Generalized logical operations among conditional events

Abstract

We generalize, by a progressive procedure, the notions of conjunction and disjunction of two conditional events to the case of n conditional events. In our coherence-based approach, conjunctions and disjunctions are suitable conditional random quantities. We define the notion of negation, by verifying De Morgan’s Laws. We also show that conjunction and disjunction satisfy the associative and commutative properties, and a monotonicity property. Then, we give some results on coherence of prevision assessments for some families of compounded conditionals; in particular we examine the Fréchet-Hoeffding bounds. Moreover, we study the reverse probabilistic inference from the conjunction $\mathcal {C}_{n + 1}$ of n + 1 conditional events to the family $\{\mathcal {C}_{n},E_{n + 1}|H_{n + 1}\}$. We consider the relation with the notion of quasi-conjunction and we examine in detail the coherence of the prevision assessments related with the conjunction of three conditional events. Based on conjunction, we also give a characterization of p-consistency and of p-entailment, with applications to several inference rules in probabilistic nonmonotonic reasoning. Finally, we examine some non p-valid inference rules; then, we illustrate by an example two methods which allow to suitably modify non p-valid inference rules in order to get inferences which are p-valid.

Appendix A

Proof of Theorem 5

We observe that $(ii)$ follows by $(i)$, by replacing $\mathcal {F}$ by $\overline {\mathcal {F}}$; indeed, by $(i)$ it holds that $\overline {\mathcal {D}(\overline {\mathcal {F}})}=\mathcal {C}(\overline {\overline {\mathcal {F}}})=\mathcal {C}(\mathcal {F})$. Then, it is enough to proof the assertion $(i)$. We will prove the assertion by induction.

• Step 1: $n = 1, \mathcal {F}=\{E_{1}|H_{1}\}$.We have $\overline {\mathcal {D}(\mathcal {F})}=\overline {E_{1}|H_{1}}= 1-E_{1}|H_{1}=\overline {E}_{1}|H_{1}=\mathcal {C}(\overline {\mathcal {F}})$.

  Thus the assertion holds when $n = 1$.

• Step 2: $n = 2$, $\mathcal {F}=\{E_{1}|H_{1},E_{2}|H_{2}\}$.

  We set

  $$\begin{array}{@{}rcl@{}} &&P(E_{1}|H_{1})=x, P(E_{2}|H_{2})=y, \mathbb{P}[(E_{1}|H_{1})\\ &&\vee (E_{2}|H_{2})]=w, \mathbb{P}[(\overline{E}_{1}|H_{1}) \wedge (\overline{E}_{2}|H_{2})]=t. \end{array} $$

  We observe that the family $\{ E_{1}H_{1}\vee E_{2}H_{2}, \overline {E}_{1}H\overline {E}_{2}H_{2}, \overline {H}_{1}\overline {E}_{2}H_{2}, \overline {E}_{1}H_{1}\overline {H}_{2}, \overline {H}_{1}\overline {H}_{2}\}$ is a partition of the sure event ${\Omega }$. Moreover, by Definitions 3 and 4 we have

  $$\begin{array}{@{}rcl@{}} \overline{\mathcal{D}(\mathcal{F})}&=&1-(E_{1}|H_{1})\vee (E_{2}|H_{2})\\ &=&\left\{\begin{array}{ll} 0, &\text{ if}~E_{1}H_{1}\vee E_{2}H_{2}~\text{is true,}\\ 1, &\text{ if}~\overline{E}_{1}H_{1}\overline{E}_{2}H_{2}~\text{is true,}\\ 1-x, &\text{ if}~\overline{H}_{1}\overline{E}_{2}H_{2}~\text{is true,}\\ 1-y, &\text{ if}~\overline{E}_{1}H_{1}\overline{H}_{2}~\text{is true,}\\ 1-w, &\text{ if}~\overline{H}_{1}\overline{H}_{2}~\text{is true}. \end{array} \right. \end{array} $$

  (39)

  and

  $$\begin{array}{@{}rcl@{}} \mathcal{C}(\overline{\mathcal{F}})&=&(\overline{E}_{1}|H_{1})\wedge (\overline{E}_{2}|H_{2})\\ &=& \left\{\begin{array}{ll} 0, &\text{ if}~E_{1}H_{1}\vee E_{2}H_{2}~\text{is true,}\\ 1, &\text{ if}~\overline{E}_{1}H\overline{E}_{2}H_{2}~\text{is true,}\\ 1-x, &\text{ if}~\overline{H}_{1}\overline{E}_{2}H_{2}~\text{is true,}\\ 1-y, &\text{ if}~\overline{E}_{1}H_{1}\overline{H}_{2}~\text{is true,}\\ t, &\text{ if}~\overline{H}_{1}\overline{H}_{2}~\text{is true}. \end{array} \right. \end{array} $$

  (40)

  We observe that $\overline {\mathcal {D}(\mathcal {F})}$ and ${\mathcal {C}(\overline {\mathcal {F}})}$ coincide when $H_{1}\vee H_{2}$ is true. Thus, by Theorem 3, $\mathbb {P}(\overline {\mathcal {D}(\mathcal {F})})=\mathbb {P}(\mathcal {C}(\overline {\mathcal {F}}))$ and hence 1 − w = t. Therefore $\overline {\mathcal {D}(\mathcal {F})}$ still coincides with $\mathcal {C}(\overline {\mathcal {F}})$ when $H_{1}\vee H_{2}$ is false, so that $\overline {\mathcal {D}(\mathcal {F})}= \mathcal {C}(\overline {\mathcal {F}})$.

• Step 3: $\mathcal {F}=\{E_{1}|H_{1},E_{2}|H_{2},\ldots , E_{n}|H_{n}\}$.

  (Inductive Hypothesis) Let us assume that for any (strict) subset $S\subset \{1,\ldots ,n\}$, by defining $\mathcal {F}_{S}=\{E_{i}|H_{i}, i\in S\}$, it holds that $\overline {{\mathcal {D}(\mathcal {F}_{S})}}=\mathcal {C}(\overline {\mathcal {F}_{S}})$. Now we will prove that $\overline {{\mathcal {D}(\mathcal {F}_{S})}}=\mathcal {C}(\overline {\mathcal {F}_{S}})$ when $S=\{1,\ldots ,n\}$, in which case $\mathcal {F}_{S}=\mathcal {F}$. By Definition 7 we have

  $$\begin{array}{@{}rcl@{}} \overline{\mathcal{D}(\mathcal{F})}\!&=&\!{\sum}_{h = 0}^{m} w_{h}C_{h}, \text{where}\\ w_{h}&=& \left\{ \begin{array}{llll} \!0, &\!\text{ if } S_{h}^{\prime}\neq \emptyset,\\ \!1, &\!\text{ if } S_{h}^{\prime\prime}=\{1,2,\ldots,n\}, \\ \!1 - y_{S^{\prime\prime\prime}_{h}},\! &\!\text{ if } S_{h}^{\prime}=\emptyset \text{ and } S_{h}^{\prime\prime\prime} \!\neq\emptyset. \end{array} \right.\\ \end{array} $$

  (41)

  We continue to use the subsets $S_{h}^{\prime }, S_{h}^{\prime \prime }$, $S_{h}^{\prime \prime \prime }$ as defined in formula (9) also with the family $\overline {\mathcal {F}}$; moreover we set $t_{S}=\mathbb {P}[\bigwedge _{i\in S} (\overline {E}_{i}|H_{i})]=\mathbb {P}[{\mathcal {C}(\overline {\mathcal {F}}_{S})}]$. Based on Definition 5, we have

  $$\begin{array}{@{}rcl@{}} \mathcal{C}(\overline{\mathcal{F}})&=&{\sum}_{h = 0}^{m} z_{h}C_{h}, \text{where}\\ z_{h}&=&\left\{ \begin{array}{llll} 0, &\text{ if } S_{h}^{\prime}\neq \emptyset,\\ 1, &\text{ if } S_{h}^{\prime\prime}=\{1,2,\ldots,n\}, \\ t_{S^{\prime\prime\prime}_{h}}, &\text{ if } S_{h}^{\prime}=\emptyset \text{ and } S_{h}^{\prime\prime\prime} \neq\emptyset. \end{array} \right.\\ \end{array} $$

  (42)

  Then, $\overline {\mathcal {D}(\mathcal {F})}-\mathcal {C}(\overline {\mathcal {F}})={\sum }_{h = 0}^{m} (w_{h}-z_{h})C_{h}$, where

  $$\begin{array}{@{}rcl@{}} &&{}w_{h}-z_{h}\\ &&{\kern2pt}= \left\{ \begin{array}{llll} 0, &\text{ if } S_{h}^{\prime}\neq \emptyset,\\ 0, &\text{ if } S_{h}^{\prime\prime}=\{1,2,\ldots,n\}, \\ 1-y_{S^{\prime\prime\prime}_{h}}-t_{S^{\prime\prime\prime}_{h}}, &\text{ if } S_{h}^{\prime}=\emptyset \text{ and } S_{h}^{\prime\prime\prime} \neq\emptyset. \end{array} \right.\\ \end{array} $$

  (43)

  By the inductive hypothesis, it holds that $1-y_{S^{\prime \prime \prime }_{h}}=\mathbb {P}[\overline {{\mathcal {D}(\mathcal {F}_{{S^{\prime \prime \prime }_{h}}})}}]=\mathbb {P}[\mathcal {C}(\overline {\mathcal {F}}_{{S^{\prime \prime \prime }_{h}}})]=t_{S^{\prime \prime \prime }_{h}}$ for h = 1,…,m, because $S^{\prime \prime \prime }_{h}\subset \{1,2,\ldots ,n\}$. Then, $\overline {\mathcal {D}(\mathcal {F})}-\mathcal {C}(\overline {\mathcal {F}})={\sum }_{h = 0}^{m} (w_{h}-z_{h})C_{h}$, where

  $$ w_{h}-z_{h}= \left\{ \begin{array}{llll} 0, & h = 1,\ldots,m,\\ 1-y_{S^{\prime\prime\prime}_{0}}-t_{S^{\prime\prime\prime}_{0}}, & h = 0. \end{array} \right. $$

  (44)

  By recalling that $S^{\prime \prime \prime }_{0}=\{1,2,\ldots ,n\}$, $\overline {\mathcal {D}(\mathcal {F})}$ and ${\mathcal {C}(\overline {\mathcal {F}})}$ coincide when $H_{1}\vee H_{2} \vee {\cdots } \vee H_{n}$ is true. Thus, by Theorem 3, $\mathbb {P}[\overline {\mathcal {D}(\mathcal {F})}]=\mathbb {P}[\mathcal {C}(\overline {\mathcal {F}}))]$, that is $1-y_{S^{\prime \prime \prime }_{0}}=t_{S^{\prime \prime \prime }_{0}}$. Therefore $\overline {\mathcal {D}(\mathcal {F})}$ still coincides with $\mathcal {C}(\overline {\mathcal {F}})$ when $H_{1}\vee H_{2} \vee {\cdots } \vee H_{n}$ is false, so that $\overline {\mathcal {D}(\mathcal {F})}= \mathcal {C}(\overline {\mathcal {F}})$.

□

Proof of Theorem 6

• (i) Assume that, for every $(\mu ,\nu )\in {\Pi }$, the values of $X|H$ and $Y|K$ associated with the constituent $C_{h}$ are such that $X|H\leq Y|K$, for each $C_{h}$ contained in H ∨ K; then for each given coherent assessment $(\mu ,\nu )$, by choosing $s_{1}= 1, s_{2}=-1$ in the random gain, we have

  $$\begin{array}{@{}rcl@{}} G\!&=&\!H(X - \mu)-K(Y - \nu) = (X|H - \mu) - (Y|K - \nu)\\ \!&=&\! (X|H - Y|K) + (\nu - \mu) . \end{array} $$

  Then, by the hypothesis, $\mathcal {G}_{H\vee K}\leq (\nu - \mu )$ and by coherence $0=\mathbb {P}(\mathcal {G}_{H\vee K}) \leq \nu - \mu$. Then $\mu \leq \nu$, $\forall (\mu ,\nu ) \in {\Pi }$.

• (ii) By hypothesis, it holds that $(XH+\mu H^{c})(H\vee K)\leq (YK+\nu K^{c})(H\vee K)$; moreover, from condition (i), $\mu \leq \nu$ for every $(\mu ,\nu )\in {\Pi }$; then

  $$\begin{array}{@{}rcl@{}} X|H\! &=&\! XH+\mu \overline{H}=(XH+\mu \overline{H})(H\vee K)\\ &&\!+(XH + \mu \overline{H})\overline{H} \overline{K} = (XH + \mu \overline{H})(H\!\vee\! K)\\ &&\!+\mu \overline{H} \overline{K}\leqslant (YK + \nu \overline{K})(H\vee K) + \nu \overline{HK}\\ &&\!=(YK + \nu \overline{K})(H\vee K) + (YK + \nu \overline{K})\overline{HK}\\ &&\!= YK + \nu \overline{K} = Y|K . \end{array} $$

  Vice versa, $X|H\leq Y|K$ trivially implies $X|H\leq Y|K$ when $H\vee K$ is true.

□

Proof of Theorem 7

We distinguish three cases: $(a)$ the value of $\mathcal {C}_{n}$ is 0, with some $E_{i}|H_{i}$ false, $i\leq n$; $(b)$ the value of $\mathcal {C}_{n}$ is 1, with $E_{i}|H_{i}$ true, $i = 1,\ldots ,n$; $(c)$ the value of $\mathcal {C}_{n}$ is $\mathbb {P}[\bigwedge _{i\in S} (E_{i}|H_{i})]=\mathbb {P}(\mathcal {C}_{S})=x_{S}$, for some subset $S\subseteq \{1,2,\ldots ,n\}$.

• Case (a). It holds that $\mathcal {C}_{n + 1}= 0=\mathcal {C}_{n}$.

• Case (b). The value of $\mathcal {C}_{n + 1}$ is 1, or 0, or $x_{n + 1}$, according to whether $E_{n + 1}|H_{n + 1}$ is true, or false, or void; thus $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$.

• Case (c). We distinguish three cases: $(i)$E_n+ 1|H_n+ 1 is true; $(ii)$E_n+ 1|H_n+ 1 is false; $(iii)$E_n+ 1|H_n+ 1 is void. In the case $(i)$ the value of $\mathcal {C}_{n + 1}$ is $x_{S}$, thus $\mathcal {C}_{n + 1}=\mathcal {C}_{n}$. In the case $(ii)$ the value of $\mathcal {C}_{n + 1}$ is 0, thus $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$. In the case $(iii)$ the value of $\mathcal {C}_{n + 1}$ is $x_{S\cup \{n + 1\}}=\mathbb {P}[\bigwedge _{i\in S\cup \{n + 1\}}(E_{i}|H_{i})]$; then, in order to prove that $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$, we need to prove that $x_{S\cup \{n + 1\}}\leq x_{S}$. We proceed by induction on the cardinality of S, denoted by s. Let be $s = 1$, with $\mathcal {C}_{S}=E_{i}|H_{i}$, for some $i\in \{1,\ldots ,n\}$. We note that $x_{S}=\mathbb {P}(E_{i}|H_{i})=x_{i}$, $x_{S\cup \{n + 1\}}=\mathbb {P}((E_{i}|H_{i})\wedge (E_{n + 1}|H_{n + 1}))=x_{\{i,n + 1\}}$ and by Theorem 4 it holds that $x_{S\cup \{n + 1\}}=x_{\{i,n + 1\}}\leq x_{i}=x_{S}$. Now, let be s ≥ 2 and $x_{S\cup \{n + 1\}}\leq x_{S}$ for every $s< n$, so that, based on Definition 5, $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$ when S is a strict subset of {1, 2,…,n}. If $S=\{1,2,\ldots ,n\}$, as $E_{i}|H_{i}$ is void for all $i = 1,\ldots ,n + 1$, it holds that $\mathcal {C}_{n}=\mathbb {P}(\mathcal {C}_{n})=x_{\{1,\ldots ,n\}}$ and $\mathcal {C}_{n + 1}=x_{\{1,\ldots ,n + 1\}}=\mathbb {P}(\mathcal {C}_{n + 1})$ and, in order to prove that $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$, it remains to prove that $\mathbb {P}(\mathcal {C}_{n + 1})\leq \mathbb {P}(\mathcal {C}_{n})$. By applying Theorem 6, with $X|H=\mathcal {C}_{n + 1}=Z_{n + 1}|(H_{1}\vee {\cdots } \vee H_{n + 1})$ and $Y|K=\mathcal {C}_{n}=Z_{n}|(H_{1}\vee {\cdots } \vee H_{n})$, as $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$ when H₁ ∨⋯ ∨ H_n+ 1 is true (i.e., $s<n$ ), it follows that $\mathbb {P}(\mathcal {C}_{n + 1})\leq \mathbb {P}(\mathcal {C}_{n})$; therefore $\mathcal {C}_{n + 1}\leq \mathcal {C}_{n}$.

□

Proof

of Theorem 9.

Case $(i)$. We proceed by induction. The property is satisfied for $n = 1$; indeed, if $\mathcal {C}_{1}=E_{1}|H_{1}\in \{1,0,x_{1}\}$, where $x_{1}=\mathbb {P}(E_{1}|H_{1}) \in [0,1]$, then $\mathcal {C}_{1}\in [0,1]$. Let us assume that the property holds for $k<n$, that is $\mathcal {C}_{k}\in [0,1]$, for every $k<n$. Based on Definition 5 we distinguish three cases: (a) the value of $\mathcal {C}_{n}$ is 0; $(b)$ the value of $\mathcal {C}_{n}$ is 1; $(c)$ the value of $\mathcal {C}_{n}$ is $\mathbb {P}[\bigwedge _{i\in S} (E_{i}|H_{i})]=x_{S}$, for some subset $S\subseteq \{1,2,\ldots ,n\}$. In the cases $(a)$ and $(b)$, $\mathcal {C}_{n}\in [0,1]$. In the case $(c)$, if $S=\{i_{1},\ldots ,i_{k}\}\subset \{1,2,\ldots ,n\}$, then $\mathcal {C}_{n} \in [0,1]$, because $x_{S}=\mathbb {P}(\bigwedge _{j = 1}^{k} (E_{i_{j}}|H_{i_{j}}))$ is a possible value of $\mathcal {C}_{k}=\bigwedge _{j = 1}^{k} (E_{i_{j}}|H_{i_{j}})$, with $k<n$. Finally, if $S=\{1,2,\ldots ,n\}$ (that is the conditioning events $H_{1},\ldots , H_{n}$ are all false), then $\mathcal {C}_{n}=\mathbb {P}(\mathcal {C}_{n})$ and $\mathbb {P}(\mathcal {C}_{n})\in [0,1]$ because the values of $\mathcal {C}_{n}$ restricted to H₁ ∨⋯ ∨ H_n all belong to $[0,1]$. Therefore $\mathcal {C}_{n}\in [0,1]$. By a similar reasoning, based on Definition 7 we can prove that $\mathcal {D}_{n}\in [0,1]$. □

Proof of Theorem 10

Let $C_{0},\ldots , C_{m}$, with $m = 3^{n}-1$ be the constituents associated with $\mathcal {F}_{n + 1}=\{E_{1}|H_{1},\ldots , E_{n + 1}|H_{n + 1}\}$, where $C_{0}=\overline {H}_{1}{\cdots } \overline {H}_{n + 1}$. With each $C_{h}$, $h = 1,\ldots ,m$, we associate the point $Q_{h}=(q_{h1},q_{h2},q_{h3})$, which represents the value of the random vector $(\mathcal {C}_{n},E_{n + 1}|H_{n + 1},\mathcal {C}_{n + 1})$ when $C_{h}$ is true, where $q_{h1}$ is the value of $\mathcal {C}_{n}$, $q_{h2}$ is the value of $E_{n + 1}|H_{n + 1}$, and $q_{h3}$ is the value of $\mathcal {C}_{n + 1}$. With $C_{0}$ it is associated the point $Q_{0}=(\mu _{n},x_{n + 1},\mu _{n + 1})=\mathcal {M}$. We observe that the set of points $\{Q_{h}, h = 1,\ldots ,m\}$ contains in particular the points

$$Q_{1} = ({\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt}) , Q_{2} = ({\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt}) , Q_{3} = ({\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt}){\kern-.5pt} , Q_{4} = ({\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt}){} , $$

which are respectively associated with the following constituents or logical disjunction of constituents

$$\begin{array}{@{}rcl@{}} && E_{1}H_{1} {\cdots} E_{n}H_{n} E_{n + 1}H_{n + 1} , E_{1}H_{1} {\cdots} E_{n}H_{n} \overline{E}_{n + 1}H_{n + 1},\\ &&(\overline{E}_{1}H_{1} \!\vee \!{\cdots} \!\vee \!\overline{E}_{n}H_{n}) \!\wedge \!E_{n + 1}H_{n + 1} ,\\ &&(\overline{E}_{1}H_{1} \!\vee \!{\cdots} \!\vee\! \overline{E}_{n}H_{n}) \wedge \overline{E}_{n + 1}H_{n + 1} . \end{array} $$

Based on Remark 7, we need to prove that the set of coherent assessments ${\Pi }$ on $\{\mathcal {C}_{n}, E_{n + 1}|H_{n + 1}, \mathcal {C}_{n + 1}\}$ coincides with the convex hull $\mathcal {I}$ of Q₁,Q₂,Q₃,Q₄. We recall that coherence of $(\mu _{n},x_{n + 1},\mu _{n + 1})$ implies coherence of all the sub-assessments on the associated subfamilies of $\{\mathcal {C}_{n}, E_{n + 1}|H_{n + 1}, \mathcal {C}_{n + 1}\}$. The coherence of the single assessments $\mu _{n}$ on $\mathcal {C}_{n}$, or $x_{n + 1}$ on $E_{n + 1}|H_{n + 1}$, or $\mu _{n + 1}$ on $\mathcal {C}_{n + 1}$, simply amounts to conditions

$$\mu_{n} \in [0,1] , x_{n + 1} \in [0,1] , \mu_{n + 1} \in [0,1] , $$

respectively. Then, by the hypothesis of logical independence, the sub-assessment $(\mu _{n},x_{n + 1})$ is coherent, for every $(\mu _{n},x_{n + 1}) \in [0,1]^{2}$. By Remark 6, the coherence of the sub-assessments $(\mu _{n},\mu _{n + 1})$ and $(x_{n + 1},\mu _{n + 1})$ amounts to the conditions $0\leq \mu _{n + 1}\leq \mu _{n} \leq 1$ and 0 ≤ μ_n+ 1 ≤ x_n+ 1 ≤ 1. Finally, assuming that the above conditions are satisfied, to prove coherence of $(\mu _{n},x_{n + 1},\mu _{n + 1})$, by Theorem 1, it is enough to show that the point $(\mu _{n},x_{n + 1},\mu _{n + 1})$ belongs to the convex hull of the points $Q_{1},\ldots ,Q_{m}$. Moreover, in order $\mathcal {M}$ belongs to the convex hull of Q₁,…,Q_m the following system $({\Sigma })$ must solvable

$$ \mathcal{M}=\sum\limits_{h = 1}^{m}\lambda_{h} Q_{h}, \sum\limits_{h = 1}^{m}\lambda_{h}= 1, \lambda_{h}\geq 0, \forall h. $$

(45)

We show that the convex hull of the points $Q_{1},\ldots ,Q_{m}$ coincides with the convex hull $\mathcal {I}$ of the points $Q_{1},Q_{2},Q_{3},Q_{4}$, described in Remark 7, because all the other points $Q_{5},\ldots ,Q_{m}$, are linear convex combinations of $Q_{1},Q_{2},Q_{3},Q_{4}$, that is $Q_{h}\in \mathcal {I}$ for each $h = 5,\ldots ,m$.

We examine the following different cases which depend on the logical value of $E_{n + 1}|H_{n + 1}$: $a)$E_n+ 1|H_n+ 1 is true; $b)$E_n+ 1|H_n+ 1 is false; $c)$E_n+ 1|H_n+ 1 is void.

• a) In this case

  $$\begin{array}{@{}rcl@{}} Q_{h}&=&(q_{h1},1,q_{h1})= q_{h1}(1,1,1)+(1-q_{h1})(0,1,0)\\ &=&q_{h1} Q_{1}+(1-q_{h1})Q_{3}. \end{array} $$

• b) In this case

  $$\begin{array}{@{}rcl@{}} Q_{h}&=&(q_{h1},0,0)= q_{h1}(1,0,0)+(1-q_{h1})(0,0,0)\\ &=&q_{h1} Q_{2}+(1-q_{h1})Q_{4}. \end{array} $$

• c) In this case $Q_{h}=(q_{h1},x_{n + 1},q_{h3})$ and we distinguish the following subcases: $(i)$$\bigwedge _{i = 1}^{n} E_{i}H_{i}$ true, so that $Q_{h}=(1,x_{n + 1},x_{n + 1})$; $(ii)$$\bigvee _{i = 1}^{n} \overline {E}_{i}H_{i}$ true, so that $Q_{h}=(0,x_{n + 1},0)$; $(iii)$E_i|H_i void, for every $i \in S$ and $E_{i}|H_{i}$ true for every i ∈{1, 2,…,n}∖ S,. for some $\emptyset \neq S \subset \{1,2,\ldots ,n\}$, so that $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})$. In subcase $(i)$ it holds that

  $$\begin{array}{@{}rcl@{}} Q_{h}\!&=&\!({\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}x_{n + 1}{\kern-.5pt},{\kern-.5pt}x_{n + 1{\kern-.5pt}}) = x_{n{\kern-.5pt}+{\kern-.5pt}1}({\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt}) + ({\kern-.5pt}1 - x_{n + 1})({\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt})\\ \!&=&\! x_{n + 1} Q_{1}+(1-x_{n + 1})Q_{2}. \end{array} $$

  In subcase $(ii)$ it holds that

  $$\begin{array}{@{}rcl@{}} Q_{h}&=&({\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}x_{n + 1}{\kern-.5pt},{\kern-.5pt}0)= x_{n + 1}({\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}1{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt})+(1\!-x_{n + 1})({\kern-.5pt}0{\kern-.5pt},{\kern-.5pt}0{\kern-.5pt},0)\\ &=& x_{n + 1} Q_{3}+(1-x_{n + 1})Q_{4}. \end{array} $$

  In subcase $(iii)$, it can be verified by a finite iterative procedure that the point $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})\in \mathcal {I}$. We examine the different cases on the cardinality s of S. We recall that $\bigwedge _{i\in S}(E_{i}|H_{i})$ is denoted by $\mathcal {C}_{S}$.

• Step 1. $s = 1$. Without loss of generality we assume $S=\{1\}$, so that $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})=(x_{1},x_{n + 1},x_{\{1,n + 1\}})$, where $x_{1}=P(E_{1}|H_{1})$, $x_{\{1,n + 1\}}=\mathbb {P}[(E_{1}|H_{1})\wedge (E_{n + 1}|H_{n + 1})]$. By Theorem 4 it holds that $\max \{x_{S}+x_{n + 1}-1,0\} \leq x_{S\cup \{n + 1\}} \leq \min \{x_{S},x_{n + 1}\}$, with (x_S,x_n+ 1) ∈ [0, 1]². In other words, $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})\in \mathcal {I}$. The reasoning is the same for $S=\{i\}$, $i = 2,\ldots ,n$.

• Step 2. $s = 2$. Without loss of generality we assume $S=\{1,2\}$, so that $x_{S}=\mathbb {P}[(E_{1}|H_{1})\wedge (E_{2}|H_{2})]$, $x_{S\cup \{n + 1\}}=\mathbb {P}[\mathcal {C}_{S\cup \{n + 1\}}]=\mathbb {P}[(E_{1}|H_{1})\wedge (E_{2}|H_{2})\wedge (E_{n + 1}|H_{n + 1})]$. We denote by $C_{0}^{*},C_{1}^{*}, \ldots ,C_{m^{*}}^{*}$, the constituents associated with $\{E_{i}|H_{i}, i \in S\cup \{n + 1\}\}$, where $C_{0}^{*}=\bigwedge _{i\in S\cup \{n + 1\}}\overline {H}_{i}$. Moreover, with $C_{h}^{*}$, $h = 0,1, \ldots ,m^{*},$ we associate the point $Q_{h}^{*}=(q_{h1}^{*},q_{h2}^{*},q_{h3}^{*})$ which represents the value of the random vector $\{\mathcal {C}_{S},E_{n + 1}|H_{n + 1},\mathcal {C}_{S\cup \{n + 1\}}\}$ when $C_{h}^{*}$ is true. We observe that $Q_{0}^{*}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})$ and that $Q_{1},Q_{2},Q_{3},Q_{4}$ still belongs to the set of points $\{Q_{h}^{*}, h = 1,\ldots ,m^{*}\}$. In order that the assessment (x_S,x_n+ 1,x_S∪{n+ 1}) on $\{\mathcal {C}_{S},E_{n + 1}|H_{n + 1},\mathcal {C}_{S\cup \{n + 1\}}\}$ be coherent, the point $Q_{0}^{*}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}})$ must belong to the convex hull of points $Q_{1}^{*}, Q_{2}^{*}, \ldots , Q_{m}^{*}$. We show that for each point $Q_{h}^{*}\neq Q_{i}$, $i = 1,2,3,4$, it holds that $Q_{h}^{*}\in \mathcal {I}$. By repeating the previous reasoning we only need to analyze the subcase $(iii)$ of case $c)$. We have to show that, for every nonempty subset $S^{\prime } \subset S$, the point $Q_{h}^{*}=(x_{S^{\prime }},x_{n + 1},x_{S^{\prime }\cup \{n + 1\}})$ belongs to the convex hull $\mathcal {I}$ of $Q_{1},\ldots ,Q_{4}$. As $S=\{1,2\}$, it holds that $S^{\prime }=\{1\}$, or $S^{\prime }=\{2\}$, so that $Q_{h}^{*}=(x_{S^{\prime }},x_{n + 1},x_{S^{\prime }\cup \{n + 1\}})=(x_{1},x_{n + 1},x_{\{1,n + 1\}})$, or $Q_{h}^{*}=(x_{2},x_{n + 1},x_{\{2,n + 1\}})$. By Step 1, in both cases $Q_{h}^{*} \in \mathcal {I}$. Thus $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}}) \in \mathcal {I}$. In other words, $\max \{x_{S}+x_{n + 1}-1,0\} \leq x_{S\cup \{n + 1\}} \leq \min \{x_{S},x_{n + 1}\}$, with $(x_{S},x_{n + 1}) \in [0,1]^{2}$. The reasoning is the same for every $S=\{i,j\} \subset \{1,2,\ldots ,n\}$.

  $$............................................................................... $$

  Step$k + 1$. $s=k + 1$, $2<k + 1<n$. By induction, assume that $(x_{S^{\prime }},x_{n + 1},x_{S^{\prime }\cup \{n + 1\}}) \in \mathcal {I}$ for every $S^{\prime }=\{i_{1},i_{2},\ldots ,i_{k}\} \subset \{1,2,\ldots ,n\}$. Then, by the previous reasoning, it follows that $Q_{h}=(x_{S},x_{n + 1},x_{S\cup \{n + 1\}}) \in \mathcal {I}$ for every $S=\{i_{1},i_{2},\ldots ,i_{k + 1}\}$. In other words, $\max \{x_{S}+x_{n + 1}-1,0\} \leq x_{S\cup \{n + 1\}} \leq \min \{x_{S},x_{n + 1}\}$, with $(x_{S},x_{n + 1}) \in [0,1]^{2}$, for every $S=\{i_{1},i_{2},\ldots ,i_{k + 1}\}$.

Thus, by this iterative procedure, also in the subcase $(iii)$ of case $c)$ it holds that $Q_{h}\in \mathcal {I}$. Then, $Q_{h}\in \mathcal {I}$, $h = 5,\ldots ,m$. Finally, the condition (45) is equivalent to $\mathcal {M} \in \mathcal {I}$, so that the assessment $\mathcal {M}$ is coherent if and only if

$$\begin{array}{@{}rcl@{}} && (\mu_{n},x_{n + 1}) \in [0,1]^{2}, \max\{\mu_{n}+x_{n + 1}-1,0\} \leq \mu_{n + 1}\\ &&\leq \min\{\mu_{n},x_{n + 1}\}. \end{array} $$

□

Proof of Theorem 15

The computation of the set ${\Pi }$ is based on Section 2.2. The constituents $C_{h}$’s and the points $Q_{h}$’s associated with $(\mathcal {F},\mathcal {M})$ are illustrated in Table 1.

We recall that $Q_{h}=(q_{h1},\ldots ,q_{h7})$ represents the value associated with $C_{h}$ of the random vector (E₁|H₁,E₂|H₂,E₃|H₃, (E₁|H₁) ∧ (E₂|H₂), $(E_{1}|H_{1})\wedge (E_{3}|H_{3}), (E_{2}|H_{2})\wedge (E_{3}|H_{3})$, ${(E_{1}|H_{1})\wedge (E_{2}|H_{2})\wedge (E_{3}|H_{3})})$, $h = 1,\ldots ,26$. With $C_{0}=\overline {H_{1}}\overline {H_{2}}\overline {H_{3}}$ it is associated $Q_{0}=\mathcal {M}$. Denoting by $\mathcal {I}$ the convex hull generated by $Q_{1},Q_{2}, \ldots ,Q_{26}$, the coherence of the prevision assessment $\mathcal {M}$ on $\mathcal {F}$ requires that the condition $\mathcal {M}\in \mathcal {I}$ be satisfied; this amounts to the solvability of the following system

$$\begin{array}{@{}rcl@{}} ({\Sigma}) \hspace{1 cm} \mathcal{M}&=&{\sum}_{h = 1}^{26} \lambda_{h}Q_{h}, {\sum}_{h = 1}^{26} \lambda_{h}= 1, \\ \lambda_{h}&\geq& 0, h = 1,\ldots,26 . \end{array} $$

We observe that

$$\begin{array}{@{}rcl@{}} Q_{3}&=&x_{3}Q_{1}+(1-x_{3})Q_{2}, Q_{6}=x_{3}Q_{4}+ (1-x_{3})Q_{5}, \\ Q_{7}&=&x_{2}Q_{1}+ (1-x_{2})Q_{4}, Q_{8}=x_{2}Q_{2}+(1-x_{2})Q_{5}, \\ Q_{9}&=&x_{23}Q_{1}+(x_{2}-x_{23})Q_{2}+(x_{3}-x_{23})Q_{4}\\ &&+(x_{23}-x_{2}-x_{3}+ 1)Q_{5},\\ Q_{12}&=&x_{3}Q_{10}+(1-x_{3})Q_{11},\\ Q_{15}&=&x_{3}Q_{13} +(1-x_{3})Q_{14}, \\ Q_{16}&=&x_{2}Q_{10}+(1-x_{2})Q_{13}, \\ Q_{17}&=&x_{2}Q_{11} +(1-x_{2})Q_{14}, \\ Q_{18}&=&x_{23}Q_{10}+(x_{2}-x_{23})Q_{11}+(x_{3}-x_{23})Q_{13}\\ &&+(x_{23}-x_{2}-x_{3}+ 1)Q_{14},\\ Q_{19}&=&x_{1}Q_{1}+(1 - x_{1})Q_{10}, Q_{20}=x_{1}Q_{2}\!+(1 - x_{1})Q_{11},\\ Q_{21}&=&x_{13} Q_{1}+(x_{1}-x_{13})Q_{2} +(x_{3}-x_{13})Q_{10}\\ &&+(x_{13}-x_{1}-x_{3}+ 1) Q_{11}, \\ Q_{22}&=&x_{1}Q_{4}+(1 - x_{1})Q_{13}, Q_{23}=x_{1}Q_{5}\!+(1 - x_{1})Q_{14}, \\ Q_{24}&=&x_{13} Q_{4}+(x_{1}-x_{13})Q_{5} +(x_{3}-x_{13})Q_{13}\\ &&+(x_{13}-x_{1}-x_{3}+ 1) Q_{14},\\ Q_{25}&=&x_{12}Q_{1}+(x_{1}-x_{12})Q_{4}+(x_{2}-x_{12})Q_{10}\\ &&+(x_{12}-x_{1}-x_{2}+ 1)Q_{13}, \end{array} $$

$$\begin{array}{@{}rcl@{}} Q_{26}&=&x_{12}Q_{2}+(x_{1}-x_{12})Q_{5}+(x_{2}-x_{12})Q_{11}\\ &&+(x_{12}-x_{1}-x_{2}+ 1)Q_{14}. \end{array} $$

Thus, $\mathcal {I}$ coincides with the convex hull of the points $Q_{1}, Q_{2}, Q_{4}, Q_{5}, Q_{10}, Q_{11}, Q_{13}, Q_{14}$. For the sake of simplicity, we set: $Q_{1}^{\prime }=Q_{1}, Q_{2}^{\prime }=Q_{2}$, $Q_{3}^{\prime }=Q_{4}$, $Q_{4}^{\prime }=Q_{5}$, $Q_{5}^{\prime }=Q_{10}$, $Q_{6}^{\prime }=Q_{11}, Q_{7}^{\prime }=Q_{13}$, $Q_{8}^{\prime }=Q_{14}$. Then, the condition $\mathcal {M}\in \mathcal {I}$ amounts to the solvability of the following system

$$\begin{array}{@{}rcl@{}} ({\Sigma}^{\prime}) \hspace{1 cm} \mathcal{M}&=&{\sum}_{h = 1}^{8} \lambda_{h}^{\prime}Q_{h}^{\prime}, {\sum}_{h = 1}^{8} \lambda_{h}^{\prime}= 1,\\&&\lambda_{h}^{\prime}\geq 0, h = 1,\ldots, 8 \end{array} $$

that is

$$({\Sigma}^{\prime})\left\{ \begin{array}{l} \lambda_{1}^{\prime}+\lambda_{2}^{\prime}+\lambda_{3}^{\prime}+\lambda_{4}^{\prime} = x_{1}, \lambda_{1}^{\prime}+\lambda_{2}^{\prime}+\lambda_{5}^{\prime}+\lambda_{6}^{\prime}=x_{2},\\ \lambda_{1}^{\prime}+\lambda_{3}^{\prime}+\lambda_{5}^{\prime}+\lambda_{7}^{\prime}=x_{3}, \lambda_{1}^{\prime}+\lambda_{2}^{\prime}=x_{12},\\ \lambda_{1}^{\prime}+\lambda_{3}^{\prime}=x_{13}, \lambda_{1}^{\prime}+\lambda_{5}^{\prime}=x_{23}, \lambda_{1}^{\prime}=x_{123},\\ {\sum}_{h = 1}^{8}\lambda_{h}^{\prime}= 1, \lambda_{h}^{\prime} \geq 0, h = 1,2,\ldots,8. \end{array} \right. $$

System $({\Sigma }^{\prime })$ can be written as

$$({\Sigma}^{\prime})\left\{ \begin{array}{ll} \lambda^{\prime}_{1}=x_{123}, \lambda^{\prime}_{2}=x_{12}-x_{123},\\ \lambda^{\prime}_{3}=x_{13}-x_{123}, \lambda^{\prime}_{4}=x_{1}-x_{12}-x_{13}+x_{123},\\ \lambda^{\prime}_{5}=x_{23}-x_{123}, \lambda^{\prime}_{6}=x_{2}-x_{12}-x_{23}+x_{123},\\ \lambda^{\prime}_{7}=x_{3}-x_{13}-x_{23}+x_{123},\\ \lambda^{\prime}_{8}= 1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}-x_{123},\\ \lambda^{\prime}_{h} \geq 0, h = 1,2,\ldots,8. \end{array} \right. $$

As it can be verified, by non-negativity of $\lambda ^{\prime }_{1},\ldots ,\lambda ^{\prime }_{8}$ it follows that $({\Sigma }^{\prime })$ is solvable (with a unique solution) if and only if

$$ \left\{ \begin{array}{l} x_{123}\geq \max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}-x_{3}\},\\ x_{123}\leq \min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}, \end{array} \right. $$

(46)

or, in a more explicit way, if and only if the following conditions are satisfied

$$ \left\{ \begin{array}{l} (x_{1},x_{2},x_{3})\in[0,1]^{3},\\ \max\{x_{1}+x_{2}-1,x_{13}+x_{23}-x_{3},0\}\leq x_{12}\leq \min\{x_{1},x_{2}\},\\ \max\{x_{1}+x_{3}-1,x_{12}+x_{23}-x_{2},0\}\leq x_{13}\leq \min\{x_{1},x_{3}\},\\ \max\{x_{2}+x_{3}-1,x_{12}+x_{13}-x_{1},0\}\leq x_{23}\leq \min\{x_{2},x_{3}\},\\ 1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\geq 0,\\ x_{123}\geq \max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}\!-x_{3}\},\\ x_{123}\leq \min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}. \end{array} \right. $$

(47)

Notice that the conditions in (47) coincide with that ones in (18). Moreover, assuming $({\Sigma }^{\prime })$ solvable, with the solution $(\lambda _{1}^{\prime },\ldots ,\lambda _{8}^{\prime })$, we associate the vector $(\lambda _{1},\lambda _{2},\ldots ,\lambda _{26})$, with $\lambda _{1}=\lambda _{1}^{\prime }, \lambda _{2}=\lambda _{2}^{\prime }, \lambda _{4}=\lambda _{3}^{\prime }, \lambda _{5}=\lambda _{4}^{\prime }, \lambda _{10}=\lambda _{5}^{\prime }, \lambda _{11}=\lambda _{6}^{\prime }, \lambda _{13}=\lambda _{7}^{\prime }, \lambda _{14}=\lambda _{8}^{\prime }, \lambda _{h}= 0, h\notin \{1,2,4,5,10,11,13,14\},$ which is a solution of (Σ). Moreover, defining $\mathcal {J}=\{ 1,2,4,5,10,11,13,14\}$, it holds that $\bigvee _{h\in \mathcal {J}}C_{h}= H_{1}\wedge H_{2}\wedge H_{3}$. Therefore, ${\sum }_{h\in \mathcal {J}}\lambda _{h} ={\sum }_{h: C_{h} \subseteq H_{1}H_{2}H_{3}} \lambda _{h} = 1$ and hence ${\sum }_{h: C_{h} \subseteq H_{i}}\lambda _{h}= 1$, $i = 1,2,3$, ${\sum }_{h: C_{h} \subseteq H_{i}\vee H_{j}}\lambda _{h}= 1$, $i\neq j$, ${\sum }_{h: C_{h} \subseteq H_{1}\vee H_{2}\vee H_{3}}\lambda _{h}=~1$; thus, by (2), $I_{0} = \emptyset$. Then, by Theorem 2, the solvability of (Σ) is also sufficient for the coherence of $\mathcal {M}$. Finally, ${\Pi }$ is the set of conditional prevision assessments $(x_{1},x_{2},x_{3},x_{12},x_{13},_{23},x_{123})$ which satisfy the conditions in (18). □

Table 1 Constituents $C_{h}$’s and corresponding points Q_h’s associated with $(\mathcal {F},\mathcal {M})$, where $\mathcal {M}=(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ is a prevision assessment on $\mathcal {F}=\{E_{1}|H_{1},E_{2}|H_{2},E_{3}|H_{3}, (E_{1}|H_{1})\wedge (E_{2}|H_{2}), (E_{1}|H_{1})\wedge (E_{3}|H_{3}), (E_{2}|H_{2})\wedge (E_{3}|H_{3}), (E_{1}|H_{1})\wedge (E_{2}|H_{2})\wedge (E_{3}|H_{3})\}$

Proof of Theorem 16

Notice that, $(E_{i}|H)\wedge (E_{j}|H)=(E_{i}E_{j})|H$, for every $\{i,j\}\subset \{1,2,3\}$, and $(E_{1}|H)\wedge (E_{2}|H)\wedge (E_{3}|H)=(E_{1}E_{2}E_{3})|H$. Then $\mathcal {F}=\{E_{1}|H,E_{2}|H,E_{3}|H, (E_{1}E_{2})|H, (E_{1}E_{3})|H, (E_{2}E_{3})|H$, $(E_{1}E_{2}E_{3})|H\}$. The computation of the set ${\Pi }$ is based on Section 2.2. The constituents $C_{h}$’s and the points $Q_{h}$’s associated with $(\mathcal {F},\mathcal {M})$ are illustrated in Table 2.

Table 2 Constituents $C_{h}$’s and corresponding points Q_h’s associated with $(\mathcal {F},\mathcal {M})$, where $\mathcal {M}=(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ is a prevision assessment on $\mathcal {F}=\{E_{1}|H,E_{2}|H,E_{3}|H, (E_{1}E_{2})|H, (E_{1}E_{3})|H, (E_{2}E_{3})|H, (E_{1}E_{2}E_{3})|H\}$

We recall that $Q_{h}=(q_{h1},\ldots ,q_{h7})$ represents the value associated with $C_{h}$ of the random vector $(E_{1}|H,E_{2}|H,E_{3}|H, (E_{1}E_{2})|H$, (E₁E₃)|H, (E₂E₃)|H, $(E_{1}E_{2}E_{3})|H)$, $h = 1,\ldots ,8$. With $C_{0}=\overline {H}$ it is associated $Q_{0}=\mathcal {M}$. Denoting by $\mathcal {I}$ the convex hull generated by Q₁,Q₂,…,Q₈, as all the conditioning events coincide with H the assessment $\mathcal {M}$ on $\mathcal {F}$ is coherent if and only if $\mathcal {M}\in \mathcal {I}$; that is, if and only if the following system is solvable

$$ \mathcal{M} = {\sum}_{h = 1}^{8} \lambda_{h}Q_{h}, {\sum}_{h = 1}^{8} \lambda_{h} = 1, \lambda_{h}\!\geq\! 0, h = 1,\ldots,8 . $$

(48)

The points $Q_{1},Q_{2}, \ldots ,Q_{8}$ coincide with the points $Q_{1}^{\prime },Q_{2}^{\prime }, \ldots ,Q_{8}^{\prime }$ in the proof of Theorem 15, respectively. Then, system (48) coincides with system $({\Sigma })^{\prime }$ in the proof of Theorem 15. Therefore, it is solvable if and only if the conditions in (18) are satisfied. In other words, the set ${\Pi }$ of all coherent assessments $\mathcal {M}$ on $\mathcal {F}$ coincides with the set of points (x₁,x₂,x₃,x₁₂,x₁₃,x₂₃,x₁₂₃) which satisfy the conditions in (18). □

Proof of of Theorem 18

In order to prove the theorem it is enough to prove the following implications: a) $(i) \Rightarrow (ii)$; b) $(ii) \Rightarrow (iii)$; c) $(iii) \Rightarrow (i)$.

• a) $(i) \Rightarrow (ii)$. We recall that $\mathcal {F}$ p-entails $E_{n + 1}|H_{n + 1}$ if and only if either $H_{n + 1} \subseteq E_{n + 1}$, or there exists a nonempty $\mathcal {F}_{{\Gamma }} \subseteq \mathcal {F}$, where ${\Gamma } \subseteq \{1,\ldots ,n\}$, such that $QC(\mathcal {F}_{{\Gamma }})$ implies $E_{n + 1}|H_{n + 1}$ (see, e.g. [24, Theorem 6]). Let us first consider the case where $H_{n + 1} \subseteq E_{n + 1}$. In this case $P(E_{n + 1}|H_{n + 1}) = 1$ and $E_{n + 1}|H_{n + 1} =H_{n + 1}+\overline {H}_{n + 1}= 1$. We have $\mathcal {C}_{n + 1}=\mathcal {C}_{n} \wedge (E_{n + 1}|H_{n + 1})$, with E_n+ 1|H_n+ 1 = 1. We distinguish two cases: $(\alpha )~H_{n + 1}$ is true; $(\beta )~H_{n + 1}$ is false. In case $(\alpha )$, by Definition 5 and Remark 4, as $E_{n + 1}|H_{n + 1}$ is true it follows that the values of $\mathcal {C}_{n + 1}$ and of $\mathcal {C}_{n}$ coincide. In case $(\beta )$, let $C_{0},\ldots ,C_{m}$ be the constituents associated with $\mathcal {F}$, where $C_{0}=\overline {H}_{1}{\cdots } \overline {H}_{n}$. Then, the constituents $C_{0}^{\prime },\ldots ,C_{m}^{\prime }$ associated with $\mathcal {F}\cup \{E_{n + 1}|H_{n + 1}\}$ and contained in $\overline {H}_{n + 1}$ are $C_{0}^{\prime }=C_{0}\overline {H}_{n + 1},\ldots ,C_{m}^{\prime }=C_{m}\overline {H}_{n + 1}$. For each constituent $C_{h}^{\prime }$, $h = 1,\ldots ,m$, by formula (10) the corresponding value of $\mathcal {C}_{n}$ is $z_{h}\in \{1,0,x_{S_{h}^{\prime \prime \prime }}\}$. We denote by $z_{h}^{\prime }$ the value of $\mathcal {C}_{n + 1}$ associated with $z_{h}$ and we recall that $C_{h}^{\prime }\subseteq \overline {H}_{n + 1}$, $h = 0,1,\ldots ,m$. For each index h, if $z_{h}= 1$, then $z^{\prime }_{h}= 1$; if $z_{h}= 0$, then $z^{\prime }_{h}= 0$; if $z_{h}=x_{S_{h}^{\prime \prime \prime }}$, then $z_{h}^{\prime }=x_{S_{h}^{\prime \prime \prime }\cup \{n + 1\}}$. We set $P(E_{n + 1}|H_{n + 1})=x_{n + 1}$; in our case $x_{n + 1}= 1$. Moreover, by Theorem 10

  $$\begin{array}{@{}rcl@{}} \max\!\left\{x_{S_{h}^{\prime\prime\prime}} + x_{n + 1} - 1,\!0\right\}\!\leq\! x_{S_{h}^{\prime\prime\prime}\cup\{n + 1\}}\!\leq\! \min\!\left\{x_{S_{h}^{\prime\prime\prime}},\!x_{n + 1}{\kern-.5pt}\right\}\!; \end{array} $$

  therefore $x_{S_{h}^{\prime \prime \prime }\cup \{n + 1\}}=x_{S_{h}^{\prime \prime \prime }}$. Then, the values of $\mathcal {C}_{n + 1}$ and of $\mathcal {C}_{n}$ coincide for every $C^{\prime }_{h}$. Thus, $\mathcal {C}_{n + 1}=\mathcal {C}_{n}$ when $H_{n + 1}\subseteq E_{n + 1}$.

  We consider now the case where there exists $\mathcal {F}_{{\Gamma }}\subseteq \mathcal {F}$, $\mathcal {F}_{{\Gamma }}\neq \emptyset$, such that $QC(\mathcal {F}_{{\Gamma }})\subseteq E_{n + 1}|H_{n + 1}$. First of all we prove that $\mathcal {C}(\mathcal {F}_{{\Gamma }}\cup \{E_{n + 1}|H_{n + 1}\} )=\mathcal {C}(\mathcal {F}_{{\Gamma }})$. For the sake of simplicity, we set $\mathcal {C}(\mathcal {F}_{{\Gamma }})=\mathcal {C}_{{\Gamma }}$ and $\mathcal {C}(\mathcal {F}_{{\Gamma }}\cup \{E_{n + 1}|H_{n + 1}\} )=\mathcal {C}_{{\Gamma }\cup \{n + 1\}}$.

  If the value of $\mathcal {C}_{{\Gamma }}$ is 1 (because all the conditional events in $\mathcal {F}_{{\Gamma }}$ are true), then $QC(\mathcal {F}_{{\Gamma }})$ is true and hence $E_{n + 1}|H_{n + 1}$ is also true; thus $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}= 1$, so that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=\mathcal {C}_{{\Gamma }}$.

  If the value of $\mathcal {C}_{{\Gamma }}$ is 0 (because some conditional event in $\mathcal {F}_{{\Gamma }}$ is false), then $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}$ is 0 too, so that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=\mathcal {C}_{{\Gamma }}$.

  If $\mathcal {C}_{{\Gamma }}$ is $x_{S}$ for some nonempty subset $\mathcal {S}\subset {\Gamma }$ (that is, all the conditional events in $\mathcal {F}_{S}$ are void and the other ones in $\mathcal {F}_{{\Gamma } \setminus S}$ are true), then $QC(\mathcal {F}_{{\Gamma }})$ is true and and hence $E_{n + 1}|H_{n + 1}$ is also true; thus $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=x_{S}$, so that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=\mathcal {C}_{{\Gamma }}$.

  If $\mathcal {C}_{{\Gamma }}$ is $x_{{\Gamma }}$ because all the conditional events in $\mathcal {F}_{{\Gamma }}$ are void, then $QC(\mathcal {F}_{{\Gamma }})$ is void and for $E_{n + 1}|H_{n + 1}$ there are two cases: 1) $E_{n + 1}|H_{n + 1}$ true; 2) $E_{n + 1}|H_{n + 1}$ void. In case 1), by also recalling Remark 4, it holds that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=x_{{\Gamma }}$ so that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=\mathcal {C}_{{\Gamma }}$.

  In case 2) it holds that $\mathcal {C}_{{\Gamma }\cup \{n + 1\}} =x_{{\Gamma } \cup \{n + 1\}}$, where $x_{{\Gamma }\cup \{n + 1\}}=\mathbb {P}(\mathcal {C}_{{\Gamma }\cup \{n + 1\}})$. Now, we observe that the random quantities $\mathcal {C}_{{\Gamma }}$ and $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}$ coincide conditionally on $\bigvee _{i\in {\Gamma }\cup \{n + 1\}} H_{i}$ being true; then by Theorem 3 it holds that $\mathbb {P}(\mathcal {C}_{{\Gamma }})=\mathbb {P}(\mathcal {C}_{{\Gamma }\cup \{n + 1\}})$, that is $x_{{\Gamma }}=x_{{\Gamma }\cup \{n + 1\}}$; thus $\mathcal {C}_{{\Gamma }\cup \{n + 1\}}=\mathcal {C}_{{\Gamma }}$.

  Finally, denoting by Γ^c the set $\{1,\ldots ,n\}\setminus {\Gamma }$, by the associative property of conjunction we obtain

  $$\begin{array}{@{}rcl@{}} \mathcal{C}_{n + 1}&=&\mathcal{C}_{n}\wedge E_{n + 1}|H_{n + 1}=\mathcal{C}_{{\Gamma}^{c}}\wedge \mathcal{C}_{{\Gamma}} \wedge E_{n + 1}|H_{n + 1}\\ &=&\mathcal{C}_{{\Gamma}^{c}}\wedge \mathcal{C}_{{\Gamma}}=\mathcal{C}_{n}. \end{array} $$

• b) $(ii) \Rightarrow (iii)$. By monotonicity property of conjunction it holds that $\mathcal {C}_{n + 1}\leq E_{n + 1}|H_{n + 1}$. Then, by assuming $\mathcal {C}_{n}=\mathcal {C}_{n + 1}$, it follows $\mathcal {C}_{n}\leq E_{n + 1}|H_{n + 1}$.

• c) $(iii) \Rightarrow (i)$. Let us assume that $\mathcal {C}_{n}\leq E_{n + 1}|H_{n + 1}$, so that $\mathbb {P}(\mathcal {C}_{n})\leq P(E_{n + 1}|H_{n + 1})$. Moreover, by assuming that $P(E_{i}|H_{i})= 1$, $i = 1,\ldots , n$, from (15) it follows $\mathbb {P}(\mathcal {C}_{n})= 1$ and hence $P(E_{n + 1}|H_{n + 1})= 1$, that is $\mathcal {F}$ p-entails $E_{n + 1}|H_{n + 1}$.

□