Proof. We can obtain the models of A with respect to $\Sigma$ by extending the models of A with respect to $\Sigma'$ by any possible valuation of $\Sigma \setminus \Sigma'$ , that is, ${Mod}\,_{\Sigma}(A) = \{(\omega^a \cdot \omega^b) \mid \omega^a \in {Mod}\,_{\Sigma'}(A), \omega^b \in {{\Omega_{\Sigma \setminus \Sigma'}}}\}$ . By definition, we have that $\kappa'(\omega') = \min \{\kappa(\omega) \mid \omega \in {{\Omega_{\Sigma}}}, {\omega}_{\mid {\Sigma'}} = \omega'\} = \min \{\kappa(\omega' \cdot \omega^b) \mid \omega^b \in {{\Omega_{\Sigma \setminus \Sigma'}}}\}$ . Therefore,

\begin{align*} \kappa(A) &= \min \{\kappa(\omega^a \cdot \omega^b) \mid \omega^a \in {Mod}\,_{\Sigma'}(A), \omega^b \in {{\Omega_{\Sigma \setminus \Sigma'}}}\} \\ &= \min \{\kappa'(\omega') \mid \omega' \in {Mod}\,_{\Sigma'}(A)\} = \kappa'(A).\end{align*}

Proof. Using Lemma 1, we have $\kappa(A) = \kappa'(A)$ and $\kappa(AB) = \kappa'(AB)$ and $\kappa(A\overline{B}) = \kappa'(A\overline{B})$ . Therefore, $\kappa \models {{(B|A)}}$ iff $\kappa' \models {{(B|A)}}$ .

Proof. Let $\Sigma = \Sigma_1 \cup \Sigma_2$ . Because $A \in {{\mathcal{L}_{\Sigma_1}}}$ and $B \in {{\mathcal{L}_{\Sigma_2}}}$ , we have that ${Mod}\,_{\Sigma}(AB) = \{(\omega^1 \cdot \omega^2) \mid \omega^1 \in {Mod}\,_{\Sigma_1}(A), \omega^2 \in {Mod}\,_{\Sigma_2}(B)\}$ . Therefore, it holds that

\begin{align*} \kappa_{\oplus}(AB) &= \min \{\kappa_{\oplus}(\omega) \mid \omega \in {Mod}\,_{\Sigma}(AB)\} \\ &= \min \{\kappa_1(\omega^1) + \kappa_2(\omega^2) \mid \omega^1 \in {Mod}\,_{\Sigma_1}(A), \omega^2 \in {Mod}\,_{\Sigma_2}(B)\} \\ &= \min \{\kappa_1(\omega^1) \mid \omega^1 \in {Mod}\,_{\Sigma_1}(A)\} + \min \{\kappa_2(\omega^2) \mid \omega^2 \in {Mod}\,_{\Sigma_2}(B)\} \\ &= \kappa_1(A) + \kappa_2(B).\end{align*}

Proof. Let $\omega^1$ be any world in ${{\Omega_{\Sigma_1}}}$ . Let $\omega^2 \in {{\Omega_{\Sigma_1}}}$ such that $\kappa_2(\omega^2) = 0$ . We have that

\begin{align*} {\kappa_{\oplus}}_{|\Sigma_1}(\omega^1) &= \min \{\kappa(\omega) \mid \omega \in {{\Omega_{\Sigma}}}, {\omega}_{\mid {\Sigma_1}} = \omega^1\}\\ &= \min \{\kappa_1(\omega^1) + \kappa_2(\omega_{\mid {\Sigma_2}}) \mid \omega \in {{\Omega_{\Sigma}}}\}\\ &\overset{*}{=} \kappa_1(\omega^1). \end{align*}

Equation $(\!*\!)$ holds because we can choose, for example, $\omega = (\omega^1 \cdot \omega^2)$ , resulting in $\kappa_2(\omega_{\mid {\Sigma_2}}) = 0$ .

Proof. Using Lemma 1 we have $\kappa(A) = \kappa'(A)$ and $\kappa(AB) = \kappa'(AB)$ and $\kappa(A\overline{B}) = \kappa'(A\overline{B})$ . Therefore, $A |\hspace{-2pt} {\sim}_{\kappa} B$ iff $A |\hspace{-2pt} {\sim}_{\kappa'} B$ .

Proof. Because $\Delta$ is weakly consistent, there is a ranking function $\kappa$ with $\kappa \models \Delta$ . Let $\omega \in \kappa^{-1}(0)$ . Towards a contradiction, assume that there is a ${{(B|A)}} \in \Delta$ that is falsified by $\omega$ , that is, $\omega \models A\overline{B}$ . For $\kappa$ to accept ${{(B|A)}}$ it must be either $\kappa(A) = \infty$ or $\kappa(AB) \lt \kappa(A\overline{B})$ . Because $\omega \models A$ and $\kappa(\omega) =0$ we have $\kappa(A) \neq \infty$ . Because $\kappa(A\overline{B}) \leqslant 0$ and there are no ranks below 0 the condition $\kappa(AB) \lt \kappa(A\overline{B})$ does not hold. This is a contradiction; hence, $\omega$ does not falsify any conditional in $\Delta$ .

Proof. Assume there is a world $\omega^*$ that does not falsify any conditional in $\Delta$ . Then

\[\kappa(\omega) = \begin{cases} 0 & \text{if } \omega = \omega^*\\ \infty & \text{otherwise} \end{cases}\]

is a model of $\Delta$ . Therefore, $\Delta$ is weakly consistent.

Proof. Direction $\Rightarrow$ : Assume that $\omega$ falsifies a conditional in $\Delta^\infty$ . Then this conditional is applicable for $\omega$ .

Direction $\Leftarrow$ : Assume that at least one conditional ${{(B|A)}} \in \Delta^\infty$ is applicable in $\omega$ . There are two possible cases: Either $\omega$ falsifies one of the other conditionals in $\Delta^\infty$ or not. In the first case the lemma holds. In the second case, towards a contradiction, we assume that $\omega$ does not falsify ${{(B|A)}}$ . If ${{(B|A)}}$ is applicable in $\omega$ and $\omega$ does not falsify ${{(B|A)}}$ then $\omega$ must verify ${{(B|A)}}$ . That implies that ${{(B|A)}}$ is tolerated by $\Delta^\infty$ which contradicts the construction of ${{{{{\mathit{EZP}}}}(\Delta)}}$ .

Proof. The claim follows from Proposition 17 because for every ranking function $\kappa$ , the relation $A |\hspace{-2pt} {\sim}_{\kappa} \bot$ iff $\kappa(A) = \infty$ holds according to Equation (1).

Proof. Because $\Delta$ is weakly consistent, there is at least one world $\omega \in \Omega_\Sigma$ that does not falsify any of the conditionals (see Lemma 9). This implies $\kappa_{\vec{\eta}}(\omega) = 0$ . Thus, $\kappa_{\vec{\eta}}$ is a ranking function.

For every ${{(B|A)}} \in \Delta$ it holds that $\kappa_{\vec{\eta}}(A\overline{B}) = \infty$ because every model of $A\overline{B}$ falsifies the conditional ${{(B|A)}}$ with impact $\infty$ . For $\kappa_{\vec{\eta}}(AB)$ we have either (1) $\kappa_{\vec{\eta}}(AB) = 0$ or (2) $\kappa_{\vec{\eta}}(AB) = \infty$ . In case (1) we have $\kappa_{\vec{\eta}}(AB) = 0 \lt \infty = \kappa_{\vec{\eta}}(A\overline{B})$ . In case (2) we have $\kappa_{\vec{\eta}}(AB) = \infty$ and $\kappa_{\vec{\eta}}(A\overline{B}) = \infty$ and therefore $\kappa_{\vec{\eta}}(A) = \infty$ because $\kappa_{\vec{\eta}}(A) = \min \{\kappa_{\vec{\eta}}(AB), \kappa_{\vec{\eta}}(A\overline{B})\}$ . In both cases, $\kappa_{\vec{\eta}}$ accepts ${{(B|A)}}$ . Thus, $\kappa_{\vec{\eta}} \models \Delta$ .

Proof. Assume that $\kappa^z_{\Delta}(\omega) = \infty$ . Let ${{{{{\mathit{EZP}}}}(\Delta)}} = \{\Delta^0, \dots, \Delta^m, \Delta^\infty\}$ be the extended Z-partition of $\Delta$ . By definition of $\kappa^z_{\Delta}$ there exists a conditional ${{(B|A)}} \in \Delta^\infty$ s.t. $\omega \models A$ . Because ${{(B|A)}} \in \Delta^\infty$ the conditional ${{(B|A)}}$ is not tolerated by $\Delta^\infty$ , so there is a conditional ${{(B'|A')}} \in \Delta^\infty$ that is falsified by $\omega$ (this can be ${{(B|A)}}$ again).

Towards a contradiction assume that there is a c-representation $\kappa_{\vec{\eta}}$ of $\Delta$ with $\kappa_{\vec{\eta}}(\omega) \lt \infty$ . As $\kappa_{\vec{\eta}}$ models $\Delta$ and thus also ${{(B'|A')}}$ there must be a world $\omega^1$ that verifies ${{(B'|A')}}$ and satisfies $\kappa_{\vec{\eta}}(\omega^1) \lt \kappa_{\vec{\eta}}(\omega)$ . With the same argumentation there must be another conditional ${{(B^1|A^1)}} \in \Delta^\infty$ that is falsified by $\omega^1$ , and another world $\omega_2$ that verifies ${{(B^1|A^1)}}$ and satisfies $\kappa_{\vec{\eta}}(\omega^2) \lt \kappa_{\vec{\eta}}(\omega^1)$ . Repeating this argumentation, we obtain an infinite chain of worlds $\omega_1, \omega_2, \dots$ s.t. $\kappa_{\vec{\eta}}(\omega_1) \gt \kappa_{\vec{\eta}}(\omega_2) \gt \dots$ . But as there are only finitely many worlds (and also because there are only finitely many ranks below $\kappa_{\vec{\eta}}(\omega_1)$ ) such a chain cannot exist. Contradiction.

Proof. Let $\Delta = \{{{(B_1|A_1)}}, \dots, {{(B_n|A_n)}}\}$ be a weakly consistent belief base. Let ${{{{{\mathit{EZP}}}}(\Delta)}} = \{\Delta^0, \dots, \Delta^m, \Delta^\infty\}$ be the extended Z-partition of $\Delta$ . Construct an impact vector $\vec{\eta}$ for $\Delta$ as follows. Let $\mu^0 = 1$ and $\mu^j = |\Delta^0 \cup \dots \cup \Delta^{j-1}| \cdot \mu^{j-1} +1$ for $j=1, \dots, m$ . For ${{(B_i|A_i)}}$ with ${{(B_i|A_i)}} \in \Delta^j$ let $\eta_{i} = \mu^j$ for $j \lt \infty$ and $\eta_{j} = \infty$ for $j = \infty$ . By construction, for worlds $\omega$ that do not falsify a conditional from $\Delta^j \cup \dots \cup \Delta^m \cup \Delta^\infty$ we have $\kappa_{\vec{\eta}}(\omega) \lt \mu^j$ .

$\kappa_{\vec{\eta}}$ is a c-representation of $\Delta$ : Let ${{(B_i|A_i)}}$ be any conditional in $\Delta$ . If ${{(B_i|A_i)}} \in \Delta^\infty$ then $\kappa^z_{\Delta}(A_i) = \infty$ by the definition of $\kappa^z_{\Delta}$ which implies with Proposition 26 that $\kappa_{\vec{\eta}}(A_i) = \infty$ and therefore $\kappa_{\vec{\eta}} \models {{(B_i|A_i)}}$ . Otherwise, we have ${{(B_i|A_i)}} \in \Delta^j$ with $j \lt \infty$ . Then for any world $\omega'$ falsifying ${{(B_i|A_i)}}$ we have $\kappa_{\vec{\eta}}(\omega) \gt \mu^j$ ; hence $\kappa_{\vec{\eta}}(A_i \overline{B_i}) \geqslant \mu_j$ . Because ${{(B_i|A_i)}} \in \Delta^j$ , there is a world $\omega'$ that verifies ${{(B_i|A_i)}}$ and does not falsify a conditional in $\Delta^j \cup \dots \cup \Delta^m \cup \Delta^\infty$ . Therefore, $\kappa_{\vec{\eta}}(A_i B_i)\lt \mu_j$ . Thus, $\kappa_{\vec{\eta}}(A_i B_i)\lt \mu_j \leqslant \kappa_{\vec{\eta}}(A_i \overline{B_i})$ and $\kappa_{\vec{\eta}} \models {{(B_i|A_i)}}$ .

Furthermore, it holds that $\kappa_{\vec{\eta}}(\omega) = \infty$ iff $\omega$ falsifies a conditional in $\Delta^\infty$ . Therefore, $\kappa_{\vec{\eta}}(\omega) \lt \infty$ for all worlds $\omega$ with $\kappa^z_{\Delta}(\omega) \lt \infty$ .

Proof. Direction $\Rightarrow$ If $\kappa(\omega) = \infty$ for all $\kappa \in {Mod}\,^{ec}_{\Delta}$ , then there is no $\kappa_{\vec{\eta}} \in {Mod}\,^{ec}_{\Delta}$ with $\kappa_{\vec{\eta}}(\omega) \lt \infty$ . With Proposition 26 this implies $\kappa^z_{\Delta}(\omega) = \infty$ . By Definition 13, this is the case if a conditional in $\Delta^\infty$ is applicable for $\omega$ .

Direction $\Leftarrow$ Assume $\omega \models A$ for some ${{(B|A)}} \in \Delta^\infty$ . Then $\kappa^z_{\Delta}(\omega) = \infty$ and with Proposition 26 we have $\kappa(\omega) = \infty$ for all $\kappa \in {Mod}\,^{ec}_{\Delta}$ .

Proof. Ad 1: Let $\omega$ be a world with $\kappa(\omega)=\infty$ . Because $\kappa(\omega)$ is the sum of the impacts of the conditionals falsified by $\omega$ and $\Delta$ is finite, $\omega$ must falsify at least one conditional ${{(B_i|A_i)}}$ with impact $\eta_{i} = \infty$ . This ${{(B_i|A_i)}}$ satisfies the first part of condition (1). Moreover, all worlds $\omega'$ that also falsify ${{(B_i|A_i)}}$ will have rank $\kappa(\omega') = \infty$ , too. Therefore, also the second part of condition (1) is satisfied by ${{(B_i|A_i)}}$ , and the condition holds.

Ad 2: The rank $\kappa(\omega)$ of a world $\omega$ only depends on which conditionals in $\Delta$ the world falsifies. Therefore, if two worlds $\omega, \omega'$ verify and falsify the same conditionals in $\Delta$ they will have the same rank.

Proof. We need to show that extended c-inference satisfies (DI) and (TV). (DI) is trivial: Every c-representation of $\Delta$ accepts the conditionals in $\Delta$ by definition. Therefore, $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ for every ${{(B|A)}} \in \Delta$ . (TV) is also clear: For $\Delta = \emptyset$ , the only c-representation is $\kappa_u$ with $\kappa_u(\omega) = 0$ for all $\omega \in \Omega$ . In this case, $\kappa_u$ accepts only conditionals ${{(B|A)}}$ with $A\overline{B} = \bot$ , which are conditionals with $A \models B$ .

Proof. Let $\Delta = \{{{(B_1|A_1)}}, \dots, {{(B_n|A_n)}}\}$ be a strongly consistent belief base and $C, D \in {{\mathcal{L}}}$ . We need to show that $C |\hspace{-2pt} {\sim}^{ec}_{\Delta} D$ iff $C |\hspace{-2pt} {\sim}^{c}_{\Delta} D$ .

Direction $\Rightarrow$ : Let $C |\hspace{-2pt} {\sim}^{ec}_{\Delta} D$ , that is, every extended c-representation models ${{(D|C)}}$ . As every c-representation is an extended c-representation (Proposition 22), every c-representation models ${{(D|C)}}$ . Thus, $D |\hspace{-2pt} {\sim}^{c}_{kb} C$ .

Direction $\Leftarrow$ : Let $C |\hspace{-2pt} {\sim}^{c}_{\Delta} D$ , that is, every c-representation models ${{(D|C)}}$ . We need to show that any extended c-representation $\kappa_{\vec{\eta}}$ models ${{(D|C)}}$ . If $\vec{\eta}$ contains only finite values, it is a c-representation and thus models ${{(D|C)}}$ by assumption.

Assume that $\vec{\eta}$ contains infinite entries. Let ${{{{{\mathit{OP}}}}(\Delta)}} = \{\Delta^0, \dots, \Delta^m, \Delta^\infty\}$ be the extended tolerance partition of $\Delta$ . Because $\Delta$ is strongly consistent, we have $\Delta^\infty = \emptyset$ . Let $\mathit{fin}(\vec{\eta}) = \{\eta_{i} \mid i \in \{0,\dots, n\}, \eta_{i} \lt \infty\}$ be the set of finite values in impact vector $\vec{\eta}$ and $f_0 = 1+ |\mathit{fin}(\vec{\eta})| \cdot \max (\mathit{fin}(\vec{\eta}))$ . Now construct $\vec{\eta}^f$ from $\vec{\eta}$ as follows. For ${{(B_i|A_i)}} \in \Delta^0$ with $\eta_{i} = \infty$ let $\eta_{i}^f = f_0$ . Let $f_1 = (f_0+1) \cdot | \{{{(B_i|A_i)}} \in \Delta^0 \mid \eta_{i} = \infty\} |$ . For ${{(B_i|A_i)}} \in \Delta^1$ with $\eta_{i} = \infty$ let $\eta_{i}^f = f_1$ . Let $f_2 = (f_1+1) \cdot \big| \{{{(B_i|A_i)}} \in \Delta^1 \mid \eta_{i} = \infty\} \big|$ . For ${{(B_i|A_i)}} \in \Delta^1$ with $\eta_{i} = \infty$ let $\eta_{i}^f = f_2$ ; and so on. By construction the sum of the impacts in $\mathit{fin}(\vec{\eta})$ is less than $f_0$ and the sum of the impacts of the conditionals in $\Delta^0 \cup \dots \cup \Delta^j$ is less than $f_j$ for $j=0, \dots, m$ .

Let $\kappa^f = \kappa_{\vec{\eta}^f}$ . Now verify that:

1. 1. $\kappa^f$ is a c-representation of $\Delta$ . For this we need to check that $\kappa^f$ models all conditionals in $\Delta$ .

2. 2. $\mathord{{|\hspace{-2pt}}{\sim}^{}_{\kappa^f}} \subseteq \mathord{{|\hspace{-2pt}}{\sim}^{}_{\kappa_{\vec{\eta}}}}$ , that is, every inference in $|\hspace{-2pt} {\sim}_{\kappa^f}$ is also an inference in $|\hspace{-2pt} {\sim}_{\kappa_{\vec{\eta}}}$ .

From (1) it follows that $\kappa^f$ is a model of ${{(D|C)}}$ because $C |\hspace{-2pt} {\sim}^{c}_{\Delta}D$ . With (2.) it follows that ${{(D|C)}}$ is modelled by $\kappa_{\vec{\eta}}$ .

Ad (1): Let ${{(B_i|A_i)}} \in \Delta$ . We distinguish three cases.

Case 1: $\kappa_{\vec{\eta}}(A_i B_i) \lt \kappa_{\vec{\eta}}(A_i \overline{B_i}) \lt \infty$

In this case $\kappa^f(A_i B_i) \lt \kappa^f(A_i \overline{B_i}) \lt f_0$ and therefore $\kappa^f \models {{(B_i|A_i)}}$ .

Case 2: $\kappa_{\vec{\eta}}(A_i B_i) \lt \infty$ and $\kappa_{\vec{\eta}}(A_i \overline{B_i}) = \infty$

In this case $\kappa^f(A_i B_i) \lt f_0 \lt \kappa^f(A_i \overline{B_i})$ and therefore $\kappa^f \models {{(B_i|A_i)}}$ .

Case 3: $\kappa_{\vec{\eta}}(A_i B_i) = \infty$ and $\kappa_{\vec{\eta}}(A_i \overline{B_i}) = \infty$

Assume that ${{(B_i|A_i)}}$ is in $\Delta^j$ . Then there is a world $\omega$ s.t. $\omega \models A_i B_i$ and $\omega$ falsifies no conditional in $\Delta^0 \cup \dots \cup \Delta^j$ . Therefore, $\kappa^f(\omega) \lt f_j$ and thus $\kappa^f(A_i B_i) \lt f_j$ . Any model of $A_i \overline{B_i}$ falsifies ${{(B_i|A_i)}}$ , therefore, $\kappa^f(A_i \overline{B_i}) \gt f_j$ . Thus, we have $\kappa^f(A_i B_i) \lt f_j \lt \kappa^f(A_i \overline{B_i})$ and therefore $\kappa^f \models {{(B_i|A_i)}}$ .

Ad (2): Assume that $X |\hspace{-2pt} {\sim}_{\kappa^f}Y$ . There are two cases.

Case 1: $\kappa^f(X \overline{Y}) \lt f_0$

In this case $\kappa^f(XY) \lt \kappa^f(X \overline{Y}) \lt f_0$ and therefore $\kappa_{\vec{\eta}}(XY) \lt \kappa_{\vec{\eta}}(X \overline{Y}) \lt \infty$ . Hence, $X |\hspace{-2pt} {\sim}_{\kappa_{\vec{\eta}}} Y$ .

Case 2: $ \kappa^f(X \overline{Y}) \geqslant f_0$

In this case $\kappa_{\vec{\eta}}(X \overline{Y}) = \infty$ and therefore $X |\hspace{-2pt} {\sim}_{\kappa_{\vec{\eta}}} Y$ .

Proof. Every ranking function, and thus every extended c-representation, induces a preferential inference relation. The intersection of two preferential inference relations is again preferential. As extended c-inference is the intersection of the inference relations induced by each extended c-representation, extended c-inference is preferential.

Proof. We need to show that $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} \bot$ iff $A |\hspace{-2pt} {\sim}^{p}_{\Delta} \bot$ . Using Lemma 14, it is sufficient to show that $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} \bot$ iff $\kappa^z(A) = \infty$ .

Direction $\Leftarrow$ : Let $\kappa^z_{\Delta}(A) = \infty$ . Then Proposition 26 states that $\kappa_{\vec{\eta}}(A) = \infty$ for every extended c-representation $\kappa_{\vec{\eta}}(A)$ of $\Delta$ . Thus, $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} \bot$ .

Direction $\Rightarrow$ : Let $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} \bot$ , that is, there is no extended c-representation $\kappa_{\vec{\eta}}$ of $\Delta$ s.t. $\kappa_{\vec{\eta}}(A) \lt \infty$ . By Proposition 26, we have $\kappa^z_{\Delta}(A) = \infty$ .

Proof. Let $\Delta$ be a belief base with $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ and $\top \mathord{{|\hspace{-2pt}}{\sim}\!\!\!\!/\;\;}^{ec}_{\Delta} \overline{A}$ . If $\Delta$ is not weakly consistent, $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ holds trivially. For the remainder of the proof assume that $\Delta$ is weakly consistent. Let $\kappa$ be any extended c-representation of $\Delta$ , and let $\Omega_{\mathit{zero}} = \kappa^{-1}(0)$ be the set of worlds with rank 0 with respect to $\kappa$ . Then, $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ implies that $\kappa(B) \lt \kappa(\overline{B})$ and therefore that for every $\omega \in \Omega_{\mathit{zero}}$ we have $\omega \models B$ . Because $\top \mathord{{|\hspace{-2pt}}{\sim}\!\!\!\!/\;\;}^{ec}_{\Delta} \overline{A}$ there is at least one $\omega \in \Omega_{\mathit{zero}}$ with $\omega \models A$ . This $\omega$ is a model of AB implying that $\kappa(AB) = 0$ . We also have that $\kappa(A\overline{B}) \gt 0$ because there are no models of B with rank 0. In summary, we have $\kappa(AB) = 0 \lt \kappa(A\overline{B})$ implying that $A |\hspace{-2pt} {\sim}_{\kappa} B$ . Because the c-representation $\kappa$ was chosen arbitrarily, we have $A |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ .

Proof. We first show that extended c-inference satisfies (wSM). Let $\Delta, \Delta'$ be belief bases with $\Delta \subseteq \Delta'$ and $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ . If $\Delta'$ is not weakly consistent, $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ holds trivially. For the remainder of the proof assume that $\Delta'$ and therefore also $\Delta$ is weakly consistent. Let $\kappa'$ be an extended c-representation of $\Delta'$ , and let $\Omega_{\mathit{zero}} = \kappa^{\prime-1}(0)$ be the set of worlds with rank 0 with respect to $\kappa'$ . Let $\omega$ be any world in $\Omega_{\mathit{zero}}$ . Because $\kappa' \models \Delta'$ , the world $\omega$ cannot falsify any conditional in $\Delta'$ . Thus, $\omega$ also does not falsify any conditional in $\Delta$ . Let $\kappa$ be an extended c-representation of $\Delta$ . Because $\omega$ does not falsify any conditional in $\Delta$ , we have $\kappa(\omega) = 0$ . Since $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta} B$ implies that $\kappa(B) \lt \kappa(\overline{B})$ , we have $\omega \models B$ . As $\omega$ was chosen arbitrarily, every world in $\Omega_{\mathit{zero}}$ is a model of B. Thus $\kappa'(B) = 0 \lt \kappa'(\overline{B})$ , implying $\top |\hspace{-2pt} {\sim}_{\kappa'} B$ . Because $\kappa'$ was chosen arbitrarily among the extended c-representations of $\Delta'$ , we have $\top |\hspace{-2pt} {\sim}^{ec}_{\Delta'} B$ .

Since c-inference and extended c-inference coincide for strongly consistent belief bases, c-inference satisfies (wSM) as well.

Proof. Let ${{(B_1|A_1)}}, \dots, {{(B_k|A_k)}}$ be the conditionals in $\Delta_1$ and let ${{(B_{k+1}|A_{k+1})}}, \dots, {{(B_n|A_n)}}$ be the conditionals in $\Delta_2$ . Let $\vec{\eta}^1$ be an impact vector inducing $\kappa_1$ and let $\vec{\eta}^2$ be an impact vector inducing $\kappa_2$ . Let $\omega$ be any world in ${{\Omega_{\Sigma}}}$ . Because of the syntax splitting, ${\omega}_{\mid {\Sigma_1}}$ falsifies the same worlds in $\Delta_1$ as $\omega$ and ${\omega}_{\mid {\Sigma_2}}$ falsifies the same worlds in $\Delta_2$ as $\omega$ . Let $\vec{\eta}$ be the impact vector for $\Delta$ that combines the impacts from $\vec{\eta}^1$ and $\vec{\eta}^2$ . We have that

(10) \begin{align} \begin{split} \kappa_{\oplus} &= \kappa_1({\omega}_{\mid {\Sigma_1}}) + \kappa_2({\omega}_{\mid {\Sigma_2}}) \\ &= \sum_{\substack{1 \leqslant i \leqslant k\\ {\omega}_{\mid {\Sigma_1}} \models A_i\overline{B_i}}} \eta_{i} + \sum_{\substack{k+1 \leqslant i \leqslant n\\{\omega}_{\mid {\Sigma_2}} \models A_i\overline{B_i}}} \eta_{i} \\ &= \sum_{\substack{1 \leqslant i \leqslant n\\ \omega \models A_i\overline{B_i}}} \eta_{i} = {{{\kappa_{\vec{\eta}}}}.} \end{split} \end{align}

Additionally, because $\kappa_1 = {\kappa_{\oplus}}_{\mid {\Sigma_1}}$ and $\kappa_2 = {\kappa_{\oplus}}_{\mid {\Sigma_2}}$ (see Lemma 4), Lemma 2 yields that $\kappa_{\oplus}$ models the conditionals in $\Delta_1$ and in $\Delta_2$ . Therefore, $\kappa_{\oplus}$ is an extended c-representation of $\Delta$ .

Proof. Let ${{(B_1|A_1)}}, \dots, {{(B_k|A_k)}}$ be the conditionals in $\Delta_1$ and let ${{(B_{k+1}|A_{k+1})}}, \dots, {{(B_n|A_n)}}$ be the conditionals in $\Delta_2$ . For any $\omega \in {{\Omega_{\Sigma}}}$ we have that

\[ \kappa(\omega) = \sum_{\substack{1 \leqslant i \leqslant n\\ \omega \models A_i\overline{B_i}}} \eta_{i} = \sum_{\substack{1 \leqslant i \leqslant k\\ \omega \models A_i\overline{B_i}}} \eta_{i} + \sum_{\substack{k+1 \leqslant i \leqslant n\\ \omega \models A_i\overline{B_i}}} \eta_{i} = {{\kappa_{\vec{\eta}^1}}}(\omega) + {{\kappa_{\vec{\eta}^2}}}(\omega). \]

Because $X \in {{\mathcal{L}_{\Sigma_1}}}$ and $Y \in {{\mathcal{L}_{\Sigma_2}}}$ we have that ${Mod}\,_{\Sigma}(XY) = \{(\omega^x \cdot \omega^y) \mid \omega^x \in {Mod}\,_{\Sigma_1}(X), \omega^y \in {Mod}\,_{\Sigma_2}(Y)\}$ . In summary, we have that

\begin{align*} \begin{split} \kappa(XY) &= \min \{\kappa(\omega) \mid \omega \in {Mod}\,_{\Sigma}(XY)\} \\ &= \min \{\kappa(\omega^x \cdot \omega^y) \mid \omega^x \in {Mod}\,_{\Sigma_1}(X), \omega^y \in {Mod}\,_{\Sigma_2}(Y)\} \\ &= \min \{{{\kappa_{\vec{\eta}^1}}}(\omega^x) + {{\kappa_{\vec{\eta}^2}}}(\omega^y) \mid \omega^x \in {Mod}\,_{\Sigma_1}(X), \omega^y \in {Mod}\,_{\Sigma_2}(Y)\} \\ &= \min \{{{\kappa_{\vec{\eta}^1}}}(\omega^x) \mid \omega^x \in {Mod}\,_{\Sigma_1}(X)\} + \min \{{{\kappa_{\vec{\eta}^2}}}(\omega^y) \mid \omega^y \in {Mod}\,_{\Sigma_2}(Y)\} \\ &= {{\kappa_{\vec{\eta}^1}}}(X) + {{\kappa_{\vec{\eta}^2}}}(Y). \end{split} \end{align*}

Proof. Let $\kappa \in {{\mathit{Mod}^{ec}_{\Sigma}({\Delta})}}$ . By Lemma 2, ${\kappa}_{|\Sigma_1}$ models the same conditionals in ${{{{(\mathcal{L}|\mathcal{L})}}_{\Sigma_1}}}$ as $\kappa$ . Especially, ${\kappa}_{|\Sigma_1}$ models all conditionals in $\Delta_1$ . It remains to be shown that ${\kappa}_{|\Sigma_1}$ can be constructed from integer impacts.

For this let ${{(B_1|A_1)}}, \dots, {{(B_k|A_k)}}$ be the conditionals in $\Delta_1$ and let ${{(B_{k+1}|A_{k+1})}}, \dots, {{(B_n|A_n)}}$ be the conditionals in $\Delta_2$ . Let $\vec{\eta}$ be an impact vector inducing $\kappa$ , and let $\vec{\eta}^1$ be the impact vector containing the impacts from $\vec{\eta}$ for $\Delta_1$ . Let $\omega^0 \in {{\Omega_{\Sigma}}}$ be a world that does not falsify any conditional in $\Delta$ (see Lemma 9). Let $\omega'$ be any world in ${{\Omega_{\Sigma_1}}}$ . All worlds $\omega$ with ${\omega}_{\mid {\Sigma_1}} = \omega'$ falsify the same conditionals in $\Delta_1$ . The world $\omega^* = (\omega' \cdot {\omega^0}_{\mid {\Sigma_2}})$ falsifies no conditional in $\Delta_2$ and is thus one of the worlds with the lowest rank in $\kappa$ that coincides with $\omega'$ on $\Sigma_1$ . We have that

\begin{align*} {\kappa}_{|\Sigma_1}(\omega') &= \min \{\kappa(\omega) \mid {\omega}_{\mid {\Sigma_1}} = \omega'\}\\ &= \min \{\sum_{\substack{1 \leqslant i \leqslant n\\\omega \models A_i\overline{B_i}}} \eta_{i} \mid {\omega}_{\mid {\Sigma_1}} = \omega'\} \\ &= \sum_{\substack{1 \leqslant i \leqslant n\\\omega^* \models A_i\overline{B_i}}} \eta_{i} = \sum_{\substack{1 \leqslant i \leqslant k\\\omega' \models A_i\overline{B_i}}} \eta_{i} = {{\kappa_{\vec{\eta}^1}}}(\omega'). \end{align*}

Therefore, ${\kappa}_{|\Sigma_1}$ is an extended c-representation of $\Delta_1$ .