Reasoning about Dependence, Preference and Coalitional Power

Abstract

This paper presents a logic of preference and functional dependence (LPFD) and its hybrid extension (HLPFD), both of whose sound and strongly complete axiomatization are provided. The decidability of LPFD is also proved. The application of LPFD and HLPFD to modelling cooperative games in strategic form is explored. The resulted framework provides a unified view on Nash equilibrium, Pareto optimality and the core. The philosophical relevance of these game-theoretical notions to discussions of collective agency is made explicit. Some key connections with other logics are also revealed, for example, the coalition logic, the logic of functional dependence and the logic of ceteris paribus preference.

Appendix

1.1 Strong Completeness of $\textsf{C}_\textsf{Nom}$

The strategic of our proof is to show that every $\textsf{C}_\textsf{Nom}$-consistent set $\Gamma$ of formulas is satisfiable. Let $\Gamma$ be a fixed consistent set. We first show the follow lemma:

Lemma 1

Let $\Gamma$ be a $\textsf{C}_\textsf{Nom}$-consistent set and $\mathsf {Nom'}=\textsf{Nom}\cup \{j_n:n\in \omega \}$. Then $\Gamma$ can be extended to a maximal $\textsf{C}_\mathsf {Nom'}$-consistent set $\Gamma ^+$ of formulas satisfying the following conditions:

• (Named) $\Gamma ^+\cap \mathsf {Nom'}\ne \varnothing$;

• (Pasted) For all $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\varphi \in \Gamma ^+$, there is a nominal $j\in \mathsf {Nom'}$ such that $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\wedge @_j\varphi \in \Gamma ^+$.

The proof of Lemma 9 is standard. Since $\textsf{Nom}$ is countable, we can assume that $\Gamma$ itself is a named and pasted $\textsf{C}_\mathsf {Nom'}$-MCS without loss of generality.

For each $i\in \textsf{Nom}$ such that $@_i\top \in \Gamma$, we define $\Delta _i=\{\varphi :@_i\varphi \in \Gamma \}$. The readers can check that $\Delta _i$ is a MCS for each $i\in \textsf{Nom}$.

Definition 1

The canonical model $\mathfrak {M}_\Gamma =(W_\Gamma ,\sim _\Gamma ,\le _\Gamma ,V_\Gamma )$ is defined by:

• $W_\Gamma =\{\Delta _i:@_i\top \in \Gamma \}$;

• for each $v\in \textsf{V}$, $\Delta _i\sim _v\Delta _j$ if and only if $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\{v\},\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Gamma$;

• for each $v\in \textsf{V}$, $\Delta _i\le _v\Delta _j$ if and only if $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\varnothing ,\{v\},\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Gamma$;

• $V(P\textbf{x})=\{\Delta _i:@_iP\textbf{x}\in \Gamma \}$ and $V(i)=\Delta _i$.

Lemma 2

$\mathfrak {M}_\Gamma =(W,\sim ,\le ,V)$ is an HRPD-model.

Proof

Since $\Gamma$ is named, $W_\Gamma \ne \varnothing$. Let $v\in \textsf{V}$. By axiom (Ord,1,2,3), $\sim _v$ is a pre-order and $\le _v$ is an equivalence relation. Note that $V(i)\in W$ for each $i\in \textsf{Nom}\cap \text {dom}(V)$. To show that $\mathfrak {M}_\Gamma$ is an HRPD-model, it suffices to show that V satisfies (Val). Let $P\in \textsf{Pred}$ and $\textbf{x}\in \textsf{V}^{\textsf{ar}(P)}$. Suppose $\Delta _i\sim _{\textsf{ran}(\textbf{x})}\Delta _j$ and $\Delta _i\in V(P\textbf{x})$. Then $P\textbf{x}\in \Delta _i$. By (Dep), $\mathbb {D}_XP\textbf{x}\in \Delta _i$, which entails $P\textbf{x}\in \Delta _j$. $\square$

Lemma 3

Let $\mathfrak {M}_\Gamma =(W,\sim ,\le ,V)$, $i\in \textsf{Nom}$ and $\Delta _i\in W$. Then

• (1) If $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Delta _i$, then $\Delta _iR(X,Y,Z)\Delta _j$;

• (2) If $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\varphi \in \Gamma$, then there is $j\in \textsf{Nom}$ with $\varphi \in \Delta _j$ and $\Delta _iR(X,Y,Z)\Delta _j$.

• (3) $D_Xs\in \Delta _i$ if and only if $\mathfrak {M}_\Gamma ,\Delta _i\models D_Xs$.

• (4) For all $\varphi \in \mathcal {L}_\textsf{Nom}$, $\varphi \in \Delta _i$ if and only if $\mathfrak {M}_\Gamma ,\Delta _i\models \varphi$.

Proof

For (1), suppose $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Delta _i$. By axiom (Ord,5), we see $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\{x\},\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j,$ $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\varnothing ,\{y\},\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j,$ $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\varnothing ,\varnothing ,\{z\}\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Delta _i$ for all $x\in X$, $y\in Y$ and $z\in Z$, which entails by axiom (Ord,4) that $\Delta _i\sim _X\Delta _j$, $\Delta _i\le _Y\Delta _j$ and $\Delta _i<_Z\Delta _j$. Thus $\Delta _iR(X,Y,Z)\Delta _j$.

For (2), suppose $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\varphi \in \Gamma$. Since $\Gamma$ is pasted, there is $j\in \textsf{Nom}$ such that $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\wedge @_j\varphi \in \Gamma$. Thus $\varphi \in \Delta _j$ and $\Delta _iR(X,Y,Z)\Delta _j$.

For (3), suppose $D_Xs\in \Delta _i$ and $\Delta _i\sim _X\Delta _j$. We show that $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\{s\},\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Delta _i$. Assume $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}\{s\},\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\not \in \Delta _i$. Then by axiom (DD,1), we see $\mathbb {D}_X\lnot j\in \Delta _i$, which contradicts to $\Delta _i\sim _X\Delta _j$. Thus $\mathfrak {M}_\Gamma ,\Delta _i\models D_Xs$. Suppose $D_Xs\not \in \Delta _i$. Then $i\wedge \lnot D_Xs\in \Delta _i$. By axiom (DD,2), we see $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\mathbb {D}_s\lnot i\in \Gamma$. Since $\Gamma$ is pasted, there is $j\in \textsf{Nom}$ such that $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,\varnothing ,\varnothing \hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\wedge @_j\mathbb {D}_s\lnot i\in \Gamma$. Thus $\Delta _i\sim _X\Delta _j$ and $\Delta _i\not \sim _s\Delta _j$. Note that $\sim _s$ is symmetric, $\Delta _j\not \sim _s\Delta _i$. Thus $\mathfrak {M}_\Gamma ,\Delta _i\not \models D_Xs$.

For (4), the proof proceeds by induction on the complexity of $\varphi$. The case when $\varphi =D_Xs$ follows from (3). The case $\varphi =P\textbf{x}$ or $\varphi \in \textsf{Nom}$ is trivial. The Boolean cases are also trivial. Let $\varphi =\llbracket {X,Y,Z}\rrbracket \psi$. Assume $\llbracket {X,Y,Z}\rrbracket \psi \not \in \Delta _i$. Then $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\lnot \psi \in \Delta _i$ and so $@_i\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\lnot \psi \in \Gamma$. By (2), $\lnot \psi \in \Delta _j$ for some $\Delta _j\in R(X,Y,Z)(\Delta _i)$. Then $\psi \not \in \Delta _j$ and by induction hypothesis, $\mathfrak {M}_\Gamma ,\Delta _j\not \models \psi$, which entails $\mathfrak {M}_\Gamma ,\Delta _i\not \models \llbracket {X,Y,Z}\rrbracket \psi$. Assume that $\mathfrak {M}_\Gamma ,\Delta _i\not \models \llbracket {X,Y,Z}\rrbracket \psi$. Then there is $\Delta _j\in R(X,Y,Z)(\Delta _i)$ such that $\mathfrak {M}_\Gamma ,\Delta _j\not \models \psi$. By induction hypothesis, $\psi \not \in \Delta _j$ and so $\lnot \psi \wedge j\in \Delta _j$. Note that $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}j\in \Delta _i$, we see $\langle \hspace{-5.0pt}\langle \hspace{1.111pt}X,Y,Z\hspace{1.111pt}\rangle \hspace{-5.0pt}\rangle \hspace{1.38885pt}\lnot \psi \in \Delta _i$, which entails $\llbracket {X,Y,Z}\rrbracket \psi \not \in \Delta _i$. $\square$

Theorem. $\textsf{C}_\textsf{Nom}$ is sound and strongly complete.

Proof

Soundness is not hard to verify. By Lemma 11, $\mathfrak {M}_\Gamma ,\Gamma \models \Gamma$. By the arbitrariness of $\Gamma$, we obtain the strong completeness. $\square$