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Reasoning about coalitional agency and ability in the logics of “bringing-it-about”

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Abstract

The logics of “bringing-it-about” have been part of a prominent tradition for the formalization of individual and institutional agency. They are the logics to talk about what states of affairs an acting entity brings about while abstracting away from the means of action. Elgesem’s proposal analyzes the agency of individual agents as the goal-directed manifestation of an individual ability. It has become an authoritative modern reference. The first contribution of this paper is to extend Elgesem’s logic of individual agency and ability to coalitions. We present a general theory and later propose several possible specializations. As a second contribution, we offer algorithms to reason with the logics of bringing-it-about and we analyze their computational complexity.

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Notes

  1. We develop the research agenda presented in the earlier extended abstract published as [46].

  2. Maybe an exception is [33]. There, the agency of a coalition is defined as adequate combinations of individual agency of its members. The additional operators with coalitions are then merely syntactic sugar, leaving the expressivity of the logic unchanged. The logic also goes beyond the Anselmian tradition by allowing explicit action terms in the language.

  3. The main reference is [15] which unfortunately is not published. The Ref. [16] is essentially the Chapter 2 of [15].

  4. Similar distinctions between different kinds of ability have later been made by Kapitan [27] and Mele [31]. Mele calls them simple ability to A and ability to A intentionally. He writes: “an agent’s \(A\)-ing at a time is sufficient for his having the simple ability to \(A\) at that time.”, and “being able to \(A\) intentionally entails having a simple ability to \(A\) and the converse is false” [31, p. 448].

  5. Notice that \(\lnot E_x\top \) becomes redundant.

  6. See [52] for a similar and even weaker account of ability.

  7. Miller uses the terminology “collective end”.

  8. This example is adapted from [5].

  9. Observe that, as pointed out in Sect. 2.5 for Constraints 5 and 7, axiom Ax3 is redundant in presence of axiom Ax7. Given its importance in \(\mathsf{ELG }\) \(^N\), we conserve it in this axiomatization.

  10. Contrarily to \(\mathsf{BIAT }\) \(^N\) and \(\mathsf{ELG }\) \(^N\), we need \(N\) to be finite. Our language refers to coalitions of agents that are subsets of \(N\), and the set of subsets of an enumerable infinite set is not enumerable. Hence, a logic \(\mathsf{COAL }\) \(^N\) with a possibly infinite enumerable set \(N\) of agents would not be finitely axiomatizable.

  11. See for instance the results of the last SAT competition. http://www.satcompetition.org/.

  12. We understand group attitude only loosely in the sense of [48].

  13. See for instance [44].

  14. It would just consist in rewriting the case of Constraint 9 in the proof of Theorem 4.

  15. Nevertheless, it should be noted that bringing-it-about logics are not inherently incompatible with a refined ontology of time, as argued in [15].

  16. A rule of thumb is that over a non-restricted language (all Boolean combinations, unbounded modal depth, etc), if a classical modal logic admits axiom Ax1 (axiom \(K\)) but does not admit \(\lnot E_x \varphi \rightarrow E_x \lnot E_x \varphi \) (axiom \(5\)), it is PSPACE-hard.

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Acknowledgments

I would like to thank Andreas Herzig and Emiliano Lorini, as well as my colleagues at the Laboratory for Applied Ontology for inspiring and commenting earlier presentations of this work. It also greatly benefited from suggestions and corrections of reviewers on previous drafts. This research was funded by a Marie Curie Actions Fellowship FP7 PEOPLE COFUND Trentino.

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Annex: proof of Theorem 3

Annex: proof of Theorem 3

1.1 Left-to-right direction

Suppose \(\varphi \) is \(\mathsf{BIAT \text{- }\mathsf sat } \,\). This means that there is a model of individual agency \(M = ( S, N, EE, v )\), and a world \(w \in S\), such that \(M, w \models \varphi \). Define the function \(\pi : {sub^{\lnot }}(\varphi ) \longrightarrow \{0,1\}\) as follows: for all \(\psi \in {sub^{\lnot }}(\varphi )\),

$$\begin{aligned} \pi (\psi ) = \left\{ \begin{array}{ll} 1 &{} \quad \text{ when } \; M,\, w \models \psi \\ 0 &{} \quad \text{ otherwise. } \end{array}\right. \end{aligned}$$

It is readily checked that \(\pi \) is a semi-valuation. It remains to show that \(\pi \) satisfies all three conditions.

  1. 1.

    if \(\pi (E_a \varphi ) = 1\) then \(M, w \models E_a \varphi \) by definition of \(\pi \). So \(|| \varphi ||^{M} \in EE_w(a)\). Since \(M\) is a model of individual agency we also have that \(w \in || \varphi ||^{M} \) by Constraint 2. Hence, \(M, w \models \varphi \), which means that \(\pi (\varphi ) = 1\).

  2. 2.

    As previously, \(\pi (E_a \varphi ) = 1\) implies that \(|| \varphi ||^{M} \in EE_w(a)\). By Constraint 1E \(|| \varphi ||^{M} \not = S\). It means that there is a world \(u \in || \lnot \varphi ||^{M} \). Hence, \(M, u \models \lnot \varphi \), which means that \(\lnot \varphi \) is \(\mathsf{BIAT \text{- }\mathsf sat } \,\).

  3. 3.

    Again, for all \(j\), \(\pi (E_a \psi _j) = 1\) implies that \(|| \psi _j ||^{M} \in EE_w(a)\). By Constraint 4 we obtain that \((\bigcap _j || \psi _j ||^{M} ) \in EE_w(a)\). By hypothesis, \(\pi (E_a\psi ) =0\). Thus, \(|| \psi ||^{M} \not \in EE_w(a)\) and clearly \(|| \psi ||^{M} \not = \bigcap _j || \psi _j ||^{M} \). It implies that there must exist a world \(u\) such that either \(u \in \bigcap _j || \psi _j ||^{M} \cap || \lnot \psi ||^{M} \) or \(u \in \bigcup _j || \lnot \psi _j ||^{M} \cap || \psi ||^{M} \). It means that the formula \((\bigwedge _j (\psi _j) \wedge \lnot \psi ) \vee (\bigvee _j (\lnot \psi _j) \wedge \psi )\) is indeed satisfiable: it is true in the world \(u\).

1.2 Right-to-left direction

Suppose now that there is a semi-valuation \(\pi \) such that we have (1), (2) and (3) like in the formulation of Theorem 3. We can rewrite (3) as: if \(E_a \psi _1, \ldots , E_a \psi _k, E_a \psi \in {sub^{\lnot }}(\varphi )\), with \(\pi (E_a \psi _j) = 1\) for all \(j\), and \(\pi (E_a \psi ) = 0\), then there is a witness pointed model of individual agency \((M_{\psi _1, \ldots \psi _k, \psi }, w_{\psi _1, \ldots \psi _k, \psi })\) such that \(M_{\psi _1, \ldots \psi _k, \psi }, w_{\psi _1, \ldots \psi _k, \psi } \models \left( \bigwedge _j (\psi _j) \wedge \lnot \psi \right) \vee \left( \bigvee _j (\lnot \psi _j) \wedge \psi \right) \). We can also rewrite (2) as: if \(E_a \psi \in {sub^{\lnot }}(\varphi )\) and \(\pi (E_a\psi ) = 1\) then there is a witness pointed model of individual agency \((M_\psi ,\, w_\psi )\) such that \(M_\psi ,\, w_\psi \models \lnot \psi \).

Let \((M^1,\, w^1) \ldots , (M^\mathsf n ,\, w^\mathsf n )\) be an enumeration of all these witness pointed models of agency \(M^i = (S^i,\, N,\, EE^i,\, v^i)\). This enumeration is finite because the length of \(\varphi \) is finite and only a finite number of \(E_a \psi \) are in \({sub^{\lnot }}(\varphi )\). We assume w.l.o.g. that \(S^i \cap S^j = \emptyset \) when \(i \not = j\).

To establish that \(\varphi \) is \(\mathsf{BIAT \text{- }\mathsf sat } \,\), we are going to construct a model of individual agency that satisfies it. Suppose a new fresh world \(\mathsf w \not \in \bigcup _i S^i\). Let \(M = (S,\,N,\,EE,\,v)\) such that: \(S = \{ \mathsf w \} \cup \bigcup _{i = 1}^\mathsf n S^i\).

Instrumental for the definition of \(v\) and \(EE\), we first introduce an intention assignment for formulas in \({sub^{\lnot }}(\varphi )\). Let \(V : {sub^{\lnot }}(\varphi ) \longrightarrow 2^S\), such that

$$\begin{aligned} V(\psi ) = \{ \mathsf w \mid \pi (\psi ) = 1\} \cup \bigcup _{i = 1}^\mathsf n || \psi ||^{M^i} . \end{aligned}$$

Our model valuation \(v\) is simply defined as the projection of \(V\) on the set \(P\) of propositional variables (we can assume w.l.o.g. that \(P \subseteq {sub^{\lnot }}(\varphi )\)). That is, \(v(p) = V(p)\) for \(p \in P\).

We finally define \(EE\). For every \(u \in S\), we let \(X \in EE_u(a)\) iff there are \(E_a \psi _1 , \ldots , E_a \psi _k \in {sub^{\lnot }}(\varphi )\), with \(X = \bigcap _{j = 1}^k V(\psi _j)\) and

$$\begin{aligned} \left\{ \begin{array}{ll} \pi (E_a \psi _j) = 1 \quad \text{ for }\; \text{ all } \; 1 \le j \le k&{} \quad \text{ when } \; u = \mathsf w ;\\ M^i,\, u \models E_a \psi _j \quad \text{ for }\; \text{ all } \; 1 \le j \le k&{} \quad \text{ when } \; u \in S^i,\; 1 \le i \le \mathsf n . \end{array}\right. \end{aligned}$$

This following claim will be useful later.

Claim 1

If \(V(\psi ) \in EE_a(u)\) and \(E_a\psi \in {sub^{\lnot }}(\varphi )\) then either \(u = \mathsf w \) and \(\pi (E_a\psi ) = 1\) or \(u \in S^c\) and \(M^c, u \models E_a \psi \) (\(1 \le c \le \mathsf n \)).

Proof

Case \(u = \mathsf w \): suppose \(V(\psi ) \in EE_a(\mathsf w )\). That is there \(E_a \psi _1, \ldots E_a \psi _k \in {sub^{\lnot }}(\varphi )\) such that \(V(\psi ) = \bigcap _{j = 1}^k V(\psi _j)\) and \(\pi (E_a\psi _j) = 1\) for all \(j\).

Now suppose that \(E_a \psi \in {sub^{\lnot }}{\varphi }\) and for contradiction, that \(\pi (E_a \psi ) = 0\). By Hypothesis (3), \(M_{\psi _1, \ldots \psi _k, \psi }, w_{\psi _1, \ldots \psi _k, \psi } \models \left( \bigwedge _j (\psi _j) \wedge \lnot \psi \right) \vee \left( \bigvee _j (\lnot \psi _j) \wedge \psi \right) \). This means that \(|| \psi ||^{M_{\psi _1, \ldots \psi _k, \psi }} \not = \bigcap _{j = 1}^k || \psi _j ||^{M_{\psi _1, \ldots \psi _k, \psi }} \). Consequently, \(V(\psi ) \not = \bigcap _{j = 1}^k V(\psi _j)\)—a contradiction.

Case \(u \in S^c\): suppose \(V(\psi ) \in EE_a(u)\). So there are \(E_a \psi _1, \ldots , E_a\psi _k \in {sub^{\lnot }}{\varphi }\) such that \(V(\psi ) = \bigcap _{j = 1}^k V(\psi _j)\) and \(M^c, u \models E_a \psi _j\) for all \(1\le j \le k\). So we have \(|| \psi _j ||^{M^c} \in EE^c_a(u)\) for all \(1\le j \le k\). By Constraint 4 on \(M^c\) we deduce that \(\bigcap _{j = 1}^k || \psi _j ||^{M^c} \in EE^c_a(u)\). Since \(V(\psi ) = \bigcap _{j = 1}^k V(\psi _j)\) and because \(S^i \cup S^j = \emptyset \) when \(i \not = j\), we also have that \(|| \psi ||^{M^c} = \bigcap _{j = 1}^k || \psi _j ||^{M^c} \). Hence, \(|| \psi ||^{M^c} \in EE^c_a(u)\), which means that \(M^c, u \models E_a \psi \).

We need to make sure that \(M\) is indeed a model of individual agency.

Claim 2

\(M\) is a model of individual agency.

Proof

Constraint 1E: Clearly by construction and because \(S^i \not \in EE^i_u(a)\), we have \(S \not \in EE_u(a)\) for \(u \not = \mathsf w \). Now, suppose for contradiction that \(S \in EE_\mathsf w (a)\). It means that there are \(E_a \psi _1 , \ldots , E_a \psi _k \in {sub^{\lnot }}(\varphi )\), with \(S = \bigcap _{j = 1}^k V(\psi _j)\) and \(\pi (E_a \psi _j) = 1\) for all \(1 \le j \le k\). Hence, \(V(\psi _j) = \{ \mathsf w \mid \pi (\psi _j) = 1\} \cup \bigcup _{i = 1}^\mathsf n || \psi _j ||^{M^i} = S\) for all \(0 \le j \le k\). As we assumed that \(S^i \cap S^j = \emptyset \) when \(i \not = j\), it means that \(|| \psi _j ||^{M^i} = S^i\) for all \(i\). In particular, \(|| \psi _j ||^{M^j} = S^j\). But since, \(\pi (E_a \psi _j) = 1\), by Hypothesis (2), it means that \(M_j, w_j \models \lnot \psi _j\), that is, \(w_j \in \emptyset \)—a contradiction.

Constraint 2: Now, suppose that \(X \in EE_u(a)\). We need to show that \(u \in X\).

By construction, there are \(E_a \psi _1 , \ldots , E_a \psi _k \in {sub^{\lnot }}(\varphi )\), such that \(X = \bigcap _{j = 1}^k V(\psi _j)\) and

$$\begin{aligned} \left\{ \begin{array}{ll} \pi (E_a \psi _j) = 1 \quad \text{ for }\; \text{ all } \; 1 \le j \le k&{} \quad \text{ if } \; u = \mathsf w ;\\ M^i, u \models E_a\psi _j \quad \text{ for }\; \text{ all } \; 1 \le j \le k&{} \quad \text{ if } \; u \in S^i, \;1 \le i \le \mathsf n . \end{array}\right. \end{aligned}$$

If \(u \in S^c\), we have \(|| \psi _j ||^{M^c} \in EE_u(a)\) for all \(1 \le j \le k\). Since \(M^c\) is a model of individual agency, we also have \(u \in || \psi _j ||^{M^c} \) for all \(1 \le j \le k\) (Constraint 2 on \(M^c\)). So \(u \in \bigcap _{j = 1}^k || \psi _j ||^{M^c} \). Then, \(u \in \bigcap _{j = 1}^k \bigcup _{i = 1}^\mathsf n || \psi _j ||^{M^i} \). Then, \(u \in \bigcap _{j = 1}^k \{ \mathsf w \mid \pi (\psi _j) = 1\} \cup \bigcap _{j = 1}^k \bigcup _{i = 1}^\mathsf n || \psi _j ||^{M^i} \) which is equivalent to \(u \in X\).

If \(u = \mathsf w \), by Hypothesis (1), we have \(\pi (\psi _j) = 1\) for all \(1 \le j \le k\). So indeed, \(u \in \bigcap _{j = 1}^k V(\psi _j)\).

Constraint 4: \(X_1 \in EE_u (a)\) and \(X_2 \in EE_u(a)\). We need to show that \((X_1 \cap X_2) \in EE_u(a)\).

We do not give the details. Again, if \(u \in S^c\) (\(1 \le c \le \mathsf n \)), it holds in virtue of \(M^a\) being a model of individual agency; this time we use Constraint 4 on \(M^a\). If \(u = \mathsf w \), the proof uses Hypothesis (3).

Finally, we are going to prove that \(\varphi \) is true in \(M\) at \(\mathsf w \). To do this, we show that for all \(\psi \in {sub^{\lnot }}(\varphi )\), we have \(V(\psi ) = || \psi ||^{M} \).

Claim 3

If \(\psi \in {sub^{\lnot }}(\varphi )\) then \(V(\psi ) = || \psi ||^{M} \).

Proof

The proof is by induction on the structure of \(\psi \). This holds for \(\psi = p,\; p \in P\), by definition of \(v(p)\) and \(|| p ||^{M} \). It also holds for \(\psi = \lnot \gamma \), and \(\psi = \gamma _1 \vee \gamma _2\), because \(V(\lnot \gamma ) = S \setminus V(\gamma )\) and \(V(\gamma _1 \vee \gamma _2) = V(\gamma _1) \cup V(\gamma _2)\) respectively.

For \(\psi = E_a\gamma \), we assume inductively that \(V(\gamma ) = || \gamma ||^{M} \) and show that for every \(u \in S\), \(u \in V(E_a\gamma )\) iff \(u \in || E_a\gamma ||^{M} \).

Suppose \(u \in V(E_a\gamma )\) then either \(u = \mathsf w \) and \(\pi (E_a\gamma ) = 1\), or \(u \in S^a\) and \(M^a, u \models E_a \gamma \). In both cases, by definition of \(EE\), we have \(V(\gamma ) \in EE_a(u)\). By induction hypothesis, we have \(|| \gamma ||^{M} \in EE_a(u)\). Hence \(M, u \models E_a \gamma \), that is, \(u \in || E_a\gamma ||^{M} \).

The other way around suppose that \(u \in || E_a\gamma ||^{M} \), that is, \(M, u \models E_a \gamma \). It means that \(|| \gamma ||^{M} \in EE_a(u)\). By induction hypothesis, \(V(\gamma ) \in EE_a(u)\). We conclude using Claim 1 that \(u \in V(E_a\gamma )\).

To sum up, we have built a structure \(M\) that is a model of individual agency (Claim 2), and \(|| \varphi ||^{M} = V(\varphi )\) (Claim 3). By definition of \(V\) and because \(\pi (\varphi ) = 1\) (\(\pi \) is a semi-valuation of \(\varphi \)), we have that \(M,\, \mathsf w \models \varphi \). So indeed, \(\varphi \) is \(\mathsf{BIAT \text{- }\mathsf sat } \,\). This concludes the proof of Theorem 3.

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Troquard, N. Reasoning about coalitional agency and ability in the logics of “bringing-it-about”. Auton Agent Multi-Agent Syst 28, 381–407 (2014). https://doi.org/10.1007/s10458-013-9229-x

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