Realism, rhetoric, and reliability

Abstract

Ockham’s razor is the characteristic scientific penchant for simpler, more testable, and more unified theories. Glymour’s early work on confirmation theory (1980) eloquently stressed the rhetorical plausibility of Ockham’s razor in scientific arguments. His subsequent, seminal research on causal discovery (Spirtes et al. 2000) still concerns methods with a strong bias toward simpler causal models, and it also comes with a story about reliability—the methods are guaranteed to converge to true causal structure in the limit. However, there is a familiar gap between convergent reliability and scientific rhetoric: convergence in the long run is compatible with any conclusion in the short run. For that reason, Carnap (1945) suggested that the proper sense of reliability for scientific inference should lie somewhere between short-run reliability and mere convergence in the limit. One natural such concept is straightest possible convergence to the truth, where straightness is explicated in terms of minimizing reversals of opinion (drawing a conclusion and then replacing it with a logically incompatible one) and cycles of opinion (returning to an opinion previously rejected) prior to convergence. We close the gap between scientific rhetoric and scientific reliability by showing (1) that Ockham’s razor is necessary for cycle-optimal convergence to the truth, and (2) that patiently waiting for information to resolve conflicts among simplest hypotheses is necessary for reversal-optimal convergence to the truth.

Proofs and Lemmas

Proof (proposition 1)

$\Rightarrow :$ Suppose that $A=O\cap C$ for $O,C^c$ open. Then

which is an intersection of two closed sets and therefore closed. The final equality follows from the fact that $\overline{O\cap C} \cap C^c \subseteq \overline{O} \cap \overline{C} \cap C^c = \overline{O} \cap C \cap C^c = \varnothing$.

$\Leftarrow :$ Suppose that is closed. Since holds, for every A, we have that A is a difference of closed sets, and is, therefore, locally closed. $\square$

Proof (proposition 2)

$\Rightarrow$ Suppose that $\lambda$ solves $\mathfrak {P}$. For each w, choose $E_w \in \mathsf {Lock}(\lambda ,w)$. Then $\{E_w : w\in W \}$ is a (countable) cover of W by locking information. Let $F_w= \bigcup \{ E \in \mathcal{I}: E\subset E_w \text { and } \lambda (E) \ne \mathcal{Q}(w)\}$. Then $E_w\setminus F_w$ is locally closed. We claim that $A = \bigcup _{w\in A} E_w\setminus F_w$ for each $A\in \mathcal{Q}$. Let $w\in A$. Then $w\in E_w$, and $w\notin F_w$, since $E_w$ is locking for w. So $w\in E_w\setminus F_w\subseteq \bigcup _{v\in A} E_v\setminus F_v$. Suppose that $v \in \bigcup _{w\in A} E_w \setminus F_w$. Then, for some $w\in A$, $v\in E_w\setminus F_w$. Suppose that $\mathcal{Q}(v) = B\ne A$. Then there is $E_v \in \mathsf {Lock}(\lambda ,v)$, and $\lambda (E_v\cap E_w) = B \ne A$. But then $v\in F_w$. Contradiction. We have shown that each $A\in \mathcal{Q}$ is a countable union of locally closed propositions.

$\Leftarrow :$ Suppose that for each answer A to $\mathcal{Q}$ there is a countable collection $\mathcal{C}_{A}$ of locally closed sets such that $A = \bigcup \mathcal{C}_{A}$. Let $\{C_{i}: i \in \mathbb {N}\}$ enumerate $\bigcup _{A \in \mathcal{Q}}\mathcal{C}_{A}$. For each i, let $E_{i}$, $F_{i}$ be open such that $C_{i} = E_{i} \setminus F_{i}$. Define:

$$\begin{aligned} \sigma (E) = {\left\{ \begin{array}{ll} \underset{i \in \mathbb {N}}{\min }\; E \subseteq E_i {\text { and }} E \nsubseteq F_i &{}\,\,\, {\text { if defined}}; \\ \omega &{}\,\,\, {\text { otherwise}}. \\ \end{array}\right. } \end{aligned}$$

Furthermore, define:

$$\begin{aligned} \lambda (E) = {\left\{ \begin{array}{ll} \mathcal{Q}(E_{\sigma (E)}\setminus F_{\sigma (E)})&{}\,\,\, {\text { if }} \sigma (E) < \omega ; \\ \mathcal{Q}(E) &{}\,\,\, {\text { otherwise}}. \\ \end{array}\right. } \end{aligned}$$

Let $w\in W$. Let i be least in $\mathbb {N}$ that $w\in E_i\setminus F_i$. Then, for each $j<i$, either $w \notin E_j$, or $w\in F_j$. Let:

$$\begin{aligned} E ~=~ \left( \underset{j<i \text { and } w\in F_j}{\bigcap } F_j\right) \cap E_i, \end{aligned}$$

under the usual convention that $\bigcap _{j < 0} F_{j} = W$. Let $F \in \mathcal{I}(w)$ such that $F \subseteq E$. Then $F \subseteq E\subseteq E_i$ and, since $F\in \mathcal{I}(w)$, we have that $F\nsubseteq F_i$. Furthermore, since $F\in \mathcal{I}(w)$, it follows that $F\nsubseteq E_j$, or $F\subseteq F_j$, for all $j<i$. So $\lambda (F) = \mathcal{Q}(E_i\setminus F_i) = \mathcal{Q}(w)$. So $E\in \mathsf {Lock}(\lambda ,w)$, as required. $\square$

Proof (proposition 3)

Recall the solution given in the proof of proposition 2. Let $E \in \mathcal{I}$ be non-empty. Suppose that $\lambda (E) = \mathcal{Q}(E)$. Then $E \cap \lambda (E) = E \cap \mathcal{Q}(E) = E \ne \varnothing$. Suppose that $\lambda (E) = \mathcal{Q}(E_{\sigma (E)}\setminus F_{\sigma (E)})$. Then $E\subseteq E_{\sigma (E)}$ and $E \not \subseteq F_{\sigma (E)}$, so $E \cap \lambda (E) = E \cap \mathcal{Q}(E_{\sigma (E)}\setminus F_{\sigma (E)}) \ne \varnothing$. $\square$

Proof (proposition 4)

The result follows immediately from proposition 2 and the fact that a set is locally closed in the subspace topology $\mathcal{I}|_{C}$ iff it is the intersection of C with a locally closed set. $\square$

Proof (proposition 5)

Let H be in $\mathcal{Q}^{*}{\Vert }_{E}$. $\Rightarrow$ Suppose that H is not downward closed in $\mathcal{Q}^{*}{\Vert }_{E}$. So there exist A, B in $\mathcal{Q}^{*}{\Vert }_{E}$ such that $A \subseteq H$ and $B \prec A$, but $B \not \subseteq H$. Since B and H are disjunctions of answers, there exists answer $C \subseteq B$ such that C is disjoint from H. Then $C \prec H$, so H is not minimal in $\mathcal{Q}^{*}{\Vert }_{E}$.

$\Leftarrow$ Suppose that H is not minimal in $\mathcal{Q}^{*}{\Vert }_{E}$. Then there exists $A \in \mathcal{Q}^{*}{\Vert }_{E}$ such that $A \prec H$. Hence, A is disjoint from H, so H is not downward closed in $\mathcal{Q}^{*}{\Vert }_{E}$. $\square$

Proof (proposition 6)

For the $\Leftarrow$ direction of the first equivalence, suppose that , for all $A\in \mathcal{Q}^*$, and for $B\in \mathcal{Q}$, . Then . But since $\mathcal{Q}$ is a partition, $B\subseteq A_\prec$. For the $\Rightarrow$ direction, suppose that $\mathcal{Q}$ is homogeneous, $A\in \mathcal{Q}^*$, and . Then by homogeneity, $w\in \mathcal{Q}(w)\prec A$. So $w\in A_\prec$.

The second equivalence is an immediate consequence of proposition 1. $\square$

Proof (proposition 7)

Let $A,B,C\in \mathcal{Q}$, $A\prec B$, and $B\prec C$. Thus, $A\subseteq \overline{B} \subseteq \overline{C}$. So it remains only to show that $A\ne C$. Each proposition is disjoint from its frontier, so . Furthermore, if B locally closed is open, by proposition 1. So, since , is an open set containing B and disjoint from A. Therefore $B\nprec A$, and $A\ne C$ as required. $\square$

Proof (proposition 8)

Let A be in $\mathcal{Q}{\Vert }_{E}$, and let B be in $\mathcal{Q}^{*}{\Vert }_{E}$, for E in $\mathcal{I}$. Then E is not disjoint from A, B. Suppose that $A \prec _{W} B$. Then, in general, $A \prec _{E} B$. Next, suppose that $A \not \prec _{W} B$ but $A \prec _{E} B$. Then $A \cap E \prec _{W} B$. So $\mathcal{Q}$ is not homogeneous for $\mathfrak {I}$. $\square$

Proof (proposition 9)

Define the method:

$$\begin{aligned} \lambda (E) = {\left\{ \begin{array}{ll} \bigcup {Simp}(\mathcal{Q}{\Vert }_{E}) &{}\,\,\, {\text { if }}{Simp}(\mathcal{Q}{\Vert }_{E})\hbox { is co-initial in }\mathcal{Q}{\Vert }_{E}, \\ \bigcup \mathcal{Q}{\Vert }_{E} &{}\,\,\, {\text { otherwise}}. \\ \end{array}\right. } \end{aligned}$$

Clearly, $\lambda$ is patient. Let $w\in W$. Since $\mathcal{Q}$ is natural, $\mathcal{Q}(w)_\prec$ is closed, by proposition 6. so $\mathcal{Q}(w)_\prec ^c$ is an open neighborhood of w. By normality, $Q(w)_\succeq$ is an open neighborhood of w as well. Let $E = Q(w)_\succeq \cap \mathcal{Q}(w)_\prec ^c$. Then $\mathcal{Q}(w) = \bigcup Simp(\mathcal{Q}{\Vert }_{E}) = \lambda (E)$. So $E\in \mathsf {Lock}(\lambda ,w)$, by proposition 8. $\square$

Proof (proposition 10)

Omitted. $\square$

Proof (proposition 11)

Suppose that $\lambda$ is a non-Ockham solution to problem $\mathfrak {P}$. Then there exists information $E_0\in \mathcal{I}$ and answer A in $\mathcal{Q}{\Vert }_{E_0}$ such that $A \prec \lambda (E_{0})$. Let $H = \lambda (E_{0})$. Let world w be in $A\cap E_0$. Since $\lambda$ is a solution, there exists $E_1$ in $\mathsf {Lock}(\lambda ,w)$. There exists world v in $E_0 \cap E_1 \cap H$, since $A \prec H$. There exists $E_2 \in \mathsf {Lock}(\lambda , v)$, because $\lambda$ is a solution, so $\lambda (E_{2}) = \mathcal{Q}(v)$. Let $B = \mathcal{Q}(v)$. Let $e = (E_0, ~E_0 \cap E_1,~ E_0\cap E_1 \cap E_2)$. Then $\lambda (e) = (H, A, B)$, since $E_1$, $E_2$ are locking. Note that $B \subseteq H$, because v is in H. Furthermore, A is disjoint from H, and therefore from B, because $A \prec H$. So (H, A, B) is a cycle sequence. $\square$

Proof (lemma 1)

Enumerate $\mathcal{Q}$ as $H_1, H_2, \ldots$. Define:

$$\begin{aligned} \preceq _0 = & {} \preceq \\ \preceq _{n+1} = & {} \{ (H_i, H_j) : H_i \preceq _n H_{n+1} \text { and } H_j |_n H_{n+1} \} \cup \preceq _n; \end{aligned}$$

where $H_i |_n H_j$ holds iff $H_i$ and $H_j$ are not ordered by $\preceq _n$.

We show, by induction on n, that $\preceq _{n}$ is a partial order on $\mathcal{Q}$ such that $H_{\prec _n}$ is closed for all $H\in \mathcal{Q}$. The base case follows from propositions 7 and 6. For induction, suppose that $\preceq _{n}$ is a partial order.

Reflexivity Relation $\preceq _{n}$ is reflexive, and $\preceq _{n+1}$ extends $\preceq _{n}$.

Antisymmetry Suppose that $H_i \preceq _{n+1} H_j$ and $H_j \preceq _{n+1} H_i$. If we have also that $H_i \preceq _n H_j$ and $H_j \preceq _n H_i$, then we are done. So suppose that $H_i \npreceq _n H_j$ and $H_j \preceq _n H_i$. Then $H_i \preceq _n H_{n+1}$, and . But $H_j\preceq _n H_{n+1}$, by the transitivity of $\preceq _n$. Contradiction. Finally, suppose that $H_i \npreceq _n H_j$ and $H_j \npreceq _n H_i$. Then $H_i \preceq _n H_{n+1}$; ; $H_j \preceq _n H_{n+1}$; and . Contradiction.

Transitivity Suppose that $H_i \preceq _{n+1} H_j$ and $H_j \preceq _{n+1} H_k$. If $H_i \preceq _{n} H_j$ and $H_j \preceq _{n} H_k$, then we are done. So suppose that $H_i\npreceq _n H_j,H_j\preceq _n H_k$. Then, $H_i\preceq _n H_{n+1}$ and . If $H_{n+1} \preceq _n H_k$ then, by transitivity of $\preceq _n$, we are done. So suppose that . But then, $H_i\preceq _{n+1} H_k$, as required. Now, suppose that $H_i\preceq _n H_j,H_j\npreceq _n H_k$. Then, $H_j\preceq _n H_{n+1}$ and . But then, $H_i,H_j\preceq _{n+1} H_k$. Finally, suppose that $H_i\npreceq _n H_j,H_j\npreceq _n H_k$. Then, $H_i,H_j\preceq _n H_{n+1}$ and . So $H_i\preceq _{n+1} H_k$, as required.

Next, we show that, if $H_{\prec _n}$ is closed, for all $H\in \mathcal{Q}$, then $H_{\prec _{n+1}}$ is also closed, for all $H\in \mathcal{Q}$. Consider arbitrary $H \in \mathcal{Q}$. If H is ordered with $H_{n+1}$ by $\preceq _n$, then $H_{\prec _{n+1}} = H_{\prec _n}$, and we are done. If , then $H_{\prec _{n+1}} = H_{\prec _n} \cup H_{{n+1}_{\preceq _n}} = H_{\prec _n} \cup H_{{n+1}_{\prec _n}} \cup \overline{H}_{n+1}.$ The final equality follows from the fact that $\overline{H}_{n+1} = {H_{n+1}}_\preceq \subseteq {H_{n+1}}_{\preceq _n}$. Then $H_{\prec _{n+1}}$ is closed, since it is a finite union of closed sets.

Take the directed union of the orders:

$$\begin{aligned} \preceq _\omega= & {} \bigcup _{i} \preceq _i. \end{aligned}$$

We show that $\preceq _\omega$ is a total order.

Reflexivity follows from the fact that $\preceq$ is extended by $\preceq _\omega$.

Anti-symmetry If $H_i \preceq _\omega H_j$ and $H_j \preceq _\omega H_i$, then $H_i \preceq _n H_j$ and $H_j\preceq _m H_i$, for some n, m. Suppose that $n\le m$. Then $H_i \preceq _m H_j$, since $\preceq _n \subseteq \preceq _m$. So $H_i = H_j$, since $\preceq _m$ is a partial order.

Transitivity Similarly, if $H_i\preceq _\omega H_j$ and $H_j \preceq _\omega H_k$, then for some m, $H_i \preceq _m H_j$ and $H_j \preceq _m H_k$. So by the transitivity of $\preceq _m$, $H_i\preceq _m H_k$. Since $\preceq _m$ is extended by $\preceq _\omega$, we have that $H_i \preceq _\omega H_k$. Totality. Note that $H_1, H_2, \ldots , H_m$ are totally ordered by $\preceq _m$. So $\preceq _\omega$ is total.

It remains to show that for $H_i \in \mathcal{Q}$, ${H_i}_{\prec _\omega }$ is closed. That follows immediately from the fact that ${H_i}_{\prec _\omega } = {H_i}_{\prec _i}$, which we already showed to be closed. $\square$

Proof (proposition 12)

Apply lemma 1 to obtain a total ordering $\preceq ^*$ of the elements of $\mathcal{Q}$, such that $H_{\prec ^*}$ is closed, for every $H\in \mathcal{Q}$. Define:

$$\begin{aligned} \lambda (E) = \left\{ \begin{array}{l@{\quad }l} \min _{\preceq ^*} (\mathcal{Q}{\Vert }_{E}) &{}\text { if } \min _{\preceq ^*} (\mathcal{Q}{\Vert }_{E}) \ne \varnothing ; \\ \bigcup \mathcal{Q}{\Vert }_{E} &{}\text { otherwise. } \end{array} \right. \end{aligned}$$

(1)

Since $H_{\prec ^*}$ is refutable, if H is true, then H eventually becomes minimal in $\preceq ^*$. Once minimal, H will remain minimal until refuted—so $\lambda$ will never perform a cycle. To see this suppose that $c=(\lambda (E_0), \lambda (E_1), \lambda (E_2))$ is an output sequence. If $\lambda (E_0) = \bigcup \mathcal{Q}{\Vert }_{E_0}$, then $\lambda (E_1) \subseteq \lambda (E_0)$, and c is not a reversal sequence. So suppose that $\lambda (E_0) = \min _{\preceq ^*} \mathcal{Q}{\Vert }_{E_0}$. But then if $\lambda (E_0)$ is disjoint from $\lambda (E_1)$, we have that $\lambda (E_0) \notin \mathcal{Q}{\Vert }_{E_1}$. Since $\lambda$ is clearly consistent, c cannot be a cycle sequence. We have shown that $\lambda$ performs no cycles of length three. Since every cycle sequence contains a cycle sub-sequence of length three, $\lambda$ performs no cycles. $\square$

Lemma 2

For arbitrary $B\subseteq W,$

$$\begin{aligned}&A_n|_B \cap \overline{A_{n-1}|_B \cap \overline{\ldots \cap \overline{A_1|_B \cap \overline{A_0|_B } } } }\\&\quad \subseteq ~ A_n|_B \cap \overline{A_{n-1} \cap \overline{\ldots \cap \overline{A_1 \cap \overline{A_0 } } } }. \end{aligned}$$

Proof (lemma 2)

$A_1|_B \subseteq A_1|_B$. By the induction hypothesis, we have that:

$$\begin{aligned}&A_{n+1}|_B \cap \overline{A_n|_B \cap \overline{A_{n-1}|_B \cap \overline{\ldots \cap \overline{A_1|_B \cap \overline{A_0|_B } } } }}\\&\quad \subseteq ~ A_{n+1}|_B \cap \overline{A_n|_B \cap \overline{A_{n-1} \cap \overline{\ldots \cap \overline{A_1 \cap \overline{A_0 } } } }}\\&\quad \subseteq ~ A_{n+1}|_B \cap \overline{ A_n \cap \overline{A_{n-1} \cap \overline{\ldots \cap \overline{A_1 \cap \overline{A_0 } } } }}, \end{aligned}$$

where, at each step, we use the fact that if $B\subseteq C$, then $A\cap \overline{B} \subseteq A\cap \overline{C}$. $\square$

Say that reversal sequence $a = (A_0, \ldots , A_n)$, with $A_i$ in $\mathcal{Q}^*$, is forcible in $\mathfrak {P}$ iff, for every solution $\lambda$ to $\mathfrak {P}$, there is an input sequence e, such that $\lambda (e)\leqslant a$.

Lemma 3

If $\mathfrak {P}$ is solvable, then reversal sequence $a = (A_0, \ldots , A_n)$ is forcible in $\mathfrak {P}$ iff:

$$\begin{aligned} A_0 \cap \overline{ A_1 \cap \overline{\ldots \cap \overline{A_{n-1}\cap \overline{A_n}}}}\ne \varnothing . \end{aligned}$$

Proof (lemma 3)

$\Leftarrow :$ Suppose that $\lambda$ is a solution to $\mathfrak {P}$, and $A_0 \cap \overline{ A_1 \cap \overline{\ldots \cap \overline{A_{n-1}\cap \overline{A_n}}}}\ne \varnothing$. Let:

$$\begin{aligned} w_0\in & {} A_0 \cap \overline{ A_1 \cap \overline{\ldots \cap \overline{A_{n-1}\cap \overline{A_n}}}};\\ w_{i+1}\in & {} \bigcap _{j\le i} E_{j} \cap A_{i+1} \cap \overline{ A_{i+2} \cap \overline{\ldots \cap \overline{A_{n-1}\cap \overline{A_n}}}}; \end{aligned}$$

where $E_j \in \mathsf {Lock}(\lambda , w_j)$. Letting $e=(\bigcap _{j\le i} E_{j})_{i=0}^n$, we have that $\lambda (e)\leqslant a$.

$\Rightarrow :$ Suppose that $\mathfrak {P}$ is solvable. We proceed by induction on n. In the base case, suppose that $A_0=\varnothing$. By proposition 3, there is a solution $\lambda$ such that $\lambda (E)\ne \varnothing$ for all nonempty $E\in \mathcal{I}$. So $a=(A_0)$ is not forcible. For the inductive case, suppose that $A_0 \cap \overline{ A_1 \cap \overline{\ldots \cap \overline{A_{n}\cap \overline{A_{n+1}}}}} = \varnothing$. Let:

$$\begin{aligned} C= & {} A_1 \cap \overline{\ldots \cap \overline{A_{n}\cap \overline{A_{n+1}}}} \end{aligned}$$

By Lemma 2:

$$\begin{aligned} A_1|_{\overline{C}^c} \cap \overline{\ldots \cap \overline{A_{n}|_{\overline{C}^c} \cap \overline{A_{n+1}|_{\overline{C}^c}}}}\subseteq & {} A_1|_{\overline{C}^c} \cap \overline{\ldots \cap \overline{A_{n} \cap \overline{A_{n+1}}}}\\= & {} \overline{C}^c \cap A_1 \cap \overline{\ldots \cap \overline{A_{n} \cap \overline{A_{n+1}}}}\\= & {} \overline{C}^c \cap C = \varnothing . \end{aligned}$$

So, by the induction hypothesis, $(A_1, ..., A_{n+1})$ is not forcible in the $\mathfrak {P}|_{\overline{C}^c}$ subproblem. Let $\lambda _2$ be a solution that never performs that reversal sequence in $\mathfrak {P}|_{\overline{C}^c}$. Furthermore, let $\lambda _{1}$ be a solution to the $\mathfrak {P}|_{\overline{C}}$ subproblem. Both solutions exist by proposition 4, since $\mathfrak {P}$ is solvable. Furthermore, we can assume that solution $\lambda _{1}$ is consistent, by proposition 3. Define the method:

$$\begin{aligned} \lambda ^*(E) = {\left\{ \begin{array}{ll} \mathcal{Q}(\lambda _1(E \cap \overline{C})) &{}\,\,\, {\text { if }}\overline{C} \cap E \ne \varnothing ; \\ \mathcal{Q}(\lambda _2(E)) &{}\,\,\, {\text { otherwise}}. \\ \end{array}\right. } \end{aligned}$$

Since $\overline{C}$ is refutable, it is possible to focus on the $\mathfrak {P}|_{\overline{C}}$ subproblem, until $\overline{C}$ is refuted. Since, by assumption, $A_0\cap \overline{C} = \varnothing$, consistent method $\lambda _1$ never outputs $A_0$. Therefore, $\lambda _1$ cannot perform the reversal sequence $(A_0, A_1, \ldots , A_{n+1})$. Furthermore, since $\lambda _2$ never performs the reversal sequence $(A_1, ..., A_{n+1})$, it follows that $\lambda ^*$ solves $\mathfrak {P}$, and never performs the reversal sequence $(A_0, A_1, \ldots , A_{n+1})$. $\square$

Proof (proposition 14)

$\Rightarrow :$ Suppose that $\lambda$ is a solution to $\mathfrak {P}$ that is not patient. Then $\lambda (E)$ is not co-initial in $\mathcal{Q}{\Vert }_{E}$, for some $E\in \mathcal{I}$. So there is $A\in \mathcal{Q}{\Vert }_{E}$ such that $\lambda (E) \cap \overline{A}=\varnothing$, by homogeneity. Let $w\in A\cap E$, and let $F\in \mathsf {Lock}(\lambda ,w)$. Then $(\lambda (E), \lambda (E\cap F))$ is a reversal sequence that is not forcible, by lemma 3. So $\lambda$ is not reversal-optimal.

$\Leftarrow :$ Suppose $\lambda$ is a patient solution and $b=(\lambda (E_0), \lambda (E_1), \ldots , \lambda (E_n))$ is a reversal sequence. Let $A_n$ be an element of $\mathcal{Q}$ such that $A_n\subseteq \lambda (E_n)$. By patience, there exists $A_{n-1} \in \mathcal{Q}$, such that $A_{n-1} \subseteq \lambda (E_{n-1})$ and $A_{n-1} \prec A_n$. Also by patience, there is an answer $A_{n-2} \subseteq \lambda (E_{n-2})$ such that $A_{n-2} \prec A_{n-1}$, and so on. That yields a recipe for constructing a reversal sequence $a=(A_0, A_1, \ldots , A_n)$, such that $A_0 \prec A_1 \prec \cdots \prec A_n$ and $a\le b$. Furthermore, it is easy to check that $A_0 \cap \overline{ A_1 \cap \overline{\ldots \cap \overline{A_{n-1}\cap \overline{A_n}}}} = A_0$. By lemma 3, the reversal sequence a is forcible. Therefore b is forcible as well. It has been shown that every reversal sequence performed by a patient method is forcible. Therefore, patient solutions are reversal-optimal. $\square$

Proof (proposition 16)

By proposition 9, every natural, normal problem has a patient Ockham solution. By proposition 14, every patient Ockham solution is reversal-optimal. It remains to show that the method described in the proof of proposition 9 is cycle-free. Recall that the method is:

$$\begin{aligned} \lambda (E) = {\left\{ \begin{array}{ll} \bigcup {Simp}(\mathcal{Q}{\Vert }_{E}) &{}\,\,\, {\text { if }} {Simp}(\mathcal{Q}{\Vert }_{E}) \hbox { is co-initial in } \mathcal{Q}{\Vert }_{E}, \\ \bigcup \mathcal{Q}{\Vert }_{E} &{}\,\,\, {\text { otherwise}}. \\ \end{array}\right. } \end{aligned}$$

It is easy to see that $\lambda$ is consistent. Suppose that $c=(\lambda (E_0), \lambda (E_1), \lambda (E_2))$ is a cycle sequence. If $\lambda (E_0) = \bigcup \mathcal{Q}{\Vert }_{E}$ then, by consistency, $\lambda (E_1) \subseteq \lambda (E_0)$, and c is not a reversal sequence. So it must be that $\lambda (E_0) = \bigcup {Simp}(\mathcal{Q}{\Vert }_{E_0})$. If c is a cycle sequence, $\lambda (E_2) \subseteq \lambda (E_0) = \bigcup {Simp}(\mathcal{Q}{\Vert }_{E_0})$. Furthermore, by consistency, $\lambda (E_2) \subseteq \mathcal{Q}{\Vert }_{E_2} \subseteq \mathcal{Q}{\Vert }_{E_1}$. By proposition 8, if $\lambda (E_2) \subseteq \bigcup {Simp}(\mathcal{Q}{\Vert }_{E_0})$, and $\lambda (E_2)\subseteq \mathcal{Q}{\Vert }_{E_1}$, then $\lambda (E_2) \subseteq \bigcup {Simp}(\mathcal{Q}{\Vert }_{E_1})$. So c is not a reversal sequence. Contradiction. We have shown that $\lambda$ performs no cycles of length three. Since every cycle sequence contains a cycle sub-sequence of length three, $\lambda$ performs no cycles. $\square$

Proof (proposition 17)

Suppose that $\lambda$ is reversal-optimal in $\mathfrak {P}$. Then $\lambda$ is a patient solution to $\mathfrak {P}$, by proposition 14. By the proof of proposition 14, it follows that every reversal sequence produced by $\lambda$ is forcible. Hence, if the simplicity order in $\mathfrak {P}$ has no path of length $\geqslant 2$, then $\lambda$ performs no reversal sequence and is, therefore, reversal-admissible. Alternatively, let $A_{0} \prec A_{1}$, for distinct $A_{0}, A_{1} \in \mathcal{Q}$. Let $\lambda '$ solve $\mathfrak {P}$. It suffices for admissibility of $\lambda$ to find information states $E_{0}, E_{1}$ such that $\lambda (E_{0}, E_{1}) = \lambda '(E_{0}, E_{1})$, for then $\lambda '(E_{0}, E_{1}) \not < \lambda (E_{0}, E_{1})$. Choose world $w_{0}$ in $A_{0}$. Let information state $F_{0}$ be locking for $\lambda$ in $w_{0}$ and let information state $G_{0}$ be locking for $\lambda '$ in $w_{0}$. Thus, information state $E_{0} = F_{0} \cap G_{0}$ is locking for both $\lambda$ and $\lambda '$ in $w_{0}$. By the definition of $\prec$, there exists world $w_{1}$ in $A_{1} \cap E_{0}$. Let $F_{1}$ be locking for $\lambda$ in $w_{1}$ and let $G_{1}$ be locking for $\lambda '$ in $w_{1}$. Then $E_{0} \cap F_{1} \cap G_{1}$ is locking for both $\lambda$ and $\lambda '$ in $w_{1}$. Let $e = (E_{0}, E_{1})$. Then $\lambda (e) = (A_{0}, A_{1}) = \lambda '(e)$, as desired. $\square$