Appendix: proofs
Theorem 1
(\(\iota \)-Bisimulation Theorem) For all pointed deontic game models (M, a) and \((M',a')\): if \((M,a)\rightleftarrows _\iota (M',a')\), then \((M,a)\equiv _{{\mathfrak {L}}_\iota }(M',a')\).
Proof
By structural induction on \(\phi \).
Basis: Suppose \((M,a)\rightleftarrows _\iota (M',a')\). We check cases p and \(\star _i\).
- p::
-
By Definition 5(i), for all \(p\in {\mathfrak {P}}\) it holds that \(a\in v(p)\) iff \(a'\in v'(p)\). Hence, for all \(p\in {\mathfrak {P}}\) it holds that \((M,a)\models p\) iff \((M',a')\models p\).
- \(\star _i\)::
-
By Definition 5(ii), for all \(i\in \mathcal {N}\) it holds that \(a^{}_i\in \text { Adm}_M(i)\) iff \(a'_i\in \text { Adm}_{M'}(i)\). Hence, for all \(i\in \mathcal {N}\) it holds that \((M,a)\models \star _i\) iff \((M',a')\models \star _i\).
Induction hypothesis For all pointed deontic game models \((M^{}_1,a^{}_1)\) and \((M'_1,a'_1)\) and all \(\psi \in {\mathfrak {L}}_\iota \) with fewer operators than \(\phi \): if \((M^{}_1,a^{}_1)\rightleftarrows _\iota (M'_1,a'_1)\), then \((M^{}_1,a^{}_1)\models \psi \) iff \((M'_1,a'_1)\models \psi \). Induction step Suppose \((M,a)\rightleftarrows _\iota (M',a')\). We check case [i]. The other cases are proved analogously.
- [i]::
-
Suppose \((M,a)\models [i]\psi \). Suppose \(b'\in A'\) such that \(a'_i=b'_i\). By Definition 5(vi), there is a \(b\in A\) such that \(a^{}_i=b^{}_i\) and \((b,b')\in R\). Then \((M,b)\rightleftarrows _\iota (M',b')\). Because \((M,a)\models [i]\psi \) and \(a^{}_i=b^{}_i\), we have \((M,b)\models \psi \). By the Induction Hypothesis, \((M',b')\models \psi \). Hence, for all \(b'\in A'\) such that \(a'_i=b'_i\) it holds that \((M',b')\models \psi \). Hence, \((M',a')\models [i]\psi \). Suppose \((M',a')\models [i]\psi \). Suppose \(b\in A\) such that \(a^{}_i=b^{}_i\). By Definition 5(v), there is a \(b'\in A'\) such that \(a'_i=b'_i\) and \((b,b')\in R\). Then \((M,b)\rightleftarrows _\iota (M',b')\). Because \((M',a')\models [i]\psi \) and \(a'_i=b'_i\), we have \((M',b')\models \psi \). By the Induction Hypothesis, \((M,b)\models \psi \). Hence, for all \(b\in A\) such that \(a^{}_i=b^{}_i\) it holds that \((M,b)\models \psi \). Hence, \((M,a)\models [i]\psi \).
Therefore, \((M,a)\equiv _{{\mathfrak {L}}_\iota }(M',a')\). \(\square \)
Lemma 1
Let \(M=\langle \mathcal {N}, (A_i), d, v\rangle \) be a deontic game model, let \({\mathcal{G}}\subseteq \mathcal {N}\), and let \(M'_{\mathcal{G}}=\langle \mathcal {N},(A'_i),d',v'\rangle \) be the unit \({\mathcal{G}}\)-transform of M. Let \(a^{}_i, b^{}_i\in A_i\) and \(x_i^{},y^{}_i\in \{+,-\}\). Then
-
(i)
\(a^{}_i\succeq _M b^{}_i\) iff \((a^{}_i,x^{}_{i})\succeq _{M'_{\mathcal{G}}}(b^{}_i,x^{}_{i})\)
-
(ii)
If \((a^{}_i,x^{}_{i})\succ _{M'_{\mathcal{G}}}\,(b^{}_i,y^{}_{i})\), then \(a^{}_{i}\succ _{M}\, b^{}_{i}\).
Proof
(i) (\(\Rightarrow \)) Suppose \(a^{}_{i}\succeq _M b^{}_{i}\). Consider an arbitrary \(c'_{-i}\in A'_{-i}\). Then \(c'_{-i}=(c^{}_{-i},y^{}_{-i})\) for a \(c^{}_{-i}\in A^{}_{-i}\) and a \(y ^{}_{-i}\in \{+,-\}^{\mathcal {N}-i}\). Consider \((x^{}_i,y^{}_{-i})\in \{+,-\}^\mathcal {N}\).
If \((x^{}_i,y^{}_{-i})^{}_{\mathcal{G}}\) is constant, then, by definition, it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=d(a^{}_i,c^{}_{-i})\) and \(d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=d(b^{}_i,c^{}_{-i})\). By supposition, it must be that \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). Hence, \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\ge d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\).
If \((x^{}_i,y^{}_{-i})^{}_{\mathcal{G}}\) is not constant, then, by definition, it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=0\) and \(d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=0\). Hence, \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\ge d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\).
Hence, for all \(c'_{-i}\in A'_{-i}\) it holds that \(d'(a^{}_i,x^{}_{i},c'_{-i})\ge d'(b^{}_i,x^{}_{i},c'_{-i})\). Therefore, \((a^{}_i,x^{}_{i})\succeq _{M'_{\mathcal{G}}}(b^{}_i,x^{}_{i})\).
(\(\Leftarrow \)) Suppose \((a^{}_i,x^{}_{i})\succeq _{M'_{\mathcal{G}}}(b^{}_i,x^{}_{i})\). Consider an arbitrary \(c^{}_{-i}\in A^{}_{-i}\). Take a \(y^{}_{-i}\in \{+,-\}^{\mathcal {N}-i}\) such that \((x^{}_i,y^{}_{-i})^{}_{\mathcal{G}}\) is constant. Then \((c^{}_{-i},y^{}_{-i})\in A'_{-i}\). By definition of \(d'\), it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=d(a^{}_i,c^{}_{-i})\) and \(d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})=d(b^{}_i,c^{}_{-i})\). By supposition, it must be that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\ge d'(b^{}_i,x^{}_{i},c^{}_{-i},y^{}_{-i})\). Hence, \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). Hence, for all \(c^{}_{-i}\in A^{}_{i}\) it holds that \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). Therefore, \(a^{}_i\succeq _M b^{}_i\).
(ii) Suppose \((a^{}_i,x^{}_{i})\succ _{M'_{\mathcal{G}}}(b^{}_i,y^{}_{i})\). If \(x^{}_i=y^{}_i\), then, by part (i) of this lemma, \(a^{}_{i}\succ _{M}\, b^{}_{i}\). Suppose \(x^{}_i\ne y^{}_i\).
We first prove \(a^{}_{i}\succeq _{M} b^{}_{i}\). Consider an arbitrary \(c^{}_{-i}\in A^{}_{-i}\). There are two cases: (a) If \(i\in {\mathcal{G}}\), then take a \(z^{}_{-i}\in \{+,-\}^{\mathcal {N}-i}\) such that \((y^{}_i,z^{}_{-i})^{}_{\mathcal{G}}\) is constant. Then \((x^{}_i,z^{}_{-i})^{}_{\mathcal{G}}\) is not constant. By definition of \(d'\), it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},z^{}_{-i})=0\). By supposition, \(d'(b^{}_i,y^{}_{i},c^{}_{-i},z^{}_{-i})=0\). By definition of \(d'\), it holds that \(d(b^{}_i,c^{}_{-i})=0\). Hence, \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). (b) If \(i\not \in {\mathcal{G}}\), then take a \(z^{}_{-i}\in \{+,-\}^{\mathcal {N}-i}\) such that \((x^{}_i,z^{}_{-i})^{}_{\mathcal{G}}\) is constant. Then \((y^{}_i,z^{}_{-i})^{}_{\mathcal{G}}\) is also constant. By definition of \(d'\), it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},z^{}_{-i})=d(a^{}_i,c^{}_{-i})\) and \(d'(b^{}_i,y^{}_{i},c^{}_{-i},z^{}_{-i})=d(b^{}_i,c^{}_{-i})\). By supposition, it holds that \(d'(a^{}_i,x^{}_{i},c^{}_{-i},z^{}_{-i})\ge d'(b^{}_i,y^{}_{i},c^{}_{-i},z^{}_{-i})\). Hence, \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). Hence, for all \(c^{}_{-i}\in A^{}_{i}\) it holds that \(d(a^{}_i,c^{}_{-i})\ge d(b^{}_i,c^{}_{-i})\). Therefore, \(a^{}_i\succeq _M b^{}_i\).
We now prove \(b^{}_{i}\not \succeq _{M} a^{}_{i}\). Because \((b^{}_i,y^{}_{i})\not \succeq _{M'_{\mathcal{G}}}(a^{}_i,x^{}_{i})\), there is a \(c^*_{-i}\in A^{}_{-i}\) and a \(z^*_{-i}\in \{+,-\}^{\mathcal {N}-i}\) such that \(d'(a^{}_i,x^{}_{i},c^*_{-i},z^*_{-i})=1\) and \(d'(b^{}_i,y^{}_{i},c^*_{-i},z^*_{-i})=0\). Then \((x^{}_i,z^*_{-i})^{}_{\mathcal{G}}\) is constant and \(d(a^{}_i,c^*_{-i})=1\). There are two cases: (a) If \(i\in {\mathcal{G}}\), take a \(z^{**}_{-i}\in \{+,-\}^{\mathcal {N}-i}\) such that \((y^{}_i,z^{**}_{-i})^{}_{\mathcal{G}}\) is constant. Then \((x^{}_i,z^{**}_{-i})^{}_{\mathcal{G}}\) is not constant. By definition of \(d'\), it holds that \(d'(a^{}_i,x^{}_{i},c^*_{-i},z^{**}_{-i})=0\). By supposition, \(d'(b^{}_i,y^{}_{i},c^*_{-i},z^{**}_{-i})=0\). By definition of \(d'\), it holds that \(d(b^{}_i,c^*_{-i})=0\). (b) If \(i\not \in {\mathcal{G}}\), then \((y^{}_i,z^*_{-i})^{}_{\mathcal{G}}\) is constant. Because \(d'(b^{}_i,y^{}_{i},c^*_{-i},z^*_{-i})=0\) and by definition of \(d'\), it must be that \(d(b^{}_i,c^*_{-i})=0\). Either way, \(d(b^{}_i,c^*_{-i})=0\). Hence, there is a \(c^*_{-i}\in A^{}_{-i}\) such that \(d(a^{}_i,c^*_{-i})=1\) and \(d(b^{}_i,c^*_{-i})=0\). Therefore, \(b^{}_{i}\not \succeq _{M} a^{}_{i}\). \(\square \)
Theorem 2
Let \(M=\langle \mathcal {N}, (A_i), d, v\rangle \) be a deontic game model, let \({\mathcal{G}}\subseteq \mathcal {N}\), and let \(M'_{\mathcal{G}}=\langle \mathcal {N},(A'_i),d',v'\rangle \) be the unit \({\mathcal{G}}\)-transform of M. Let \(R_1\subseteq A\times A'\) be given by \(\{(a,(a,x)):a\in A \text{ and } x\in \{+,-\}^\mathcal {N}\}\). Then \(R_1\) is an \(\iota \)-bisimulation between M and \(M'_{\mathcal{G}}\).
Proof
Suppose that \((a,a')\in R_1\) for \(a\in A\) and \(a'\in A'\). Then \(a'=(a,x)\) for some \(x\in \{+,-\}^\mathcal {N}\). Note that \(a'_i=(a^{}_i,x^{}_i)\).
(i) Because \(a'=(a,x)\), for all \(p\in {\mathfrak {P}}\) it holds that \(a\in v(p)\) iff \(a'\in v'(p)\).
(ii) Suppose \(a^{}_i\not \in \text { Adm}_M(i)\). Then there is a \(b^{}_i\in A^{}_i\) such that \(b^{}_{i}\succ _M a^{}_{i}\), that is, \(b^{}_{i}\succeq _M a^{}_{i}\) and \(a^{}_{i}\not \succeq _M b^{}_{i}\). By Lemma 1(i), it holds that \((b^{}_{i},x^{}_i)\succeq _{M'_{\mathcal{G}}} (a^{}_{i},x^{}_i)\) and \((a^{}_{i},x^{}_i)\not \succeq _{M'_{\mathcal{G}}} (b^{}_{i},x^{}_i)\), that is, \((b^{}_{i},x^{}_i)\succ _{M'_{\mathcal{G}}} (a^{}_{i},x^{}_i)\). Hence, \((a^{}_i,x^{}_i)\not \in \text { Adm}_{M'_{\mathcal{G}}}(i)\) and therefore, \(a'_i\not \in \text { Adm}_{M'_{\mathcal{G}}}(i)\).
Suppose \(a'_i\not \in \text { Adm}_{M'_{\mathcal{G}}}(i)\). Then \((a^{}_i,x^{}_i)\notin \text { Adm}_{M'_{\mathcal{G}}}(i)\). Then there is a \((b^{}_i,y^{}_i)\in A'_i\) such that \((b^{}_i,y^{}_i)\succ _{M'_{\mathcal{G}}}(a^{}_i,x^{}_i)\). By Lemma 1(ii), we have \(b^{}_{i}\succ _{M} \, a_{i}\). Therefore, \(a^{}_i\not \in \text { Adm}_M(i)\).
(iii) Consider an arbitrary \(b\in A\). Let \(b'=(b,x)\). Obviously, \(b'\in A'\) and \((b,b')\in R_1\). Hence, there is a \(b'\in A'\) such that \((b,b')\in R_1\).
(iv) Consider an arbitrary \(b'\in A'\). It holds that \(b'=(b,y)\) for some \(b\in A\) and some \(y\in \{+,-\}^\mathcal {N}\). Obviously, \((b,b')\in R_1\). Hence, there is a \(b\in A\) such that \((b,b')\in R_1\).
(v) Suppose \(a^{}_i=b^{}_i\) for an arbitrary \(b\in A\). Let \(b'=(b,x)\). Obviously, \(b'\in A'\) and \((b,b')\in R_1\). Because \(a^{}_i=b^{}_i\) and \(x^{}_i=x^{}_i\), it holds that \(a'_i=b'_i\). Hence, there is a \(b'\in A'\) such that \(a'_i=b'_i\) and \((b,b')\in R_1\).
(vi) Suppose \(a'_i=b'_i\) for an arbitrary \(b'\in A'\). It holds that \(b'=(b,y)\) for some \(b\in A\) and some \(y\in \{+,-\}^\mathcal {N}\). Obviously, \((b,b')\in R_1\). Because \(a'_i=(a^{}_i,x^{}_i)\) and \(b'_i=(b^{}_i,y^{}_i)\), it must be that \(a^{}_i=b^{}_i\) and \(x^{}_i=y^{}_i\). Hence, there is a \(b\in A\) such that \(a^{}_i=b^{}_i\) and \((b,b')\in R_1\).
Therefore, \((M,a)\rightleftarrows _\iota (M'_{\mathcal{G}},(a,x))\). \(\square \)
Lemma 2
Let \(M=\langle \mathcal {N}, (A_i), d, v\rangle \) be a deontic game model, let \({\mathcal{G}}\subseteq \mathcal {N}\), and let \(M''_{\mathcal{G}}=\langle \mathcal {N},(A''_i),d'',v''\rangle \) be the zero \({\mathcal{G}}\)-transform of M. Let \(a^{}_i, b^{}_i\in A_i\) and \(x^{}_i,y^{}_i\in \{+,-\}\). Then
-
(i)
\(a^{}_i\succeq _M b^{}_i\) iff \((a^{}_i,x^{}_i)\succeq _{M''_{\mathcal{G}}}(b^{}_i,x^{}_i)\)
-
(ii)
If \((a^{}_i,x^{}_i)\succ _{M''_{\mathcal{G}}}\,(b^{}_i,y^{}_i)\), then \(a^{}_{i}\succ _{M}\, b^{}_{i}\).
Proof
Analogous to the proof of Lemma 1. \(\square \)
Theorem 3
Let \(M=\langle \mathcal {N}, (A_i), d, v\rangle \) be a deontic game model, let \({\mathcal{G}}\subseteq \mathcal {N}\), and let \(M''_{\mathcal{G}}=\langle \mathcal {N},(A''_i),d'',v''\rangle \) be the zero \({\mathcal{G}}\)-transform of M. Let \(R_2\subseteq A\times A''\) be given by \(\{(a,(a,x)):a\in A \text{ and } x\in \{+,-\}^\mathcal {N}\}\). Then \(R_2\) is an \(\iota \)-bisimulation between M and \(M''_{\mathcal{G}}\).
Proof
Analogous to the proof of Theorem 2. \(\square \)