c'} { Premise}(r')$. $\square$

Observation 2 Let $\Gamma$ be a consistent case base and X a new fact situation confronting the court, and suppose the result model of constraint requires a decision for the side s in the situation X. Then the reason model of constraint likewise requires a decision for s in this situation.

Proof

This result has already been established in the standard setting, using slightly different terminology, as Observation 5 from Horty (2011), and so it is shown here only for the dimensional setting.

Consider, then, a dimensional case base $\Gamma$ and fact situation X, where the result model of constraint requires a decision for s. Then there is some case $c = \langle Y, r, s \rangle$ from $\Gamma$ such that $Y \le ^s X$, which means, in the dimensional setting, that $Y(d) \le ^s X(d)$ for each dimension d. Now suppose that the result model does not require a decision for s in the situation X. Then it must be possible to consistently decide X for $\overline{s}$—that is, there must be some rule $r'$ favoring $\overline{s}$ such that $\Gamma \cup \{c'\}$ is consistent where $c' = \langle X, r', \overline{s}\rangle$.

We can verify that by showing that Y satisfies each magnitude factor from ${ Premise}(r')$, as follows. Suppose $M_{d,p}^{\overline{s}}$ is a magnitude factor from ${ Premise}(r')$. Then since $c'$ is a case, we have , that is, $p \le ^{\overline{s}} X(d)$. By assumption, we have $Y(d) \le ^s X(d)$, which yields $X(d) \le ^{\overline{s}} Y(d)$ by duality of the ordering relation on dimension values. By transitivity, this and the previous inequality then tell us that $p \le ^{\overline{s}} Y(d)$, or that . In the same way, we can verify that by showing that X satisfies each magnitude factor from ${ Premise}(r)$. Suppose $M_{d,p}^{s}$ is a magnitude factor from ${ Premise}(r)$. Then since c is a case, we have , or $p \le ^s Y(d)$. We again have $Y(d) \le ^s X(d)$ by assumption, and then $p \le ^s X(d)$ by transitivity, so that .

Since , and of course , we have ${ Premise}(r') <_c { Premise}(r)$. And since , and of course , we have ${ Premise}(r) <_{c'} { Premise}(r')$. But together, these two conclusions tell us that $\Gamma \cup \{c'\}$ is inconsistent, contrary to assumption. $\square$

Observation 3 Let $\Gamma$ be a consistent dimensional case base and X a new dimensional fact situation confronting the court, and suppose the reason model of constraint requires a decision for the side s in the situation X. Then the result model of constraint likewise requires a decision for the side s in this situation.

Proof

We reason by contraposition. Consider a dimensional case base $\Gamma$ and fact situation X, where the result model of constraint does not require a decision for s. Then there is no case c in $\Gamma$ such that $Outcome (c) = s$ and $Facts (c) \le ^s X$. In other words, for every c from $\Gamma$ with $Outcome (c) = s$, it is not the case that $Facts (c)(d) \le ^s X(d)$ for each dimension d—that is, for each such case c, there is some dimension d for which $Facts (c)(d) \le ^s X(d)$ fails. We show that the reason model cannot require a decision for s in the situation X either, by constructing a rule r favoring $\overline{s}$ such that $\Gamma \cup \{ c \}$ is consistent where $c = \langle X, r, \overline{s}\rangle$.

If there are no cases in $\Gamma$ that have been decided for s, then r can be any rule at all favoring $\overline{s}$ whose premise is satisfied by X, so we focus on the more interesting situation in which there are, in fact, cases c from $\Gamma$ such that $Outcome (c) = s$. For each such case c, let us define $d_c$ as a representative dimension for which $Facts (c)(d) \le ^s X(d)$. (It follows from the argument in the previous paragraph that there is at least one such dimension; if there are more than one, $d_c$ can be chosen arbitrarily.) We know, therefore, that (*) $Facts (c)(d_c) \le ^s X(d_c)$ fails for each c from $\Gamma$ such that $Outcome (c) = s$.

Now consider the magnitude factor

$$\begin{aligned} M_{d_c,\; X(d_c)}^{\overline{s}}, \end{aligned}$$

which holds in any situation in which the value of that situation along the dimension $d_c$ favors the side $\overline{s}$ at least as strongly as $X(d_c)$—that is, at least as strongly as the value of the situation X along the dimension $d_c$. We form the rule r by collecting together all the factors of this form for each case c from $\Gamma$ such that $outcome(c) = s$. More precisely, we take r as the rule

$$\begin{aligned} \{ M_{d_c,\; X(d_c)}^{\overline{s}} : \; c \in \Gamma \, {\mathrm{and}} \, Outcome (c) = s \} \rightarrow \overline{s}. \end{aligned}$$

In order to establish that $c = \langle X, r, \overline{s}\rangle$ is a case, we verify that by showing that X satisfies each magnitude factor from ${ Premise}(r)$. But this is trivial, since just in case $X(d_c) \le ^{\overline{s}} X(d_c)$, which is an instance of the reflexivity property of the value ordering.

Next, we establish that $\Gamma \cup \{c\}$ is consistent. Suppose otherwise. In that case, Observation 1 tells us that there is some $c' = \langle Y, r', s \rangle$ belonging to $\Gamma$ such that ${ Premise}(r') <_c { Premise}(r)$ and ${ Premise}(r) <_{c'} { Premise}(r')$. But ${ Premise}(r) <_{c'} { Premise}(r')$ requires that , which is impossible. Why? Because, since $c'$ is a case from $\Gamma$ with $Outcome (c') = s$, we know that ${ Premise}(r)$ contains a magnitude factor of the form $M_{d_{c'},\; X(d_{c'})}^{\overline{s}}$, which Y would have to satisfy. But just in case $X(d_{c'}) \le ^{\overline{s}} Y(d_{c'})$, which is equivalent by duality of the value ordering to $Y(d_{c'}) \le ^s X(d_{c'})$, which is equivalent, since $Facts (c') = Y$, to $Facts (c')(d_{c'}) \le ^s X(d_{c'})$, which we know to be false by (*) above. $\square$

Observation 4 Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the result model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, the result model of constraint requires a decision for s in the dimensional fact situation $\mathcal{D}(X)$.

Proof

We begin by verifying that, if X and Y are standard fact situations, then (1) $X \le ^s Y$ holds in the standard setting just in case, moving to the dimensional setting, we have (2) $\mathcal{D}(X) \le ^s \mathcal{D}(Y)$. In the standard setting, (1) is equivalent to (3) $X^s \subseteq Y^s$ and (4) $Y^{\overline{s}} \subseteq X^{\overline{s}}$, while in the dimensional setting, (2) means that $\mathcal{D}(X)(d) \le ^s \mathcal{D}(Y)(d)$ for each dimension d. But based on our interpretation of standard information into the dimensional setting, there are only two dimensions to consider, $d_s$ and $d_{\overline{s}}$, so that (2) is equivalent to (5) $\mathcal{D}(X)(d_s) \le ^s \mathcal{D}(Y)(d_s)$ and (6) $\mathcal{D}(X)(d_{\overline{s}}) \le ^s \mathcal{D}(Y)(d_{\overline{s}})$. Since, according to our interpretation, $\mathcal{D}(X)(d_s)$ is $X^s$ and $\mathcal{D}(Y)(d_s)$ is $Y^s$, (5) is just the statement (7) $X^s \le ^s Y^s$, where $X^s$ and $Y^s$ are to be interpreted as values along the dimension $d_s$, which holds, according to our dimensional value ordering, just in case $X^s \subseteq Y^s$, which is simply (3). And since, according to our interpretation, $\mathcal{D}(X)(d_{\overline{s}})$ is $X^{\overline{s}}$ and $\mathcal{D}(Y)(d_{\overline{s}})$ is $Y^{\overline{s}}$, (6) is just the statement (8) $X^{\overline{s}} \le ^s Y^{\overline{s}}$, where $X^{\overline{s}}$ and $Y^{\overline{s}}$ are to be interpreted as values along the dimension $d_{\overline{s}}$. By duality of the ordering relation on dimension values, (8) is equivalent to (9) $Y^{\overline{s}} \le ^{\overline{s}} X^{\overline{s}}$, which holds, according to our dimensional value ordering, just in case $Y^{\overline{s}} \subseteq X^{\overline{s}}$, which is simply (4).

Next, we turn to the result itself. Reasoning from left to right (the other direction is similar), with $\Gamma$ a standard case base and X a standard fact situation, suppose the result model requires a decision for s in X. Then there is a case c in $\Gamma$ with $Outcome (c) = s$ such that $Facts (c) \le ^s X$. But by our definition of the dimensionalization function, the case $\mathcal{D}(c)$ belongs to $\mathcal{D}(\Gamma )$, we still have $Outcome (\mathcal{D}(c)) = s$, and as we have just seen, $\mathcal{D}( Facts (c)) \le ^s \mathcal{D}(X)$, so that the result model requires a decision for s in the dimensional setting as well. $\square$

Lemma 1

LetXbe a standard fact situation andWa standard reason. Thenif and only if .

Proof

This fact is verified in Section 5.1 of the text, immediately after the definition of the $\mathcal{D}'$ function. $\square$

Lemma 2

Let$\Gamma$be a standard case base and$c = \langle X, r, s \rangle$and$c' = \langle Y, r', \overline{s}\rangle$cases belonging to$\Gamma$. Then${ Premise}(r') <_c { Premise}(r)$if and only if$\mathcal{D}'({ Premise}(r)) <_{\mathcal{D}(c)} \mathcal{D}'({ Premise}(r))$.

Proof

${ Premise}(r') <_c { Premise}(r)$ is equivalent to (1) and (2) , while $\mathcal{D}'({ Premise}(r')) <_{\mathcal{D}(c)} \mathcal{D}'({ Premise}(r))$ is equivalent to (3) and (4) . But (2) and (4) are obvious and (1) is equivalent to (3) by Lemma 1. Therefore ${ Premise}(r') <_c { Premise}(r)$ is equivalent to $\mathcal{D}'({ Premise}(r')) <_{\mathcal{D}(c)} \mathcal{D}'({ Premise}(r))$. $\square$

Lemma 3

Let$\Gamma$be a consistent standard case base,Xa standard fact situation, andra standard rule supporting the outcomes. Then if$\mathcal{D}(\Gamma ) \cup \{ \langle \mathcal{D}(X), \mathcal{D}(r), s \rangle \}$is consistent, so is$\Gamma \cup \{ \langle X, r, s \rangle \}$.

Proof

Suppose for contraposition that $\Gamma \cup \{ c \}$ is inconsistent where $c = \langle X, r, s \rangle$. Then $\Gamma$ must contain some case $c' = \langle Y, r', \overline{s}\rangle \}$ allowing us to show that ${ Premise}(r') <_c { Premise}(r)$ and ${ Premise}(r) <_{c'} { Premise}(r')$. But then by Lemma 2 we will also have $\mathcal{D}'({ Premise}(r')) <_{\mathcal{D}(c)} \mathcal{D}'({ Premise}(r))$ and $\mathcal{D}'({ Premise}(r)) <_{\mathcal{D}(c')} \mathcal{D}'({ Premise}(r'))$, and since $\mathcal{D}(c')$ must belong to $\mathcal{D}(\Gamma )$, it follows that $\mathcal{D}(\Gamma ) \cup \{ \mathcal{D}(c) \}$ is inconsistent as well. $\square$

Observation 5 Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the reason model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, a decision for s in $\mathcal{D}(X)$ is also required by the reason model with the constraint on rule selection subject to the first refinement.

Proof

Since the two directions are similar, we prove only the left to right direction. Assume, then, that the reason model requires a decision for s in the situation X. Then there must be some standard rule r supporting s such that $\Gamma \cup \{ \langle X, r, s \rangle \}$ is consistent, and there can be no standard rule consistently supporting the opposite side—that is, no $r'$ supporting $\overline{s}$ such that $\Gamma \cup \{ \langle X, r', \overline{s}\rangle \}$ is consistent.

Now suppose that, moving to the dimensional setting, the reason model subject to the first refinement does not require a decision for s in the situation $\mathcal{D}(X)$. Then there must be some dimensional rule $r''$ supporting $\overline{s}$ such that $\mathcal{D}(\Gamma ) \cup \{ \langle \mathcal{D}(X), r'', \overline{s}\rangle \}$ is consistent. Since , we know from Lemma 1 that , and of course $Outcome (\mathcal{D}(r)) = s$. By the refinement of the reason model, $r''$ must therefore be separated from $\mathcal{D}(r)$.

Since r is a standard rule, it has the form $W \rightarrow s$ for some $W \subseteq F^{s}$, with the consequence that $\mathcal{D}(r)$ has the form $\{M_{d_s,W}^{s}\} \rightarrow s$. The rule $\mathcal{D}(r)$, therefore, addresses the dimension $d_s$, so that, by separation, $r''$ must address the dimension $d_{\overline{s}}$. The rule $r''$ must therefore have the form $\{ M_{d_{\overline{s}},V}^{\overline{s}} \} \rightarrow \overline{s}$ for some $V \subseteq F^{\overline{s}}$, so that $r''$ is $\mathcal{D}(r''')$ where $r'''$ is the standard rule $V \rightarrow \overline{s}$. But since $\mathcal{D}(\Gamma ) \cup \{ \langle \mathcal{D}(X), r'', \overline{s}\rangle \}$ is consistent, we can conclude from Lemma 3 that $\Gamma \cup \{X, r''', \overline{s}\}$ is consistent as well, contrary to our assumption that the standard model requires a decision for s in X. $\square$

Observation 6 Let $\Gamma$ be a standard case base and X a standard fact situation confronting the court. Then the reason model of constraint requires a decision for the side s in the situation X if and only if, moving to the dimensional setting and working against the background of the dimensional case base $\mathcal{D}(\Gamma )$, a decision for s in $\mathcal{D}(X)$ is also required by the reason model with the constraint on rule selection subject to the second refinement.

Proof

Again we prove only the left to right direction. Assume, then, that the reason model requires a decision for s in the situation X, and suppose that, moving to the dimensional setting, the reason model subject to the second refinement does not require a decision for s in the situation $\mathcal{D}(X)$. Then there must be a dimensional rule r such that (1) $\mathcal{D}(\Gamma ) \cup \{ \langle \mathcal{D}(X), r, \overline{s}\rangle \}$ is consistent and (2) r is the dimensionalization of some standard rule $r'$. It then follows from Lemma 3 that $\Gamma \cup \{ \langle X, r', \overline{s}\rangle \}$ is consistent as well, contrary to the assumption that the reason model in the standard setting requires a decision for s. $\square$