Guilty Minds and Collective Know-how

Abstract

Many philosophers agree that moral responsibility often includes an epistemic requirement. Inspired by the dictum that ‘ought implies can’, this epistemic perspective induces the requirement that an agent can only be held morally responsible if she knows how to fulfil her subjective obligation yet decided not to. The central theorems of this chapter concern the characterization of both individual and collective know-how. On the one hand, I show that an individual agent knows how to do something if and only if the agent knows that it is possible for her to knowingly do it. Collective know-how, on the other hand, requires common knowledge of an effective division of tasks among the group’s members and of the fact that each member knows how to carry out her part. Accordingly, no communication is needed once the group collectively knows how to achieve some collective goal: if each member knowingly plays her part (which she knows how to do), then the group jointly achieves their collective goal. I explore some connections to the literature on knowledge, action, and ability in philosophy, economics, and artificial intelligence, and analyse the problem of responsibility voids and gaps from the perspective of informational responsibility systems.

The “epistemic condition” corresponds to the excuse of ignorance. It captures the intuition that an agent is responsible only if he both knows the particular facts surrounding his action, and acts with the proper sort of beliefs and intentions.

Fischer and Ravizza (1998, p. 13)

Parts of this chapter draw from (Duijf et al., 2021).

Appendices

Appendix C: Guilty Minds and Collective Know-how

C.1 Epistemic Stit Theory

Theorem 4.1 (Completeness Epistemic Stit)

The following axiom schemas, in combination with a standard axiomatization for propositional logic, and the standard rules ( such as necessitation)for the normal modal operators, provide a complete Hilbert system for the validities on epistemic stit models:

(S5 Historical Necessity)	S5 for
(S5 Agency)	for each group $\mathcal {G}$: S5 for $[\mathcal {G}\;\mathsf {stit}]$
(Agent Monotonicity)	for all groups $\mathcal {F}$ and $\mathcal {G}$ satisfying $\mathcal {F}\subseteq \mathcal {G}$: $[\mathcal {F}\;\mathsf {stit}] \varphi \to [\mathcal {G}\;\mathsf {stit}] \varphi$
(Independence of Agency)	for all groups $\mathcal {F}$ and $\mathcal {G}$ satisfying $\mathcal {F}\cap \mathcal {G}=\emptyset$: $\Diamond [\mathcal {F}\;\mathsf {stit}] \varphi \wedge \Diamond [\mathcal {G}\;\mathsf {stit}] \psi \to \Diamond ([\mathcal {F}\;\mathsf {stit}] \varphi \wedge [\mathcal {G}\;\mathsf {stit}] \psi )$
(S5 Knowledge)	for each i ∈Ags: S5 for K i
(Public Knowledge)	for each group $\mathcal {G}$: $\mathsf {C}_{\mathcal {G}}\varphi \to (\varphi \land \mathsf {E}_{\mathcal {G}}\mathsf {C}_{\mathcal {G}}\varphi )$
(Induction)	for each group $\mathcal {G}$: $\varphi \land \mathsf {C}_{\mathcal {G}}(\varphi \to \mathsf {E}_{\mathcal {G}}\varphi ) \to \mathsf {C}_{\mathcal {G}}\varphi$

Proof (Sketch)

The epistemic stit logic is a so-called fusion of epistemic logic and non-epistemic stit theory. The complete logical system for epistemic stit theory is therefore given by the simple combination of the logical systems for epistemic logic and (non-epistemic) stit logic (see Carnielli & Coniglio, 2020, Sect. 4.1).

It is well-established that the non-epistemic fragment is complete with respect to stit models, for instance, by using Sahlqvist correspondence (see the standard textbook treatment of Blackburn et al. (2001)). The completeness of the epistemic logic is standard (see, e.g. Meyer & van der Hoek, 1995). (Broersen, 2011, Theorem 2.1 proves a similar result for Xstit, the difference is that his logic concerns Xstit and excludes common knowledge.)□

C.2 Individual Practical Knowledge

Proposition 4.1

(OAC):

The own-action condition , schematically expressed by K _i φ →K _i[i s t i t]φ, corresponds to the condition:

For all indices m∕h, $m^{\prime }/h^{\prime }_1$ , and $m^{\prime }/h^{\prime }_2$ , if $m/h\sim _i m^{\prime }/h^{\prime }_1$ and $h^{\prime }_2\in \mathit {Act}_{i}(m^{\prime }/h^{\prime }_1)$ then $m/h\sim _i m^{\prime }/h^{\prime }_2$.

(Unif-H):

The uniformity of historical possibility property , schematically expressed by ♢K _i φ →K _i♢φ, corresponds to the confluency condition:

For all indices m∕h ₁ , m∕h ₂ , and $m^{\prime }/h^{\prime }_1$ , if $m/h_1\sim _i m^{\prime }/h^{\prime }_1$ then there is an index $m^{\prime }/h^{\prime }_2$ such that $m/h_2\sim _i m^{\prime }/h^{\prime }_2$.

Proof

These correspondences can be checked using the algorithm SQEMA (Conradie et al., 2006).□

Theorem 4.2 (Impossibility Results)

Implicit individual know-how and explicit individual know-how cannot be characterized in the current logic:

• 1. Given an agent i and a formula φ, there is noformula ψ such that for every model $\mathcal {M}$ and every index m∕h it holds that, relative to m∕h, agent i implicitlyknows how to φ if and only if $\mathcal {M},m/h\vDash \psi$.

• 2. Given an agent i and a formula φ, there is noformula ψ such that for every model $\mathcal {M}$ and every index m∕h it holds that, relative to m∕h, agent i explicitlyknows how to φ if and only if $\mathcal {M},m/h\vDash \psi$.

Proof

Implicit individual know-how would be characterizable in $\mathfrak {L}_{\mathit {estit}}$ if and only if, given an agent i ∈Ags and a formula φ, there is a formula ψ available such that for any model $\mathcal {M}$ and any index m∕h it holds that $\mathcal {M},m/h\vDash \psi$ iff i implicitly knows how to φ.

We argue by contradiction. Suppose ψ characterizes that agent i implicitly knows how to p. Let P ^′⊂ P denote the set of propositional variables occurring in either ψ or p, and let q∉P ^′. Consider a model $\mathcal {M}$ and an index m∕h in which agent i has two options, that is, $\mathit {Act}_{i}^{m}$ contains two acts. Let the epistemic indistinguishability relation ∼_i be given by m ₁∕h ₁ ∼_i m ₂∕h ₂ if and only if m ₂∕h ₂ ∈Act _i(m ₁∕h ₁). Consider the valuation function V  which assigns to each propositional variable the set of all indices in the model. It is now easy to see that i implicitly knows how to p, because for every formula χ we have or . The assumption implies that $\mathcal {M},\vDash \psi$ holds. See Fig. 4.11 for the initial model in which agent i implicitly knows how to p.

Fig. 4.11

The initial model $\mathcal {M}$

We can construct a countermodel $\mathcal {M}^{\prime }$ from $\mathcal {M}$ by adjusting the propositional valuation to V ^′ in the following way: one choice guarantees q, the other guarantees ¬q, and both guarantee any p ^′∈ P ^′ (see the constructed model in Fig. 4.12). Note that V ^′ only differs from V  on q. It is easy to see that in $\mathcal {M}^{\prime }$, i does not implicitly know how to p, because, in $\mathcal {M}^{\prime }$, q is a proper refinement of p for agent i. However, because we did not alter the truth values of the propositions occurring in ψ, it is the case that ψ also holds in $\mathcal {M}^{\prime }$. Contradiction.

Fig. 4.12

The constructed model $\mathcal {M}^{\prime }$

In a similar way, one can prove that explicit knowledge is also not characterizable.□

Theorem 4.3 (Characterization of Individual Know-how)

Individual know-how is characterized by

$$\displaystyle \begin{aligned}\mathsf{K}_i \Diamond\mathsf{K}_i [i\;\mathsf{stit}]\varphi.\end{aligned}$$

That is, for every model $\mathcal {M}$ , every index m∕h, every individual agent i, and every formula φ it holds that, relative to m∕h, agent i knows how to φ if and only if $\mathcal {M},m/h\vDash \mathsf {K}_i\Diamond \mathsf {K}_i [i\;\mathsf {stit}]\varphi$.

Proof

(If.) Follows immediately by letting ψ := φ in Definition 4.5.

(Only if.) It is immediate for implicit know-how. For explicit know-how it follows from two facts. First, is monotonic: , which entails . Replacing χ ₁ with K _i[i s t i t]ψ and χ ₂ with K _i[i s t i t]φ yields

Second, by K _i-necessitation, and two applications of the K-axiom for K _i it holds that:

By assumption, both the antecedent of the principal implication and the antecedent of the subordinate implication obtain. Hence, K _i♢K _i[i s t i t]φ holds, as desired. □

C.3 Collective Know-how

Theorem 4.4

Let $\mathcal {M}$ be an epistemic stit model, and let m∕h be an index. Assume that $\mathcal {M}$ satisfies (Unif-H). Suppose $\mathcal {G}$ collectively knows how to φ, as witnessed by $(\varphi _i)_{i\in \mathcal {G}}$ . Then

(*)

and

(**)

Proof

Assume all the stated assumptions. To clarify, the crucial assumptions are (1) and (2) $\mathsf {C}_{\mathcal {G}}\bigwedge _{i\in \mathcal {G}} \mathsf {K}_{i}\Diamond \mathsf {K}_{i}[i\;\mathsf {stit}]\varphi _i$. It is important to note that (2) entails (2’) $\mathsf {C}_{\mathcal {G}}\bigwedge _{i\in \mathcal {G}} \Diamond [i\;\mathsf {stit}]\varphi _i$.

First, consider the following subformula of (1), which results from ignoring the outermost modality $\mathsf {C}_{\mathcal {G}}$: (1’) . Let us take an arbitrary $j\in \mathcal {G}$. The right-to-left implication of (1’) entails that at any index m∕h where (1’) obtains, we get that $m/h\vDash \langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi \to \langle j\;\mathsf {stit}\rangle [j\;\mathsf {stit}]\varphi _j$. Note that $\vDash \langle j\;\mathsf {stit}\rangle [j\;\mathsf {stit}]\varphi _j\to [j\;\mathsf {stit}]\varphi _j$ and, therefore, at any index m∕h where (1’) obtains, we get $m/h\vDash \langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi \to [j\;\mathsf {stit}]\varphi _j$. Since j was arbitrary, is an S5 modality and $\mathsf {C}_{\mathcal {G}}$ is a normal modality, it follows that (1) entails .

For the converse, consider the subformula of (2’), which results from ignoring the outermost modality $\mathsf {C}_{\mathcal {G}}$: (2”) $\bigwedge _{i\in \mathcal {G}} \Diamond [i\;\mathsf {stit}]\varphi _i$. The independence of agency condition entails that at any index where (2”) holds also $\Diamond \bigwedge _{i\in \mathcal {G}} [i\;\mathsf {stit}]\varphi _i$ obtains. Let us take an arbitrary $j\in \mathcal {G}$. Then, at any index m∕h where (2”) obtains, we get that $m/h\vDash [j\;\mathsf {stit}]\varphi _j\to \langle j\;\mathsf {stit}\rangle \bigwedge _{i\in \mathcal {G}} [i\;\mathsf {stit}]\varphi _i$. The left-to-right implication in (1’) entails that at any index m∕h where (1’) obtains, we get that $m/h\vDash \langle j\;\mathsf {stit}\rangle \bigwedge _{i\in \mathcal {G}} [i\;\mathsf {stit}]\varphi _i\to \langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$. Hence, at any index m∕h where (2”) and (1’) hold, we get $m/h\vDash [j\;\mathsf {stit}]\varphi _j\to \langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$. Since j was arbitrary and and $\mathsf {C}_{\mathcal {G}}$ are normal modalities, we have , as desired.

Second, (∗∗) follows from (∗) and two facts. First, the fact that $\vDash \mathsf {C}_{\mathcal {G}}\chi \to \mathsf {C}_{\mathcal {G}}\mathsf {K}_i\chi$, for each $i\in \mathcal {G}$. Second, the fact that (Unif-H) entails that . □

Observation 4.5 (Interchangeability & Effectivity)

Let S be a game model, let $\mathcal {G}$ be a group of agents, and let φ be a formula. The following are equivalent:

• 1. ;

• 2. $P_{\mathcal {G}}:=\{a_{\mathcal {G}}\in A_{\mathcal {G}}\mid S,a\vDash [\mathcal {G}\;\mathsf {stit}]\varphi \}$ is interchangeable.

Proof

(1. ⇒ 2.) Assume 1. Let b, c ∈ A be such that $b_{\mathcal {G}},c_{\mathcal {G}}\in P_{\mathcal {G}}$, and let $i\in \mathcal {G}$. To prove 2., we take an arbitrary d ∈ A such that $d_{\mathcal {G}}=(b_{\mathcal {G} -i},c_{i})$, and prove that $d_{\mathcal {G}}\in P_{\mathcal {G}}$. For any $j\in \mathcal {G}-i$ it holds that $S,d\vDash \langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$, because $S,b\vDash [\mathcal {G}\;\mathsf {stit}]\varphi$ and d _j = b _j. Hence, $S,d\vDash \bigwedge _{j\in \mathcal {G}-i}\langle j\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$. In a similar way, we can prove that $S,d\vDash \langle i\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$ (note the ‘i’ here). By the right-to-left implication in 1. it follows that $S,d\vDash [\mathcal {G}\;\mathsf {stit}]\varphi$. Hence $d_{\mathcal {G}}\in P_{\mathcal {G}}$.

(2. ⇒ 1.) Assume 2. Take any a ∈ A. We need to show that $S,a\vDash [\mathcal {G}\;\mathsf {stit}]\varphi \leftrightarrow \bigwedge _{i\in \mathcal {G}}\langle i\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$. The left-to-right implication follows immediately from the fact that [i s t i t] is reflexive, i.e. satisfies $\vDash \chi \to \langle i\;\mathsf {stit}\rangle \chi$. To establish the right-to-left implication, assume $S,a\vDash \bigwedge _{i\in \mathcal {G}} \langle i\;\mathsf {stit}\rangle [\mathcal {G}\;\mathsf {stit}]\varphi$. That is, for every $i\in \mathcal {G}$ there is a $b^{i}_{\mathcal {G}}\in P_{\mathcal {G}}$ such that $b^{i}_{i}=a_{i}$. Using the fact that $P_{\mathcal {G}}$ is interchangeable, one can inductively show that for any subgroup $\mathcal {F}\subseteq \mathcal {G}$ there is a $b^{\mathcal {F}}_{\mathcal {G}}\in P_{\mathcal {G}}$ such that $b^{\mathcal {F}}_{\mathcal {F}}=a_{\mathcal {F}}$. Hence, $a_{\mathcal {G}}=b^{\mathcal {G}}_{\mathcal {G}}\in P_{\mathcal {G}}$ and therefore there is an a ^′∈ A such that $a^{\prime }_{\mathcal {G}}=a_{\mathcal {G}}$ and $S,a^{\prime }\vDash [\mathcal {G}\;\mathsf {stit}]\varphi$. Since $\vDash [\mathcal {G}\;\mathsf {stit}]\varphi \leftrightarrow [\mathcal {G}\;\mathsf {stit}][\mathcal {G}\;\mathsf {stit}]\varphi$ and $a^{\prime }_{\mathcal {G}}=a_{\mathcal {G}}$, it holds that $S,a\vDash [\mathcal {G}\;\mathsf {stit}]\varphi$, as desired.□