Local Fact Change Logic

Abstract

We investigate a modal logic of model change, introducing a new operator which allows for changing the valuation at a particular state in a model. After investigating some properties of the logic, and aspects of its expressive power, we show it to be undecidable by way of a reduction using memory logic.

This chapter is in its final form and it is not submitted to publication anywhere else.

Appendix

1.1 Proof of Fact 4

Fact 8

Let $A\subseteq \mathsf {Prop}$. If $\psi \in \mathsf {Bool}(\mathsf {At}(\varphi ))$ and $\mathfrak {F}, V, s \models \psi$ and $\mathfrak {F}, V^s_A, s \models \psi$ then

$$ \mathfrak {F}, V, s \models \varphi \quad \text {iff} \quad \mathfrak {F}, V^s_A, s \models \varphi . $$

Proof

Recall that $\psi$ is of the form $\pm p_0 \wedge \dots \pm p_n$, where $p_0, \dots p_n \in \mathsf {At}(\varphi )$. Since $\psi$ has no modal operators and its evaluation only depends on the values of propositional variables in $\mathsf {At}(\varphi )$, we know $V(s) \cap \mathsf {At}(\varphi ) = V^s_A(s) \cap \mathsf {At}(\varphi )$. From this, we conclude $V(t) \cap \mathsf {At}(\varphi ) = V^s_A(t) \cap \mathsf {At}(\varphi )$ for all t. Observing that $\varphi$ also only depends on the values of propositional variables in $\mathsf {At}(\varphi )$ gives us the result. $\square$

Fact 9

iff .

Proof

For simplicity, we will omit $\mathfrak {F}$ below. We have the following equivalences:

(1)

(2)

(3)

(4)

$$\begin{aligned}&\text {iff}\quad V^s_B, s \models \lnot p \vee \lnot \psi _{V,s} \vee \varphi \text { for all }B \end{aligned}$$

(5)

$$\begin{aligned}&\text {iff}\quad \text {for all }B, \mathrm{either }V^s_B, s\not \models p~\mathrm{or }V^s_B, s\not \models \psi _{V,s}~\mathrm{or }V^s_B, s\models \varphi . \end{aligned}$$

(6)

Equation (1) follows from the definition of , and (2) via basic logic. Taking $\varphi _{V,s}$ to characterise V at s, equivalence (3) holds, and since $V\models \psi _{V,s}$ by the definition of $\psi _{V,s}$, we get (4). Equations (5) and (6) then follow from the truth definitions. In the remainder of the proof, we will establish that (6) holds iff .

($\Rightarrow$) Suppose $V^s_B, s\not \models p$ or $V^s_B, s\not \models \psi _{V,s}$ or $V^s_B, s\models \varphi$ for all B. Let $A = V(s) \cup \{p\}$. We have that $V^s_A, s \models p$ trivially, and $V^s_A, s \models \psi _{V,s}$ since $V, s \models \psi _{V,}$ and $p \notin \mathsf {At}(\psi _{V,s})$, so we must have $V^s_A, s \models \varphi$. Hence $V^s_A, s \models p \wedge \psi _{V,s} \wedge \varphi$ and so . From this it quickly follows that from the definition of .

($\Leftarrow$) Suppose . By similar reasoning to the move from (2) to (4), we know that . Thus, $V^s_C, s \models p \wedge \psi _{V,s} \wedge \varphi$ for some C. Take arbitrary B and suppose $V^s_{B}, s \models p$ and $V^s_{B}, s \models \psi _{V,s}$. Since $V^s_B$ and $V^s_C$ agree on p and $\psi _{V,s}$ at s, we know that $V^s_{B}, s \models \varphi$ iff $V^s_C, s \models \varphi$, by Fact 8. Since $V^s_C, s\models \varphi$, we have $V^s_B, s \models \varphi$. Since B was arbitrary, we can conclude that for all B, either $V^s_B, s\not \models p$ or $V^s_B, s\not \models \psi _{V,s}$ or $V^s_B, s\models \varphi$. $\square$

Fact 4

iff $\mathfrak {F}, V, s \models [+p] \varphi$, for every formula $\varphi$ and all frames $\mathfrak {F}$, valuations V and points s.

Proof

By induction on the complexity of $\varphi$. For simplicity, we will omit reference to $\mathfrak {F}$ in what follows. Let $A = V(s) \cup \{p\}$.

For $\varphi$ propositional there are two cases. If $\varphi = p$, then iff , which is the case since . It’s trivial to see that $V^s_A, s \models p$. The second case is when $\varphi = q \ne p$. In this case, we conclude that iff and . Since we know and , we conclude that iff $V,s\models q$. By the definition of truth, $V,s\models q$ iff $q\in V(s)$ iff $q\in V(s)\cup \{p\}$ (since $q\ne p$). This holds iff $V^s_A, s \models q$ iff $V, s \models [+p]q$.

Next consider the case where $\varphi = \lnot \psi$ for some $\psi$. Using Fact 9, iff . The required result follows quickly using the inductive hypothesis. Similar arguments to those given in Fact 9 can be used to establish the case for $\varphi = \psi \wedge \chi$ for some $\psi$ and $\chi$.

Now we take for some $\chi$. Using a similar argument to Fact 9, we know that iff , where $\psi _{V,s} \in \mathsf {Bool}(\mathsf {At}(\chi )\setminus \{p\})$ characterises V at s. By definition, this holds iff for some B. This immediately establishes that since we can move to $V^s_B$. Hence, . For the other direction, notice that $V^s_A, s \models p$ and $V^s_A, s \models \psi _{V,s}$ since $V, s \models \psi _{V,s}$ and $p \notin \mathsf {At}(\psi _{V,s})$.

Finally,⁴ suppose $\varphi = \lozenge \chi$. Unpacking definitions, we know that iff $V^s_B, s \models p \wedge \psi _{V,s} \wedge \lozenge \chi$ for some B, where $\psi _{V,s} \in \mathsf {Bool}(\mathsf {At}(\chi )\setminus \{p\})$ characterises V at s. So $V^s_B, s \models p \wedge \psi _{V,s}$ and by the definition of $V^s_A$ and $\psi _{V,s}$ we know $V^s_A, s \models p \wedge \psi _{V,s}$. It follows that $V^s_B$ and $V^s_A$ agree on all propositional variables in $\mathsf {At}(\lozenge \chi )$, and so by Fact 8 we can conclude that $V^s_B, s \models \lozenge \chi$ iff $V^s_A, s \models \lozenge \chi$.

Initially, we found that iff $V^s_B, s \models p \wedge \psi _{V,s} \wedge \lozenge \chi$, and now we have shown that $V^s_B, s \models \lozenge \chi$ iff $V^s_A, s \models \lozenge \chi$. From these, we establish that iff $\mathfrak {F}, V, s \models [+p]\lozenge \chi$. $\square$

1.2 Bisimulation

We can define a natural notion of bisimulation for LFC. Like the Ehrenfeucht–Fraïssé games, this is similar to the approach for ML in Areces et al. (2011).

Definition 20

Let $\mathfrak {F}_1 = (W_1, R_1)$ and $\mathfrak {F}_2 = (W_2, R_2)$.

A relation Z between $W_1 \times \mathsf {Prop}^{W_1}$ and $W_2 \times \mathsf {Prop}^{W_2}$ is a bisimulation if the following clauses hold:    

Non-empty:

$Z \ne \emptyset$.

Agree:

If $(s_1, V_1) Z (s_2, V_2)$ then $V_1(s_1) = V_2(s_2)$.

Zig:

If $(s_1, V_1) Z (s_2, V_2)$ and $R_1s_1t_1$ then there is $t_2\in A_2$ with $R_2s_2t_2$ and $(t_1, V_1) Z (s_t, V_2)$.

Zag:

If $(s_1, V_1) Z (s_2, V_2)$ and $R_2s_2t_2$ then there is $t_1\in A_1$ with $R_1s_1t_1$ and $(t_1, V_1) Z (s_t, V_2)$.

Change:

If $(s_1, V_1) Z (s_2, V_2)$ then $(s_1, {V_1}^{s_1}_A) Z (s_2, {V_2}^{s_2}_A)$ for every $A \subseteq \mathsf {Prop}$.

This is the standard definition of bisimulation, but with valuations included. This is because we must keep track not only of which state we’re in, but what the current valuation is. The last clause simply ensures that a change of valuation at bisimilar points preserves bisimilarity (assuming the valuation is changed uniformly).

Fact 5

(Correctness of bisimulation). Let $\mathfrak {F}_1 = (W_1, R_1)$ and $\mathfrak {F}_2 = (W_2, R_2)$. For every formula $\varphi \in \mathcal {L}_{\mathsf {LFC}}$, $s_1 \in W_1$ and $s_2 \in W_2$, if there is a bisimulation Z with $(s_1, V_1) Z (s_2, V_2)$ then $\mathfrak {F}_1, V_1, s_1 \models \varphi$ iff $\mathfrak {F}_2, V_2, s_2 \models \varphi$.

Proof

By induction on $\varphi$. The Boolean and $\lozenge$ cases are standard.

The final case is where . iff $\mathfrak {F}_1, {V_1}^{s_1}_A, s_1 \models \psi$ for some $A \subseteq \mathsf {Prop}$. Since $(s_1, V_1) Z (s_2, V_2)$, by Change we know $(s_1, {V_1}^{s_1}_A) Z (s_2, {V_2}^{s_2}_A)$. Applying the inductive hypothesis, $\mathfrak {F}_1, {V_1}^{s_1}_A, s_1 \models \psi$ iff $\mathfrak {F}_2, {V_2}^{s_2}_A, s_2 \models \psi$. But this holds exactly when , as required. $\square$