A Stochastic Model of Mathematics and Science

Abstract

We introduce a framework that can be used to model both mathematics and human reasoning about mathematics. This framework involves stochastic mathematical systems (SMSs), which are stochastic processes that generate pairs of questions and associated answers (with no explicit referents). We use the SMS framework to define normative conditions for mathematical reasoning, by defining a “calibration” relation between a pair of SMSs. The first SMS is the human reasoner, and the second is an “oracle” SMS that can be interpreted as deciding whether the question–answer pairs of the reasoner SMS are valid. To ground thinking, we understand the answers to questions given by this oracle to be the answers that would be given by an SMS representing the entire mathematical community in the infinite long run of the process of asking and answering questions. We then introduce a slight extension of SMSs to allow us to model both the physical universe and human reasoning about the physical universe. We then define a slightly different calibration relation appropriate for the case of scientific reasoning. In this case the first SMS represents a human scientist predicting the outcome of future experiments, while the second SMS represents the physical universe in which the scientist is embedded, with the question–answer pairs of that SMS being specifications of the experiments that will occur and the outcome of those experiments, respectively. Next we derive conditions justifying two important patterns of inference in both mathematical and scientific reasoning: (i) the practice of increasing one’s degree of belief in a claim as one observes increasingly many lines of evidence for that claim, and (ii) abduction, the practice of inferring a claim’s probability of being correct from its explanatory power with respect to some other claim that is already taken to hold for independent reasons.

Appendices

Appendix 1: Additional Facets of the SMS Framework

1.1 Claim Trajectories

Although it was not needed for the results or discussion in this paper, we can also use the SMS framework to assign probabilities to trajectories of claims produced at different steps of an SMS.

Definition 11.1

A claim vector trajectory $\vec {C}^{n}$ is a map taking any $i \in 1, \ldots , n$ to a claim vector if n is finite, and taking any $i \in \mathbbm {Z}^+$ to a claim vector if $n = \infty$.

As shorthand, if a claim vector trajectory contains only a single claim vector, then we will sometimes write that trajectory as that single claim vector. In the sequel the spaces ${\mathcal {Q}}$ and ${\mathcal {V}}$ will often be implicit. We will also cavalierly use |.| to indicate cardinality. So for example, |C| is the number of components of a claim vector C, $|\hat{C}|$ is the number of elements in the claim set $\hat{C}$, etc.

It will occasionally be useful to consider probability distributions concerning the event that the claim vector generated at the n-th step of a particular SMS contains the answer v in response to the question q. We present some definitions that will be useful when working with such distributions for the case where ${\mathcal {C}}$ is countable; their extensions to other spaces is straight-forward.

Definition 11.2

Suppose we are given an SMS $\varphi =(\mathfrak {C},X)$ and a claim vector trajectory $\vec {C}^{n-1} = (C^1, C^2, \ldots , C^{n-1})$ for some finite $n > 1$. The associated question semi-distribution is the function mapping all $q\in {{\mathcal {Q}}}$ to the value

$$\begin{aligned} P^n(q, \vec {C}^{n-1}) := \sum _{v^{\prime }\in {{\mathcal {V}}}} P_{\mathfrak {C}} \left( X(1)=C^{1},\dots , X(n-1) = C^{n-1}, (q, v') \in \hat{X}(n)\right) . \end{aligned}$$

$P^n(q, \vec {C}^{n-1})$ is the joint probability that the question q occurs in a claim in the step-n claim vector and that the earlier claim vectors were $C^1, C^2, \ldots , C^{n-1}$. Intuitively, one can think of this as the probability that a reasoner considers a particular question at step n and also outputs a particular set of claims at previous steps. Note that in general, for any fixed $\vec {C}^{n-1}$, the associated question semi-distribution is not normalized when considered as a function of questions q.

We define claim semi-distributions $P^n((q,v),\vec {C}^{n-1})$ analogously to question semi-distributions. We combine these definitions as follows:

Definition 11.3

For an SMS $\varphi =(\mathfrak {C},X)$, question q, and claim vector trajectory $\vec {C}^{n-1}$ where $P^n(q, \vec {C}^{n-1}) \ne 0$, the associated response distribution is the map from all $v\in {{\mathcal {V}}}$ to the value

$$\begin{aligned} P^n(v | q,\vec {C}^{n-1})=\frac{P^n((q, v), \vec {C}^{n-1})}{P^n(q, \vec {C}^{n-1})}. \end{aligned}$$

The response distribution for a given SMS $\varphi$, question q, and claim vector trajectory $\vec {C}^{n-1}$ specifies the probability that q is given an answer v in the n-th step of $\varphi$, conditional on $\varphi$ having produced the earlier claims specified in the partial claim vector trajectory $\vec {C}^{n-1}$ and having produced some claim that includes the question q in the n-th claim vector.¹⁷ This conditional character of the response probability distribution will be exploited below to represent path-dependent mathematical reasoning, in which the answers given by an SMS to questions earlier in the process of generating mathematical claims affect the answers given to questions posed later in the process.

1.2 An Alternative Distribution over Claim Sets

In the body of this paper, we considered the following distribution over claim sets:

$$\begin{aligned} P^{n}(\hat{C}):=\sum _{\{C\in {\mathcal {C}}^{*}:\hat{C}\subseteq U(C)\}}P_{\mathfrak {C}}(X(n)=C). \end{aligned}$$

(20)

This is the probability that an SMS outputs, at step n, a claim vector whose unordering is a superset of $\hat{C}$. But we may also want to consider the probability that an SMS outputs, at a step n, a vector whose unordering simply is the claim set $\hat{C}$. Such a distribution, which we denote $\underline{P}^{n}(\hat{C})$, can be defined in the obvious way:

$$\begin{aligned} \underline{P}^{n}(\hat{C}):=\sum _{\{C\in {\mathcal {C}}^{*}:\hat{C}= U(C)\}}P_{\mathfrak {C}}(X(n)=C). \end{aligned}$$

(21)

We can then use this distribution to define a related question semi-distribution and response distribution:

$$\begin{aligned} \underline{P}^{n}((q,v),\hat{C}):= & {} \sum _{\{C\in {\mathcal {C}}^{*}:\hat{C}\cup \{(q,v)\}= U(C)\}}P_{\mathfrak {C}}(X(n)=C), \end{aligned}$$

(22)

$$\begin{aligned} \underline{P}^{n}(q,\hat{C}):= & {} \sum _{v\in {\mathcal {V}}}\underline{P}^{n}((q,v),\hat{C}), \end{aligned}$$

(23)

$$\begin{aligned} \underline{P}^{n}(v|q,\hat{C}):= & {} \frac{\underline{P}^{n}((q,v),\hat{C})}{\underline{P}^{n}(q,\hat{C})}. \end{aligned}$$

(24)

Finally, we can extend this distribution to apply to trajectories of claim sets, rather than claim vectors, e.g.:

$$\begin{aligned} \underline{P}^n(q,\vec {\hat{C}}^{n-1})&:= \sum _{\vec {C}^{n-1} : U(C^i) = \hat{C}^i \, \forall 1 \le i \le n-1} P^n(q,\vec {C}^{n-1}), \end{aligned}$$

(25)

where $\vec {\hat{C}}^{n-1}$ is a trajectory of unordered claim sets $(\hat{C}^{1},\dots ,\hat{C}^{n-1})$.

Appendix 2: Proofs of Propositions

1.1 Proof of Lemma 3.3

Proof

Choose some SMS that is backward-consistent after step $\kappa$, and some claim set $\hat{C}$. For all steps $j > \kappa$ and $j> i > \kappa$,

$$\begin{aligned} P^j(\hat{C})&= \sum _{\hat{C}' : \hat{C} \subseteq \hat{C}'} {\underline{P}}^j(\hat{C}') \nonumber \\&= \sum _{{C}' : \hat{C} \subseteq U({C}')} P_{\mathfrak {C}} (X(j, .) = {C}') \nonumber \\&= \sum _{{C}' : \hat{C} \subseteq U(C')} \left[ \sum _{{C}'' : \hat{C} \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}', X(i, .) = {C}'')\right. \nonumber \\&\quad \left. + \sum _{{C}'' : \hat{C} \not \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}', X(i, .) = {C}'') \right] \nonumber \\&= \sum _{{C}' : \hat{C} \subseteq U(C')} \bigg [ \sum _{{C}'' : \hat{C} \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}' \, \vert \, X(i, .) = {C}'') P_{\mathfrak {C}} (X(i, .) = {C}'') \nonumber \\&\quad + \sum _{{C}'' : \hat{C} \not \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}', X(i, .) = {C}'') \bigg ]. \end{aligned}$$

(26)

Due to backward-consistency, $U(C^{\prime \prime })\subseteq U(C^{\prime })$ and so if $\hat{C}\subseteq U(C^{\prime \prime })$, then $\hat{C}\subseteq U(C^{\prime })$. This means that for all $C^{\prime \prime }$ such that $\hat{C}\subseteq U(C^{\prime \prime })$,

$$\begin{aligned} P_{\mathfrak {C}}(X(j,.)\in \{C^{\prime }: \hat{C}\subseteq U(C^{\prime })\}|X(i,.)=U(C^{\prime \prime }))=1. \end{aligned}$$

(27)

Thus, the sum over $C'$ of the conditional distribution in the summand in the last line equals 1, for all $C''$ in the associated sum. Therefore we get

$$\begin{aligned} P^j(\hat{C})&= \sum _{{C}'' : \hat{C} \subseteq U(C'')} P_{\mathfrak {C}} (X(i, .) = {C}'')\nonumber \\&\quad + \sum _{{C}' : \hat{C} \subseteq U(C')} \sum _{{C}'' : \hat{C} \not \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}', X(i, .) = {C}'') \nonumber \\&= P^i(\hat{C}) + \sum _{{C}' : \hat{C} \subseteq U(C')} \sum _{{C}'' : \hat{C} \not \subseteq U(C'')} P_{\mathfrak {C}} (X(j, .) = {C}', X(i, .) = {C}'') \nonumber \\&\ge P^i(\hat{C}). \end{aligned}$$

(28)

So for any claim set $\hat{C}$, $P^j(\hat{C})$ is a monotonically increasing function of j. In addition, that probability is upper-bounded as one varies over all j, by 1. By the completeness axiom of the reals, that means it has a least upper bound. Therefore the limit of that probability goes to infinity is well-defined. Replacing j with m establishes the claim. $\square$

1.2 Proof of Proposition 6.2

Proof

The divergence function D in the definition of embed-calibrated can only equal 0 if its two arguments are identical. Therefore for all embedded prediction triples $(v, q, \hat{C})$, and $v^m_1$,

$$\begin{aligned} \Psi (\psi (q), v)(v_1^m) = \overline{P}_1\left( v_1^m \;|\; \psi (q), E^{-1}[\{(q, v)\} \cup {\hat{C}}] \right) . \end{aligned}$$

(29)

Define

$$\begin{aligned} \hat{C}_2(v^m_1)&:= E \left[ \{(\psi (q), v^m_1)\} \cup E^{-1}[\{(q, v)\} \cup {\hat{C}}]\right] . \end{aligned}$$

(30)

Since $(q, \hat{C})$ is a discriminating pair,

$$\begin{aligned} E^{-1}(\hat{{C}}_2(v^m_1)) = \{(\psi (q), v^m_1)\} \cup E^{-1}[\{(q, v)\} \cup {\hat{C}}]. \end{aligned}$$

(31)

By the definition of embedding function, this means that

$$\begin{aligned} \overline{P}_{1} \left( \{(\psi (q), v^m_1)\} \cup E^{-1}[\{(q, v)\} \cup {\hat{C}}]\right)&= P^{n}_{2} \left( \hat{C}_2(v^m_1)\right) . \end{aligned}$$

(32)

Since this is true for all $v^m_1$, it also is true when we sum over such $v^m_1$. Therefore

$$\begin{aligned} \overline{P}_{1} \left( v^m_1 \,|\, \psi (q), E^{-1}[\{(q, v)\} \cup \hat{C}] \right)&= \frac{P^{n}_{2} (\hat{C}_2(v^m_1))}{\sum _{v^m_1} P^{n}_{2} (\hat{C}_2(v^m_1))}. \end{aligned}$$

(33)

(See discussion just below Definition 3.1.) This completes the proof. $\square$

1.3 Derivation of Definition 7.1

We begin with an alternative definition, that a collection of evidence paths ${\mathcal {B}}$ is a set of claim sets $\{B(i)\}$ such that:

$$\begin{aligned}&\forall \, i \in \{2, \ldots , n\} \qquad \dfrac{P\left( B(1), \ldots , B(i) \,|\, (q, v^*), \beta \right) }{P\left( B(1), \ldots , B(i-1) \,|\, \beta , (q, v^*)\right) \times P\left( B(i) \,|\, \beta , (q, v^*)\right) } \nonumber \\&\quad \ge \dfrac{P\left( B(1), \ldots , B(i) \,|\, q, \beta \right) }{P\left( B(1), \ldots , B(i-1) \,|\, q, \beta ) \times P(B(i) \,|\, q, \beta \right) }. \end{aligned}$$

(34)

Intuitively speaking, Eq. (10) tells us that each B(i) is a line of evidence for the claim $(q, v^*)$, when considered in isolation of all the others. Equation (34) goes on to tell us that those different lines of evidence do not “work at cross-purposes”, thwarting one another. (The precise indexing of the n sets in ${\mathcal {B}}$ is not important, so long as we can find such an indexing for which Eq. (34) holds.)

To illustrate Eq. (34), suppose that B(1) and B(2) are two lines of reasoning that each establish for a given mathematician-SMS that (conditional on $\beta$) the answer to q is $v^{*}$. However, suppose that both lines of reasoning depend on steps such that, if the answer to q did indeed turn out to be $v^{*}$ rather than something else, it would imply to the mathematician-SMS that B(1) and B(2) are very unlikely to both be true. Thus, from the perspective of the mathematician-SMS, $P\left( B(1), B(2) \,|\, (q, v^*), \beta \right) \ll P\left( B(1), B(2) \,|\, q, \beta \right)$. This is one way of formalizing the notion that B(1) and B(2) thwart one another, conditioned on the answer $v^*$ in response to the question q (in addition to the foundational claim set $\beta$), but not if that answer is unspecified.

We still need to quantify the strength of that thwarting though. To see how to do this, suppose that although the supposition that q has the answer $v^{*}$ might increase the mathematician-SMS’s degree of belief that B(1) is true, the overall effect is mild: $P(B(1)|(q,v^{*}), \beta )$ is not much greater than $P(B(1)|q,\beta )$. Suppose that similarly, $P(B(2)|(q,v^{*}), \beta )$ is not much greater than $P(B(2)|q,\beta )$. Under these conditions, Eq. (34) may not hold for B(1) and B(2), so that they are not both evidence paths for the claim $(q,v^{*})$. Intuitively, they thwart each other too much. Thus, enforcing Eq. (34) ensures that such thwarting does not occur.

If ${\mathcal {B}}$ is a collection of evidence paths, then not only is each B(i) a line of evidence for the answer $v^*$ to the question q when considered in isolation, but furthermore, as we iteratively combine more and more of those evidence paths, we strictly increase the probability of the response $v^*$ to the question q:

Proposition 12.1

Let ${\mathcal {B}}$ be a collection of N evidence paths for the claim $(q, v^*)$. Then for all $i: N \ge i \ge 1$,

$$\begin{aligned} P(v^*\,|\, q, \beta , B(1), \ldots , B(i))&> P(v^*\,|\, q, \beta , B(1), \ldots , B(i-1)), \end{aligned}$$

(35)

(where we interpret the $i=1$ version of this statement to mean $P(v^*\,|\, q, \beta , B(1)) \ge P(v^*\,|\, q, \beta )$).

Proof

Evaluating Eq. (10) for $i = 1$ directly establishes the $i=1$ version of Eq. (11):

$$\begin{aligned} P(v^*\,|\, q, \beta , B(1)) \;>\; P(v^*\,|\, q, \beta ). \end{aligned}$$

Next, for any $i: 2 \le i \le n$, we can use Bayes’ theorem to expand

$$\begin{aligned}&\dfrac{P(v^*\,|\, q, \beta , B(1), \ldots , B(i))}{P(v^*\,|\, q, \beta , B(1), \ldots , B(i-1))} \nonumber \\&\quad = \dfrac{P(v^*\,|\, q, {\beta , B(i))}}{P(v^*\,|\, q, {\beta })} \nonumber \\&\qquad \times \dfrac{P({B(i)} \,|\, q, {\beta }) \times P({B(1), \ldots , B(i-1)} \,|\, q, {\beta })}{P({B(1), \ldots , B(i)} \,|\, q, {\beta })} \nonumber \\&\qquad \times \dfrac{P({B(1), \ldots , B(i)} \,|\, (q, v^*), {\beta })}{P({B(i)} \,|\, (q, v^*), {\beta }) \times P({B(1), \ldots , B(i-1)} \,|\, (q, v^*), {\beta })}. \end{aligned}$$

(36)

Using Eq. (10) again establishes that the first term on the RHS of Eq. (36) is $> 1$. Equation (34) establishes that the product of the second and third terms on the RHS is also $> 1$. Therefore the entire RHS of Eq. (36) is $> 1$. This establishes the claim. $\square$

So this alternative definition of an evidence path implies the one we adopted in the main text.

1.4 Proof of Proposition 7.3

Proof

By hypothesis, $\varphi _2$ is calibrated with the community-SMS $\varphi _1$ at step n for any $\hat{C}_{2}\subseteq \beta \cup \bigcup _i B(i)$. So for all $1 \le i \le n$,

$$\begin{aligned}&\sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))D[ \Psi (\psi (q),v)({\mathcal {V}}_{1}), \nonumber \\\;&\quad \overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), \beta , B(1), \ldots , B(i) \right)] \le \epsilon. \end{aligned}$$

(37)

The convexity of D[., .] over its first argument then establishes

$$D\left[ \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))\Psi (\psi (q),v)({\mathcal {V}}_{1}), \;\overline{P}_1( {\mathcal {V}}_1 \;|\; \psi (q), \beta , B(1), \ldots , B(i)) \right] \le \epsilon$$

(38)

i.e.,

$$\begin{aligned} D\left[ F^n({\mathcal {V}}_1 \,|\, \psi (q), {\beta , B(1), \ldots , B(i)}), \; \overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), {\beta , B(1), \ldots , B(i)} \right) \right]&\le \epsilon . \end{aligned}$$

(39)

Next, since all of the B(i) are evidence paths for $(\psi (q), v^*)$ under the distribution $F^n$ and conditioned on $\beta$, by Eq. (11), for all $1 \le i \le N$,

$$\begin{aligned} F^n(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i)})&> F^n(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i-1)}). \end{aligned}$$

(40)

Finally, by hypothesis D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (39) and (40). Then since a divergence equals zero only if its arguments are identical, for all $1 \le i \le n$, for small enough $\epsilon$,

$$\begin{aligned} \overline{P}_1(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i)})&> \overline{P}_1(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i-1)}), \end{aligned}$$

(41)

as claimed. $\square$

1.5 Proof of Proposition 7.4

Proof

By hypothesis $\varphi _2$ is embed-calibrated with the universe-SMS $\varphi _1$ at step n for any $\hat{C}_{2}\subseteq \beta \cup \bigcup _i B(i)$. So for all $1 \le i \le n$,

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))D\Big [\Psi (\psi (q),v)({\mathcal {V}}_{1}), \; \nonumber \\{} & {} \quad \overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup \beta , B(1), \ldots , B(i)] \right) \Big ] \le \epsilon . \end{aligned}$$

(42)

The convexity of D[., .] then establishes

$$\begin{aligned}{} & {} D\Big [\sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))\Psi (\psi (q),v)({\mathcal {V}}_{1}), \;\nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))\overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup \beta , B(1), \ldots , B(i)] \right) \Big ] \le \epsilon , \end{aligned}$$

(43)

i.e.,

$$\begin{aligned}{} & {} D\Big [F^{n}_{2}(v|q,\beta , B(1), \ldots , B(i)), \; \nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))\overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup \beta , B(1), \ldots , B(i)] \right) \Big ] \le \epsilon . \end{aligned}$$

(44)

Next, since all of the B(i) are evidence paths for $(\psi (q), v^*)$ under the distribution $F^n$ and conditioned on $\beta$, by Eq. (11), for all $1 \le i \le N$,

$$\begin{aligned} F^n(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i)})&> F^n(v^*\,|\, \psi (q), {\beta , B(1), \ldots , B(i-1)}). \end{aligned}$$

(45)

Finally, by hypothesis D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (44) and (45). Then since a divergence equals zero only if its arguments are identical, for all $1 \le i \le n$, for small enough $\epsilon$,

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i))\overline{P}_1(v^*\,|\, \psi (q), E^{-1}[\{(q,v)\}\cup \beta , B(1), \ldots , B(i)) \nonumber \\{} & {} \quad > \ \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,\beta , B(1), \ldots , B(i-1))\overline{P}_1(v^*\,|\, \psi (q), E^{-1}[\{(q,v)\}\cup \beta , B(1), \ldots , B(i-1)]), \end{aligned}$$

(46)

as claimed. $\square$

1.6 Proof of Proposition 7.5

Proof

By hypothesis, $(q, E[\hat{C}_1])$ is an embedding prediction pair for step n for any claim set $\hat{C}_{1}\subseteq \beta \cup \bigcup _i B(i)$. By hypothesis it is also true that $\varphi _2$ is embed-calibrated with SMS $\varphi _1$ at step n for the prediction pair $(q, E[\hat{C}_1])$ for any such $\hat{C}_1$. So for all $1 \le i \le n$,

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,E[\beta , B(1), \ldots , B(i)])D\Big [\Psi (\psi (q),v)({\mathcal {V}}_{1}), \nonumber \\{} & {} \quad \overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i)]] \right) \Big ] \le \epsilon . \end{aligned}$$

(47)

Due to the convexity of D[., .], this means that

$$\begin{aligned}{} & {} D\Big [\sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,E[\beta , B(1), \ldots , B(i)])\Psi (\psi (q),v)({\mathcal {V}}_{1}), \nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,E[\beta , B(1), \ldots , B(i)])\overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i)]] \right) \Big ]\le \epsilon , \end{aligned}$$

(48)

or

$$\begin{aligned}{} & {} D\Big [F^n({\mathcal {V}}_1 \,|\, \psi (q), E[\beta , B(1), \ldots , B(i)]), \nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,E[\beta , B(1), \ldots , B(i)])\overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i)]] \right) \Big ]\le \epsilon . \end{aligned}$$

(49)

The same line of reasoning for $E[\beta , B(1), \ldots , B(i-1)]$ yields:

$$\begin{aligned}{} & {} D\Big [F^n({\mathcal {V}}_1 \,|\, \psi (q), E[\beta , B(1), \ldots , B(i-1)]), \nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|q,E[\beta , B(1), \ldots , B(i-1)])\overline{P}_1\left( {\mathcal {V}}_1 \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i-1)]] \right) \Big ] \le \epsilon . \end{aligned}$$

(50)

Since, by hypothesis, all of the B(i) are evidence paths for $(\psi (q), v^*)$ under the distribution $\overline{P}_{1}$ and conditioned on $\beta$, for all $1 \le i \le N$, Eq. (11) allows us to write

$$\begin{aligned} \overline{P}_1(v^*\,|\, \psi (q), \beta , B(1), \ldots , B(i)) \;>\; \overline{P}_1(v^*\,|\, \psi (q), \beta , B(1), \ldots , B(i-1)). \end{aligned}$$

(51)

Via the fifth condition of the proposition, this implies that

$$\begin{aligned}{} & {} \overline{P}_1\left( v^{*} \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i)]] \right) \nonumber \\{} & {} \quad > \overline{P}_1\left( v^{*} \;|\; \psi (q), E^{-1}[\{(q,v)\}\cup E[\beta , B(1), \ldots , B(i-1)]] \right) , \end{aligned}$$

(52)

for all $v\in {\mathcal {V}}_{2}$. By hypothesis D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (49) to (51). Then since a divergence equals zero only if its arguments are identical, for all $1 \le i \le n$, for small enough $\epsilon$,

$$\begin{aligned}{} & {} F^n({\mathcal {V}}_1 \,|\, \psi (q), E[\beta , B(1), \ldots , B(i)])\nonumber \\{} & {} \quad >F^n({\mathcal {V}}_1 \,|\, \psi (q), E[\beta , B(1), \ldots , B(i-1)]), \end{aligned}$$

(53)

as claimed. $\square$

1.7 Proof of Proposition 8.1

Proof

By hypothesis, ${F^{n}}$ satisfies the abductive premise,

$$\begin{aligned} { {F^{n}}(v^*\;|\; q^*, (q^\dagger , v^\dagger ),\hat{C}_{2}) = \alpha {F^{n}}(v^*\;|\; q^*, \hat{C}_{2}), }\end{aligned}$$

(54)

which means the abduction implication must hold,

$$\begin{aligned} {F^{n}}(v^\dagger \;|\; q^\dagger , (q^*, v^*),\hat{C}_{2}) = \alpha {F^{n}}(v^\dagger \;|\; q^\dagger ,\hat{C}_{2}). \end{aligned}$$

(55)

Since hypothesis $F^{n}_{2}(v^*\,|\, q^*, q^\dagger , \hat{C}_{2}):= \sum _{v^\dagger } F^{n}_{2}(v^*, v^\dagger \,|\, q^*, q^\dagger , \hat{C}_{2}) = F^{n}_{2}(v^*\,|\, q^*, \hat{C}_{2})$, Eq. (55) implies

$$\begin{aligned} {F^{n}}(v^*, v^\dagger \;|\; q^*, q^\dagger , \hat{C}_{2} ) = \alpha {F^{n}}(v^\dagger \;|\; q^\dagger , \hat{C}_{2}) {F^{n}}(v^*\;|\; q^*, \hat{C}_{2}). \end{aligned}$$

(56)

Also by hypothesis, $\varphi _2$ is calibrated with $\varphi _{1}$ at step n for $\hat{C}_{2}$, for each of the three questions $\psi ^{-1}(q^{*})$, $\psi ^{-1}(q^{\dagger })$, and $\psi ^{-1}(q^{*},q^{\dagger })$, and for $m=1,2$. Then choosing $m=2$, we get

$$\begin{aligned}&\sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2})D\left[ \Psi (\psi (\psi ^{-1}(q^{*},q^{\dagger })),v)({\mathcal {V}}_{1},{\mathcal {V}}_{1}), \overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , \hat{C}_{2}\right) \right] \le \epsilon . \end{aligned}$$

(57)

The convexity of D[., .] establishes

$$\begin{aligned} D\left[ \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }), \hat{C})\Psi (\psi (\psi ^{-1}(q^{*},q^{\dagger })),v)({\mathcal {V}}_{1},{\mathcal {V}}_{1}), \; \overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , \hat{C}_{2}\right) \right] \le \epsilon , \end{aligned}$$

(58)

or

$$\begin{aligned}&D\left[ {F^{n}}({\mathcal {V}}_{1},{\mathcal {V}}_{1} \,|\, q^*, q^\dagger ,\hat{C}_{2}),\overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , \hat{C}_{2}\right) \right] \le \epsilon , \end{aligned}$$

(59)

where with abuse of notation we write $\overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , \hat{C}_{2}\right)$ for a fixed value of the triple $(q^*, q^\dagger , \hat{C}_{2})$ for the associated distribution over an event space ${\mathcal {V}}_{1} \times {\mathcal {V}}_{1}$.

Go through the analogous reasoning for $m=1$ twice, once for each of the two distinct questions $q' \ne q, q'' \ne q$ that we assume exist, questions which (via $\psi (.)$) specify the single question $q^*$ and the single question $q^\dagger$, respectively. In these two cases calibration means that:

$$\begin{aligned}&D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^*,\hat{C}_{2}), \; \overline{P}_1 \left( {\mathcal {V}}_{1} \,|\, q^*, \hat{C}_{2}\right) \right] \le \epsilon , \end{aligned}$$

(60)

$$\begin{aligned}&D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^\dagger ,\hat{C}_{2}), \; \overline{P}_1 \left( {\mathcal {V}}_{1} \,|\, q^\dagger , \hat{C}_{2}\right) \right] \le \epsilon , \end{aligned}$$

(61)

(where we have extended the definition of ${F^{n}}$ in the obvious way to the case where it has one claim as an argument rather than two).

By hypothesis, D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (59) to  (61). Then since a divergence equals zero only if its arguments are identical, for small $\epsilon$ Eq. (59) implies:

$$\begin{aligned} \overline{P}_1\left( v^*, v^\dagger \,|\, q^*, q^\dagger , \hat{C}_{2}\right)&\simeq {F^{n}}(v^*, v^\dagger \,|\, q^*, q^\dagger ,\hat{C}_{2}) \nonumber \\&= \alpha {F^{n}}(v^\dagger \;|\; q^\dagger ,\hat{C}_{2}) P(v^*\;|\; q^*, \hat{C}_{2}). \end{aligned}$$

(62)

Equation (60) implies:

$$\begin{aligned} \overline{P}_1\left( v^*\,|\, q^*,\hat{C}_{2}\right)&\simeq {F^{n}}(v^*\,|\, q^*,\hat{C}_{2}), \end{aligned}$$

(63)

and Eq. (61) implies:

$$\begin{aligned} \overline{P}_1\left( v^\dagger \,|\, q^\dagger ,\hat{C}_{2}\right)&\simeq {F^{n}}(v^\dagger \,|\, q^\dagger , \hat{C}_{2}). \end{aligned}$$

(64)

Together, Eqs. (62) through  (64) imply:

$$\begin{aligned} \overline{P}_1\left( v^*, v^\dagger \,|\, q^*, q^\dagger , \hat{C}_{2}\right)&\simeq \alpha \overline{P}_1 \left( v^*\,|\, q^*, \hat{C}_{2}\right) \overline{P}_1 \left( v^\dagger \,|\, q^\dagger , \hat{C}_{2}\right) . \end{aligned}$$

(65)

By hypothesis,

$$\begin{aligned} \overline{P}_1 \left( v^*\,|\, q^*, q^\dagger , \hat{C}_{2}\right)&= \overline{P}_1 \left( v^*\,|\, q^*, \hat{C}_{2}\right) \end{aligned}$$

(66)

and so Eq. (65) implies that

$$\begin{aligned} \overline{P}_{1}(v^{\dagger }|q^{\dagger },(q^{*},v^{*}), \hat{C}_{2})\simeq \alpha \overline{P}_{1}(v^{\dagger }|q^{\dagger }, \hat{C}_{2})>\overline{P}_{1}(v^{\dagger }|q^{\dagger }, \hat{C}_{2}), \end{aligned}$$

(67)

as claimed. $\square$

1.8 Proof of Proposition 8.2

Proof

By hypothesis, ${F^{n}}$ satisfies the abductive premise. Following the same steps as at the beginning of the proof of Proposition 8.1, we get

$$\begin{aligned} {F^{n}}(v^*, v^\dagger \;|\; q^*, q^\dagger , \hat{C}_{2} ) = \alpha {F^{n}}(v^\dagger \;|\; q^\dagger , \hat{C}_{2}) {F^{n}}(v^*\;|\; q^*, \hat{C}_{2}). \end{aligned}$$

(68)

By hypothesis, $\varphi _{2}$ is embed-calibrated with $\varphi _{1}$ for $\hat{C}_{2}$ for each of the three questions $\psi ^{-1}(q^*)$, $\psi ^{-1}(q^\dagger )$, and $\psi ^{-1}(q^*,q^\dagger )$, and for $m=1,2$. Then choosing $m=2$, we get

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2}])D\Big [ \Psi (\psi (\psi ^{-1}(q^{*},q^{\dagger })),v)({\mathcal {V}}_{1},{\mathcal {V}}_{1})\overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup \hat{C}_{2}]]\right) \Big ]\le \epsilon . \end{aligned}$$

(69)

The convexity of D[., .] establishes

$$\begin{aligned}{} & {} D\Big [{F^{n}}({\mathcal {V}}_{1}|\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2}]),\sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2}])\overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup \hat{C}_{2}]]\right) \Big ]\le \epsilon . \end{aligned}$$

(70)

Go through the analogous reasoning for $m=1$ twice, once for each of the two distinct questions $q' \ne q, q'' \ne q$ that we assume exist, questions which (via $\psi (.)$) specify the single question $q^*$ and the single question $q^\dagger$, respectively. In these two cases calibration means that:

$$D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^*,\hat{C}_{2}), \; \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*}),\hat{C}_{2})\overline{P}_1\left( {\mathcal {V}}_{1} \,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup \hat{C}_{2}]\right) \right] \le \epsilon$$

(71)

$$D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^\dagger ,\hat{C}_{2}), \; \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),\hat{C}_{2})\overline{P}_1\left( {\mathcal {V}}_{1} \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup \hat{C}_{2}]\right) \right] \le \epsilon$$

(72)

(where we have extended the definition of ${F^{n}}$ in the obvious way to the case where it has one claim as an argument rather than two).

By hypothesis, D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (70) to (72). Then since a divergence equals zero only if its arguments are identical, for small $\epsilon$ Eq. (70) implies:

$$\begin{aligned}{} & {} {F^{n}}(v^*,v^\dagger |\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2}) \nonumber \\{} & {} \quad ={F^{n}}(v^\dagger \;|\; q^\dagger , \hat{C}_{2}) {F^{n}}(v^*\;|\; q^*, \hat{C}_{2}) \nonumber \\{} & {} \quad \simeq \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),\hat{C}_{2})\overline{P}_1\nonumber \\{} & {} \qquad \times \left( v^*,v^\dagger \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup \hat{C}_{2}]\right) . \end{aligned}$$

(73)

Equation (71) implies:

$$\begin{aligned} {F^{n}}(v^*\,|\, q^*,\hat{C}_{2})&\simeq \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*}),\hat{C}_{2})\overline{P}_1 \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup \hat{C}_{2}]\right) , \end{aligned}$$

(74)

and Eq. (72) implies:

$$\begin{aligned} {F^{n}}(v^\dagger \,|\, q^\dagger ,\hat{C}_{2})&\simeq \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),\hat{C}_{2})\overline{P}_1 \left( v^\dagger \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup \hat{C}_{2}]\right) . \end{aligned}$$

(75)

Together, Eqs. (73) through  (75) imply:

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }), \hat{C}_{2})\overline{P}_1\left( v^*,v^\dagger \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup \hat{C}_{2}]\right) \nonumber \\{} & {} \quad \simeq \ \alpha \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*}),\hat{C}_{2})\overline{P}_1 \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup \hat{C}_{2}\right) \nonumber \\{} & {} \qquad \times \ \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),\hat{C}_{2})\overline{P}_1 \left( v^\dagger \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup \hat{C}_{2}]\right) . \end{aligned}$$

(76)

By hypothesis, for all $v\in {\mathcal {V}}_{2}$,

$$\begin{aligned}{} & {} \overline{P}_1 \left( v^*\,|\, q^*, q^{\dagger }, E^{-1}[\{(\psi ^{-1}(q^*,q^\dagger ),v)\}\cup \hat{C}_{2}]\right) =\overline{P}_1 \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup \hat{C}_{2}]\right) , \end{aligned}$$

(77)

and so Eq. (76) implies that

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),(q^{*},v^{*}), \hat{C}_{2})\overline{P}_{1}(v^{\dagger }|q^{\dagger },(q^{*},v^{*}), E^{-1}[\{(\psi ^{-1}(q^{\dagger }),v)]\}\cup \hat{C}_{2}]) \nonumber \\{} & {} \quad > \sum _{v\in {\mathcal {V}}_{2}}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }), \hat{C}_{2})\overline{P}_{1}(v^{\dagger }|q^{\dagger }, E^{-1}[\{\psi ^{-1}(q^{\dagger }),v)\}\cup \hat{C}_{2}]), \end{aligned}$$

(78)

as claimed. $\square$

1.9 Proof of Proposition 8.3

Proof

By hypothesis, $\overline{P}_{1}$ satisfies the abductive premise for any $\hat{C}_{1}\in {\hat {\mathcal{C}}}_{1}$, meaning that for any $v\in {\mathcal {V}}_{2}$,

$$\begin{aligned}&\overline{P}_1\left( v^*\,|\, q^*, (q^\dagger ,v^{\dagger }), E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \nonumber \\&\quad = \alpha \overline{P}_1 \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup E[\hat{C}_{1}]]\right) , \end{aligned}$$

(79)

which allows us to derive

$$\begin{aligned}&\overline{P}_1\left( v^\dagger \,|\, q^\dagger , (q^*,v^{*}), E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \nonumber \\&\quad = \alpha \overline{P}_1 \left( v^\dagger \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup E[\hat{C}_{1}]]\right) . \end{aligned}$$

(80)

Since, also by hypothesis,

$$\begin{aligned}&\overline{P}_{1}\left( v^*\,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \nonumber \\&\quad = \overline{P}_{1} \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup E[\hat{C}_{1}]]\right) , \end{aligned}$$

(81)

we are able to rewrite Eq. (80) as

$$\begin{aligned}{} & {} \overline{P}_1\left( v^\dagger , v^*\,|\, q^\dagger , q^*, E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \nonumber \\{} & {} \quad = \alpha \overline{P}_1 \left( v^\dagger \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup E[\hat{C}_{1}]\right) \overline{P}_1 \nonumber \\{} & {} \qquad \times \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup E[\hat{C}_{1}]]\right) . \end{aligned}$$

(82)

By hypothesis, $\varphi _{2}$ is embed-calibrated with $\varphi _{1}$ for $E[\hat{C}_{1}]$ for each of the three questions $\psi ^{-1}(q^*)$, $\psi ^{-1}(q^\dagger )$, and $\psi ^{-1}(q^*,q^\dagger )$, and for $m=1,2$. Then choosing $m=2$ and applying Eq. (82), we get

$$\begin{aligned}{} & {} \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),E[\hat{C}_{1}])D\Big [ \Psi (\psi (\psi ^{-1}(q^{*},q^{\dagger })),v)({\mathcal {V}}_{1},{\mathcal {V}}_{1}), \nonumber \\{} & {} \quad \overline{P}_1\left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \Big ]\le \epsilon . \end{aligned}$$

(83)

The convexity of D[., .] establishes

$$\begin{aligned}{} & {} D\Big [{F^{n}}({\mathcal {V}}_{1},{\mathcal {V}}_{1}|\psi ^{-1}(q^{*},q^{\dagger }),E[\hat{C}_{1}]), \nonumber \\{} & {} \quad \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }),E[\hat{C}_{1}]) \overline{P}_1\nonumber \\{} & {} \quad \left( {\mathcal {V}}_{1}, {\mathcal {V}}_{1} \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) \Big ] \le \epsilon . \end{aligned}$$

(84)

Go through the analogous reasoning for $m=1$ twice, once for each of the two distinct questions $q' \ne q, q'' \ne q$ that we assume exist, questions which (via $\psi (.)$) specify the single question $q^*$ and the single question $q^\dagger$, respectively. In these two cases calibration means that:

$$D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^*,E[\hat{C}_{1}]), \; \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*}),E[\hat{C}_{1}])\overline{P}_1\left( {\mathcal {V}}_{1} \,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup E[\hat{C}_{1}]]\right) \right] \le \epsilon ,$$

(85)

$$D\left[ {F^{n}}({\mathcal {V}}_{1} \,|\, q^\dagger ,E[\hat{C}_{1}]), \; \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),E[\hat{C}_{1}])\overline{P}_1\left( {\mathcal {V}}_{1} \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup E[\hat{C}_{1}]]\right) \right] \le \epsilon$$

(86)

where we have extended the definition of ${F^{n}}$ in the obvious way to the case where it has one claim as an argument rather than two.

By hypothesis, D[., .] is a locally Lipschitz continuous function of its probability distribution arguments (where those distributions are considered as vectors in a Euclidean metric space) when evaluated for the distributions specified in  Eqs. (84) to (86). Then since a divergence equals zero only if its arguments are identical, for small $\epsilon$ Eq. (84) implies:

$$\begin{aligned}{} & {} {F^{n}}(v^*,v^\dagger |\psi ^{-1}(q^{*},q^{\dagger }),E[\hat{C}_{1}]) \nonumber \\{} & {} \quad ={F^{n}}(v^\dagger \;|\; q^\dagger , E[\hat{C}_{1}]) {F^{n}}(v^*\;|\; q^*, E[\hat{C}_{1}]) \nonumber \\{} & {} \quad \simeq \ \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*},q^{\dagger }), E[\hat{C}_{1}])\overline{P}_1\nonumber \\{} & {} \qquad \times \left( v^*,v^\dagger \,|\, q^*, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^{*},q^{\dagger }),v)\}\cup E[\hat{C}_{1}]]\right) . \end{aligned}$$

(87)

Equation (85) implies:

$$\begin{aligned} {F^{n}}(v^*\,|\, q^*,E[\hat{C}_{1}])&\simeq \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{*}),E[\hat{C}_{1}])\overline{P}_1\nonumber \\&\quad \times \left( v^*\,|\, q^*, E^{-1}[\{(\psi ^{-1}(q^*),v)\}\cup E[\hat{C}_{1}]]\right) , \end{aligned}$$

(88)

and Eq. (86) implies:

$$\begin{aligned} {F^{n}}(v^\dagger \,|\, q^\dagger ,E[\hat{C}_{1}])&\simeq \sum _{v\in {\mathcal {V}}_2}P^{n}_{2}(v|\psi ^{-1}(q^{\dagger }),E[\hat{C}_{1}])\overline{P}_1\nonumber \\&\quad \times \left( v^\dagger \,|\, q^\dagger , E^{-1}[\{(\psi ^{-1}(q^\dagger ),v)\}\cup E[\hat{C}_{1}]]\right) . \end{aligned}$$

(89)

Together, Eqs. (87) through (89) imply:

$$\begin{aligned} {F^{n}}(v^*,v^\dagger |\psi ^{-1}(q^{*},q^{\dagger }),E[\hat{C}_{1}])&\simeq \alpha {F^{n}}(v^*\,|\, q^*,E[\hat{C}_{1}]) {F^{n}}(v^\dagger \,|\, q^\dagger ,E[\hat{C}_{1}]). \end{aligned}$$

(90)

By hypothesis,

$$\begin{aligned} {F^{n}}\left( v^*\,|\, q^*, q^\dagger , E[\hat{C}_{1}]\right)&= {F^{n}}\left( v^*\,|\, q^*, E[\hat{C}_{1}]\right) , \end{aligned}$$

(91)

and so Eq. (90) implies that

$$\begin{aligned} {F^{n}}(v^{\dagger }|q^{\dagger },(q^{*},v^{*}),E[\hat{C}_{1}])>{F^{n}}(v^{\dagger }|q^{\dagger }, E[\hat{C}_{1}]), \end{aligned}$$

(92)

as claimed.

$\square$