Continuous Accessibility Modal Logics

Abstract

In classical modal semantics, a binary accessibility relation connects worlds. In this paper, we present a uniform and systematic treatment of modal semantics with a continuous accessibility relation alongside the continuous accessibility modal logics that they model. We develop several such logics for a variety of philosophical applications. Our main conclusions are as follows. Modal logics with a continuous accessibility relation are sound and complete in their natural classes of models. The class of Kripke frames where a continuous accessibility relation has a magnitude characterizing its degree of accessibility is not modally definable, and this has unappreciated significance to completeness proofs for such logics, revealing a methodological advantage of using classical multimodal semantics over fuzzy modal semantics. There is a pseudometric space modal logic that is complete in the class of pseudometric spaces, a natural semantic setting for quantitative modal reasoning about similarity. There is a metric space modal logic that is complete in the class of metric spaces, a natural semantic setting for quantitative modal reasoning about neighborhoods and counterfactual stability. There is a real line continuous temporal logic that is canonical for real lines, a natural semantic setting for quantitative modal reasoning about time.

Appendix A: Plantability Proofs

Recall the definition of a plantable property from Section 3.3:

Definition 5

Let π be a frame property. Let $\mathfrak {F} = (M, \{R_x: {x} \in [0,\infty ]\})$ be an arbitrary frame that has π. Let $\left (\mathfrak {M}, {a}\right )$ be an arbitrary pointed model extending $\mathfrak {F}$. Let $\left (\mathfrak {M}^{\dagger },\langle {a}\rangle \right )$ be its planted model, with $\mathfrak {M}^{\dagger } = \big ({M}^{\dagger },\{R_{x}^{\dagger }: x \in [0,\infty ]\}, {V}^{\dagger }\big )$.

Property π is plantable if and only if there exists a set of accessibility relations $\{R_{x}^{\prime } : {x} \in [0, \infty ]\}$ such that:

• 1. for all ${x}\in [0, \infty ]$, $R_{x}^{\prime } \supseteq R_{x}^{\dagger }$;

• 2. $\big ({M}^{\dagger },\{R_{x}^{\prime } : {x} \in [0, \infty ]\}\big )$ has π;

• 3. $\big ({M}^{\dagger },\{R_{x}^{\prime } : {x} \in [0, \infty ]\}\big )$ is well-founded; and

• 4. for all w^‡,u^‡∈ M^‡ corresponding to w and u, respectively, and for all ${x} \in [0, \infty ]$, $R_{x}^{\prime }(w^{\dagger }, {u}^{\dagger })$ holds only if R_x(w,u) holds.

Below we show that various properties considered in this paper are plantable.

1.1 A.1 Proof of Lemma 2

We will show that upwardly closed accessibility is plantable.

Let $\left ({\mathfrak {M}}, {a}\right )$ be an arbitrary pointed model with upwardly closed accessibility, with $\mathfrak {M}=\left ({M}, \{R_{x} : {x} \in [0, \infty ]\}, {V}\right )$. Let $\left (\mathfrak {M}^{\dagger },\langle {a}\rangle \right )$ be its planted model, with $\mathfrak {M}^{\dagger }=({M}^{\dagger }, \{R_{x}^{\dagger } : {x} \in [0, \infty ]\},{{V}^{\dagger }})$. Define $\mathfrak {M}^{\prime }:=\left ({M}^{\dagger },\{R_{x}^{\prime } : {x} \in [0, \infty ]\},{V}^{\dagger }\right )$, where $\{R_{x}^{\prime } : {x} \in [0, \infty ]\}$ is constructed as follows.

• Definition of $R_{x}^{\prime }$: For w^‡,u^‡∈ M^‡ and ${x,y} \in [0, \infty ]$,

  $$ R_{x}^{\prime} := \{({w}^{\dagger},{u}^{\dagger}) : R_{y}^{\dagger}({w}^{\dagger},{u}^{\dagger}) \text{ holds for some} y \leq x\}. $$

In other words, the x-accessibility relations of $\mathfrak {M}^{\prime }$ are defined to be the upwardly closed accessibility relations of $\mathfrak {M}^{\dagger }$.

By construction, $\{R_{x}^{\prime } : {x} \in [0, \infty ]\}$ satisfies the first three conditions of being a plantable model. For the fourth condition, fix arbitrary w^‡,u^‡∈ M^‡, corresponding to w,u ∈ M, respectively. Suppose that $R_{x}^{\prime }({w}^{\dagger }, {u}^{\dagger })$ holds for some ${x} \in [0, \infty ]$. By definition, there is some y ≤ x such that $R_{y}^{\dagger }({w}^{\dagger }, {u}^{\dagger })$ holds. Since $(\mathfrak {M},{a})$ is bisimilar to $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$, R_y(w,u) holds. Since $(\mathfrak {M},{a})$ has upwardly closed accessibility, R_x(w,u) holds.

1.2 A.2 Proof of Lemma 3

We will show that upwardly closed accessibility, reflexivity, symmetry, and additively transitive accessibility are jointly plantable.

Let $(\mathfrak {M},{a})$ be an arbitrary pointed model that has upwardly closed accessibility, reflexivity, symmetry, and additively transitive accessibility, with $\mathfrak {M}=({M}, \{R_{x} : {x} \in [0, \infty ]\}, {V})$. Let $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ be its planted model, with $\mathfrak {M}^{\dagger }=({M}^{\dagger }, \{R_{x}^{\dagger } : {x} \in [0, \infty ]\},{V}^{\dagger })$. Define $\mathfrak {M}^{\prime }:=({M}^{\dagger },\{R_{x}^{\prime } : {x} \in [0, \infty ]\}, {V}^{\dagger })$, where $\{{R}_{x}^{\prime } : {x} \in [0, \infty ]\}$ is constructed via transfinite recursion as follows.

• Definition of $R_{x}^{\prime }$:

  $$ {R_{x}^{\prime}} := \bigcup\limits_{{\alpha}\in{\text{ORD}}} R_{x}^{{\prime}{\alpha}}. $$

• Base case: For w^‡,u^‡∈ M^‡ and ${x,y}\in [0, \infty ]$,

  $$ R_{x}^{{\prime}0} := \Big\{({w}^{\dagger},{u}^{\dagger}) : R_{y}^{\dagger}({w}^{\dagger},{u}^{\dagger}) \text{or} R_{y}^{\dagger}({u}^{\dagger},{w}^{\dagger}) \text{holds for some} y \leq x \Big\}. $$

• Successor case: For w^‡,u^‡,v^‡∈ M^‡ and ${x,y} \in [0, \infty ]$,

  $$ R_{{x}+{y}}^{{\prime}{\alpha}+1} := \left\{({w}^{\dagger},{v}^{\dagger}) : R_{x}^{{\prime}{\alpha}}({w}^{\dagger},{u}^{\dagger}) \text{and} R_{y}^{{\prime}{\alpha}}({u}^{\dagger},{v}^{\dagger}) \text{hold for some} {u}^{\dagger}\right\}. $$

• Limit case:

  $$ R_{x}^{{\prime}{\alpha}} := \bigcup\limits_{\beta<\alpha} R_{x}^{{\prime}{\beta}}. $$

In other words, the x-accessibility relations of $\mathfrak {M}^{\prime }$ are defined to be the upwardly closed and symmetrized accessibility relations of $\mathfrak {M}^{\dagger }$, successively made additively transitive (which preserves upwardly closed accessibility and symmetry).

By construction, $R_{x}^{\prime } {\supseteq } R_{x}^{\dagger }$ for all ${x} \in [0, \infty ],$ satisfying the first condition of being a plantable property. We show that $\{R_{x}^{\prime } : {x} \in [0, \infty ]\}$ satisfies the remaining three conditions.

Lemma 1

$\mathfrak {M}^{\prime }$ has all of upwardly closed accessibility, reflexivity, symmetry, and additively transitive accessibility.

Proof

To show that $\mathfrak {M}^{\prime }$ has upwardly closed accessibility, we perform a simple proof by transfinite induction. For the base case, by construction, $\{R_{x}^{{\prime }0} : {x} \in [0, \infty ]\}$ is upwardly closed. For the successor case, the inductive hypothesis is that, for fixed ${\alpha } \in {\text {ORD}}, \{R_{x}^{{\prime }{\alpha }} : {x} \in [0, \infty ]\}$ is upwardly closed. Fix w^‡,v^‡∈ M^‡ and ${z^{\prime } \geq z}\in [0, \infty ]$. Suppose that $R_{z}^{{\prime }{\alpha }+1}({w}^{\dagger },{v}^{\dagger })$ holds. Then there is some u^‡∈ M^‡ and $x, y \in [0, \infty ]$ such that $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{v}^{\dagger })$ and $R_{y}^{{\prime }{\alpha }}({u}^{\dagger },{v}^{\dagger })$ hold and x + y = z. Since $\{R_{x}^{{\prime }{\alpha }} : {x} \in [0, \infty ]\}$ is upwardly closed, there is an ${{x}^{\prime }} \geq {{x,y}^{\prime }}\geq {y} \in [0, \infty ]$ such that $R_{{x}{^{\prime }}}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger })$ and $R_{{y}{^{\prime }}}^{{\prime }{\alpha }}({u}^{\dagger },{v}^{\dagger })$ hold and $x^{\prime }+y^{\prime }=z^{\prime }$. It follows from the definition of the successor case that $R_{{z}^{\prime }}^{{\prime }{\alpha }+ 1}({w}^{\dagger },{v}^{\dagger })$ holds. Therefore, $\{R_{x}^{{\prime }{\alpha }+1} : {x} \in [0, \infty ]\}$ is upwardly closed. For the limit case, the proof is transparent.

To show that $\mathfrak {M}^{\prime }$ is reflexive, note that $(\mathfrak {M}, a)$ is reflexive. By the definition of planted models, $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is thus reflexive in 0-accessibility. It follows that $\mathfrak {M}^{\prime }$ is reflexive in 0-accessibility. Since $\mathfrak {M}^{\prime }$ also has upwardly closed accessibility, $\mathfrak {M}^{\prime }$ is reflexive.

To show that $\mathfrak {M}^{\prime }$ is symmetric, we again proceed by transfinite induction. For the base case, by construction, $\{R_{x}^{{\prime }0} : {x} \in [0, \infty ]\}$ is symmetric. For the successor case, the inductive hypothesis is that, for fixed ${\alpha } \in {\text {ORD}}, \{R_{x}^{{\prime }{\alpha }} : {x} \in [0, \infty ]\}$ is symmetric. Fix w^‡,v^‡∈ M^‡ and ${z}\in [0, \infty ]$. Suppose that $R_{z}^{{\prime }{\alpha }+1}({w}^{\dagger },{v}^{\dagger })$ holds. Then there is some u^‡∈ M^‡ and $x, y \in [0, \infty ]$ such that $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger }) \text {and} R_{y}^{{\prime }{\alpha }}({{u}^{\dagger }},{{v}^{\dagger }})$ hold and x + y = z. Since $\{R_{x}^{{\prime }{\alpha }} : {x} \in [0, \infty ]\}$ is symmetric, $R_{x}^{{\prime }{\alpha }}({u}^{\dagger },{w}^{\dagger })$ and $R_{y}^{{\prime }{\alpha }}({v}^{\dagger },{u}^{\dagger })$ also hold. It follows from the definition of the successor case that $R_{x+y}^{{\prime }{\alpha }+1}({v}^{\dagger },{w}^{\dagger })$ holds, which means $R_{z}^{{\prime }{\alpha }+1}({v}^{\dagger },{w}^{\dagger })$ holds. Therefore, $\{R_{x}^{{\prime }{\alpha }+1} : {x} \in [0, \infty ]\}$ is symmetric. For the limit case, the proof is transparent.

To show that $\mathfrak {M}^{\prime }$ has additively transitive accessibility, fix w^‡,u^‡,v^‡∈ M^‡ and ${x,y} \in [0,\infty ]$. Suppose that $R_{x}^{\prime }({w}^{\dagger },{u}^{\dagger })$ and $R_{y}^{\prime }({u}^{\dagger },{v}^{\dagger })$ hold. By the definition of $R_{x}^{\prime }$, it follows that for some α,β ∈ORD, $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger })$ and $R_{y}^{_{{\prime }{\beta }}}({u}^{\dagger },{v}^{\dagger })$ hold. Fix one such α and β, and let γ be the larger of the two. By the definitions of the successor case and the limit case, it follows from reflexivity in 0-accessibility that if $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger })$ holds, $R_{x}^{_{{\prime }{\gamma }}}({w}^{\dagger },{u}^{\dagger })$ holds, and if $R_{y}^{_{{\prime }{\beta }}}({u}^{\dagger },{v}^{\dagger })$ holds, $R_{y}^{_{{\prime }{\gamma }}}({u}^{\dagger },{v}^{\dagger })$ holds. Thus, by the definition of the successor case, $R_{x+y}^{_{{\prime }{\gamma }+1}}({w}^{\dagger },{v}^{\dagger })$ holds. From this it follows by the definition of $R_{x}^{\prime }$ that $R_{x+y}^{{\prime }}({w}^{\dagger },{v}^{\dagger })$ holds. $\mathfrak {M}^{\prime }$ therefore has additively transitive accessibility. □

Lemma 2

$\mathfrak {M}^{\prime }$ is well-founded.

Proof

$(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is well-founded. Here is the concern for $\mathfrak {M}^{\prime }$. Planted models ensure well-foundedness by expanding one world’s continuum-many accessibility relations to a single world into one world’s single accessibility relation to continuum-many worlds. But the construction of $R_{x}^{\prime }$ allows one world to access another through continuum-many paths. We must ensure that these paths don’t generate a set of accessibility relations with no minimum.

Definition 14

For all w^‡,v^‡∈ M^‡, a direct path from w^‡ tov^‡ is a path from w^‡ to v^‡ such that:

• 1. for each element of the path ${u_{1}^{\dagger }}$ before v^‡, the next element ${u}_{2}^{\dagger }$ is such that, for some ${x} \in [0,\infty ]$, either ${R_{x}^{\dagger }}({u_{1}^{\dagger }},{u}_{2}^{\dagger })$ holds or ${R_{x}^{\dagger }}({u}_{2}^{\dagger },{u_{1}^{\dagger }})$ holds; and

• 2. no element appears in the path more than once.

The planted model $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is just the tree-unraveling of $(\mathfrak {M}, a)$, save for self-0-accessibility. Since, by definition, no direct path has a world that appears more than once, self-0-accessibility cannot connect two elements of a direct path. So for the construction of direct paths, $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is a rooted tree-like structure. From this it follows that, for all w^‡,v^‡∈ M^‡, there is exactly one direct path from w^‡ to v^‡, and the path has a finite number of elements. We will henceforth call it the direct path from w^‡ to v^‡.

Definition 15

For all w^‡,v^‡∈ M^‡, the length of the direct path from w^‡ tov^‡ is the sum of the indices of the accessibility relations connecting each pair of successive elements of the path.

More formally, let $\langle {u_{1}^{\dagger }},{u}_{2}^{\dagger },\ldots ,{{u}_{n}^{\dagger }}\rangle$ be the direct path from w^‡ to v^‡. Define {x_i : 1 ≤ i ≤ n} as follows.

$$ x_{i} := \begin{cases} x \text{such that } R_{x}^{\dagger}(u_{i}^{\dagger},u_{i+1}^{\dagger}) \text{ or } R_{x}^{\dagger}(u_{i+1}^{\dagger},u_{i}^{\dagger}) \text{ holds} & \text{ if } i<n; \\ 0 & \text{ if } i=n. \end{cases} $$

The length of the path is then:

$$ \lambda({w}^{\dagger},{v}^{\dagger}) := \sum\limits_{i=1}^{n} x_{i}. $$

Consider arbitrary worlds w^‡,u^‡,v^‡∈ M^‡, and the direct paths from w^‡ to u^‡, from u^‡ to v^‡, and from w^‡ to v^‡. Construct a possibly indirect path from w^‡ to v^‡ by concatenating the direct paths from w^‡ to u^‡ and from u^‡ to v^‡, merging the last element of the former with the first element of the latter. The direct path from w^‡ to v^‡ is a subpath of the possibly indirect path from w^‡ to v^‡, for otherwise there would be more than one direct path from w^‡ to v^‡. Therefore, for all w^‡,u^‡,v^‡∈ M^‡,

$$ \lambda({w}^{\dagger},{v}^{\dagger})\leq\lambda({w}^{\dagger},{u}^{\dagger})+\lambda({u}^{\dagger},{v}^{\dagger}). $$

To prove that $\mathfrak {M}^{\prime }$ is well-founded, we will prove that, for all w^‡,v^‡∈ M^‡, λ(w^‡,v^‡) is the minimum x such that $R_{x}^{\prime }({w}^{\dagger },{v}^{\dagger })$ holds. Since $\mathfrak {M}^{\prime }$ is reflexive in 0-accessibility and has additively transitive accessibility, for all w^‡,v^‡∈ M^‡, $R_{{\lambda }({w}^{\dagger },{v}^{\dagger })}^{\prime }{({w}^{\dagger },{v}^{\dagger })}$ holds. We thus need to show that, for all x such that $R_{x}^{\prime }({w}^{\dagger },{v}^{\dagger })$ holds, x ≥ λ(w^‡,v^‡). We show this by performing a proof by transfinite induction.

For the base case, fix w^‡,v^‡∈ M^‡ and ${x} \in [0,\infty ]$. Suppose that $R_{x}^{{\prime }0}({w}^{\dagger },{v}^{\dagger })$ holds. By construction, there is a y ≤ x such that either $R_{y}^{\dagger }({w}^{\dagger },{v}^{\dagger })$ holds or $R_{y}^{\dagger }({v}^{\dagger },{w}^{\dagger })$ holds. Since $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is a rooted tree-like structure, it follows that the direct path between w^‡ and v^‡ is 〈w^‡,v^‡〉, which has length λ(w^‡,v^‡) = y. Therefore, x ≥ λ(w^‡,v^‡).

For the successor case, the inductive hypothesis is that, for fixed α ∈ord: for all w^‡,v^‡∈ M^‡ and ${x} \in [0,\infty ]$, if $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{v}^{\dagger })$ holds, then x ≥ λ(w^‡,v^‡). Now fix w^‡,v^‡∈ M^‡ and ${z} \in [0,\infty ]$. Suppose that $R_{z}^{{\prime }{\alpha }+1}({w}^{\dagger }{v}^{\dagger })$ holds. Then there is some u^‡∈ M^‡ and $x, y \in [0, \infty ]$ such that $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger })$ and $R_{y}^{{\prime }{\alpha }}({{u}^{\dagger }},{{v}^{\dagger }})$ hold and x + y = z. By hypothesis, x ≥ λ(w^‡,u^‡) and y ≥ λ(u^‡,v^‡). Therefore, z = x + y ≥ λ(w^‡,u^‡) + λ(u^‡,v^‡) ≥ λ(w^‡,v^‡).

For the limit case, the proof is straightforward. □

Lemma 3

For all w^‡,v^‡∈ M^‡ corresponding to w,v ∈ M, respectively, and for all ${x} \in [0,\infty ]$, if $R_{x}^{\prime }$(w^‡,v^‡) holds, then R_x(w,v) holds.

Proof

We perform a proof by transfinite induction.

For the base case, fix w^‡,v^‡∈ M^‡, corresponding to w,v ∈ M, respectively, and fix ${x} \in [0,\infty ]$. Suppose that $R_{x}^{{\prime }0}({w}^{\dagger },{v}^{\dagger })$ holds. By construction, there is a y ≤ x such that either $R_{y}^{\dagger }({w}^{\dagger },{v}^{\dagger })$ holds or $R_{y}^{\dagger }({v}^{\dagger },{w}^{\dagger })$ holds. Since $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is bisimilar to $(\mathfrak {M}, a)$, either R_y(w,v) or R_y(v,w) holds. Since $\mathfrak {M}$ is symmetric, if R_y(v,w) holds, then R_y(w,v) holds. Thus, R_y(w,v) holds. Since $\mathfrak {M}$ has upwardly closed accessibility, R_x(w,v) holds.

For the successor case, the inductive hypothesis is that, for fixed α ∈ord: for all w^‡,v^‡∈ M^‡ corresponding to w,v ∈ M, respectively, and all ${x} \in [0,\infty ]$, if $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{v}^{\dagger })$ holds, then R_x(w,v) holds. Now fix w^‡,v^‡∈ M^‡ corresponding to w,v ∈ M, respectively, and fix ${z} \in [0,\infty ]$. Suppose that $R_{z}^{{\prime }{\alpha }+1}({w}^{\dagger },{v}^{\dagger })$ holds. Then there is some u^‡∈ M^‡ corresponding to u ∈ M and $x, y \in [0,\infty ]$ such that $R_{x}^{{\prime }{\alpha }}({w}^{\dagger },{u}^{\dagger })$ and $R_{y}^{{\prime }{\alpha }}({{u}^{\dagger }},{{v}^{\dagger }})$ hold and x + y = z. By hypothesis, R_x(w,u) and R_y(u,v) hold. Since $\mathfrak {M}$ has additively transitive accessibility, R_x+y(w,v) holds, which means R_z(w,v) holds.

For the limit case, the proof is straightforward. □

1.3 A.3 Proof of Lemma 5

We will show that upwardly closed accessibility, reflexivity, symmetry, additively transitive accessibility, and D-bounded accessibility are jointly plantable for each ${D} \in [0, \infty ]$. The proof is the same as for Lemma 3 in Appendix A.2, except as follows.

Fix ${D} \in [0, \infty ]$. Stipulate that $(\mathfrak {M}, a)$ also has D-boundedness. Define $\mathfrak {M}^{\prime }:=({M}^{\dagger },\{R_{x}^{D} : {x} \in [0, \infty ]\}, {V}^{\dagger })$, with:

• Definition of $R_{x}^{D}$: For w^‡,u^‡∈ M^‡ and ${x,y} \in [0,\infty ]$,

  $$ R_{x}^{D}{} :={} \{({w}^{\dagger}{\kern-.5pt},{{u}^{\dagger}}):{\kern-.5pt} R_{x}^{\prime}({w}^{\dagger},{u}^{\dagger}) \text{holds, or} R_{y}^{\prime}({w}^{\dagger}{\kern-.5pt},{{u}^{\dagger}}) \text{holds for some} y \geq x \geq D\}. $$

In other words, the accessibility relations of $\mathfrak {M}^{\prime }$ are defined to be the same as in Lemma 3 in Appendix A.2, except that they are D-bounded.

By construction, $R_{x}^{D} \supseteq R_{x}^{\dagger }$ for all ${x} \in [0,\infty ]$. Also by construction, $\mathfrak {M}^{\prime }$ will have D-bounded accessibility, preserving upwardly closed accessibility, reflexivity, symmetry, and additive transitivity. Again by construction, since $R_{x}^{\prime }$ is well-founded, $R_{x}^{D}$ is well-founded. So the first three conditions of being a plantable property are satisfied.

For the fourth condition, fix w^‡,u^‡,∈ M^‡, corresponding to w,u ∈ M, respectively, and fix ${x} \in [0,\infty ]$. Suppose $R_{x}^{D}({w}^{\dagger },{{u}^{\dagger }})$ holds. Then either $R_{x}^{\prime }({w}^{\dagger },{u}^{\dagger })$ holds or there is a y ≥ x ≥ D such that $R_{y}^{\prime }({w}^{\dagger },{{u}^{\dagger }})$ holds. In case $R_{x}^{\prime }({w}^{\dagger },{u}^{\dagger })$ holds, then by Lemma 15, R_x(w,u) holds. In case there is a y ≥ x ≥ D such that $R_{y}^{\prime }({w}^{\dagger },{{u}^{\dagger }})$ holds, then by Lemma 15, R_y(w,u). Since $\mathfrak {M}$ has upwardly closed accessibility, $R_{\infty }(w,u)$ holds. Since $\mathfrak {M}$ has D-bounded accessibility, R_D(w,u) holds. Since $\mathfrak {M}$ has upwardly closed accessibility, R_x(w,u) holds.

1.4 A.4 Proof of Lemma 9

We will show that upwardly closed accessibility, reflexivity, symmetry, additively transitive accessibility, D-bounded accessibility, and coreflexive 0-accessibility are jointly plantable for each ${D} \in [0, \infty ]$. The proof is the same as for Lemma 5 in Appendix A.3, except as follows.

Stipulate that $(\mathfrak {M}, a)$ is coreflexive in 0-accessibility. By the definition of planted models, $(\mathfrak {M}^{\dagger },\langle {a}\rangle )$ is also coreflexive in 0-accessibility. From this it follows that $\mathfrak {M}^{\prime }$ is coreflexive in 0-accessibility.