Belief and Expertise Paper¶
Note: the paper was accepted at KR 2022.
Misc results¶
Impossibility result showing that a report of \(A\) from \(i\) leading to new belief in \(A\) should not lead to belief in expertise of \(i\) on \(A\). Some of the postulates are specifically chosen so that our score-based example satisfies them; we can strengthen them to make it more presentable while maintaining the impossibility.
Proposition 1
The following postulates cannot hold simultaneously:
If \(A \notin B^\sigma_c\) and \(A \in B^{\sigma \concat \tuple{i, c, A}}\), then \(E_i(A) \in B^{\sigma \concat \tuple{i, c, A}}_c\)
\(K^\sigma = \cn(K^\sigma)\) and \(B^\sigma = \cn(B^\sigma)\)
\(K^\sigma \sqsubseteq B^\sigma\)
If \(\tuple{i, c, A} \in \sigma\) then \(S_i(A) \in K^\sigma_c\)
Suppose \(|\propmods(A)| = |\propmods(A')| = 1\), \(A \ne A'\), \(A \in K^\sigma_c\) and \(i \notin \S(\sigma)\). Then \(E_i(A \wedge A') \in B^{\sigma \concat \tuple{i, c, A'}}_c\)
Suppose \(A, A' \in \lprop\) are such that \(\propmods(A) = \propmods(A') \cup \{v\}\), \(E_i(A) \in B^\sigma_c\), \(\neg E_i(A') \in K^\sigma_c\) and \(\neg A' \notin B^\sigma_c\). Then \(E_i(A) \in B^{\sigma \concat \tuple{i, c, A'}}_c\)
If \(\sigma\) is \(\ast\)-consistent then \(B^\sigma\) is consistent
If \(\S(\sigma; \hat{c}) = \{\ast\}\) and \(\sigma\) is \(\ast\)-consistent, then \([B^\sigma_{\hat{c}}] \subseteq \cnprop(\Gamma)\), where \(\Gamma = \{A \mid \tuple{\ast, \hat{c}, A} \in \sigma\}\)
Moreover, the scored based operator with \(r_0 \equiv 0\) and \(d(W, \tuple{i, c, A}) = \#(\Pi^W_i[A] \setminus \propmods(A))\) for \(W, c \models S_i(A)\) (and \(d(W, \tuple{i, c, A}) = \infty\) otherwise) satisfies all but the first postulate.
A postulate which could be considered a defining property of belief in expertise follows from some others.
Proposition 2
The following postulate
If \(E_i(a) \in B^\sigma_c\) and \(\neg A \notin B^\sigma_c\), then \(A \in B^{\sigma \concat \tuple{i, c, A}}_c\)
is implied by the conjunction of
\(B^\sigma = \cn(B^\sigma)\) (half of M1)
\(B^{\sigma \concat \rho} \sqsubseteq \cn(B^\sigma \sqcup K^\rho)\), with equality if \(B^\sigma \sqcup K^\rho\) is consistent (C5 from conditioning postulates; paper notation)
If \(\tuple{i, c, A} \in \sigma\) then \(S_i(A) \in K^\sigma_c\)
If \(\sigma = (\tuple{i, c, A})\) and \(\propmods(A) \ne \emptyset\) then \(K^\sigma_c \subseteq \cnprop(A)\) and \(K^\sigma_d \subseteq \cnprop(\emptyset)\) for \(d \ne c\)
Note that
The results still holds if the postulates are weakened to refer only to one specific \(i \in \S\)
The “\(\neg A \in B^\sigma_c\)” clause is important. Given the closure postulate, \(E_i(\falsum)\) is always in \(B^\sigma_c\) (since \(E_i(\falsum)\) is valid), but we would not want \(\falsum \in B^{\sigma \concat \tuple{i, c, \falsum}}\)
This result can be applied to any conditioning operator with \(\X_\sigma\) set to satisfy soundness constraints (see handwritten notes for verification of the final postulate)
A possible disadvantage of simple conditioning.
Proposition 3
Suppose a conditioning operator has the following property
\(If A \rightarrow A' \in \cnprop(\emptyset)\) then \(\X_{\tuple{i, c, A}} \subseteq \X_{\tuple{i, c, A'}}\)
Then it satisfies
\(If A \rightarrow A' \in \cnprop(\emptyset)\) and
\[\begin{aligned} \rho_1 &= \sigma \concat \tuple{i, c, A} \\ \rho_2 &= \sigma \concat \tuple{i, c, A'} \concat \tuple{i, c, A} \end{aligned}\]then \(B^{\rho_1} = B^{\rho_2}\) and \(K^{\rho_1} = K^{\rho_2}\)
In general this second property might be undesirable: it says that the weaker report \(A'\) has no affect once it is trumped by a stronger sentence \(A\). For instance, consider
There is some argument that we should believe \(p\), but this can only be so if we believe \(p\) on
which does not seem reasonable.
Score-based operators with flat prior rankings are characterised by tautological beliefs (resp., knowledge) on the empty sequence.
Proposition 4
A score-based operator has the property
\(B^\emptyset = K^\emptyset = \cn(\emptyset,\ldots,\emptyset)\)
if and only if \(r_0\) is finite and constant
Note that the elementatity of \(\X_\sigma\) and \(\Y_\sigma\) really is needed here for the “only if” direction.
Revision by reliable information behaves like AGM revision for a wide class of score-based operators.
Proposition 5
Say an operator is AGM-like if it has the following property
If \(\sigma\) is \(\ast\)-consistent, for any \(c\) there is a tpo \(\le\) on \(\vals\) such that
\(\propmods([B^\sigma_c]) = \min_{\le}\vals\)
\(\propmods([B^{\sigma \concat \tuple{\ast, c, A}}_c]) = \min_{\le}\propmods(A)\)
whenever \(\neg A \notin K^\sigma_c\)
Then any score-based operator satisfying
\(K^\sigma\) is consistent whenever \(\sigma\) is \(\ast\)-consistent
\(d(W, \tuple{\ast, c, A}) = 0\) if \(W, c \models A\) and \(d(W, \tuple{\ast, c, A}) = \infty\) otherwise
is AGM-like.
See Richard’s email for an improved version of the AGM-like property.
The elementary condition in the definition of conditioning operators is actually required for at least one of the postulates in the representation result.
Proposition 6
There is an operator which fails the following postulate (from the characterisation of conditioning operators)
\(B^{\sigma \concat \rho} \sqsubseteq \cn(B^\sigma \sqcup K^\rho)\), with equality if \(B^\sigma \sqcup K^\rho\) is consistent
but which satisfies all but the elementariness conditions required for conditioning operators.
We present some results which are useful elsewhere. First: beliefs about expertise come exactly from the closed sets in the join of the maxiplausible partitions (for model-based operators).
Lemma 1
For a set of worlds \(S \subseteq \W\) and \(i \in \S\), write \(\Pi^S_i = \bigvee \{\Pi^W_i \mid W \in S\}\) for the join of the \(i\)-partitions of worlds in \(S\).
Then for any model-based operator,
Proof.
A sketch of the proof is as follows:
Observe that \(\Pi\) refines \(\Pi'\) iff \(R \subseteq R'\), where \(R, R'\) are the equivalence relations corresponding to \(\Pi, \Pi'\)
Observe that if \(\tilde{\Pi} = \bigvee_{t \in T}\Pi_t\) is a join of a collection, we have \(\tilde{R} = \left(\bigcup_{t \in T}R_t\right)^*\), where \(\cdot^*\) denotes the transitive closure operation. This follows from (1) together with the observation that \(R^*\) is the smallest transitive relation containing \(R\).
For the left-to-right direction of the result: we need to show \(\Pi^{\Y_\sigma}_i[A] \subseteq \propmods(A)\). If \(v\) lies in the LHS then \(v\) is related to some \(w \in \propmods(A)\) in \(R^{\Y_\sigma}_i\). By (2), there are \(u_0,\ldots,u_N\) and \(W_1,\ldots,W_{N-1} \in \Y_\sigma\) such that \(u_0 = w\), \(u_N = v\) and \(u_{n+1} \in \Pi^{W_n}_i(u_n)\). One can show \(u_n \in \propmods(A)\) by induction; this holds for \(n = 0\) by assumption, and the inductive step follows from \(W_n \in \Y_\sigma\) and \(E_i(A) \in B^\sigma_c\). In particular, \(v = u_N \in \propmods(A)\) as required.
For the right-to-left direction, take \(W \in \Y_\sigma\). Then \(\Pi^W_i[A] \subseteq \Pi^{\Y_\sigma}_i[A] = \propmods(A)\) since \(\Pi^W_i\) refines \(\Pi^{\Y_\sigma}_i\). Hence \(W, c \models E_i(A)\).
The next lemma shows that the models of the propositional beliefs in case \(c\) (for a model-based operator) are exactly those appearing as the \(c\)-valuation in some maxiplausible world.
Lemma 2
For any model-based operator,
Proof.
The \(\supseteq\) inclusion is clear from the definition of \(B^\sigma_c\) for model-based operators. For the other inclusion, take \(v \in \propmods([B^\sigma_c])\). Let \(Q_v\) be any propositional formula with \(\propmods(Q_v) = \{v\}\). Then \(\neg Q_v \notin B^\sigma_c\), since otherwise we would have \(v \in \propmods([B^\sigma_c]) \subseteq \propmods(\neg Q_v) = \vals \setminus \{v\}\) which is clearly false. Hence there is \(W \in Y_\sigma\) such that \(W, c \not\models \neg Q_v\), i.e. \(W, v \models Q_v\), so \(v^W_c = v\).
We come to selective revision properties.
Proposition 7
For \(i \in \S\), \(A \in \lprop\) and \(\Gamma \subseteq \lext\), write
i.e. \(x_i(A; \Gamma)\) is the result of “expanding” \(A\) according to the expertise statements (of source \(i\)) in \(\Gamma\). 1
Consider the following properties
S: If \(\tuple{i, c, A} \in \sigma\) then \(S_i(a) \in K^\sigma_c\)
X1: \([B^\sigma_c] \supseteq \cnprop(X_{\sigma, c})\), where \(X_{\sigma, c} = \{x_i(A; B^\sigma_c) \mid \tuple{i, c, A} \in \sigma\})\)
X2: If \(\sigma\) is \(\ast\)-consistent, \([B^\sigma_c] \subseteq \cnprop(X_{\sigma, c})\)
Then
For a model-based operator, S implies X1
X2 is not satisfied by the score-based operator given by \(r_0 \equiv 0\) and
\[d(W, \tuple{i, c, A}) = \begin{cases} |\Pi^W_i[A] \setminus \propmods(A)|,& W, c \models S_i(A) \\ \infty,& \text{ otherwise } \end{cases}\]
Proof.
(Sketch) We show \(\propmods([B^\sigma_c]) \subseteq \propmods(X_{\sigma, c})\). Let \(v \in \propmods([B^\sigma_c])\). By Lemma 2 there is \(W \in \Y_\sigma\) with \(v^W_c = v\). Suppose \(\tuple{i, c, A} \in \sigma\). To show \(v \in \propmods(x_i(A; B^\sigma_c))\), take any \(A'\) in the conjunction which defines \(x_i\). Then \(A' \in \cnprop(A)\) and \(E_i(A') \in B^\sigma_c\). Since \(W \in \Y_\sigma\) we have \(W, c \models E_i(A')\), so \(\Pi^W_i[A'] = \propmods(A')\). Now S and M2 imply \(W, c \models S_i(A)\), so \(v^W_c \in \Pi^W_i[A]\). Hence
\[v = v^W_c \in \Pi^W_i[A] \subseteq \Pi^W_i[A'] = \propmods(A')\]This shows \(v \in \propmods(x_i(A; B^\sigma_c))\) and we are done.
Suppose \(\C = \{c_1, c_2, c_3\}\), \(\S = \{i, \ast\}\) and \(\propvars = \{p, q\}\). Consider the input sequence \(\sigma\) containing the following reports
\[\begin{aligned} &\tuple{i, c_1, p}, \\ &\tuple{i, c_1, q}, \\ &\tuple{\ast, c_1, \neg (p \wedge q)}, \\ &\tuple{i, c_2, (p \vee q) \wedge \neg (p \wedge q)}, \\ &\tuple{\ast, c_2, \neg p \wedge \neg q}, \\ &\tuple{i, c_3, p \wedge q}, \end{aligned}\]Note that \(\sigma\) is \(\ast\)-consistent. According to the Haskell implementation of the score based operator in question, \(\Y_\sigma\) contains 7 worlds, and we have
\[\{v^W_{c_3} \mid W \in \Y_\sigma\} = \{ pq, \bar{p}q, p\bar{q} \} = \propmods(p \vee q)\]By Lemma 2 and the fact that \([B^\sigma_c] = \cnprop([B^\sigma_c])\) we get
\[[B^\sigma_{c_3}] = \cnprop(p \vee q)\]The maxiplausible partitions are
\[\{\Pi^W_i \mid W \in \Y_\sigma\} = \left\{ \{\bar{p}\bar{q}, p\bar{q} \mid \bar{p}q, pq\}, \{\bar{p}\bar{q}, p\bar{q}, \bar{p}q \mid pq\}, \{\bar{p}\bar{q}, \bar{p}q \mid p\bar{q}, pq\} \right\}\]so
\[\Pi^{\Y_\sigma}_i = \{\bar{p}\bar{q}, p\bar{q}, \bar{p}q, \bar{p}\bar{q}\}\]i.e. their join is the trivial (coarsest) partition. By Lemma 1, the only formulas of the form \(E_i(A)\) in \(B^\sigma_{c_3}\) are equivalent to \(E_i(\true)\) and \(E_i(\falsum)\). Hence
\[\cnprop(X_{\sigma, c_3}) = \cnprop( \underbrace{x_i(p \wedge q; B^\sigma_{c_3})}_{ \equiv \true } ) = \cnprop(\emptyset)\]In particular, \(p \vee q \in [B^\sigma_{c_3}] \setminus \cnprop(X_{\sigma, c_3})\), so X2 does not hold.
A lemma concerning consistency of propositional beliefs.
Lemma 3
For any model-based operator, any sequence \(\sigma\) and \(c, d \in \C\),
If \([B^\sigma_c]\) is inconsistent iff \([B^\sigma_c] = \lprop\).
\([B^\sigma_c]\) is consistent iff \([B^\sigma_d]\) is.
Proof.
It follows from (M1) that \([B^\sigma_c]\) is closed under \(\cnprop\), and the result follows.
If \([B^\sigma_c]\) is inconsistent, then \([B^\sigma_c] = \lprop\) by (1). Recall that, by definition of a model-based operator,
\[[B^\sigma_c] = \{ A \in \lprop \mid \forall W \in \Y_\sigma: W, c \models A \}\]So for \([B^\sigma_c] = \lprop\) to hold we must have \(\Y_\sigma = \emptyset\). But then clearly \([B^\sigma_d] = \lprop\) too, so \([B^\sigma_d]\) is inconsistent.
Selective revision¶
We consider operators whose propositional beliefs are formed by weakening each input report according to some selection function (i.e. “selecting” only part of the information present in each report).
Notation. For a sequence \(\sigma\) and case \(c \in \C\), write \(\sigma \rs c = \{\tuple{i, A} \mid \tuple{i, c, A} \in \sigma\}\).
Definition 1
A selection scheme is a mapping \(f\) which assigns to each input sequence \(\sigma\) a function \(f_\sigma: \S \times \C \times \lprop \to \lprop\) such that \(f_\sigma(i, c, A) \in \cnprop(A)\).
An operator is selective if there is a selection scheme \(f\) such that, for all \(\ast\)-consistent \(\sigma\) and \(c \in \C\),
Note that this definition already requires that an operator have no non-trivial prior propositional beliefs, since we must have \([B^\emptyset_c] = \cnprop(\emptyset)\). Also note that since only \(\ast\)-consistent sequences \(\sigma\) are considered in the definition, we may define a selection scheme by specifying \(f_\sigma\) only for \(\ast\)-consistent \(\sigma\).
If a selective (model-based) operator satisfies Soundness, we get a (more standard?) notion of selective revision where reports are replaced by their weakenings.
Proposition 8
Suppose a selective model-based operator with scheme \(f\) satisfies
(Soundness) If \(\tuple{i, c, A} \in \sigma\) then \(S_i(A) \in K^\sigma_c\)
Take any \(\ast\)-consistent \(\sigma = (\tuple{i_n, c_n, A_n})_{n=1}^N\), and write
Then for all \(c \in \C\) we have \([B^\rho_c] = [B^\sigma_c]\).
Proof.
First note that whenever \(\tuple{i, A} \in \sigma \rs c\), we have \(\tuple{\ast, c, f_\sigma(i, c, A)} \in \rho\). By Soundness, closure, the fact that \(S_\ast(A) \equiv A\) and \(K^\rho \sqsubseteq B^\rho\), we have
Monotony of \(\cnprop\) and deductive closure of \([B^\rho_c]\) therefore gives
But by selectivity, the expression on the LHS is just \([B^\sigma_c]\). Hence \([B^\sigma_c] \subseteq [B^\rho_c]\).
For the other inclusion, first suppose \(\rho\) is \(\ast\)-consistent. Then by selectivity applied to \(\rho\) we have
so \([B^\rho_c] \subseteq [B^\sigma_c]\). Now suppose \(\rho\) is not \(\ast\)-consistent. By construction of \(\rho\), this means there is a case \(d\) such that \(\{f_\sigma(i, d, A) \mid \tuple{i, A} \in \sigma \rs d\}\) is inconsistent. But, since \(\sigma\) is \(\ast\)-consistent, we have that \([B^\sigma_d]\) is equal to the consequences of this set; therefore \([B^\sigma_d]\) is inconsistent. By Lemma 3 \([B^\sigma_c] = \lprop\), so clearly \([B^\rho_c] \subseteq [B^\sigma_c]\), and we are done.
[TODO: would it be better to just define selective revision as above? This would remove the restriction that initial propositional beliefs have to consist of just tautologies. Then the case in Definition 1 is recovered if we have the (quite natural) postulate
If \(\S(\sigma) = \{\ast\}\), then \([B^\sigma_c] = \cnprop(\{A \mid \tuple{\ast, c, A} \in \sigma\})\)
i.e. if the only source is \(\ast\), then we just believe all reports (and nothing else). But we may no longer have the characterisation results below.]
Selectivity is characterised by the following postulate:
(S1) For any \(\ast\)-consistent \(\sigma\) and \(c \in \C\), write \(\Gamma_c = \{A \mid \exists i \in \S: \tuple{i, c, A} \in \sigma\}\). Then \([B^\sigma_c] \subseteq \cnprop(\Gamma_c)\).
(S1) simply says that propositional beliefs for case \(c\) should not go beyond the reports given for case \(c\). This expresses the lack of prior propositional beliefs together with independence between different cases.
Proposition 9
A model-based operator satisfies (S1) if and only if it is selective.
Proof.
“if”: Suppose a model-based operator is selective according to a scheme \(f\). Take any \(\ast\)-consistent \(\sigma\) and \(c \in \C\). Write
so that \([B^\sigma_c] = \cnprop(\Delta)\). For \(\tuple{i, A} \in \sigma \rs c\), we have \(f_\sigma(i, c, A) \in \cnprop(A) \subseteq \cnprop(\Gamma_c)\) from the definition of a selection function and the fact that \(A \in \Gamma_c\). Hence \(\Delta \subseteq \cnprop(\Gamma_c)\), so
as required for (S1).
“only if”: Suppose an operator is model-based and satisfies (S1). Take any \(\ast\)-consistent sequence \(\sigma\). For \(c \in \C\), set
By (S1), we have \(M_c \supseteq \propmods(\Gamma_c)\). Now set
Define a selection function \(f_\sigma\) by letting \(f_\sigma(i, c, A)\) be any formula with \(\propmods(f_\sigma(i, c, A) = F_\sigma(i, c, A)\). Since \(F_\sigma(i, c, A)\) contains the models of \(A\), clearly \(f_\sigma(i, c, A) \in \cnprop(A)\). Therefore \(f\) is indeed a selection function.
We claim that, for any \(c\), \(M_c = \bigcap_{\tuple{i, A} \in \sigma \rs c}F_\sigma(i, c, A)\). The “\(\subseteq\)” inclusion is clear since each \(F_\sigma(i, c, A)\) contains \(M_c\) by definition. For the “\(\supseteq\)” inclusion, suppose \(v \in \bigcap_{\tuple{i, A} \in \sigma \rs c}F_\sigma(i, c, A)\) but \(v \notin M_c\). For all \(A \in \Gamma_c\) there is some \(i\) with \(\tuple{i, A} \in \sigma \rs c\), so we have \(v \in F_\sigma(i, c, A)\). But \(v \notin M_c\), so we must have \(v \in \propmods(A)\). This shows that \(v \in \propmods(\Gamma_c)\). But \(\propmods(\Gamma_c) \subseteq M_c\) by (S1); contradiction.
By the claim, we have
Since \([B^\sigma_c]\) is deductively closed (as a consequence of the model-based postulate (M1)) we get that
as required for selectivity.
Independence properties¶
So far we have allowed the weakening of a report \(A\) from a source \(i\) to depend on the case \(c\): Definition 1 says nothing about how \(f_\sigma(i, c, A)\) and \(f_\sigma(i, d, A)\) are related for \(c \ne d\). But since our framework assumes expertise is independent of case, it is natural to require that a report of \(A\) from source \(i\) has the same effect in \(c\) as in \(d\). We make this precise.
Definition 2
A selection scheme \(f\) is case independent if for every \(\sigma\), \(i \in \S\), \(c, d \in \C\) and \(A \in \lprop\),
(where \(\equiv\) denotes logical equivalence of propositional sentences).
A selective operator is said to be case independent if it has a case independent scheme.
Case independent selectivity is characterised by a stronger version of (S1):
(S2) For any \(\ast\)-consistent \(\sigma\), \(c \in \C\) and \(K \subseteq \C\), write
\[\Gamma^K_c = \{ A \mid \exists i \in \S: \left( \tuple{i, c, A} \in \sigma \text{ and } \forall d \in K: \tuple{i, d, A} \notin \sigma \right) \}\]Then
\[[B^\sigma_c] \subseteq \cnprop\left( \Gamma^K_c \cup \bigcup_{d \in K}[B^\sigma_d] \right)\]
Note that (S1) is just the special case of (S2) where \(K = \emptyset\). While (S2) arises as a purely technical condition, attempts can be made to rationalise it. Whereas (S1) says that the reports in case \(c\) provide an upper bound on \([B^\sigma_c]\), (S2) goes further to say that we can remove some \(c\)-reports from this upper bound and add in beliefs about other cases, if a source makes the same report in multiple cases. For example, consider
(S1) requires that \([B^\sigma_c] \subseteq \cnprop(\{p, q\})\). But (S2) with \(K = \{d\}\) notices that \(j\) reports \(q\) in both \(c\) and \(d\), and requires \([B^\sigma_c] \subseteq \cnprop(\{p\} \cup [B^\sigma_d])\). This fails, for instance, for the selection function given by \(f_\sigma(j, d, q) = \top\) and \(f_\sigma(i', c', A) = A\) for all other \(i', c', A\).
Finally, note that the constraint on \([B^\sigma_c]\) in (S2) becomes trivial if \(c \in K\).
Proposition 10
A model-based operator satisfies (S2) if and only if it is selective and case independent.
Proof.
“if”: Suppose a model-based operator is selective according to some case independent scheme \(f\). Take any \(\ast\)-consistent \(\sigma\) and \(K \subseteq \C\). Write
for any \(c \in \C\), and write
where \(\hat{c}\) is an arbitrary fixed case. By case independence, the choice of \(\hat{c}\) is irrelevant, and we have \(F(i, A) = \propmods(f_\sigma(i, c, A))\) for any report \(\tuple{i, c, A}\). Consequently, by selectivity, we have that for any \(c \in \C\),
Now fix some \(c \in \C\). Let \(\Gamma^K_c\) be as in the statement of (S2). Write \(\Lambda = \Gamma^K_c \cup \bigcup_{d \in K}[B^\sigma_d]\). We need to show that \([B^\sigma_c] \subseteq \cnprop(\Lambda)\); or equivalently, \(M_c \supseteq \propmods(\Lambda)\). Take \(v \in \propmods(\Lambda)\). Then
Recall that \(M_c = \bigcap_{\tuple{i, A} \in \sigma \rs c}{F(i, A)}\). Take \(\tuple{i, A} \in \sigma \rs c\). If \(A \in \Gamma^K_c\), then we have \(v \in \propmods(\Gamma^K_c) \subseteq \propmods(A) \subseteq F(i, A)\), where the last inclusion follows from \(f_\sigma(i, c, A) \in \cnprop(A)\). Otherwise, \(A \notin \Gamma^K_c\). Since \(\tuple{i, c, A} \in \sigma\), from the definition of \(\Gamma^K_c\) we must have \(\tuple{i, d, A} \in \sigma\) for some \(d \in K\). This means \(\tuple{i, A} \in \sigma \rs d\). Combining \(v \in M_d\) and \((\ast)\), we have \(v \in F(i, A)\). This shows that \(v \in M_c\), and (S2) is shown.
“only if”: Suppose a model-based operator satisfies (S2). Take any \(\ast\)-consistent sequence \(\sigma\). As before, for \(c \in \C\) write \(M_c = \propmods([B^\sigma_c])\). For \(i \in \S\) and \(A \in \lprop\), write
Set
and define a scheme \(f\) by setting \(f_\sigma(i, c, A)\) to be any formula with models \(F_\sigma(i, A)\). Then \(f\) is a selection scheme since \(F_\sigma(i, A)\) contains the models of \(A\), and \(f\) is case independent since \(\propmods(f_\sigma(i, c, A)) = F_\sigma(i, A)\) does not depend on \(c\). We need to show that, for any \(c\), \([B^\sigma_c] = \cnprop(\{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\})\). By deductive closure of \([B^\sigma_c]\) (which follows from the model-based postulate (M1)), it is sufficient to show that
To that end, fix \(c \in \C\). The “\(\subseteq\)” inclusion is clear since if \(\tuple{i, A} \in \sigma \rs c\) then \(c \in \C(i, A)\), so \(F_\sigma(i, A) \supseteq M_c\). For the “\(\supseteq\)” inclusion, let \(v \in \bigcap_{\tuple{i, A} \in \sigma \rs c}{F_\sigma(i, A)}\). Set
Applying (S2) and taking the models of both sides, we get
To show \(v \in M_c\) it is therefore sufficient to show that \(v\) lies in the RHS above. If \(d \in K\) then clearly \(v \in M_d\) (by definition of \(K\)), so we only need to show that \(v \in \propmods(\Gamma^K_c)\). Let \(A \in \Gamma^K_c\). Then there is \(i\) such that \(\tuple{i, A} \in \sigma \rs c\), so \(v \in F_\sigma(i, A)\). Suppose for contradiction that \(v \notin \propmods(A)\). Then, by definition of \(F_\sigma(i, A)\), there is \(d \in \C(i, A)\) such that \(v \in M_d\). On the on hand, \(d \in \C(i, A)\) means \(\tuple{i, d, A} \in \sigma\), so from the definition of \(\Gamma^K_c\) we must have \(d \notin K\). But on the other hand \(v \in M_d\) gives \(d \in K\) directly from the definition of \(K\): contradiction. Therefore we must have \(v \in \propmods(A)\) and consequently \(v \in \propmods(\Gamma^K_c)\). This completes the proof.
[TODO:
Is there a less complicated combination of postulates equivalent to (S2)?
Can we characterise other interesting properties of \(f\)? E.g. ones from Richard’s JAIR paper? An analogue of Indempotence would be particularly interesting.
Does our cardinality-based score-based operator satisfy (S2)? If so, what does the selection function look like?]
It follows from the proof of Proposition 10 that every model-based, selective, case independent operator admits a scheme \(f\) of a specific form.
Proposition 11
Any model-based, selective, case independent operator admits a scheme \(f\) such that
for all \(\ast\)-consistent sequences \(\sigma\), \(i \in \S\) and \(A \in \lprop\).
Proof.
Consider an operator with the stated properties. By the “if” direction of Proposition 10, the operator in question satisfies (S2). Then in the proof of the “only if” direction of Proposition 10, a selection scheme \(f\) is constructed such that, for \(\ast\)-consistent \(\sigma\),
But since this is shown to be a scheme for the original operator, we have
and the result now follows.
For examples, taking \(\sigma = \left(\tuple{i, c, p}, \tuple{j, c, q}, \tuple{j, d, q}, \tuple{k, d, r} \right)\) as before, we get
[TODO: Are all CI \(f\)s of this form? The RHS implies \(f_\sigma(i, c, A)\) for any CI \(f\) when \(\tuple{i, c, A} \in \sigma\): what about the other direction?]
For completeness, we can also consider independence with respect to the other parameters of a selection function. Say a scheme \(f\) is…
sequence independent if for all \(\ast\)-consistent \(\sigma\), \(\rho\), we have \(f_\sigma(i, c, A) \equiv f_\rho(i, c, A)\)
source independent if for all \(\ast\)-consistent \(\sigma\), \(f_\sigma(i, c, A) \equiv f_\sigma(j, c, A)\)
report independent if for all \(\ast\)-consistent \(\sigma\), \(f_\sigma(i, c, A_1) = f_\sigma(i, c, A_2)\)
Proposition 12
A model-based operator is selective and has a sequence independent scheme if and only if it satisfies (S1) and \([B^{\sigma \concat \rho}_c] = \cnprop([B^\sigma_c] \cup [B^\rho_c])\) for all \(\ast\)-consistent \(\sigma\), \(\rho\) such that \(\sigma \concat \rho\) is \(\ast\)-consistent, and \(c \in \C\)
Every model-based selective operator has a source independent scheme
A selective operator has a report independent scheme if and only if \([B^\sigma_c] = \cnprop(\emptyset)\) for all \(\ast\)-consistent \(\sigma\)
Proof.
“if”: Set
\[F(i, c, A) = \propmods([B^{\tuple{i, c, A}}_c])\]Note that by (S1), \([B^{\tuple{i, c, A}}_c] \subseteq \cnprop(A)\), so \(F(i, c, A) \supseteq \propmods(A)\).
Define a scheme \(f\) by letting \(f_\sigma(i, c, A)\) be any sentence with models \(F(i, c, A)\). Clearly \(f\) is sequence independent.
Let \(\sigma\) be \(\ast\)-consistent and \(c \in \C\). We need to show that \([B^\sigma_c] = \cnprop(\{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\})\). Note that since \(\sigma\) is \(\ast\)-consistent, any subsequences is too. In particular, each singleton sequence \(\tuple{i, d, A}\) – for \(\tuple{i, d, A} \in \sigma\) – is \(\ast\)-consistent, as is any concatentation of such singleton sequences. Clearly \(\sigma\) is the concatenation of all these singleton sequences. Using the assumed property and basic properties of \(\cnprop\), 2 we find
\[ [B^\sigma_c] = \cnprop\left( \bigcup_{\tuple{i, d, A} \in \sigma}{ [B^{\tuple{i, d, A}}_c] } \right)\]Now by (S1), for \(d \ne c\) we have
\[[B^{\tuple{i, d, A}}_c] \subseteq \cnprop(\emptyset)\]Hence the terms with \(d \ne c\) do not contribute to the consequences of the union above, so
\[[B^\sigma_c] = \cnprop\left( \bigcup_{\tuple{i, A} \in \sigma \rs c}{ [B^{\tuple{i, c, A}}_c] } \right)\]Since \(\propmods(f_\sigma(i, c, A)) = F(i, c, A) = \propmods([B^{\tuple{i, c, A}}_c])\), we get
\[\begin{aligned} [B^\sigma_c] &= \cnprop\left( \bigcup_{\tuple{i, A} \in \sigma \rs c}{ \{f_\sigma(i, c, A)\} } \right) \\ &= \cnprop(\{ f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c \}) \end{aligned}\]as required.
“only if”: Suppose a model-based selective operator has a sequence independent scheme \(f\). Let \(\sigma\) and \(\rho\) be \(\ast\)-consistent sequences such that \(\sigma \concat \rho\) is also \(\ast\)-consistent, and take \(c \in \C\).
By selectivity, we have
\[[B^{\sigma \concat \rho}_c] = \cnprop(\{ f_{\sigma \concat \rho}(i, c, A) \mid \tuple{i, A} \in (\sigma \concat \rho) \rs c \})\]But \((\sigma \concat \rho) \rs c = (\sigma \rs c) \cup (\rho \rs c)\). Using this together with the fact that \(f_{\sigma \concat \rho}(i, c, A) \equiv f_\sigma(i, c, A) \equiv f_\rho(i, c, A)\) by sequence independence, we have
\[[B^{\sigma \concat \rho}_c] = \cnprop( \underbrace{\{ f_{\sigma}(i, c, A) \mid \tuple{i, A} \in \sigma \rs c \}}_{(1)} \cup \underbrace{\{ f_{\rho}(i, c, A) \mid \tuple{i, A} \in \rho \rs c \}}_{(2)} )\]By selectivity again, \([B^\sigma_c]\) is the consquences of \((1)\), and \([B^\rho_c]\) is the consequences of \((2)\). Using basic properties of the consequence operator, 3 we get \([B^{\sigma \concat \rho}_c] = \cnprop([B^\sigma_c] \cup [B^\rho_c])\) as required.
This is proved in a similar way to Proposition 11. By the “if” direction of Proposition 9, any model-based selective operator satisfies (S1). So we may apply the “only if” direction of the same result to construct a selection scheme corresponding to the operator in question. According to this construction, we may take \(f\) such that
\[\propmods(f_\sigma(i, c, A)) = \propmods(A) \cup \propmods([B^\sigma_c])\]for \(\ast\)-consistent \(\sigma\). Note that the RHS does not depend on \(i\), so this scheme is source independent.
“if”: Set \(f_\sigma(i, c, A) = \top\). For any \(\ast\)-consistent \(\sigma\), we have
\[\cnprop(\{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\}) = \cnprop(\{\top,\ldots,\top\}) = \cnprop(\emptyset) = [B^\sigma_c] \]so the operator in question if selective according to \(f\). Since \(f\) is clearly report independent, we are done.
“only if”: Consider a selective operator with a report independent scheme \(f\). Since \(f_\sigma(i, c, A) \in \cnprop(A)\) by definition of a selection scheme, we have by report independence
\[f_\sigma(i, c, A) \equiv f_\sigma(i, c, \top) \in \cnprop(\top) = \cnprop(\emptyset)\]That is, \(f_\sigma(i, c, A) \equiv \top\) for all \(\sigma\), \(i\), \(c\) and \(A\). In particular, for \(\ast\)-consistent \(\sigma\) and \(c \in \C\) we have by selectivity that
\[[B^\sigma_c] = \cnprop(\{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\}) = \cnprop(\{\top,\ldots,\top\}) = \cnprop(\emptyset)\]as required.
Selectivity and conditioning¶
(S1) characterises selectivity among the model-based operators. Since conditioning operators are model-based, (S1) also characterises selectivity among the conditioning operators. But given that conditioning operators are defined by a fixed tpo \(\le\), it is natural to look instead for a property of \(\le\) which does the job of (S1). In this section we provide such a property, which characterises selectivity among the conditioning operators with a specific choice of the knowledge collection \(K^\sigma\).
We need some preliminary definitions and lemmas.
Definition 3
For worlds \(W, W'\) and a case \(c\), say that \(W\) refines \(W'\) at \(c\) if for all sources \(i\)
Intuitively, \(W\) refines \(W'\) at \(c\) if each source is more knowledgable in case \(c\) at \(W\) then they are at \(W'\). Note that taking \(i = \ast\) we in fact have that \(W\) and \(W'\) share the same \(c\) valuation. This notion of “local refinement” can be captured by sequences of reports by soundness. In what follows, write
i.e. \(\X^\snd_\sigma\) is the set of worlds making all reports in \(\sigma\) sound at their respective cases. We have the following.
Lemma 4
Let \(W \in \W\) and \(c \in \C\). Then there is a \(\ast\)-consistent sequence \(\sigma_{W, c}\) – containing only reports for case \(c\) – such that \(W' \in \X^\snd_{\sigma_{W, c}}\) iff \(W\) refines \(W'\) at \(c\).
Proof.
For a valuation \(u \in V\), let \(Q(u)\) be any propositional formula with \(\propmods(Q(u)) = \{u\}\). Let \(\sigma_{W, c}\) be any sequence consisting of all (and only) reports of the form
for \(i \in \S\) and \(u \in \Pi^W_i(v^W_c)\). Note that such \(\sigma\) exists since there are only finitely many sources and valuations \(u\). Also note that \(\sigma_{W, c}\) is \(\ast\)-consistent since the only report from \(\ast\) is \(\tuple{\ast, c, Q(v^W_c)}\). We show the desired equivalence.
“if”: Suppose \(W\) refines \(W'\) at \(c\). We need to show \(W' \in \X^\snd_{\sigma_{W, c}}\). Any report in \(\sigma_{W, c}\) is of the form \(\tuple{i, c, Q(u)}\) for \(i \in \S\) and \(u \in \Pi^W_i(v^W_c)\). By refinement, \(u \in \Pi^{W'}_i(v^{W'}_c)\), so \(v^{W'}_c \in \Pi^{W'}_i(u) = \Pi^{W'}_i[Q(u)]\). Hence \(W', c \models S_i(Q(u))\), so \(W' \in \X^\snd_{\sigma_{W, c}}\).
“only if”: Suppose \(W' \in \X^\snd_{\sigma_{W, c}}\). Let \(i \in \S\) and \(u \in \Pi^W_i(v^W_c)\). Then, by construction, \(\tuple{i, c, Q(u)} \in \sigma_{W, c}\). By soundness, \(W', c \models S_i(Q(u))\), i.e. \(v^{W'}_c \in \Pi^{W'}_i[Q(u)] = \Pi^{W'}_i(u)\). Equivalently, \(u \in \Pi^{W'}_i(v^{W'}_c)\). This shows \(\Pi^W_i(v^W_c) \subseteq \Pi^{W'}_i(v^{W'}_c)\) as required for refinement at \(c\).
Local refinement also preserves soundness of reports.
Lemma 5
Suppose \(W\) refines \(W'\) at \(c\). Then for any source \(i\) and \(A \in \lprop\),
Proof.
Suppose \(W, c \models S_i(A)\). Then \(v^W_c \in \Pi^W_i[A]\), i.e \(\propmods(A) \cap \Pi^W_i(v^W_c) \ne \emptyset\). But by refinement, \(\Pi^W_i(v^W_c) \subseteq \Pi^{W'}_i(v^{W'}_c)\). Hence \(\propmods(A) \cap \Pi^{W'}_i(v^{W'}_c) \ne \emptyset\) also, i.e. \(v^{W'}_c \in \Pi^{W'}_i[A]\), so \(W', c \models S_i(A)\).
We can now introduce the promised property of tpos \(\le\) over \(\W\). In what follows, write \(\W_{c\ :\ v} = \{W \in \W \mid v^W_c = v\}\) for the set of worlds whose \(c\)-valuation is \(v\):
(CS) For all \(W \in \W\), \(c \in \C\) and \(v \in \vals\) there is \(W' \in \W_{c\ :\ v}\) such that \(W' \le W\) and \(W\) refines \(W'\) at all cases \(d \ne c\)
The characterisation result is as follows.
Proposition 13
Let \(\le\) be a tpo on \(\W\). Then the model-based operator defined by \(\X_\sigma = \X^\snd_\sigma\) and \(\Y_\sigma = \min_{\le}\X_\sigma\) is selective if and only if \(\le\) satisfies (CS).
Proof.
“if”: By Proposition 9 it is sufficient to show that our operator satisfies (S1). To that end, let \(\sigma\) be \(\ast\)-consistent and take \(c \in \C\). We need \([B^\sigma_c] \subseteq \cnprop(\Gamma^\sigma_c)\), i.e. \(\propmods([B^\sigma_c]) \supseteq \propmods(\Gamma^\sigma_c)\).
Take any \(v \in \propmods(\Gamma^\sigma_c)\). Since \(\sigma\) is \(\ast\)-consistent, \(\X_\sigma = \X^\snd_\sigma\) is non-empty. 4 Hence \(\Y_\sigma\) is also non-empty. Take \(W \in \Y_\sigma\). By (CS), there is \(W' \in \W_{c\ :\ v}\) such that \(W' \le W\) and \(W\) refines \(W'\) at all \(d \ne c\). We claim \(W' \in \X_\sigma\). Indeed, take \(\tuple{i, d, A} \in \sigma\).
Case 1: \(d = c\). Here \(\tuple{i, c, A} \in \sigma\), so \(A \in \Gamma^\sigma_c\). Hence \(v \in \propmods(\Gamma^\sigma_c) \subseteq \propmods(A)\). Since \(W' \in \W_{c\ :\ v}\), we have \(v^{W'}_c = v \in \propmods(A)\), i.e. \(W', c \models A\). It follows that \(W', c \models S_i(A)\).
Case 2: \(d \ne c\). In this case we have that \(W\) refines \(W'\) at \(d\). Since \(W \in \Y_\sigma \subseteq \X_\sigma\), we have \(W, d \models S_i(A)\). By Lemma 5, \(W', d \models S_i(A)\) also.
Hence \(W' \in \X_\sigma\). But \(W' \le W\) and \(W \in \Y_\sigma = \min_{\le}\X_\sigma\) means \(W'\) must also be \(\le\)-minimal in \(\X_\sigma\). Hence \(W' \in \Y_\sigma\). Since \(v^{W'}_c = v\), we have by Lemma 2 that \(v \in \propmods([B^\sigma_c])\), as required.
“only if”: Suppose our operator is selective, i.e. satisfies (S1). To show (CS) holds, take any \(W\), \(c\) and \(v\). Write \(\C \setminus \{c\} = \{d_1, \ldots, d_N\}\). By Lemma 4, for each \(1 \le n \le N\) there is a sequence \(\sigma_n\) (\(= \sigma_{W, d_n}\)) such that \(\X_{\sigma_n}\) consists of exactly those \(W'\) refined by \(W\) at \(d_n\). Moreover, each \(\sigma_n\) is \(\ast\)-consistent and only includes reports for case \(d_n\). Now let \(\sigma\) be the concatenation
where, as in an earlier proof, \(Q(u)\) denotes some propositional formula with \(\propmods(Q(u)) = \{u\}\). Note that \(\sigma\) is \(\ast\)-consistent, so we may apply (S1) for case \(c\). Taking models of both sides of (S1), we get
In particular, \(v \in \propmods([B^\sigma_c])\). By Lemma 2, there is \(W' \in \Y_\sigma\) with \(v^{W'}_c = v\). That is, \(W' \in \Y_\sigma \cap \W_{c\ :\ v}\). Recall that \(\X_{\sigma_n}\) consists exactly of those worlds refined by \(W\) at \(d_n\). Since
we have that \(W\) refines \(W'\) at each \(d_n\). But the \(d_n\) cover all of \(\C \setminus \{c\}\). Hence \(W\) refines \(W'\) at all cases \(d \ne c\).
It only remains to show that \(W' \le W\). This is now straightforward: since \(W\) clearly refines itself at each \(d_n\) and \(W \in \X_{\tuple{\ast, c, Q(v) \vee Q(v^W_c)}}\) by construction, we have \(W \in \X_\sigma\). Consequently, \(W' \in \Y_\sigma = \min_{\le}{\X_\sigma}\) gives \(W' \le W\), and we are done.
Note that the above result applies (slightly) more generally than to just conditioning operators, since we do not require that \(\Y_\sigma = \min_{\le}{X^\snd_\sigma}\) is elementary.
Does (CS) have any intuitive interpretation? As with (S2), it arises as a technical condition, but we can attempt some rough post hoc justification. First note that when passing from an input sequence \(\sigma\) to \(\X^\snd_\sigma\), information about the reports in a particular case \(c\) are lost. Using the notation of (S1), \(\Gamma_c\) is not recoverable from \(\X^\snd_\sigma\). Since the tpo \(\le\) does not depend on \(\sigma\), it follows that \(\Gamma_c\) cannot be recovered from \(\Y_\sigma\) either. But (S1) – which characterises selectivity – places some constraint on \(\Y_\sigma\) which depends on \(\Gamma_c\). In particular, (S1) requires that the worlds in \(\Y_\sigma\) contain “enough” valuations for case \(c\) (namely, all models of \(\Gamma_c\) are included). Intuitively, this suggests that in order for selectivity to hold, \(\le\) should not be too “fussy” about valuations (otherwise some models of \(\Gamma_c\) might be excluded in \(\Y_\sigma = \min_{\le}\X^\snd_\sigma\)). With this said, consider what it means for (CS) to fail. There must be some \(W\), \(c\) and \(v\) such that, whenever \(W\) refines \(W'\) at all \(d \ne c\) and \(v^{W'}_c = v\), we have \(W < W'\). That is, \(\le\) discriminates against those \(W'\) which satisfy the same soundness sentences at cases \(d \ne c\) as \(W\), and which have \(v\) for the \(c\)-valuation. In this sense \(\le\) is too “fussy”: one can construct a sequence \(\sigma\) where \(v\) is a model of \(\Gamma_c\) but no world with \(v\) at its \(c\)-valuation will ever be minimal in \(X^\snd_\sigma\), and thus selectivity fails.
However, while (CS) is difficult to interpret, there is a very natural postulate which implies it:
(PE) If \(W\) and \(W'\) are partition equivalent – that is, \(\Pi^W_i = \Pi^{W'}_i\) for all \(i \in \S\) – then \(W \simeq W'\).
(PE) clearly implies (CS): 5 for \(W\), \(c\), \(v\), let \(W'\) be obtained from \(W\) by setting the \(c\)-valuation to \(v\) and leaving other valuations and partitions the same. Then \(W \in \W_{c\ :\ v}\) by construction, \(W\) and \(W'\) are partition equivalent, so \(W \simeq W'\) and in particular \(W' \le W\), and it is easily checked that \(W\) refines \(W'\) at all \(d \ne c\) (in fact, \(W'\) refines \(W\) at such \(d\) too). It follows that all conditioning operators with \(\X_\sigma = \X^\snd_\sigma\) and whose tpo \(\le\) respects partition equivalence are selective (for instance, example 3 in the paper-in-progress at the time of writing).
[TODO: Try to generalise this result to other choices of \(\X_\sigma\). This might be possible based on the following observation. For any mapping \(\sigma \mapsto \X_\sigma\), define a binary relation \(R_c\) on \(\W\), for each case \(c\), by
If \(X_\sigma = X^\snd_\sigma\) then \(R_c\) is refinement at \(c\), so this does in a sense generalise the above. It seems plausible that under some conditions, we will also have a version of Lemma 4 referring to \(R_c\) instead of \(c\)-refinement, and that the statement and proof of Proposition 13 can be adapted to replace all instances of \(c\)-refinement with relation under \(R_c\). But the fine details need to be checked, if this is something worth spending time on.]
Implementability¶
Every scheme \(f\) is trivially implemented by some operator: set \(B^\sigma_c = \cnprop(\{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\})\) and set \(K^\sigma_c\) arbitrarily. However, not all schemes are implementable by a model-based operator. We have the following.
Proposition 14
Let \(f\) be a selection scheme. Then the following are equivalent:
There is a selective, model-based operator corresponding to \(f\).
For any \(\ast\)-consistent \(\sigma\), writing \(\Delta_c = \{f_\sigma(i, c, A) \mid \tuple{i, A} \in \sigma \rs c\}\) we have that \(\Delta_c\) is consistent if and only if \(\Delta_d\) is consistent, for all \(c, d \in \C\).
Proof.
(1) \(\implies\) (2): Consider some model-based, selective operator corresponding to \(f\) and let \(\sigma\) be a \(\ast\)-consistent sequence. Take \(c, d \in \C\). It is sufficient to show that \(\Delta_c\) inconsistent implies \(\Delta_d\) is also inconsistent. So suppose \(\Delta_c\) is inconsistent. By definition of selectivity, \([B^\sigma_c] = \cnprop(\Delta_c)\), so \([B^\sigma_c]\) is inconsistent. Since the operator in question is model-based, we have by Lemma 3 that \([B^\sigma_d] = \cnprop(\Delta_d)\) is also inconsistent, and therefore \(\Delta_d\) is too.
(2) \(\implies\) (1): Suppose \(f\) satisfies the condition in (2). If \(\sigma\) is not \(\ast\)-consistent, set \(\X_\sigma = \Y_\sigma = \emptyset\). Otherwise, set
Consider the corresponding model-based operator. We need to show it is selective and corresponds to the scheme \(f\). So, let \(\sigma\) be \(\ast\)-consistent. We need \([B^\sigma_c] = \cnprop(\Delta_c)\) for each \(c\).
First note that, by (2), either \(\Delta_c\) is inconsistent for all \(c\) or \(\Delta_c\) is consistent for all \(c\). In the first case, clearly \(\Y_\sigma = \emptyset\). By the definition of a model-based operator each \([B^\sigma_c]\) is then inconsistent, so
In the second case, each \(\Delta_c\) is consistent. Take any \(c \in \C\). We will show that \(\propmods([B^\sigma_c]) = \propmods(\Delta_c)\), which implies \([B^\sigma_c] = \cnprop(\Delta_c)\) since each \([B^\sigma_c]\) is deductively closed. First suppose \(v \in \propmods([B^\sigma_c])\). By Lemma 2, there is \(W \in \Y_\sigma\) such that \(v = v^W_c\). By our choice of \(\Y_\sigma\), \(v \in \propmods(\Delta_c)\). Hence \(\propmods([B^\sigma_c]) \subseteq \propmods(\Delta_c)\). Now suppose \(v \in \propmods(\Delta_c)\). For every other \(d \in \C\), take some \(v_d \in \propmods(\Delta_d)\); this is possible since each \(\Delta_d\) is consistent. Letting \(\Pi_\ast\) be the unit partition of \(\vals\), define a world \(W\) by
Then clearly \(W \in \Y_\sigma\), so \(v = v^W_c \in \propmods([B^\sigma_c])\) by Lemma 2. Hence \(\propmods(\Delta_c) \subseteq \propmods([B^\sigma_c])\), and we are done.
For an example of how (2) may fail, consider \(f\) given by \(f_\sigma(i, c, A) = A\). Then on the sequence \(\sigma = (\tuple{i, c, p}, \tuple{j, c, \neg p}, \tuple{k, d, q})\) (which is clearly \(\ast\)-consistent) we have that \(\Delta_c = \{p, \neg p\}\) is inconsistent, while \(\Delta_d = \{q\}\) is consistent.
[TODO: What about conditions on \(f\) which ensure there is a conditioning (or score-based) selective operator corresponding to \(f\)?]
Selectivity and expertise¶
The discussion of selectivity so far has only referred to propositional beliefs. In this section we look at how a selection scheme may relate to beliefs about expertise.
[Note: notation change: we use \(\phi\) etc for propositional formulas instead of \(A\) etc.]
Definition 4
A selection scheme \(f\) is expertise-compatible (EC) with an operator \(\sigma \mapsto \tuple{B^\sigma, K^\sigma}\) if for all \(\ast\)-consistent \(\sigma\) and \(\tuple{i, c, \phi} \in \sigma\) we have \(E_i(f_\sigma(i, c, \phi)) \in B^\sigma_c\).
Say an EC scheme \(f\) is tight (for a given operator) if we additionally have that for all \(\psi \in \cnprop(\phi)\), if \(\psi \notin \cnprop(f_\sigma(i, c, \phi))\) then \(E_i(\psi) \notin B^\sigma_c\).
Say an operator is EC selective if it selective via some EC scheme, and tight-EC selective if it is selective via some tight EC scheme. Since selective operators form propositional beliefs by weakening each report \(\tuple{i, c, \phi}\) to \(f_\sigma(i, c, \phi)\), EC selectivity says that reports are weakened to some formula on which the reporting source is trusted (i.e. believed to be an expert). Tight-EC selectivity requires that the reporting source is not trusted on any strictly stronger formula which is still weaker than \(\phi\).
Tight schemes are unique up to logical equivalence.
Proposition 15
Suppose an operator satsifies Closure and has tight EC schemes \(f, f'\). Then \(f_\sigma(i, c, \phi) \equiv f'_\sigma(i, c, \phi)\) for all \(\tuple{i, c, \phi} \in \sigma\).
Proof.
Let \(\tuple{i, c, \phi} \in \sigma\). Write
Then, by definition of a selection scheme, \(\theta \land \theta' \in \cnprop(\phi)\). By the EC property of \(f\) and \(f'\), we have \(E_i(\theta) \in B^\sigma_c\) and \(E_i(\theta') \in B^\sigma_c\). Since \(\models (E_i(\theta) \land E_i(\theta')) \rightarrow E_i(\theta \land \theta')\), by Closure we have \(E_i(\theta \land \theta') \in B^\sigma_c\). By tightness of \(f\), \(\theta \land \theta' \in \cnprop(\theta)\). In particular, \(\theta' \in \cnprop(\theta)\). Similarly, by tightness of \(f'\) we have \(\theta \land \theta' \in \cnprop(\theta')\) and \(\theta \in \cnprop(\theta')\). Hence \(\theta \equiv \theta'\), as required.
If we assume Soundness, EC selectivity actually implies tight-EC selectivity, and the tight EC scheme takes a particularly simple form. Using the notation \(\Pi^{\Y_\sigma}_i = \bigvee_{W \in \Y_\sigma}{\Pi^W_i}\) from Lemma 1, we have the following.
Proposition 16
If a model-based operator is EC selective and satisfies Soundness, then it is tight-EC selective via a scheme \(f\) such that
Proof.
Let \(f\) be any selection scheme with \(\propmods(f_\sigma(i, c, \phi)) = \Pi^{\Y_\sigma}_i[\phi] \supseteq \propmods(\phi)\). Since \(\Pi^{\Y_\sigma}_i[\Pi^{\Y_\sigma}_i[\phi]] = \Pi^{\Y_\sigma}_i[\phi]\), we have by Lemma 1 that \(E_i(f_\sigma(i, c, \phi)) \in B^\sigma_c\). Hence \(f\) is expertise-compatible. To show \(f\) is tight, suppose \(\psi \in \cnprop(\phi)\) and \(E_i(\psi) \in B^\sigma_c\). Then \(\propmods(\phi) \subseteq \propmods(\psi)\) and, by Lemma 1 again, \(\Pi^{\Y_\sigma}_i[\psi] = \propmods(\psi)\). By monotonicity of expansion by \(\Pi^{\Y_\sigma}_i\), we have
i.e. \(\psi \in \cnprop(f_\sigma(i, c, \phi))\). Hence \(f\) is tight.
It only remains to show that the operator is selective according to \(f\). By assumption, there exists some EC scheme \(g\) according to which the operator is selective. Write \(G_\sigma(i, c, \phi)\) for \(\propmods(g_\sigma(i, c, \phi))\). Let \(\sigma\) be a \(\ast\)-consistent sequence and \(c \in \C\). By selectivity,
Let \(\tuple{i, \phi} \in \sigma \rs c\). Since \(g_\sigma(i, c, \phi) \in \cnprop(\phi)\) by definition of a selection scheme and \(E_i(g_\sigma(i, c, \phi)) \in B^\sigma_c\) by the EC property for \(g\), we have by tightness of \(f\) that \(g_\sigma(i, c, \phi) \in \cnprop(f_\sigma(i, c, \phi))\). Hence \(G_\sigma(i, c, \phi) \supseteq \Pi^{\Y_\sigma}_i[\phi]\). From \((1)\) we get
We show the reverse inclusion also holds. Let \(v \in \propmods\proppart{B^\sigma_c}\). By Lemma 2, there is \(W \in \Y_\sigma\) such that \(v^W_c = v\). Let \(\tuple{i, \phi} \in \sigma \rs c\). Then \(\tuple{i, c, \phi} \in \sigma\), so by Soundness (and Containment, which holds for all model-based operators) we have \(W, c \models S_i(\phi)\). That is, \(v = v^W_c \in \Pi^W_i[\phi] \subseteq \Pi^{\Y_\sigma}_i[\phi]\), where the last inclusion follows since \(W \in \Y_\sigma\) implies that \(\Pi^W_i\) refines \(\Pi^{\Y_\sigma}_i\). Hence the left-to-right inclusion holds; together with \((2)\) we get
By deductive closure of \(\proppart{B^\sigma_c}\) we get \(\proppart{B^\sigma_c} = \cnprop(\{f_\sigma(i, c, \phi) \mid \tuple{i, \phi} \in \sigma \rs c\})\), i.e. our operator is selective according to \(f\). This completes the proof.
Note that in Proposition 16, \(f_\sigma(i, c, \phi)\) does not depend on \(c\). We have the following corollary.
Corollary 1
For model-based operators, Soundness and expertise-compatible selectivity implies case independent selectivity.
The scheme \(f\) in Proposition 16 also hints that the propositional beliefs of case \(c\) only depend on the \(c\)-reports in \(\sigma\) and the expertise statements believed on the basis of \(\sigma\). Formally, we can show that the propositional belief set \(\proppart{B^\sigma_c}\) consists of exactly the propositional \(c\)-consequences of \(K^\sigma\) and \(E(B^\sigma)\) together (see further down this document for the definition of \(E(B^\sigma)\)). That is, the propositional beliefs in \(B^\sigma\) can be recovered just from the formulas of the form \(E_i(\phi)\), together with the knowledge \(K^\sigma\).
Proposition 17
If a model-based operator is EC selective and satisfies Soundness, then for any \(\ast\)-consistent \(\sigma\) and \(c \in \C\),
Proof.
“\(\supseteq\)”: Since \(K^\sigma \sqsubseteq B^\sigma\) by Containment and \(E(B^\sigma) \sqsubseteq B^\sigma\) by definition, \(K^\sigma \sqcup E(B^\sigma) \sqsubseteq B^\sigma\). Hence \(\cn(K^\sigma \sqcup E(B^\sigma)) \sqsubseteq \cn(B^\sigma) = B^\sigma\) by Closure. In particular, \(\cn_c(K^\sigma \sqcup E(B^\sigma)) \subseteq B^\sigma_c\), and the desired inclusion follows.
“\(\subseteq\)”: Since \(B^\sigma\) and \(\cn(K^\sigma \sqcup E(B^\sigma))\) are deductively closed collections, it is sufficient to show that
To that end, let \(v \in \propmods\proppart{ \cn_c(K^\sigma \sqcup E(B^\sigma))}\). By an argument almost identical to the proof of Lemma 2, there is \(W \in \mods(K^\sigma \sqcup E(B^\sigma))\) such that \(v^W_c = v\).
Now, by Proposition 16 we have
We need to show \(v \in \propmods\proppart{B^\sigma_c}\), so let \(\tuple{i, \phi} \in \sigma \rs c\). Let \(\psi\) be a propositional formula such that \(\propmods(\psi) = \Pi^{\Y_\sigma}_i[\phi]\). Then \(E_i(\psi) \in B^\sigma_c\) by Lemma 1, so \(E_i(\psi) \in E(B^\sigma)_c\). Hence \(W, c \models E_i(\psi)\), i.e. \(\Pi^W_i[\psi] = \propmods(\psi) = \Pi^{\Y_\sigma}_i[\phi]\). On the other hand we have \(\propmods(\psi) \supseteq \propmods(\phi)\), so by monotonicity of expansion by a partition we get
Now, since \(\tuple{i, c, \phi} \in \sigma\), Soundness gives \(S_i(\phi) \in K^\sigma_c\), so \(W \in \mods(K^\sigma)\) means \(W, c \models S_i(\phi)\). That is, \(v \in \Pi^W_i[\phi]\). By \((2)\), \(v \in \Pi^{\Y_\sigma}_i[\phi]\). Since \(\tuple{i, \phi} \in \sigma \rs c\) was arbitrary, we have by \((1)\) that \(v \in \propmods\proppart{B^\sigma_c}\), which completes the proof.
If we additionally assume Consistency and (OS) (see further down this document…), the condition in Proposition 17 becomes a characterisation of EC selectivity.
Proposition 18
Any model-based operator satisfying Soundness, Consistency and (OS) is EC selective if and only if \(\proppart{B^\sigma_c} = \proppart{\cn_c(K^\sigma \sqcup E(B^\sigma))}\) for all \(\ast\)-consistent \(\sigma\) and \(c \in \C\).
Proof.
The “only if” statement was shown in Proposition 17. For the “if” statement, take any model-based operator satisfying Soundness, Consistency and (OS), and suppose the stated condition holds.
Let \(\sigma\) be \(\ast\)-consistent and take \(c \in \C\). We aim to show that
This will imply EC selectivity since we may take any \(f\) such that \(\propmods(f_\sigma(i, c, \phi)) = \Pi^{\Y_\sigma}_i[\phi]\) (selectivity will then follow by Closure, and the EC property will follow from Lemma 1).
“\(\subseteq\)”: Let \(v \in \propmods\proppart{B^\sigma_c}\). Then there is \(W \in \Y_\sigma\) such that \(v^W_c = v\). Take \(\tuple{i, \phi} \in \sigma \rs c\). By Soundness and Containment, \(W, c \models S_i(\phi)\), i.e. \(v \in \Pi^W_i[\phi]\). Since \(\Pi^W_i\) refines \(\Pi^{\Y_\sigma}_i\), we have \(\Pi^W_i[\phi] \subseteq \Pi^{\Y_\sigma}_i[\phi]\) and consequently \(v \in \Pi^{\Y_\sigma}_i[\phi]\), as required.
“\(\supseteq\)”: Let \(v \in \bigcap_{\tuple{i, \phi} \in \sigma \rs c}{ \Pi^{\Y_\sigma}_i[\phi] }\). By Consistency, there is some \(W_0 \in \Y_\sigma\). Define a world \(W\) by
First we show \(W \in \mods(K^\sigma)\). By Soundness and (OS), \(K^\sigma = \cn(G^\snd_\sigma)\), so \(\mods(K^\sigma) = \mods(G^\snd_\sigma)\). Let \(\tuple{i, d, \phi} \in \sigma\). If \(d = c\), then \(\tuple{i, \phi} \in \sigma \rs c\), so by assumption we have \(v \in \Pi^{\Y_\sigma}_i[\phi]\); hence
so \(W, c \models S_i(\phi)\). If \(d \ne c\), then since \(W_0, d \models S_i(\phi)\) (by Soundness and Containtment again) and \(\Pi^{W_0}_i\) refines \(\Pi^{\Y_\sigma}_i\), we have
so \(W, d \models S_i(\phi)\). This shows \(W \in \mods(G^\snd_\sigma) = \mods(K^\sigma)\).
Next we show \(W \in \mods(E(B^\sigma))\). Supose \(E_i(\phi) \in B^\sigma_d\). Then, by Lemma 1, \(\Pi^{\Y_\sigma}_i[\phi] = \propmods(\phi)\). By definition of \(W\), we get \(\Pi^W_i[\phi] = \propmods(\phi)\), so \(W, d \models E_i(\phi)\). Hence \(W \in \mods(E(B^\sigma))\).
Together, this means \(W \in \mods(K^\sigma \sqcup E(B^\sigma))\). Now let \(\phi \in \proppart{B^\sigma_c} = \proppart{\cn_c(K^\sigma \sqcup E(B^\sigma))}\). By definition of \(\cn_c(\cdot)\), this means \(\mods(K^\sigma \sqcup E(B^\sigma)) \subseteq \mods_c(\phi)\). In particular, \(W, c \models \phi\). Since \(v^W_c = v\), we get \(v \in \propmods(\phi)\). Since \(\phi \in \proppart{B^\sigma_c}\) was arbitrary, we get \(v \in \propmods\proppart{B^\sigma_c}\) as required.
Now, is EC selectivity actually a desirable property? Equivalently (if we accept Soundness, Consistency and (OS)), do we want \(\proppart{B^\sigma_c} = \proppart{\cn_c(K^\sigma \sqcup E(B^\sigma))}\)? In some situations this property forces the operator to be very conservative with its propositional beliefs (or to violate some other natural properties). Suppose we have only one propositional variable \(p\), and consider the following sequence:
Ideas of symmetry suggest that we should not believe either \(E_i(p)\) or \(E_j(p)\). Ideas of trusting sources as much as possible suggest we should, however, believe \(E_i(p) \lor E_j(p)\). Note that
so that by Soundness and Closure, we should believe \(p\) in case \(c\). 6 But this is immediately at odds with EC selectivity; if \(E_i(p)\) and \(E_j(p)\) are not believed then any EC scheme \(f\) must have \(f_\sigma(i, c, p) \equiv f_\sigma(j, c, p) \equiv \top\), and selectivity gives \(\proppart{B^\sigma_c} = \cnprop(\top)\). In particular, \(p\) cannot be believed in case \(c\).
More generally, \(\proppart{B^\sigma_c} = \proppart{\cn_c(K^\sigma \sqcup E(B^\sigma))}\) says that propositional beliefs can only go beyond knowledge by believing some formulas of the form \(E_i(\phi)\), and this rules out the kind of reasoning by which \(p\) was accepted above.
Example 1. Our running score-based and conditioning operators are not EC selective (the sequence \(\sigma\) above provides the counterexample).
The weak soundness operator given by \(\X_\sigma = \Y_\sigma = \{W \mid \forall \tuple{i, c, \phi} \in \sigma: W, c \models S_i(\phi)\}\) is EC selective.
For non-trivial examples, take any model-based operator \(\sigma \mapsto \tuple{B^\sigma, K^\sigma}\) satisfying Soundness, Consistency and (OS). Then set
One can show that that the corresponding operator \(\sigma \mapsto \tuple{\tilde{B}^\sigma, \tilde{K}^\sigma}\) is EC selective and has \(\tilde{B}^\sigma \sqsubseteq B^\sigma\), \(E(\tilde{B}^\sigma) = E(B^\sigma)\) and \(\tilde{K}^\sigma = K^\sigma\). That is, we can construct a new weaker operator which preserves knowledge and expertise beliefs, but which satsifes EC selectivity.
In summary: EC selectivity appears natural at first glance but may be too restrictive. We may need to rethink what \(f_\sigma(i, c, \phi)\) really represents, if it is not in general some formula on which \(i\) has expertise.
Modifications of Conditional-success¶
(R0) If \(E_i(\phi) \in B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\), then \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
(R1) If \(\neg E_i(\phi) \notin B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\), then \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
(R2) If \(E_i(\phi) \in B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\), then \(E_i(\phi) \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
(R1+2) If \(\neg E_i(\phi) \notin B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\), then \(E_i(\phi) \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
Note that (R0) is just Conditional-success. (R1) and (R2) are modifications where the antecedent and consequent respectively are modified, and (R1+2) makes both modifications together.
Lemma 6
For any operator with Closure, (R1) implies (R0), and (R1+2) implies (R2)
For any operator with Closure, Containment and Soundness, (R2) implies (R0), and (R1+2) implies (R1)
Proof.
We show that (R1) implies (R0); showing (R1+2) implies (R2) is similar. Suppose \(E_i(\phi) \in B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\). For the sake of contradiction, suppose \(\neg E_i(\phi) \in B^\sigma_c\). Then clearly \(B^\sigma\) is inconsistent, so \(\mods(B^\sigma) = \emptyset\). But then Closure gives \(B^\sigma_c = \cn_c(B^\sigma) = \lext\). In particular, \(\neg\phi \in B^\sigma_c\). This contradicts our assumption.
Therefore we have \(\neg E_i(\phi) \notin B^\sigma_c\). By (R1), \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\) as required.
Again, we only show that (R2) implies (R0), since an identical argument shows (R1+2) implies (R1). Suppose \(E_i(\phi) \in B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\). By (R2) we have \(E_i(\phi) \in B^{\sigma \concat \tuple{i, c, \phi}}_c\). By Soundness and Containment we have \(S_i(\phi) \in B^{\sigma \concat \tuple{i, c, \phi}}_c\). Recall that \(\models (E_i(\phi) \land S_i(\phi)) \rightarrow \phi\). By Closure, we get that \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\) as required.
We get the following.
Proposition 19
For any operator satisfying the basic postulates, (R0) is weaker than the other three properties, and (R1+2) is stronger than the other three properties.
I claim that (R1+2) (and maybe even (R1) alone) is too strong. Consider the following sequence
Then, both intuitively and according to our example operators, we have \(\neg p \notin B^\sigma_c\) and \(\neg E_i(p) \notin B^\sigma_c\). Instantiating (R1+2) with \(\phi = p\) and writing
we get that \(E_i(p) \in B^\rho_c\). But this conflicts with the basic postulates. By Soundness and Containment we have \(S_i(p), S_i(\neg p) \in B^\rho_c\); since \(\models (S_i(p) \land S_i(\neg p)) \rightarrow \neg E_i(p)\) we have by Closure that \(\neg E_i(p) \in B^\rho_c\). Thus \(B^\rho\) is inconsistent, contradicting Consistency.
The problem with (R1+2) is that while both \(E_i(\phi)\) and \(\phi\) might be separately consistent with \(B^\sigma_c\), their conjunction \(\phi \land E_i(\phi)\) need not be. This suggests modified postulates:
(R3) If \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\), then \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
(R4) If \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\), then \(E_i(\phi) \in B^{\sigma \concat \tuple{i, c, \phi}}_c\)
Again, there are interrelations between the postulates.
Lemma 7
For any operator with Closure, (R3) implies (R0) and (R4) implies (R2)
For any operator with Closure, Containment and Soundness, (R4) implies (R3)
For any operator with Closure, (R1) implies (R3) and (R1+2) implies (R4)
Proof.
We show that (R3) implies (R0); the other statement can be shown to hold by a similar argument. Suppose \(E_i(\phi) \in B^\sigma_c\) and \(\neg\phi \notin B^\sigma_c\). Suppose for contradiction that \(\neg(\phi \land E_i(\phi)) \in B^\sigma_c\). Note that \(\neg\phi \in \cnprop(\neg(\phi \land E_i(\phi)), E_i(\phi))\). By Closure, \(\neg\phi \in B^\sigma_c\); clearly this is a contradiction.
Hence \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). By (R3), \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\) and (R0) is shown.
This follows by an argument identical to the one in the proof of Lemma 6.
Again, we just prove the first statement, i.e. that (R1) implies (R3). Suppose \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). By Closure, \(\neg\phi \lor \neg E_i(\phi) \notin B^\sigma_c\). Thus by Closure, neither \(\neg\phi\) nor \(\neg E_i(\phi)\) lie in \(B^\sigma_c\). By (R1), \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\) as required for (R3).
A pair of reasonable constraints on score-based operators ensure (R4) holds.
Lemma 8
Any score-based operator with properties
and
satisfies (R4).
Proof.
Suppose \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). Then there is some \(W \in \Y_\sigma\) such that \(W, c \models \phi \land E_i(\phi)\). By property (i), \(d(W, \tuple{i, c, \phi}) = 0\). Now write \(\rho = \sigma \concat \tuple{i, c, \phi}\). Then
so \(W \in \X_\rho\).
Now take any \(W' \in \Y_\rho\). Since \(\Y_\rho \subseteq \X_\rho \subseteq \X_\sigma\) and \(W \in \Y_\sigma\), we have
On the other hand \(W' \in \Y_\rho\) and \(W \in \X_\rho\), so
This means
and consequently – since \(d \ge 0\) – \(d(W', \tuple{i, c, \phi}) = 0\). By property (ii), we get \(W', c \models E_i(\phi)\). Since \(W'\) was an arbitrary member of \(\Y_\rho\), this shows that \(E_i(\phi) \in B^\rho_c\), as required for (R4).
Note that our running-example score-based operator with \(r_0 \equiv 0\) and
satisfies these conditions (in fact, we have \(d(W, \tuple{i, c, \phi}) = 0\) iff \(W, c \models \phi \land E_i(\phi)\)), so must also satisfy (R4). Given that this operator also satisfies the basic postulates, it also satisfies (R0), (R2) and (R3). By the example sequence above, it fails (R1+2). [TODO: what about R1?]
For conditioning operators and (R3), we have an impossibility result of sorts. First, some new notation: write \(W \preceq W'\) iff for all \(i \in \S\) we have that \(\Pi^W_i\) refines \(\Pi^{W'}_i\). For a permutation \(\pi: \S \to \S\) of which \(\ast\) is a fixpoint, write \(\pi(W)\) for the world with the same valuations as \(W\), and where \(\Pi^{\pi(W)}_i = \Pi^W_{\pi(i)}\).
Proposition 20
No conditioning operator satisfies the following properties simultaneously:
\(X_\sigma = \X^\snd_\sigma\)
If \(W \preceq W'\) then \(W \le W'\)
For any permutation \(\pi: \S \to \S\) with \(\pi(\ast) = \ast\), \(W \simeq \pi(W)\)
(R3)
However, any proper subset of (1) – (4) is satisfiable.
Proof.
First we show (1) – (4) are incompatible. For contradiction, suppose all properties hold. Take any distinct sources \(i_1, i_2 \in \S \setminus \{\ast\}\), distinct cases \(c, d \in \C\), and distinct valuations \(v_1, v_2 \in \vals\). Let \(\phi_1, \phi_2\) be propositional formulas with models \(\{v_1\}, \{v_2\}\) respectively. Consider the sequence
Let \(\Pi_\bot\) denote the finest partition \(\{\{u\} \mid u \in \vals\}\). Let \(\widehat{\Pi}\) denote the partition
Consider worlds \(W_1, W_2\) with
That is, \(W_1\) has \(v_1\) as the valuation for each case, and all sources but \(i_2\) have the finest partition, and \(W_2\) has \(v_2\) as the valuation for each case with all sources but \(i_1\) having the finest partition.
First we show that \(W_1 \simeq W_2\). Let \(\pi\) denote the permutation of \(\S\) which swaps \(i_1\) and \(i_2\). It is easily seen that \(\pi(W_1)\) is partition-equivalent to \(W_2\), so both \(\pi(W_1) \preceq W_2\) and \(W_2 \preceq \pi(W_1)\) by reflexivity of partition refinement. Hence \(\pi(W_1) \simeq W_2\) by (2), and \(W_1 \simeq W_2\) by (3).
Next, note that \(W_1, W_2 \in \X^\snd_\sigma\), so by (1) we have \(W_1, W_2 \in \X_\sigma\). We claim that, in fact, \(W_1, W_2 \in \Y_\sigma\). Indeed, take any \(W \in \X_\sigma\). By (1), \(W, c \models S_\ast(\phi_1 \lor \phi_2)\), so \(W, c \models \phi_1 \lor \phi_2\). We consider two cases.
Case 1: \(W, c \models \phi_1\): In this case \(v^W_c = v_1\). By (1) again, we have \(W, c \models S_{i_2}(\phi_2)\). Hence \(v_1 \in \Pi^W_{i_2}[v_2]\), and so \(\{v_1, v_2\} \subseteq \Pi^W_{i_2}[v_2]\). It follows that \(\widehat{\Pi}\) refines \(\Pi^W_{i_2}\). Since \(\widehat{\Pi}\) is the partition of \(i_2\) in \(W_1\) and all \(i \ne i_2\) have the most-refined partition \(\Pi_\bot\) in \(W_1\), we have \(W_1 \preceq W\). By (2), \(W_1 \le W\). Since \(W_1 \simeq W_2\), we have \(W_1, W_2 \le W\).
Case 2: \(W, c \models \phi_2\): The argument is similar to case 1. Here \(v^W_c = v_2\), so \(W, c \models S_{i_1}(\phi_1)\) (from (1)) gives \(v_2 \in \Pi^W_{i_1}[v_1]\) and \(\{v_1, v_2\} \subseteq \Pi^W_{i_1}[v_1]\). Hence \(\widehat{\Pi}\) refines \(\Pi^W_{i_1}\) and \(W_2 \preceq W\). By (2), \(W_2 \le W\). Again, \(W_1 \simeq W_2\) gives \(W_1, W_2 \le W\).
Since \(W\) was an arbitrary member of \(\X_\sigma\), this shows \(W_1\) and \(W_2\) are \(\le\)-minimal in \(\X_\sigma\), so \(W_1, W_2 \in \Y_\sigma\). Hence, since \(W_1, d \models \phi_1 \land E_{i_1}(\phi_1)\), we have \(\neg(\phi_1 \land E_{i_1}(\phi_1)) \notin B^\sigma_d\). Now write \(\rho = \sigma \concat \tuple{i_1, d, \phi_1}\). By (R3), we have \(\phi_1 \in B^\rho_d\).
Next, note that \(W_2 \in \X^\snd_\rho = \X_\rho\) since \(W_2 \in \X^\snd_\sigma\) and \(W_2, d \models S_{i_1}(\phi_1)\). Then – since \(\X_\rho \subseteq \X_\sigma\) and \(W_2\) is \(\le\)-minimal in \(\X_\sigma\) – we have \(W_2 \le W\) for any \(W \in \X_\rho\) and consequently \(W_2 \in \Y_\rho\). From \(\phi_1 \in B^\rho_d\) we get \(W_2, d \models \phi_1\). But \(\propmods(\phi_1) = \{v_1\}\) and \(v^{W_2}_d = v_2\), so this means \(v_1 = v_2\). This is a contradiction since \(v_1\) and \(v_2\) were assumed to be distinct. This shows that (1) – (4) cannot hold simultaneously.
For the next part, we show that any set of 3 properties from (1) – (4) are satisfiable. We do this by taking each property in turn and finding a conditioning operator which only satisfies the other 3.
Without (1):
Let \(\Pi_\top = \{V\}\) be the coarsest partition, consisting of a single cell. Write
\[\W_\top = \{W \mid \forall i \in \S \setminus \{\ast\}: \Pi^W_i = \Pi_\top\}\]Define a conditioning operator by
\[\begin{aligned} \X_\sigma &= \X^\snd_\sigma \cap \W_\top \\ {\le} &= \W \times \W \end{aligned}\]Clearly (1) fails while (2) and (3) hold. We need to show that (4) holds, i.e. the operator satisfies (R3). Take some sequence \(\sigma\) and suppose \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). Then there is \(\widehat{W} \in \Y_\sigma\) such that \(\widehat{W}, c \models \phi \land E_i(\phi)\). Set \(\rho = \sigma \concat \tuple{i, c, \phi}\). Take any \(W \in \Y_\rho\). We consider two cases.
Case 1: \(i = \ast\): Since \(\Y_\rho \subseteq \X_\rho \subseteq \X^\snd_\rho \subseteq \X^\snd_{\tuple{\ast, c, \phi}}\), we have \(W, c \models S_\ast(\phi)\). Since \(S_\ast(\phi) \equiv \phi\), we have \(W, c \models \phi\).
Case 2: \(i \ne \ast\): Here \(\widehat{W} \in \W_\top\) and \(i \ne \ast\) imply \(\Pi^{\widehat{W}}_i = \Pi_\top\). Hence \(\widehat{W}, c \models E_i(\phi)\) can only mean \(\phi \equiv \top\) or \(\phi \equiv \falsum\). But \(\widehat{W}, c \models \phi\) too, so \(\phi \not\equiv \falsum\). Hence \(\phi \equiv \top\), so clearly \(W, c \models \phi\).
In either case we have \(W, c \models \phi\). Since \(W\) was an arbitrary member of \(\Y_\rho\), we have \(\phi \in B^\rho_c\), and (R3) is shown.
Without (2):
Set
\[\begin{aligned} \X_\sigma &= \X^\snd_\sigma \\ W \le W' &\iff W \in \W_\top \text{ or } W' \notin \W_\top \end{aligned}\]It is easily checked that \(\le\) defines a total preorder. Clearly (1) holds, and (3) holds since both \(\W_\top\) and \(\W \setminus \W_\top\) are closed under permutations fixing \(\ast\). We need to show (R3) holds.
But we do not need to show it from scratch. Note that if a sequence \(\sigma\) is \(\ast\)-consistent, \(\X^\snd_\sigma \cap \W_\top \ne \empty\); just take the valuation for case \(c\) to be any model of the \(c\)-reports from \(\ast\). It follows that, for \(\ast\)-consistent \(\sigma\), we have \(\Y_\sigma = \min_{\le}{\X^\snd_\sigma} = \X^\snd_\sigma \cap \W_\top\). Note that this is exactly the same as \(\Y_\sigma\) for the operator above which satisfied all postulates except (1). Consequently, the belief sets of the two operators coincide on \(\ast\)-consistent inputs.
Now, suppose \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). Write \(\rho = \sigma \concat \tuple{i, c, \phi}\). If \(\rho\) is \(\ast\)-inconsistent then \(\X_\rho = \X^\snd_\rho = \emptyset\). Hence \(\Y_\rho = \emptyset\), so \(B^\rho_c = \lext \ni \phi\). Otherwise \(\rho\) is \(\ast\)-consistent, so \(\sigma\) is too. Thus the beliefs of our operator coincide with the operator satisfiying (R3) above, so \(\phi \in B^\rho_c\). Hence (R3) holds here too.
Without (3):
Set \(\X_\sigma = \X^\snd_\sigma\) and let \(\le\) be any linear extension of \(\W\). Clearly (1) and (2) hold. For (R3), take any \(\sigma\) and suppose \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). Then there is some \(W \in \Y_\sigma\) such that \(W, c \models \phi \land E_i(\phi)\).
As usual, write \(\rho = \sigma \concat \tuple{i, c, \phi}\). Since \(W, c \models \phi\) and \(\models \phi \rightarrow S_i(\phi)\), we have \(W \in \X^\snd_\sigma \cap \X_{\tuple{i, c, \phi}} = \X_\rho\). Now take any \(W' \in \X_\rho\). Then \(W' \in \X_\sigma\), so \(W \in \Y_\sigma\) implies \(W \le W'\). Hence \(W \in \Y_\rho\).
But recall \(\le\) is a linear order, and therefore satisfies anti-symmetry. Consequently, any non-empty set has a unique \(\le\)-minimal element. Hence \(\Y_\rho = \{W\}\). Since \(W, c \models \phi\), we have \(\phi \in B^\rho_c\) as required.
Without (4):
Set \(\X_\sigma = \X^\snd_\sigma\) and \({\le} = \W \times \W\). Clearly (1), (2) and (3) hold.
Note that properties (1) – (3) are fairly natural, and are satisfied by the example conditioning operators we have investigated. (1) says that knowledge consists only of the consequences of the soundness constraints. For an understanding of (2), note that \(W \preceq W'\) means each source has greater expertise across the board in \(W\) as they do in \(W'\): we have \(W', c \models E_i(\phi) \implies W, c \models E_i(\phi)\) for all \(\phi\). Therefore (2) says we prefer to believe all sources have as much expertise as possible (and valuations do not play a role in relative plausibility). Note also that (2) implies the partition-equivalence property we encountered in the section on selective change. Finally, (3) is a simple source-symmetry property.
If one accepts these justifications and views (1) – (3) as natural properties, Proposition 20 says that no reasonable conditioning operator satisfies (R3). Given that score-based operators can easily satisfy (R3) – see Lemma 7 and Lemma 8 – this could be a property which motivates the need for score-based operators in the first place.
We also have a “syntactic” version of this impossibility result, in which the postulates do not refer directly to the plausibility tpo \(\le\). In fact, the syntactic version applies beyond conditioning operators, so that there need not be any such tpo: the only property of conditioning which plays a role is Inclusion-vacuity.
First, for a permutation \(\xi: \propvars \to \propvars\), extend \(\xi\) in the natural way to formulas of \(\lprop\) and then \(\lext\), and let \(\xi(\sigma)\) denote the sequence in which each \(\tuple{i, c, \phi}\) is replaced with \(\tuple{i, c, \xi(\phi)}\). We can now introduce a variable-symmetry property:
(VS) \(\Phi \in B^\sigma_c\) iff \(\xi(\Phi) \in B^{\xi(\sigma)}_c\)
Next, for a collection \(G = \{\Gamma_c\}_{c \in \C}\), write \(E(G) = \{\Gamma'_c\}_{c \in \C}\) with \(\Gamma'_c = \{E_i(\phi) \mid E_i(\phi) \in \Gamma_c\}\) for the collection of expertise statements in \(G\). Say that \(G\) is maximally consistent if it is consistent and any \(G'\) with \(G \sqsubset G'\) is inconsistent. The following postulate roughly says that collections of formulas which are “expertise-maximal” are consistent with \(B^\sigma\) (provided they are consistent with \(K^\sigma\)):
(M) For maximally consistent \(G, G'\), if \(B^\sigma \sqcup G\) and \(K^\sigma \sqcup G'\) are consistent and \(E(G) \sqsubseteq E(G')\), then \(B^\sigma \sqcup G'\) is consistent
(M) is only a slight generalisation of property (2) for conditioning operators in Proposition 20.
Lemma 9
\(G = \{\Gamma_c\}_{c \in \C}\) is maximally consistent if and only if there is \(W \in \W\) such that
Moreover, for such \(G\) and \(W\) we have \(\mods(G) = \{W\}\).
Lemma 10
For any operator, (M) holds if and only if \(W \in \mods(B^\sigma)\), \(W' \in \mods(K^\sigma)\) and \(W' \preceq W\) implies \(W' \in \mods(B^\sigma)\)
Proof.
“if”: Suppose the stated property holds. Take any \(\sigma\), maximally consistent \(G = \{\Gamma_c\}_{c \in \C}\), \(G' = \{\Gamma'_c\}_{c \in \C}\) and suppose that \(B^\sigma \sqcup G\), \(K^\sigma \cup G'\) are consistent and \(E(G) \sqsubseteq E(G')\). By Lemma 9, there are \(W, W' \in \W\) such that \(\mods(G) = \{W\}\) and \(\mods(G') = \{W'\}\). Hence \(W \in \mods(B^\sigma)\) and \(W' \in \mods(K^\sigma)\).
We show \(W' \preceq W\). Let \(i \in \S\). We use the following characterisation of refinement: \(\Pi_1\) refines \(\Pi_2\) iff every cell in \(\Pi_2\) is a disjoint union of cells in \(\Pi_1\). Let \(U \in \Pi^W_i\). Let \(\phi\) be any formula with \(\propmods(\phi) = U\). Then \(\Pi^W_i[\phi] = U = \propmods(\phi)\), so \(W, c_0 \models E_i(\phi)\) (where \(c_0 \in \C\) is arbitrary). By Lemma 9 again, \(E_i(\phi) \in \Gamma_{c_0}\). Hence \(E_i(\phi) \in E(G)_{c_0} \subseteq E(G')_{c_0}\), i.e. \(E_i(\phi) \in \Gamma'_{c_0}\). By Lemma 9 once more, we get \(W', c_0 \models E_i(\phi)\). This means
i.e. \(U\) is a union of cells of \(\Pi^{W'}_i\). This shows that \(\Pi^{W'}_i\) refines \(\Pi^W_i\). Hence \(W' \preceq W\).
Applying the stated property, we have \(W' \in \mods(B^\sigma)\). Since \(\mods(G') = \{W'\}\), this shows \(B^\sigma \sqcup G'\) is consistent. Hence (M) holds.
“only if”: Suppose (M) holds. Take any \(\sigma\) and \(W, W'\) such that \(W \in \mods(B^\sigma)\), \(W' \in \mods(K^\sigma)\) and \(W' \preceq W\). Let \(G\) and \(G'\) be maximally consistent sets corresponding to \(W\) and \(W'\) respectively; this is possible due to Lemma 9. Then \(\mods(G) = \{W\}\) and \(\mods(G') = \{W'\}\). Hence \(B^\sigma \sqcup G\) and \(K^\sigma \sqcup G'\) are consistent. We show \(E(G) \sqsubseteq E(G')\). Let \(c \in \C\) and suppose \(E_i(\phi) \in \Gamma_c\). By the form of \(\Gamma_c\) given in Lemma 9, we have \(W, c \models E_i(\phi)\). Since \(\Pi^{W'}_i\) refines \(\Pi^W_i\), this means
and consequently \(W', c \models E_i(\phi)\) also. Hence \(E_i(\phi) \in \Gamma'_c\). This shows \(E(G) \sqsubseteq E(G')\). We can now apply (M) to get that \(B^\sigma \sqcup G'\) is consistent. But \(\mods(G') = \{W'\}\), so this implies \(W' \in \mods(B^\sigma)\) as required.
Finally, for any sequence \(\sigma\) write \(G^\snd_\sigma = \{\Gamma_c\}_{c \in \C}\) where \(\Gamma_c = \{S_i(\phi) \mid \tuple{i, \phi} \in \sigma \rs c\}\). Then Soundness requires that \(G^\snd_\sigma \sqsubseteq K^\sigma\). We introduce the reverse inclusion as a new postulate, which says we only know the consequences of soundness:
(OS) \(K^\sigma \sqsubseteq \cn(G^\snd_\sigma)\)
Proposition 21
No operator satisfying the basic postulates can simultaneously satisfy Inclusion-vacuity, (VS), (M), (OS) and (R3).
Proof.
Suppose, for contradiction, that all properties hold. Pick distinct variables \(p, q \in \propvars\). Set \(\theta = \bigwedge_{r \in \propvars \setminus \{p, q\}}r\) (where the conjunction is taken in any arbitrary order. Write
Then there are distinct valuations \(v_1, v_2 \in \vals\) such that \(\propmods(\phi_1) = \{v_1\}\) and \(\propmods(\phi_2) = \{v_2\}\).
Take distinct \(i_1, i_2 \in \S \setminus \{\ast\}\) and cases \(c, d \in \C\). Set
As before, let \(\Pi_\bot\) denote the finest partition \(\{\{u\} \mid u \in \vals\}\), and let \(\widehat{\Pi}\) denote the partition
Consider worlds \(W_1, W_2\) with
It is readily verified that \(W_1, W_2 \in \X^\snd_\sigma = \mods(G^\snd_\sigma)\). By (OS), \(W_1, W_2 \in \mods(K^\sigma)\).
Now, \(\sigma\) is clearly \(\ast\)-consistent, so \(B^\sigma\) is consistent by Consistency. Thus there is some \(W \in \mods(B^\sigma)\). By Soundness and Containment, \(S_\ast(\phi_1 \lor \phi_2) \in B^\sigma_c\), so \(W, c \models \phi_1 \lor \phi_2\). We consider cases.
Case 1: \(W, c \models \phi_1\). Here we have \(v^W_c = v_1\). By Soundness and Containment again, \(W, c \models S_{i_2}(\phi_2)\), i.e. \(v_1 \in \Pi^W_{i_2}[v_2]\). That is, \(\{v_1, v_2\} \subseteq \Pi^W_i[v_2]\). It follows that \(\widehat{\Pi}\) refines \(\Pi^W_{i_2}\). Since \(\widehat{\Pi}\) is the partition associated with \(i_2\) in \(W_1\) and all other sources have the finest partition \(\Pi_\bot\), we have \(W_1 \preceq W\). By the equivalent characterisation of (M) in Lemma 10, we have \(W_1 \in \mods(B^\sigma)\).
Case 2: \(W, c \models \phi_2\). By a near-identical argument with the roles of \(W_1\)/\(W_2\), \(\phi_1\)/\(\phi_2\) and \(i_1\)/\(i_2\) reversed, we get \(W_2 \in \mods(B^\sigma)\).
So, either \(W_1 \in \mods(B^\sigma)\) or \(W_2 \in \mods(B^\sigma)\). Write
for \(k \in \{1, 2\}\). Turning now to case \(d\), note that \(W_k, d \not\models \Psi_k\). Since at least one \(W_k\) lies in \(\mods(B^\sigma)\), we have either \(\Psi_1 \notin B^\sigma_d\) or \(\Psi_2 \notin B^\sigma_d\). Now, let \(\pi\) be the permutation of \(\S\) which swaps \(i_1\) and \(i_2\), and let \(\xi\) be the permutation of \(\propvars\) which swaps \(p\) and \(q\). Referring back to the definition of \(\phi_1\) and \(\phi_2\), we see that \(\xi(\phi_1) \equiv \phi_2\) and \(\xi(\phi_2) \equiv \phi_1\). It follows that \(\xi(\pi(\sigma))\) is equivalent to a permutation of \(\sigma\). By Rearrangement and Equivalence, \(B^\sigma = B^{\xi(\pi(\sigma))}\). Using with (VS) and Source-symmetry, we find
But \(\xi(\pi(\Psi_1)) \equiv \Psi_2\). By Closure we get \(\Psi_1 \in B^\sigma_d\) iff \(\Psi_2 \in B^\sigma_d\). From earlier, we know at least one side of this equivalence must be false. Hence both sides are. That is, \(\Psi_1 \notin B^\sigma_d\) and \(\Psi_2 \notin B^\sigma_d\). In particular, \(\Psi_1 = \neg(\phi_1 \land E_{i_1}(\phi_1)) \notin B^\sigma_d\). Writing \(\rho = \sigma \concat \tuple{i_1, d, \phi_1}\), we may apply (R3) to get \(\phi_1 \in B^\rho_d\).
On the other hand, \(\Psi_2 = \neg(\phi_2 \land E_{i_2}(\phi_2)) \notin B^\sigma_d\), so by Closure there is some \(W \in \mods(B^\sigma)\) such that \(W, d \models \phi_2 \land E_{i_2}(\phi_2)\). Hence \(v^W_d = v_2\). By Containment, \(W \in \mods(K^\sigma)\). Then by Soundness, \(W, c \models S_{i_2}(\phi_2)\). Since expertise formulas hold independently of case and \(\models (E_i(\psi) \land S_i(\psi)) \rightarrow \psi\), we get \(W, c \models \phi_2\). That is, \(v^W_c = v_2 = v^W_d\), i.e. \(W\) has \(v_2\) for both its \(c\) and \(d\) valuations. Using Soundness again, \(W, c \models S_{i_1}(\phi_1)\) and \(v^W_c = v^W_d\) means \(W, d \models S_{i_1}(\phi_1)\). Hence \(W \in \mods\left(G^\snd_{\tuple{i_1, d, \phi_1}}\right)\). By (OS), \(W \in \mods(K^{\tuple{i_1, d, \phi_1}})\). Hence \(B^\sigma \sqcup K^{\tuple{i_1, d, \phi_1}}\) is consistent. By Inclusion-vacuity,
In particular, \(W \in \mods(B^\rho)\). Finally, recall that from (R3) we have \(\phi_1 \in B^\rho_d\), so \(W, d \models \phi_1\). But this means \(W, d \models \phi_1 \land \phi_2\). Since \(\propmods(\phi_1 \land \phi_2) = \{v_1\} \cap \{v_2\} = \emptyset\), we have reached the desired contradiction.
Note that the cardinality-based score-based operator satisfies all the postulates of Proposition 21 except Inclusion-vacuity. [TODO: what about excluding the other postulates? Can any proper subset of Inclusion-vacuity, (VS), (M), (OS) and (R3) be satisfied together with the basic postulates?] We show (M) holds (the others are more-or-less straightforward…).
Proposition 22
The score-based operator given by \(r_0 \equiv 0\) and
satisfies (M).
Proof.
We show that the equivalent formulation of (M) in Lemma 10 holds. Suppose \(W \in \mods(B^\sigma)\), \(W' \in \mods(K^\sigma)\) and \(W' \preceq W\). Since the sets \(\X_\sigma = \{W \mid r_\sigma(W) < \infty\}\) and \(\Y_\sigma = \argmin_{\X_\sigma}{r_\sigma}\) are elementary for this operator, 7 we have \(\mods(B^\sigma) = \Y_\sigma\) and \(\mods(K^\sigma) = \X_\sigma\). So we need to show that \(W' \in \Y_\sigma\), i.e. \(r_\sigma(W')\) is minimal (across \(\X_\sigma\)). Since \(W \in \Y_\sigma\), it is sufficient to show \(r_\sigma(W') \le r_\sigma(W)\).
Let \(\tuple{i, c, \phi} \in \sigma\). Since \(W' \in \X_\sigma\) we have \(r_\sigma(W') < \infty\), so \(d(W', \tuple{i, c, \phi}) < \infty\). Hence \(d(W', \tuple{i, c, \phi}) = |\Pi^{W'}_i[\phi] \setminus \propmods(\phi)|\). Now, since \(W' \preceq W\), the partition \(\Pi^{W'}_i\) is a refinement of \(\Pi^W_i\). Hence \(\Pi^{W'}_i[v] \subseteq \Pi^W_i[v]\) for every valuation \(v \in \vals\). Consequently, \(\Pi^{W'}_i[\phi] \subseteq \Pi^W_i[\phi]\). Hence
so
Hence
and we are done.
There may be other approaches to argue that (R3) is incompatible with conditioning operators. The following result shows that any operator with (a small selection of) the basic postulates, Inclusion-vacuity, (OS) and (R3) implies an undesirable property.
Proposition 23
Suppose an operator satisfies Closure, Containment, Soundness, Inclusion-vacuity, (OS) and (R3). Then for any \(\sigma\) and \(\tuple{i, c, \phi} \in \sigma\), either \(\phi \in B^\sigma_c\) or \(\neg E_i(\phi) \in B^\sigma_c\).
Proof.
Take \(\tuple{i, c, \phi} \in \sigma\). We consider two cases.
Case 1: \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\). We aim to show that \(\phi \in B^\sigma_c\). By (R3) we have \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\).
Now, since \(\tuple{i, c, \phi} \in \sigma\), by Soundness and Containment we have \(S_i(\phi) \in B^\sigma_c\). That is, \(G^\snd_{\tuple{i, c, \phi}} \sqsubseteq B^\sigma\). With Closure this implies \(\cn(G^\snd_{\tuple{i, c, \phi}}) \sqsubseteq B^\sigma\). But from Closure, Soundness and (OS) we get \(K^{\tuple{i, c, \phi}} = \cn\left(G^\snd_{\tuple{i, c, \phi}}\right)\). Therefore \(B^\sigma \sqcup K^{\tuple{i, c, \phi}} = B^\sigma\). Since \(B^\sigma\) is consistent (otherwise Closure would imply \(B^\sigma_c = \lext\), contradicting \(\neg(\phi \land E_i(\phi)) \notin B^\sigma_c\)), Inclusion-vacuity gives \(B^{\sigma \concat \tuple{i, c, \phi}} = \cn(B^\sigma \sqcup K^{\tuple{i, c, \phi}}) = \cn(B^\sigma) = B^\sigma\). Finally, since \(\phi \in B^{\sigma \concat \tuple{i, c, \phi}}_c\) we get \(\phi \in B^\sigma_c\).
Case 2: \(\neg(\phi \land E_i(\phi)) \in B^\sigma_c\). Here we will show \(\neg E_i(\phi) \in B^\sigma_c\). From \(\tuple{i, c, \phi} \in \sigma\), Soundness and Containment, we have \(S_i(\phi) \in B^\sigma_c\). Writing \(\Phi = \neg(\phi \land E_i(\phi)) \land S_i(\phi)\), we have \(\Phi \in B^\sigma_c\) by Closure.
Now, note that
\[\begin{aligned} \Phi &= \neg(\phi \land E_i(\phi)) \land S_i(\phi) \\ &\equiv (\neg\phi \lor \neg E_i(\phi)) \land S_i(\phi) \\ &\equiv \underbrace{(\neg\phi \land S_i(\phi))}_{=\Psi_1} \lor \underbrace{(\neg E_i(\phi) \land S_i(\phi))}_{=\Psi_2} \end{aligned}\]It is easily verified that \(\models \Psi_1 \rightarrow \neg E_i(\phi)\), and clearly \(\models \Psi_2 \rightarrow \neg E_i(\phi)\). Hence \(\models (\Psi_1 \lor \Psi_2) \rightarrow \neg E_i(\phi)\). Hence \(\models \Phi \rightarrow \neg E_i(\phi)\). By Closure again, \(\neg E_i(\phi) \in B^\sigma_c\).
In particular, Proposition 23 says that if an elementary conditioning operator with \(X_\sigma = \X^\snd_\sigma\) satisfies (R3), every report \(\tuple{i, c, \phi}\) is either believed (\(\phi \in B^\sigma_c\)), or the source is distrusted (\(\neg E_i(\phi) \in B^\sigma_c\)). This leaves no room for abstaining until further information comes in; for example in
it seems reasonable to neither believe \(p\) nor \(\neg E_j(p)\), but this is ruled out by Proposition 23 (in fact, this is more or less the approach in the proof of Proposition 21, where (VS), (M) and Source-symmetry are used to derive a contradiction from a sequence similar to \(\sigma\) above).
- 1
This is sloppy, as this is an infinitary conjunction. Really I mean that \(x_i(A; \Gamma)\) is some formula (unique up to equivalence) with models \(\bigcap \{\propmods(A') \mid A' \in \cnprop(A), E_i(A) \in \Gamma\}\).
- 2
Namely, repeated use of the fact that \(\cnprop(\cnprop(\Gamma_1 \cup \Gamma_2) \cup \Gamma_3) = \cnprop(\Gamma_1 \cup \Gamma_2 \cup \Gamma_3)\).
- 3
Namely \(\cnprop(\Gamma \cup \Delta) = \cnprop(\cnprop(\Gamma) \cup \cnprop(\Delta))\).
- 4
For example, take \(W\) whose \(c\)-valuation is a model of the \(c\)-reports from \(\ast\) (such models exist by definition of \(\ast\) consistency), and where \(\Pi^W_i = \{V\}\) is the coarsest partition for all \(i \ne \ast\).
- 5
(PE) is in fact strictly stronger than (CS). To see this, take any partition \(\hat{\Pi}\) not equal to the coarsest partition \(\Pi_\top = \{V\}\). Let \(S\) be the set of worlds \(W\) such that \(\Pi^W_i = \hat{\Pi}\) for all \(i \ne \ast\). Let \(\le_1\) be the flat preorder on \(\W \setminus S\), and let \(\le_2\) be an arbitrary linear order of \(S\). Then set \({\le} = {\le_1} + {\le_2}\), i.e. \(\le\) ranks everything not in \(S\) minimally, and ranks everything in \(S\) linearly above \(\W \setminus S\). Then \(\le\) fails (PE), since any distinct elements of \(S\) are partition equivalent, but cannot rank equally. But \(\le\) does satisfy (CS). Indeed, take \(W \in \W\), \(c \in \C\) and \(v \in \vals\). If \(W \notin S\), then we can take \(W' = W[c\ :\ v]\); it is easily checked that the conditions in (CS) are satisfied. If \(W \in S\), take \(W'\) given by
\[\begin{aligned} v^{W'}_c &= v \\ v^{W'}_d &= v^W_d \qquad & (d \ne c) \\ \Pi^{W'}_i &= \Pi_\top \qquad & (i \ne \ast) \end{aligned} \]Provided that \(\S\) contains some source other than \(\ast\), \(W' \notin S\). It is again easily checked that the conditions in (CS) are satisfied.
- 6
This is a kind of credulous acceptance of \(p\), in argumentation terms. We know \(S_i(p)\) and \(S_j(p)\), and in two plausible worlds we have \(E_i(p)\) and \(E_j(p)\) respectively. Therefore there is always some argument for \(p\), in each plausible world. However, there is no single argument (for \(p\)) which is universally accepted in these worlds, since by the reports in case \(d\) we cannot believe both \(i\) and \(j\) have expertise on \(p\). Therefore \(p\) is not skeptically accepted.
- 7
\(\X_\sigma\) is elementary since \(\X_\sigma = \mods(G^\snd_\sigma)\). One can show \(\Y_\sigma\) is elementary using the characterisation of elementary sets in the paper.