## 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 KL_{3} 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 KL_{3}. 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.

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## Notes

For other inference rules it is not always the case that

*d**e**f*({*r*_{1},*r*_{2},*r*}) =*d**e**f*({*r*_{1},*r*_{2}}). However, for the first half of the proof it is sufficient to show that for every*D*∈*d**e**f*({*r*1,*r*, 2,*r*}) there exists \(D^{\prime } \in \mathit {def}(\{r1,r,2\})\) such that*D*and \(D^{\prime }\) have the same models (and therefore the same minimal model). This is straightforward for all the inference rules of Fig. ??.

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## Acknowledgements

We are very grateful to Michiel van Lambalgen for extensive feedback, and for a number of suggestions that have been incorporated into the text. Thanks also to Barnaby Evans for feedback and suggestions.

## Author information

### Authors and Affiliations

### Contributions

Richard Evans designed the logic and wrote the first draft of the paper. Marek Sergot developed the alternative semantics for KL_{1}, developed the semantics for KL_{2}, and improved the semantics for KL_{3}. Andrew Stephenson improved the philosophical discussion and added further discussion of Kant. All three authors edited, revised, and polished the final draft.

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## Appendix: Proofs

### Appendix: Proofs

**Proposition 5** (Soundness)

*KL*_{1}*is 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* ∈ *d**e**r**i**v*_{1}(*R*) then, for all *A*, *o**u**t*_{1}(*R*, *A*) = *o**u**t*_{1}(*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 *r*_{1} and *r*_{2}. 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*:

By Proposition 1, if *X* ∈ *o**u**t*_{1}(*R* ∪{*r*_{1}, *r*_{2}, *r*}, *A*) then *X* = M(*D*, *A*) for some definite program *D* ∈*d**e**f*(*R* ∪{*r*_{1}, *r*_{2}, *r*}) and *X*⊧*R* ∪{*r*_{1}, *r*_{2}, *r*}.

Suppose *r* was derived from *r*_{1} and *r*_{2} using MUST-UNION. (The other cases are similar and are omitted^{Footnote 1}). Let *r*_{1} = , *r*_{2} = , *r* = ∪ *C*. It can be checked that *d**e**f*({ ∪ *C*}) = *d**e**f*({ , }), and so:

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* ∪{*r*_{1}, *r*_{2}} then *X*⊧*R* ∪{*r*_{1}, *r*_{2}, *r*}. To see this, note that if *X*⊧ then *X*⊧ ∪ *C*. □

**Proposition 6** (Decomposition)

*A set R**of rules semantically entails a rule r**in KL*_{2}*if and only if*\(R \cup \mathcal {C} \cup aux(R)\)*semantically entails r**in KL*_{1}. *That is, for all rule sets**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 *R*_{1} and *R*_{2} are strongly equivalent in KL_{2}, *k**l*_{2}(*R*_{1}) = *k**l*_{2}(*R*_{2}), iff \(R_{1} \cup \mathcal {C} \cup aux(R_{1})\) and \(R_{2} \cup \mathcal {C} \cup aux(R_{2})\) are strongly equivalent in KL_{1}, \(kl_{1}(R_{1} \cup \mathcal {C} \cup aux(R_{1})) = kl_{1}(R_{2} \cup \mathcal {C} \cup aux(R_{2}))\).

□

###
**Proposition 7**

*Let R**be a set of rules and**A**a (finite) set of literals*:

###
*Proof*

We need to show that *o**u**t*_{1}(*R*, *A*) = *∅* iff
for all sets \(A^{\prime }\) of literals.

For left-to-right: suppose *o**u**t*_{1}(*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 *d**e**f*(*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 *d**e**f*_{r}= {*∅*}, *D* ∈*d**e**f*(*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 *o**u**t*_{1}(*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 *o**u**t*_{1}(*R*, *A*)≠*∅*. We need to show that there is some \(A^{\prime }\) such that . Take \(A^{\prime } = A\): clearly *o**u**t*_{1}(*R* ∪ , *A*) = *∅*. □

###
**Proposition 8**

*Let R**be a set of rules. If X**is a finite violating set of R**then*:

###
*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 KL_{1} 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 KL_{1}. We can also show that if \(R \cup \mathcal {C}\) is strongly inconsistent in KL_{1} 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: *o**u**t*_{1}(*R*, *∅*) = {*∅*}≠*∅*.

###
**Proposition 9**

*Let R**be a set of rules and X**a 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 ∈ *k**l*_{2}(*R*). It remains to prove that if ∈ *k**l*_{2}(*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 *d**e**f*(*R*) of *R*. Let *d**e**f*(*a**u**x*(*R*)) = {*D*_{aux}} (all rules in *a**u**x*(*R*) are rules with singleton heads and so there is a single definite program encoding *a**u**x*(*R*)). M(*D*_{R} ∪ *D*_{aux}, *X*)⊧*a**u**x*(*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 R**be a set of rules and A**a 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, *r**i**g**h**t*. 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 *a**u**x*_{m}(*R* ∪ ) = *a**u**x*_{m}(*R*) ∪ . □

**Proposition 15** (Soundness of KL_{2})

*For all sets**R**of rules*:

###
*Proof*

We need to show soundness of the inference rules of KL_{1}, \({\sim }\mspace {1mu}\)-LEFT and \({\sim }\mspace {1mu}\)-RIGHT with respect to semantic entailment in KL_{2}. Since \(kl_{2}(R) = kl_{1}(R \cup \mathcal {C} \cup aux(R))\) (Proposition 6) and KL_{1} is sound with respect to *k**l*_{1}, the soundness of KL_{1} 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* ∈ *r**i**g**h**t*(*R*) then \(R \models _{\text {KL}_{2}} r\), or more generally, that \(right(R) \subseteq kl_{2}(R)\). We want to show:

We will show that \(right(R) \subseteq aux(R)\) (which implies the above). In full: if *r* ∈ *r**i**g**h**t*(*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 *a**u**x*(*R*) contains the rule . □

**Proposition 16** (Conditional completeness of KL_{2})

*If KL*_{1}*is complete with respect to o**u**t*_{1}*then KL*_{2}*is complete with respect to o**u**t*_{2}. *That is: if, for all sets R**of rules*\(kl_{1}(R) \subseteq deriv_{1}(R)\)*then, for all sets R**of rules*\(kl_{2}(R) \subseteq deriv_{2}(R)\).

###
*Proof*

The final step is because all the inference rules used in the construction of *a**u**x*_{e}(*R*) in Proposition 14 are inference rules of KL_{2}. □

**Proposition 17** (Decomposition of KL_{2})

*If KL*
_{1}
*is complete with respect to o*
*u*
*t*
_{1}
*then*

###
*Proof*

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

□

###
**Proposition 19**

*Let**R**be a set of rules and A**a set of assumptions, both containing no negative literals. Then*:

###
*Proof*

We can see this again by looking at the rules in *a**u**x*(*R*): if *R* contains no negative literals, all rules in *a**u**x*(*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} ∈*d**e**f*(*R*) and *d**e**f*(*a**u**x*(*R*)) = {*D*_{aux}} (all rules in *a**u**x*(*R*) are rules with singleton heads and so there is a single definite program encoding *a**u**x*(*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 \(\mathcal {C}\). □

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### Cite this article

Evans, R., Sergot, M. & Stephenson, A. Formalizing Kant’s Rules.
*J Philos Logic* **49**, 613–680 (2020). https://doi.org/10.1007/s10992-019-09531-x

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DOI: https://doi.org/10.1007/s10992-019-09531-x