Formalizing Kant’s Rules

Abstract

This paper formalizes part of the cognitive architecture that Kant develops in the Critique of Pure Reason. The central Kantian notion that we formalize is the rule. As we interpret Kant, a rule is not a declarative conditional stating what would be true if such and such conditions hold. Rather, a Kantian rule is a general procedure, represented by a conditional imperative or permissive, indicating which acts must or may be performed, given certain acts that are already being performed. These acts are not propositions; they do not have truth-values. Our formalization is related to the input/ output logics, a family of logics designed to capture relations between elements that need not have truth-values. In this paper, we introduce KL₃ as a formalization of Kant’s conception of rules as conditional imperatives and permissives. We explain how it differs from standard input/output logics, geometric logic, and first-order logic, as well as how it translates natural language sentences not well captured by first-order logic. Finally, we show how the various distinctions in Kant’s much-maligned Table of Judgements emerge as the most natural way of dividing up the various types and sub-types of rule in KL₃. Our analysis sheds new light on the way in which normative notions play a fundamental role in the conception of logic at the heart of Kant’s theoretical philosophy.

Appendix: Proofs

Proposition 5 (Soundness)

KL₁is sound: $R \vdash _{\text {KL}_{1}} r$implies$R \models _{\text {KL}_{1}} r$. That is, $deriv_{1}(R) \subseteq kl_{1}(R)$.

Proof

We show that if r ∈ deriv₁(R) then, for all A, out₁(R, A) = out₁(R ∪{r}, A). The proof is by induction on the length of the derivation. The base case is trivial. For the inductive step, suppose r was derived in one step from r₁ and r₂. By the inductive hypothesis $out_{1}(R \cup \{r_{1}, r_{2}\}, A) = out_{1}(R, A)$ for all A. We must show that, for all A:

$$ out_{1}(R \cup \{r_{1}, r_{2}, r\},\ A) = out_{1}(R \cup \{r_{1}, r_{2}\},\ A) $$

By Proposition 1, if X ∈ out₁(R ∪{r₁, r₂, r}, A) then X = M(D, A) for some definite program D ∈def(R ∪{r₁, r₂, r}) and X⊧R ∪{r₁, r₂, r}.

Suppose r was derived from r₁ and r₂ using MUST-UNION. (The other cases are similar and are omitted¹). Let r₁ = , r₂ = , r = ∪ C. It can be checked that def({ ∪ C}) = def({ , }), and so:

$$ \mathit{def}(R \cup \{r_{1}, r_{2}, r\}) = \mathit{def}(R \cup \{r_{1}, r_{2}\}) $$

This establishes that $out_{1}(R \cup \{r_{1}, r_{2}, r\},\ A) \subseteq out_{1}(R \cup \{r_{1}, r_{2}\},\ A)$ for all A. To show the other inclusion, that $out_{1}(R \cup \{r_{1}, r_{2}\}\ A) \subseteq out_{1}(R \cup \{r_{1}, r_{2}, r\},\ A)$, it remains to show that if X⊧R ∪{r₁, r₂} then X⊧R ∪{r₁, r₂, r}. To see this, note that if X⊧ then X⊧ ∪ C. □

Proposition 6 (Decomposition)

A set Rof rules semantically entails a rule rin KL₂if and only if$R \cup \mathcal {C} \cup aux(R)$semantically entails rin KL₁. That is, for all rule setsR:

$$ kl_{2}(R) = kl_{1}(R \cup \mathcal{C} \cup aux(R)) $$

Proof

$kl_{2}(R) = kl_{1}(R \cup \mathcal {C} \cup aux(R))$ for all R is equivalent to saying that two rules sets R₁ and R₂ are strongly equivalent in KL₂, kl₂(R₁) = kl₂(R₂), iff $R_{1} \cup \mathcal {C} \cup aux(R_{1})$ and $R_{2} \cup \mathcal {C} \cup aux(R_{2})$ are strongly equivalent in KL₁, $kl_{1}(R_{1} \cup \mathcal {C} \cup aux(R_{1})) = kl_{1}(R_{2} \cup \mathcal {C} \cup aux(R_{2}))$.

$$ \begin{array}{@{}rcl@{}} kl_{2}(R_{1}) &=& kl_{2}(R_{2}) \\ &&\text{iff}\quad out_{2}(R_{1},A) = out_{2}(R_{2},A) \quad \text{for all \textit{A}} \\ &&\text{iff}\quad out_{1}(R_{1} \cup \mathcal{C} \cup aux(R_{1}),\ A) = out_{1}(R_{2} \cup \mathcal{C} \cup aux(R_{2}),\ A) \quad \text{for all \textit{A}} \\ &&\text{iff}\quad kl_{1}(R_{1} \cup \mathcal{C} \cup aux(R_{1})) = kl_{1}(R_{2} \cup \mathcal{C} \cup aux(R_{2})) \end{array} $$

□

Proposition 7

Let Rbe a set of rules andAa (finite) set of literals:

Proof

We need to show that out₁(R, A) = ∅ iff for all sets $A^{\prime }$ of literals.

For left-to-right: suppose out₁(R, A) = ∅. First observe that for all $A^{\prime }$. (Because if then X is computed from assumptions $A^{\prime }$ using the non-constraint rules of R and X⊧R ∪ . Since X⊧R, that means $X \in out_{1}(R,A^{\prime })$ also.)

It remains to show that for all $A^{\prime }$. Assume $X \in out_{1}(R, A^{\prime })$; we shall show . Since $X \in out_{1}(R, A^{\prime })$, there is a D in def(R) such that $X = \mathrm {M}(D, A^{\prime })$. We will prove, first, that D is one of the definite programs of R ∪ , and second, that X⊧R ∪ . First, since def_r= {∅}, D ∈def(R ∪ ). So D, as well as being one of the definite programs of R, is also one of the definite programs of R ∪ . Second, since $X \in out_{1}(R, A^{\prime })$, X⊧R. We just need to show X⊧ . Since out₁(R, A) = ∅, $A \nvDash R$ by Proposition 2. Now, since X⊧R, $A \nsubseteq X$, hence X⊧ . These two claims entail, using Proposition 1, that .

For the other direction: suppose out₁(R, A)≠∅. We need to show that there is some $A^{\prime }$ such that . Take $A^{\prime } = A$: clearly out₁(R ∪ , A) = ∅. □

Proposition 8

Let Rbe a set of rules. If Xis a finite violating set of Rthen:

Proof

If X = ∅ (∅ is a violating set of R) then {a} and $\{{\sim }\mspace {1mu} a\}$ are also violating sets of R, any atom a, and . Clearly $aux(R) \cup \mathcal {C}$ is strongly inconsistent in KL₁ and so (Proposition 4).

Suppose X≠∅. If X is a (finite) violating set of R then . We show that by showing that .

Consider any rule X −{a} , a ∈ X. Suppose, for contradiction, that . Then $X^{\prime } \supseteq X$ and . In order to satisfy X −{a} , $X^{\prime }$ must contain $\overline {a}$. But a ∈ X so $X^{\prime }$ also contains a, and $X^{\prime } \not \models \mathcal {C}$. □

Remark

The proof above shows that if ∅ is a violating set of R then $aux(R) \cup \mathcal {C}$ is strongly inconsistent in KL₁. We can also show that if $R \cup \mathcal {C}$ is strongly inconsistent in KL₁ then ∅ is a violating set of R. (Because then $out_{1}(R \cup \mathcal {C},X) = \emptyset$ for every set X of literals including every maximal consistent set X, and ∅ is thus a violating set of R.) The converse however is not true. Consider . ∅ is a violating set of R but R is not strongly inconsistent: out₁(R, ∅) = {∅}≠∅.

Proposition 9

Let Rbe a set of rules and Xa finite set of literals.

Proof

One half follows from the preceding result: if X is a (finite) violating set of R then ; $kl_{1}(\mathcal {C} \cup aux(R)) \subseteq kl_{1}(R \cup \mathcal {C} \cup aux(R))$ and so ∈ kl₂(R). It remains to prove that if ∈ kl₂(R) then X is a violating set of R. We will prove that if $out_{1}(R \cup \mathcal {C} \cup aux(R),\ X) = \emptyset$ then X is a violating set of R.

Consider any definite logic program D_R in the encoding def(R) of R. Let def(aux(R)) = {D_aux} (all rules in aux(R) are rules with singleton heads and so there is a single definite program encoding aux(R)). M(D_R ∪ D_aux, X)⊧aux(R) so it must be that $\mathrm {M}(D_{R} \cup D_{aux},\ X) \not \models R \cup \mathcal {C}$, i.e., either M(D_R ∪ D_aux, X) is inconsistent or $B \subseteq \mathrm {M}(D_{R} \cup D_{aux},\ X)$ for some rule in R.

If X is inconsistent then X is a violating set of R. If X is consistent then consider any maximal consistent $X_{m} \supseteq X$. $\mathrm {M}(D_{R} \cup D_{aux},\ X_{m}) \supseteq \mathrm {M}(D_{R} \cup D_{aux},\ X)$, and since X_m is maximal, $X_{m} \supseteq \mathrm {M}(D_{R} \cup D_{aux},\ X_{m}) \supseteq \mathrm {M}(D_{R} \cup D_{aux},\ X)$. If M(D_R ∪ D_aux, X) is inconsistent then so is X_m, and that cannot be. So $B \subseteq \mathrm {M}(D_{R} \cup D_{aux},\ X)$ for some rule in R. But then $B \subseteq X_{m}$, and X_m⊮R. □

Proposition 11

Let Rbe a set of rules and Aa set of literals.

Proof

The result follows from the previous minimality result. It is enough to consider a singleton set of assumptions A = {a}. The general result follows by repeated application.

If is strongly inconsistent the result holds trivially. Suppose it is not strongly inconsistent. We need to show that:

We will show that .

Clearly $\{\overline {a}\}$ and all violating sets of R are violating sets of . Further, right. So . Now (assuming is not strongly inconsistent) $\{\overline {a}\}$ is a minimal consistent violating set of . If X is a minimal consistent violating set of R and $\overline {a} \in X$ then X is not a minimal consistent violating set of ; if $\overline {a} \notin X$ then X is a minimal consistent violating set of . So aux_m(R ∪ ) = aux_m(R) ∪ . □

Proposition 15 (Soundness of KL₂)

For all setsRof rules:

$$ deriv_{2}(R) \subseteq kl_{2}(R) $$

Proof

We need to show soundness of the inference rules of KL₁, ${\sim }\mspace {1mu}$-LEFT and ${\sim }\mspace {1mu}$-RIGHT with respect to semantic entailment in KL₂. Since $kl_{2}(R) = kl_{1}(R \cup \mathcal {C} \cup aux(R))$ (Proposition 6) and KL₁ is sound with respect to kl₁, the soundness of KL₁ inference rules is immediate. Soundness of ${\sim }\mspace {1mu}$-LEFT is just $\mathcal {C} \subseteq kl_{2}(R)$ which also follows trivially.

It remains to show that ${\sim }\mspace {1mu}$-RIGHT is sound: if r ∈ right(R) then $R \models _{\text {KL}_{2}} r$, or more generally, that $right(R) \subseteq kl_{2}(R)$. We want to show:

$$ right(R) \subseteq kl_{1}(R \cup \mathcal{C} \cup aux(R)) $$

We will show that $right(R) \subseteq aux(R)$ (which implies the above). In full: if r ∈ right(R) then r is a rule of the form where is a rule in R. In that case B ∪{c} is a (not necessarily minimal or consistent) violating set of R, and aux(R) contains the rule . □

Proposition 16 (Conditional completeness of KL₂)

If KL₁is complete with respect to out₁then KL₂is complete with respect to out₂. That is: if, for all sets Rof rules$kl_{1}(R) \subseteq deriv_{1}(R)$then, for all sets Rof rules$kl_{2}(R) \subseteq deriv_{2}(R)$.

Proof

$$ \begin{array}{@{}rcl@{}} r \in kl_{2}(R) \ &\Rightarrow&\ r \in kl_{1}(R \cup \mathcal{C} \cup aux(R)) \\ &\Rightarrow&\ r \in kl_{1}(R \cup \mathcal{C} \cup aux_{e}(R)) \quad\qquad\qquad\qquad\qquad\hfill(\text{Proposition~13}) \\ &\Rightarrow&\ r \in deriv_{1}(R \cup \mathcal{C} \cup aux_{e}(R)) \qquad\qquad\quad\hfil(\text{completeness of KL}_{1}) \\ &\Rightarrow&\ r \in deriv_{2}(R) \end{array} $$

The final step is because all the inference rules used in the construction of aux_e(R) in Proposition 14 are inference rules of KL₂. □

Proposition 17 (Decomposition of KL₂)

If KL ₁ is complete with respect to o u t ₁ then

$$ deriv_{2}(R) = deriv_{1}(R \cup \mathcal{C} \cup aux_{e}(R)) $$

Proof

Right-in-left inclusion is noted in the proof of Proposition 16. The other inclusion is similar:

$$ \begin{array}{@{}rcl@{}} r \in deriv_{2}(R) \ &\Rightarrow&\ r \in kl_{2}(R)\qquad\qquad\qquad\qquad\qquad\qquad \quad\hfil(\text{soundness of KL}_{2}) \\ &\Rightarrow&\ r \in kl_{1}(R \cup \mathcal{C} \cup aux(R)) \\ &\Rightarrow&\ r \in kl_{1}(R \cup \mathcal{C} \cup aux_{e}(R))\qquad\qquad\qquad\qquad \hfil(\text{Proposition~13}) \\ &\Rightarrow&\ r \in deriv_{1}(R \cup \mathcal{C} \cup aux(R))\qquad\qquad \hfil(\text{completeness of KL}_{1}) \end{array} $$

□

Proposition 19

LetRbe a set of rules and Aa set of assumptions, both containing no negative literals. Then:

$$ out_{2}(R,A)^{+} = out_{1}(R,A) $$

Proof

We can see this again by looking at the rules in aux(R): if R contains no negative literals, all rules in aux(R) are of the form where c and all of B are atoms. These rules have no effect on the outcomes computed from R except possibly to add negative literals. This is clear if we look at the translation to definite logic programs: every definite program D in the encoding $\mathit {def}(R \cup \mathcal {C} \cup aux(R))$ has the form D_R ∪ D_aux where D_R ∈def(R) and def(aux(R)) = {D_aux} (all rules in aux(R) are rules with singleton heads and so there is a single definite program encoding aux(R)). Moreover, since none of the heads of clauses in D_aux appear in D_R the least model M(D_R ∪ D_aux, A) = M(D_R,M(D_aux, A)) (indeed that is true whether the set A of assumptions contains negative literals or not). If A contains no negative literals, then this least model also satisfies the constraints $\mathcal {C}$. □