A Logical Theory for Conditional Weak Ontic Necessity Based on Context Update

Abstract

Weak ontic necessity is the ontic necessity expressed by “should/ought to” in English. An example of it is “I should be dead by now”. A feature of this necessity is that whether it holds at the present world is irrelevant to whether its prejacent holds at the present world. In this paper, by combining premise semantics and update semantics for conditionals, we present a logical theory for conditional weak ontic necessity based on context update. A context is a set of ordered defaults, determining expected possible states of the present world. Sentences are evaluated with respect to contexts. When evaluating the conditional weak ontic necessity with respect to a context, we first update the context with the antecedent, then check whether the consequent holds with respect to the updated context. We compare this theory to some related works. The logic’s expressivity is studied.

Appendices

A Proofs About Comparisons to the Conditional Logic ${\textsf {VN}}$

1.1 A.1 Finite Universal Relational Models for $\Phi _{{\textsf {VN}}}$

Note that $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$ contains no nested conditionals. This implies that we can just consider universal relational models without changing the set of valid formulas in $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$.

Definition 20

(Universal relational models for $\Phi _{{\textsf {VN}}}$) A tuple $M = (W, <, V)$ is a universal relational model for $\Phi _{{\textsf {VN}}}$ if

• W and V are as usual;

• < is a well-founded, irreflexive, transitive, almost connected binary relation on W.

Lemma 6

The class of universal relational models and the class of relational models determine the same set of valid formulas in $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$.

1.2 A.2 Rephrasing the Semantics for $\Phi _{{\textsf {VN}}}$

Fact 3

Let $M = (W, <, V)$ be a universal relational model for $\Phi _{\textsf {VN}}$. Define a relation $\equiv$ on W as follows: For all w and u, $w \equiv u$ if and only if $w \not < u$ and $u \not < w$. Then, $\equiv$ is an equivalence relation. Let $\Delta _W$ be the partition of W under $\equiv$. Define a relation $\ll$ on $\Delta _W$ as follows: For all X and Y, $X \ll Y$ if and only if for all $x \in X$ and $y \in Y$, $x < y$. Then, $\ll$ is a well-founded strict well-ordering on $\Delta _W$.

Definition 21

(Sphere models for $\Phi _{\textsf {VN}}$) A tuple $M = (W, \Delta _W, \ll , V)$ is a sphere model for $\Phi _{\textsf {VN}}$ if

• W and V are as usual;

• $\Delta _W$ is a partition of W and $\ll$ is a well-founded strict well-ordering on $\Delta _W$.

Definition 22

(Sphere semantics for $\Phi _{\textsf {VN}}$) Let $M = (W, \Delta _W, \ll , V)$ be a sphere model.

where $|\phi | = \{x \mid M,x \Vdash \phi \}$ and $|\psi | = \{x \mid M,x \Vdash \psi \}$.

It can be verified that the following result holds:

Lemma 7

Sphere semantics is equivalent to relational semantics for $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$.

1.3 A.3 Rephrasing the Semantics for $\Phi _{{\textsf {VN}}}$ Again

Definition 23

(Pseudo sphere models for $\Phi _{\textsf {VN}}$) A tuple $M = (W, \Pi , V)$ is a pseudo sphere model for $\Phi _{\textsf {VN}}$ if

• W and V are as usual;

• $\Pi = (X_0, \dots , X_n, \dots )$ is a sequence of pairwise disjoint (possibly empty) subsets of W such that the union of them is W.

Definition 24

(Pseudo sphere semantics for $\Phi _{\textsf {VN}}$) Let $M = (W, \Pi , V)$ is a pseudo sphere model.

where $|\phi | = \{x \mid M,x \Vdash \phi \}$ and $|\psi | = \{x \mid M,x \Vdash \psi \}$.

It can be verified that the following result holds:

Lemma 8

Pseudo sphere semantics is equivalent to sphere semantics for $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$.

By the following result, which is easy to show, empty elements in $\Pi$ do not matter in pseudo sphere semantics.

Lemma 9

Let (M, w) and $(M',w)$ be two pointed pseudo sphere models, where $M = (W, \Pi , V)$ and $M' = (W, \Pi ', V)$. Assume that $\Pi$ and $\Pi '$ are identical if we remove all the empty elements in them. Then, (M, w) and $(M',w)$ are equivalent for $\Phi _{{\textsf {VN}}\text {-}\textsf {1}}$.

1.4 A.4 Two Transformation Lemmas

Lemma 10

Let W be a nonempty set of states. Let $Y_0, \dots , Y_k$ be a sequence of pairwise disjoint nonempty subsets of W such that $Y_0 \cup \dots \cup Y_k = W$. Define a sequence $X_0, \dots , X_k$ as follows:

• $X_0 = Y_0 \cup \dots \cup Y_k$

• $X_1 = Y_1 \cup \dots \cup Y_k$

  $\vdots$

• $X_k = Y_k$

• 1. Let $Z \subseteq W$. Let l be the greatest number such that $l \le k$ and $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$ and l is the greatest number such that $Z \cap Y_l \ne \emptyset$.

• 2. Let $Z \subseteq W$. Let l be the greatest number such that $l \le k$ and $Z \cap Y_l \ne \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$ and l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$.

Proof

1. Assume that $l = k$. Note that $X_0 \cap \dots \cap X_k= X_k = Y_k$. It is easy to see that the result holds.

Assume that $l < k$.

We first show that l is the greatest number such that $Z \cap Y_l \ne \emptyset$.

Note that $X_0 \cap \dots \cap X_l= X_l$. Then, $Z \cap X_l \ne \emptyset$. Note that $Z \cap X_0 \cap \dots \cap X_{l+1} = \emptyset$ and $X_0 \cap \dots \cap X_{l+1} = X_{l+1}$. Then, $Z \cap X_{l+1} = \emptyset$. Note that $X_l = Y_l \cup \dots \cup Y_k$ and $X_{l+1} = Y_{l+1} \cup \dots \cup Y_k$. Then, $Z \cap Y_{l+1} = \emptyset , \dots , Z \cap Y_k = \emptyset$. Then, $Z \cap Y_l \ne \emptyset$. Then, l is the greatest number such that $Z \cap Y_l \ne \emptyset$.

Second, we show $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$.

Let $a \in Z \cap X_0 \cap \dots \cap X_l$. Then, $a \in Z \cap X_l$. Note that $X_l = Y_l \cup \dots \cup Y_k$. Then, $a \in Z \cap (Y_l \cup \dots \cup Y_k)$. We claim $a \notin Y_{l+1}, \dots , a \notin Y_k$. Why? Suppose that $a \in Y_{l+1}$. Note that $X_{l+1} = Y_{l+1} \cup \dots \cup Y_k$. Then, $a \in X_{l+1}$. Then, $a \in Z \cap X_0 \cap \dots \cap X_{l+1}$. Then, l is not the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. We have a contradiction. Similarly, we know $a \notin Y_{l+2}, \dots , a \notin Y_k$. Then, $a \in Y_l$. Then, $a \in Z \cap Y_l$.

Let $a \in Z \cap Y_l$. By the definitions of $X_0, \dots , X_l$, we know $a \in X_0, \dots , a \in X_l$. Then, $a \in Z \cap X_0 \cap \dots \cap X_l$.

2. Note that $X_0 \cap \dots \cap X_l= X_l = Y_l \cup \dots \cup Y_k$ and $X_0 \cap \dots \cap X_{l+1} = X_{l+1} = Y_{l+1} \cup \dots \cup Y_k$. Also note that $Z \cap Y_{l+1} = \emptyset$, ..., $Z \cap Y_k = \emptyset$.

We first show that l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$.

Note that l is the greatest number such that $Z \cap Y_l \ne \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. Then, $Z \cap X_{l+1} = \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l\cap X_{l+1} = \emptyset$. Then, l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$.

Second, we show $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$.

Let $a \in Z \cap X_0 \cap \dots \cap X_l$. Then, $a \in Z \cap X_l$. Then, $a \in Z \cap (Y_l \cup \dots \cup Y_k)$. Note that l is the greatest number such that $Z \cap Y_l \ne \emptyset$. Then, $a \in Z \cap Y_l$.

Let $a \in Z \cap Y_l$. Then, $a \in Y_l \cup \dots \cup Y_k = X_l = X_0 \cap \dots \cap X_l$. Then, $a \in Z \cap X_0 \cap \dots \cap X_l$. $\square$

Lemma 11

Let W be a nonempty set of states. Let $X_0, \dots , X_k$ be a sequence of subsets of W, where $X_0 = W$. Define a sequence $Y_0, \dots , Y_k$ as follows:

• $Y_0 = X_0-X_1$

• $Y_1 = X_0 \cap X_1 - X_2$

  $\vdots$

• $Y_k = X_0 \cap \dots \cap X_k$

• 1. Then, $Y_0, \dots , Y_k$ are pairwise disjoint and $Y_0 \cup \dots \cup Y_k = W$.

• 2. Let $Z \subseteq W$. Let l be the greatest number such that $l \le k$ and $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$ and l is the greatest number such that $Z \cap Y_l \ne \emptyset$.

• 3. Let $Z \subseteq W$. Let l be the greatest number such that $l \le k$ and $Z \cap Y_l \ne \emptyset$. Then, $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$ and l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$.

Proof

• 1. Let $i, j \le k$ be such that $i \ne j$. We want to show $Y_i \cap Y_j = \emptyset$. Without loss of any generality, Assume that $i < j$. Assume that $Y_i \cap Y_j \ne \emptyset$. Then, there is a such that $a \in Y_i$ and $a \in Y_j$. Note that $Y_i = X_0 \cap \dots \cap X_i - X_{i+1}$ and $Y_j = X_0 \cap \dots \cap X_j - X_{j+1}$. Then, $a \notin X_{i+1}$ and $a \in X_{i+1}$. There is a contradiction. Let $a \in W$. Let l be the the greatest number such that $a \in X_0 \cap \dots \cap X_l$. Note that l exists, as $X_0 = W$. Suppose that $l = k$. Then, $a \in Y_k$. Suppose that $l < k$. Then, $a \notin X_{l+1}$. Then, by the definition of $Y_l$, $a \in Y_l$.

• 2. We first show $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$. Let $a \in Z \cap X_0 \cap \dots \cap X_l$. As l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$, $a \notin X_{l+1}$. Note that $Y_l = X_0 \cap \dots \cap X_k- X_{l+1}$. Then, $a \in Y_l$. Then, $a \in Z \cap Y_l$. Let $a \in Z \cap Y_l$. By the definition of $Y_l$, $a \in X_0 \cap \dots \cap X_l$. Then, $a \in Z \cap X_0 \cap \dots \cap X_l$. Second, we show that l is the greatest number such that $Z \cap Y_l \ne \emptyset$. Assume that there is a natural number $l' > l$ such that $l' \le k$ and $Z \cap Y_{l'} \ne \emptyset$. Let $a \in Z \cap Y_{l'}$. By the definition of $Y_{l'}$, $a \in Z \cap X_0 \cap \dots \cap X_{l'}$. Then, l is not the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. There is a contradiction.

• 3. We first show that l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$. Assume that there is a natural number $l' > l$ such that $l' \le k$ and $l'$ is the greatest number such that $Z \cap X_0 \cap \dots \cap X_{l'} \ne \emptyset$. Assume that $l' = k$. Note that $Y_k = X_0 \cap \dots \cap X_k$. Then, $Z \cap Y_k \ne \emptyset$. Then, l is not the greatest number such that $Z \cap Y_l \ne \emptyset$. We have a contradiction. Assume that $l' < k$. Let $a \in Z \cap X_0 \cap \dots \cap X_{l'}$. As $l'$ is the greatest number such that $Z \cap X_0 \cap \dots \cap X_{l'} \ne \emptyset$, we know $a \notin X_{l' + 1}$. By the definition of $Y_{l'}$, $a \in Z \cap Y_{l'}$. Then, l is not the greatest number such that $Z \cap Y_l \ne \emptyset$. We have a contradiction. Second, we show $Z \cap X_0 \cap \dots \cap X_l= Z \cap Y_l$. Let $a \in Z \cap X_0 \cap \dots \cap X_l$. As l is the greatest number such that $Z \cap X_0 \cap \dots \cap X_l\ne \emptyset$, $a \notin X_{l+1}$. By the definition of $Y_{l}$, $a \in Y_l$. Then, $a \in Z \cap Y_l$. Let $a \in Z \cap Y_l$. By the definition of $Y_{l}$, $a \in X_0 \cap \dots \cap X_l$. Then, $a \in Z \cap X_0 \cap \dots \cap X_l$.

$\square$

1.5 A.5 The Equivalence Theorem

Theorem 4

For every $\Phi _{\textsf{ConWON}\text {-}\textsf {1}}$, $\phi$ is valid in $\textsf{ConWON}$ if and only if $\textbf{tr}(\phi )$ is valid in ${\textsf {VN}}$.

Proof

Let $\alpha$ and $\beta$ be in $\Phi _\textsf{PL}$. It suffices to show that $\alpha \rhd \beta$ is satisfiable in ${\textsf {VN}}$ if and only if $[\alpha ] \beta$ is satisfiable in $\textsf{ConWON}$.

Assume that $\alpha \rhd \beta$ is satisfiable in ${\textsf {VN}}$. Then, $\alpha \rhd \beta$ is true at a pointed sphere model for $\Phi _{\textsf {VN}}$. As $\Phi _{\textsf {VN}}$ has the finite model property, $\alpha \rhd \beta$ is true at a pointed finite sphere model (M, w) for $\Phi _{\textsf {VN}}$. Let $M = (W,\Pi ,V)$, where $\Pi = (Y_k, \dots , Y_0)$.

Define $\texttt{M}' = (W, V)$, which is a model for $\textsf{ConWON}$. From the sequence $(Y_0, \dots , K_k)$, define a context $\texttt{C}= (X_0, \dots , X_k)$ for $\texttt{M}'$ as follows:

• $X_0 = Y_0 \cup \dots \cup Y_k$

• $X_1 = Y_1 \cup \dots \cup Y_k$

  $\vdots$

• $X_k = Y_k$

Assume that $|\alpha | = \emptyset$. Let $x \in W$. Then, $\texttt{M}',W,x \Vdash [\alpha ] \beta$. Here, W indicates a special context. Then, $[\alpha ] \beta$ is satisfiable in $\textsf{ConWON}$.

Assume that $|\alpha | \ne \emptyset$. As $Y_0 \cup \dots \cup Y_k = W$, there is $Y_i$ such that $|\alpha | \cap Y_i \ne \emptyset$. Let l be the greatest number such that $l \le k$ and $|\alpha | \cap Y_l \ne \emptyset$. Note that $M,w \Vdash \alpha \rhd \beta$. Then, $|\alpha | \cap Y_l \subseteq |\beta |$. By Lemma 10, $|\alpha | \cap X_0 \cap \dots \cap X_l= |\alpha | \cap Y_l$ and l is the greatest number such that $|\alpha | \cap X_1 \cap \dots \cap X_l\ne \emptyset$. Then, $\Vert \texttt{C} + \alpha \Vert = |\alpha | \cap X_1 \cap \dots \cap X_l$. Then, $\Vert \texttt{C} + \alpha \Vert \subseteq |\beta |$. Let $x \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M}',\texttt{C},x \Vdash [\alpha ] \beta$. Then, $[\alpha ] \beta$ is satisfiable in $\textsf{ConWON}$.

Assume that $[\alpha ] \beta$ is satisfiable in $\textsf{ConWON}$. Then, $[\alpha ] \beta$ is true at a contextualized pointed model $(\texttt{M},\texttt{C},w)$ for $\textsf{ConWON}$. Then, $\Vert \texttt{C} + \alpha \Vert \subseteq |\beta |$. Let $\texttt{M}= (W,V)$. Define $\texttt{C}' = \texttt{C} \oplus W$. Let $\texttt{C}' = (X_0, \dots , X_k)$. Note that $X_0 = W$. It can be verified that $\Vert \texttt{C} + \alpha \Vert = \Vert \texttt{C}' + \alpha \Vert$. Then, $\Vert \texttt{C} + \alpha ' \Vert \subseteq |\beta |$.

Define a sequence $(Y_0, \dots , Y_k)$ as follows:

• $Y_0 = X_0-X_1$

• $Y_1 = X_0 \cap X_1 - X_2$

  $\vdots$

• $Y_k = X_0 \cap \dots \cap X_k$

By Lemma 11, $Y_0, \dots , Y_k$ are pairwise disjoint and $Y_0 \cup \dots \cup Y_k = W$. Let $\Pi = (Y_k, \dots , Y_0)$. Define $M' = (W,\Pi ,V)$, which is a pseudo sphere model for ${\textsf {VN}}$.

Assume that $|\alpha | = \emptyset$. Then, there is no $Y_i$ in $\Pi$ such that $|\alpha | \cap Y_i \ne \emptyset$. Let $x \in W$. Then, $M',x \Vdash \alpha \rhd \beta$. Then, $\alpha \rhd \beta$ is satisfiable in ${\textsf {VN}}$.

Assume that $|\alpha | \ne \emptyset$. Then, $\Vert \texttt{C}' + \alpha \Vert \ne \emptyset$. Let $\Vert \texttt{C}' + \alpha \Vert = |\alpha | \cap X_0 \cap \dots \cap X_l$. By Lemma 11, $|\alpha | \cap X_0 \cap \dots \cap X_l= |\alpha | \cap Y_l$ and l is the greatest number such that $l \le k$ and $|\alpha | \cap Y_l \ne \emptyset$. Note that $\Vert \texttt{C} + \alpha ' \Vert \subseteq |\beta |$. Then, $|\alpha | \cap Y_l \subseteq |\beta |$. Let $x \in W$. Then, $M',x \Vdash \alpha \rhd \beta$. Then, $\alpha \rhd \beta$ is satisfiable in ${\textsf {VN}}$. $\square$

B Proofs About Expressivity with Respect to Validity

Lemma 12

Fix a model $\texttt{M}$.

• 1. $\Vert \texttt{C} + \alpha \Vert = \emptyset$ if and only if $|\alpha | = \emptyset$.

• 2. $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$, where $|\alpha \wedge \beta | \ne \emptyset$.

• 3. $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \theta + \beta \Vert$, where $|\alpha \wedge \beta | = \emptyset$.

• 4. $\Vert \texttt{C} + \alpha \Vert \subseteq |\alpha |$.

This result is easy to show.

Lemma 13

Let $\chi$ be a closed formula. Assume that $\Vert \texttt{C} + \alpha \Vert \ne \emptyset$. Then, for all w and u of $\texttt{M}$, $\texttt{M}, \texttt{C}, w \Vdash [\alpha ] \chi$ if and only if $\texttt{M}, \texttt{C} + \alpha , u \Vdash \chi$.

Proof

Assume that $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ]\chi$. Then, $\texttt{M}, \texttt{C} + \alpha , x \not \Vdash \chi$ for some $x \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M}, \texttt{C} + \alpha , u \not \Vdash \chi$. Assume that $\texttt{M}, \texttt{C} + \alpha , u \not \Vdash \chi$. Let $x \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M}, \texttt{C} + \alpha , x \not \Vdash \chi$. Then, $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ]\chi$. $\square$

Lemma 4

The following formulas are valid, where $\alpha , \beta$ and $\gamma$ are in $\Phi _\textsf{PL}$:

• 1. $[\alpha ] (\phi \wedge \psi ) \leftrightarrow ([\alpha ] \phi \wedge [\alpha ] \psi )$

• 2. $[\alpha ] (\phi \vee \chi ) \leftrightarrow ([\alpha ] \phi \vee [\alpha ] \chi )$, where $\chi$ is a closed formula

• 3. $[\alpha ] [\beta ] \gamma \leftrightarrow (\textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge [\alpha \wedge \beta ] \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{A}(\beta \rightarrow \gamma ))))$

• 4. $[\alpha ] \langle \beta \rangle \gamma \leftrightarrow (\textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge \langle \alpha \wedge \beta \rangle \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{E}(\beta \wedge \gamma ))))$

Proof

1. This item is easy.

2. Assume that $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ] \phi \vee [\alpha ] \chi$. Then, $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ] \phi$ and $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ] \chi$. Then, there is $u \in \Vert \texttt{C} + \alpha \Vert$ such that $\texttt{M}, \texttt{C} + \alpha , u \not \Vdash \phi$ and there is $v \in \Vert \texttt{C} + \alpha \Vert$ such that $\texttt{M}, \texttt{C} + \alpha , v \not \Vdash \chi$. As $\chi$ is a closed formula, $\texttt{M}, \texttt{C} + \alpha , u \not \Vdash \chi$. Then, $\texttt{M}, \texttt{C} + \alpha , u \not \Vdash \phi \vee \chi$. Then, $\texttt{M}, \texttt{C}, w \not \Vdash [\alpha ] (\phi \vee \chi )$. The other direction is easy.

3. Assume that $\texttt{M}, \texttt{C}, w \not \Vdash \textbf{E}\alpha$. Then, both sides of the equivalence hold at $(\texttt{M},\texttt{C},w)$ trivially.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha$ and $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}(\alpha \wedge \beta )$. Note that $\Vert \texttt{C} + \alpha \Vert \ne \emptyset$ by item 1 in Lemma 12. Also note that $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$ by Item 2 in Lemma 12.

Assume that $\texttt{M}, \texttt{C}, w \Vdash [\alpha ][\beta ] \gamma$. Let $u \in \Vert \texttt{C} + \alpha \Vert$. By Lemma 13, $\texttt{M}, \texttt{C} + \alpha , u \Vdash [\beta ]\gamma$. Then, for every $v \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$, $\texttt{M},(\texttt{C} + \alpha ) + \beta ,v \Vdash \gamma$. Then, for every $v \in \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$, $\texttt{M},\texttt{C} + (\alpha \wedge \beta ),v \Vdash \gamma$. Then, $\texttt{M}, \texttt{C}, w \Vdash [\alpha \wedge \beta ] \gamma$. Then, $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge [\alpha \wedge \beta ] \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{A}(\beta \rightarrow \gamma )))$.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge [\alpha \wedge \beta ] \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{A}(\beta \rightarrow \gamma )))$. Then, $\texttt{M}, \texttt{C}, w \Vdash [\alpha \wedge \beta ] \gamma$. Then, for every $u \in \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$, $\texttt{M},\texttt{C} + (\alpha \wedge \beta ),u \Vdash \gamma$. Then, for every $u \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$, $\texttt{M},(\texttt{C} + \alpha ) + \beta ,u \Vdash \gamma$. Let $v \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M},\texttt{C} + \alpha ,v \Vdash [\beta ]\gamma$. By Lemma 13, $\texttt{M},\texttt{C},w \Vdash [\alpha ][\beta ]\gamma$.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha$ and $\texttt{M}, \texttt{C}, w \not \Vdash \textbf{E}(\alpha \wedge \beta )$. Note that $\Vert \texttt{C} + \alpha \Vert \ne \emptyset$ by item 1 in Lemma 12. Also note that $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \theta + \beta \Vert$ by Item 3 in Lemma 12.

Assume that $\texttt{M}, \texttt{C}, w \Vdash [\alpha ][\beta ] \gamma$. Let $u \in \Vert \texttt{C} + \alpha \Vert$. By Lemma 13, $\texttt{M}, \texttt{C} + \alpha , u \Vdash [\beta ]\gamma$. Then, for every $v \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$, $\texttt{M},(\texttt{C} + \alpha ) + \beta ,v \Vdash \gamma$. Then, for every $v \in \Vert \theta + \beta \Vert$, $\texttt{M},\theta + \beta ,v \Vdash \gamma$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{A}(\beta \rightarrow \gamma )$. Then, $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge [\alpha \wedge \beta ] \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{A}(\beta \rightarrow \gamma )))$.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge [\alpha \wedge \beta ] \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{A}(\beta \rightarrow \gamma )))$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{A}(\beta \rightarrow \gamma )$. Then, for every $u \in \Vert \theta + \beta \Vert$, $\texttt{M},\theta + \beta ,u \Vdash \gamma$. Then, for every $u \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$, $\texttt{M},(\texttt{C} + \alpha ) + \beta ,u \Vdash \gamma$. Let $v \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M},\texttt{C} + \alpha ,v \Vdash [\beta ]\gamma$. By Lemma 13, $\texttt{M}, \texttt{C}, w \Vdash [\alpha ][\beta ] \gamma$.

4. Assume that $\texttt{M}, \texttt{C}, w \not \Vdash \textbf{E}\alpha$. Then, both sides of the equivalence hold at $(\texttt{M},\texttt{C},w)$ trivially.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha$ and $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}(\alpha \wedge \beta )$. Note that $\Vert \texttt{C} + \alpha \Vert \ne \emptyset$ by item 1 in Lemma 12. Also note that $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$ by Item 2 in Lemma 12.

Assume that $\texttt{M}, \texttt{C}, w \Vdash [\alpha ]\langle \beta \rangle \gamma$. Let $u \in \Vert \texttt{C} + \alpha \Vert$. By Lemma 13, $\texttt{M}, \texttt{C} + \alpha , u \Vdash \langle \beta \rangle \gamma$. Then, there is $v \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$ such that $\texttt{M},(\texttt{C} + \alpha ) + \beta ,v \Vdash \gamma$. Then, $v \in \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$ and $\texttt{M},\texttt{C} + (\alpha \wedge \beta ),v \Vdash \gamma$. Then, $\texttt{M}, \texttt{C}, w \Vdash \langle \alpha \wedge \beta \rangle \gamma$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge \langle \alpha \wedge \beta \rangle \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{E}(\beta \wedge \gamma )))$.

Assume that $\texttt{M},\texttt{C},w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge \langle \alpha \wedge \beta \rangle \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{E}(\beta \wedge \gamma )))$. Then, $\texttt{M}, \texttt{C}, w \Vdash \langle \alpha \wedge \beta \rangle \gamma$. Then, there is $u \in \Vert \texttt{C} + (\alpha \wedge \beta ) \Vert$ such that $\texttt{M},\texttt{C} + (\alpha \wedge \beta ),u \Vdash \gamma$. Then, $u \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$ and $\texttt{M},(\texttt{C} + \alpha ) + \beta ,u \Vdash \gamma$. Let $v \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M},\texttt{C} + \alpha ,v \Vdash \langle \beta \rangle \gamma$. By Lemma 13, $\texttt{M}, \texttt{C}, w \Vdash [\alpha ]\langle \beta \rangle \gamma$.

Assume that $\texttt{M}, \texttt{C}, w \Vdash \textbf{E}\alpha$ and $\texttt{M}, \texttt{C}, w \not \Vdash \textbf{E}(\alpha \wedge \beta )$. Note that $\Vert \texttt{C} + \alpha \Vert \ne \emptyset$ by item 1 in Lemma 12. Also note that $\Vert (\texttt{C} + \alpha ) + \beta \Vert = \Vert \theta + \beta \Vert$ by Item 3 in Lemma 12.

Assume that $\texttt{M}, \texttt{C}, w \Vdash [\alpha ]\langle \beta \rangle \gamma$. Let $u \in \Vert \texttt{C} + \alpha \Vert$. By Lemma 13, $\texttt{M}, \texttt{C} + \alpha , u \Vdash \langle \beta \rangle \gamma$. Then, there is $v \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$ such that $\texttt{M},(\texttt{C} + \alpha ) + \beta ,v \Vdash \gamma$. Then, $v \in \Vert \theta + \beta \Vert$ and $\texttt{M},\theta + \beta ,v \Vdash \gamma$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{E}(\beta \wedge \gamma )$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge \langle \alpha \wedge \beta \rangle \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{E}(\beta \wedge \gamma )))$.

Assume that $\texttt{M},\texttt{C},w \Vdash \textbf{E}\alpha \rightarrow ((\textbf{E}(\alpha \wedge \beta ) \wedge \langle \alpha \wedge \beta \rangle \gamma ) \vee (\lnot \textbf{E}(\alpha \wedge \beta ) \wedge \textbf{E}(\beta \wedge \gamma )))$. Then, $\texttt{M},\texttt{C},w \Vdash \textbf{E}(\beta \wedge \gamma )$. Then, there is $u \in \Vert \theta + \beta \Vert$ such that $\texttt{M},\theta + \beta ,u \Vdash \gamma$. Then, $u \in \Vert (\texttt{C} + \alpha ) + \beta \Vert$ and $\texttt{M},(\texttt{C} + \alpha ) + \beta ,u \Vdash \gamma$. Let $v \in \Vert \texttt{C} + \alpha \Vert$. Then, $\texttt{M},\texttt{C} + \alpha ,v \Vdash \langle \beta \rangle \gamma$. By Lemma 13, $\texttt{M}, \texttt{C}, w \Vdash [\alpha ]\langle \beta \rangle \gamma$. $\square$

Theorem 5

There is an effective function $\sigma$ from $\Phi _{\textsf{ConWON}}$ to $\Phi _{\textsf{ConWON}\text {-}\textsf {1}}$ such that for every $\phi \in \Phi _{\textsf{ConWON}}$, $\phi \leftrightarrow \sigma (\phi )$ is valid.

Proof

We define the modal depth of formulas of $\Phi _{\textsf{ConWON}}$ with respect to $[\cdot ]$ in the usual way.

Pick a formula $\phi$ in $\Phi _{\textsf{ConWON}}$. Repeat the following steps until we cannot proceed.

• Pick a sub-formula $[\alpha ] \psi$ of $\phi$ whose modal depth with respect to $[\cdot ]$ is 2 if $\phi$ has such a sub-formula.

• Transform $\psi$ to $\chi _1 \wedge \dots \wedge \chi _n$, where all $\chi _i$ is in the form of $(\beta _1 \vee \dots \vee \beta _k) \vee ([\gamma _1] \lambda _1 \vee \dots \vee [\gamma _l] \lambda _l) \vee (\langle \eta _1 \rangle \theta _1 \vee \dots \vee \langle \eta _m \rangle \theta _m)$, where all $\beta _i, \gamma _i, \lambda _i, \eta _i$ and $\theta _i$ are in $\Phi _\textsf{PL}$.

• Note that $[\alpha ] \psi \leftrightarrow ([\alpha ]\chi _1 \wedge \dots \wedge [\alpha ]\chi _n)$ is valid by Item 1 in Lemma 4. Repeat the following steps until we cannot proceed:

  • From $[\alpha ]\chi _1 \wedge \dots \wedge [\alpha ]\chi _n$, pick a conjunct $[\alpha ] \chi _i = [\alpha ] ((\beta _1 \vee \dots \vee \beta _k) \vee ([\gamma _1] \lambda _1 \vee \dots \vee [\gamma _l] \lambda _l) \vee (\langle \eta _1 \rangle \theta _1 \vee \dots \vee \langle \eta _m \rangle \theta _m))$.

  • By Item 2 in Lemma 4, $[\alpha ] \chi _i \leftrightarrow \xi$ is valid in $\textsf{ConWON}$, where $\xi = [\alpha ](\beta _1 \vee \dots \vee \beta _k) \vee ([\alpha ][\gamma _1] \lambda _1 \vee \dots \vee [\alpha ][\gamma _l] \lambda _l) \vee ([\alpha ]\langle \eta _1 \rangle \theta _1 \vee \dots \vee [\alpha ]\langle \eta _m \rangle \theta _m)$. In the ways specified by Items 3 and 4 in Lemma 4, transform $\xi$ to $\xi '$, whose modal depth with respect to $[\cdot ]$ is 1.

  • Replace $[\alpha ] \chi _i$ by $\xi '$ in $[\alpha ]\chi _1 \wedge \dots \wedge [\alpha ]\chi _n$.

Define $\sigma (\phi )$ as the result. It is easy to see that $\sigma (\phi )$ is in $\Phi _{\textsf{ConWON}\text {-}\textsf {1}}$ and $\phi \leftrightarrow \sigma (\phi )$ is valid. $\square$