A Spatial Logic with Time and Quantifiers

Abstract

Spatial logics are formalisms for expressing topological properties of structures based on geometrical entities and relations. In this paper we consider SLCS, the Spatial Logic for Closure Spaces, recently used for describing features of images and video frames. We extend SLCS in two directions. We first introduce first-order quantifiers, ranging on both individuals and atomic propositions. We then equip the logic with temporal operators, and provide a linear-time semantics over finite traces. The resulting formalism allows to state properties about geometrical entities whose attributes change along time. For both extensions, we prove the equivalence of their operational semantics with a denotational one.

Supported by University of Pisa project PRA_2022_99 “FM4HD”, MUR project PRIN 20228KXFN2 “STENDHAL”, CNR (Italy) and SRNSFG (Georgia) bilateral project CNR-22-010 “Model Checking for Polyhedral Logic”, and European Union - Next Generation EU - MUR project PNRR PRI ECS00000017 PRR.AP008.003 “THE - Tuscany Health Ecosystem”. The authors thank Diego Latella and Mieke Massink for fruitful discussions on spatio-temporal logics and their applications.

A Some Hints from Quantified Modal Algebras

This appendix recalls basic notions of (quantified) modal and conjugate algebras, which inspired the way we provided our logics with a denotational semantics.

1.1 A.1 Boolean and Modal Algebra

We recall the basics of boolean and modal algebras and discuss some axioms.

Definition 15

A Boolean algebra $\mathcal {A}$ is a 6-tuple $\langle A, \vee , 0, \wedge , 1, \lnot \rangle$ such that the triples $\langle A, \vee , 0 \rangle$ and $\langle A, \wedge , 1 \rangle$ are ACI (associative, commutative and with identity) monoids satisfying the usual distributivity and negation rules.

The usual negation rule means that $a \vee \lnot a = 1$ and $a \wedge \lnot a = 0$. A Boolean algebra is equivalently described as a complemented distributive lattice. In particular $a \vee b = a$ iff $a \wedge b = b$ and $a \le b$ iff $\lnot b \le \lnot a$. The partial order on A is induced by $a \le b$ if $a \vee b = b$, so that 0 is bottom and 1 is top. A well-known example of such a structure is the boolean algebra of powersets of a set, that gives rise to the algebra $\mathcal {A} = \langle \mathcal {P}(A), \cup , \emptyset , \cap , A, ^c \rangle$. We say that a Boolean algebra $\mathcal {A}$ is complete if every subset of A has a least upper bound (LUB).

Definition 16

A modal algebra $\mathcal {M}$ is a 7-tuple $\langle A, \vee , 0, \wedge , 1, \lnot , \lozenge \rangle$ such that the 6-tuple $\langle A, \vee , 0, \wedge , 1, \lnot \rangle$ is a Boolean algebra and $\lozenge : A \rightarrow A$ is a function satisfying $\lozenge 0 = 0$ and $\lozenge (a \vee b) = \lozenge a \vee \lozenge b$.

A modal algebra is complete if the underlying Boolean algebra is complete and $\lozenge (\bigvee _i a_i) = \bigvee _i \lozenge a_i$ for any i.

Monotonicity of $\lozenge$ is implied by the second axiom, which also yields that $\lozenge 1 = 1$. If $\mathcal {M}$ is finite (i.e. the set A is finite), then $\mathcal {M}$ is obviously complete.

We define the usual derived operator $\square a = \lnot \lozenge \lnot a$. Note that $\square 1 = 1$, $\square a \wedge b = \square a \wedge \square b$, and $\square$ is monotone with respect to the induced partial order

Remark 9

Modal algebras provide denotational models for propositional modal logics. Assuming a semantical function $[\cdot ]$ mapping a formula into an element of the modal algebra chosen as model, the formula $\phi$ is valid in the logics if $[\phi ] = 1$. Also, note that $[\phi \implies \rho ] = 1$ is equivalent to prove that $[\phi ] \le [\rho ]$, assuming that $[\cdot ]$ preserves the operators $\lnot$ and $\vee$ (hence, all the operators).

It is immediate that the axiom K, i.e. $\square (\phi \implies \rho ) \implies (\square \phi \implies \square \rho )$, holds in any modal algebra. By Boolean manipulation the formula is equivalent to $(\square \phi \wedge (\square (\phi \implies \rho )) \implies \square \rho$. Hence, it suffices to prove that in a modal algebra it holds $(\square a \wedge \square (a \implies b)) \le \square b$. Due to the distributivity of $\square$, this is equivalent to prove that $\square (a \wedge b) \le \square b$, which holds by monotonicity.

Also, note that what is called the necessitation rule for modal logics based on K holds, since $a = 1$ implies $\square a = \square 1 = 1$.

Definition 17

Let $\mathcal {M}$ be a modal algebra whose partial order is $\le$. Its necessity and iteration axioms are $M = a \le \lozenge a$, $4 = \lozenge \lozenge a \le \lozenge a$, and $B = a \le \square \lozenge a$.

Axioms are given in terms of the $\lozenge$ operator, but they can be rewritten using the $\square$ operator, with the reversed inequality. Hence, M and 4 can be equivalently expressed in terms of $\square$ as $\square a \le a$ and $\square a \le \square \square a$, respectively, as well as B is equivalent to $\lozenge \square a \le a$. Note that assuming M and 4 implies that $\lozenge \lozenge a = \lozenge a$.

Remark 10

Axioms M, 4, and B are known as reflexivity, transitivity, and symmetry axioms, respectively, since for modal algebras arising from Kripke frames those are the properties imposed on the underlying relation [18]. Modal algebras satisfying M and 4 are called closure algebras and are models of S4, while those satisfying all three axioms are called monadic algebras and are models of S5.

1.2 A.2 Quantified Modal Algebras

While modal algebras represent models for propositional modal logics, moving to first order quantification require the introduction of cylindric operators, a well-known abstraction for existential quantifiers  [21].

1.2.1 Cylindric Operators.

We fix a Boolean algebra $\mathcal {A}$ and a set of variables V.

Definition 18 (cylindric Boolean algebras)

A cylindric operator $\exists$ over $\mathcal {A}$ and V is a family of monotone operators $\exists _x : A \rightarrow A$ indexed by elements in V such that for all $a, b \in A$ and $x, y \in V$ it holds $a \le \exists _x a$, $\exists _x \exists y a = \exists _y \exists _x a$, and $\exists _x (a \wedge \exists _x b) = \exists _x a \wedge \exists _x b$.

Let $a \in A$. The support of a is the set of variables $sv(a) = \{x \mid \exists _x a \ne a \}$.

An element of the algebra stands for a formula possibly containing free variables. We restrict our attention to elements a with finite support, i.e., such that sv(a) is finite: this means that a is a formula containing a finite set of variables.

Now we fix a modal algebra $\mathcal {M}$ with underlying Boolean algebra $\mathcal {A}$.

Definition 19 (cylindric modal algebras)

A cylindric operator $\exists$ over $\mathcal {M}$ and V is a cylindric operator over $\mathcal {A}$ and V such that for all $a \in A$ and $x \in V$ it holds $\exists _x \lozenge a = \lozenge \exists _x a$.

Remark 11

The inequalities $\exists _x \lozenge a \ge \lozenge \exists _x a$ and $\exists _x \lozenge a \le \lozenge \exists _x a$ are known as Barcan formula and converse Barcan formula in the literature [6]. The axiom in Definition 19 is thus only one of the possible choices, and it boils down to require what is called “domain preservation”, namely, the domain is preserved along the evolution. Instead, $\exists _x \lozenge a \le \lozenge \exists _x a$ witnesses a possible domain restriction, while analogously we may have a domain increase with the reverse $\exists _x \lozenge a \ge \lozenge \exists _x a$.

The axiom implies $sv(\lozenge a) \subseteq sv(a)$, since $\exists _x a = a$ implies $\exists _x \lozenge a = \lozenge \exists _x a = \lozenge a$.

1.2.2 Soft Modal Algebras.

We now show how to build a modal algebra that admits cylindric operators. Let us fix a modal algebra $\mathcal {M}$ with underlying Boolean algebra $\mathcal {A}$ and a set of variables V.

Proposition 8

Let D be a set of elements, F the set of functions $\eta : V \rightarrow D$, and $\varGamma$ the set of functions $\gamma : F \rightarrow A$. The 7-tuple $\mathcal {F} = \langle \varGamma , \vee , 0, \wedge , 1, \lnot , \lozenge \rangle$ is a modal algebra, whose operators and constants are lifted from $\mathcal {M}$. If $\mathcal {M}$ is complete, so is $\mathcal {F}$.

For example, 0 in $\mathcal {F}$ is the function such that $0(\eta ) = 0$ for all $\eta$, and so on. In particular, note that $\gamma _1 \le \gamma _2$ means that $\gamma _1(\eta ) \le \gamma _2(\eta )$ for all $\eta$.

Let us now additionally fix a set D, and given $\eta : V \rightarrow D$, we denote as $\eta [^d/_x]$ the function coinciding with $\eta$ except for x, where $\eta [^d/_x](x) = d$.

Proposition 9

Let D be finite. The cylindric operator $\exists$ over $\mathcal {F}$ and V is defined as $(\exists _x \gamma )(\eta ) = \bigvee _{d \in D} \gamma (\eta [^d/_x])$.

If $\mathcal {M}$ is complete, the finiteness of D can be dropped.

Remark 12

By definition, $\exists _x \gamma = \gamma$ means that for all $\eta$ we have $\bigvee _{d \in D} \gamma (\eta [^d/_x]) = \gamma (\eta )$, which is equivalent to say that for all d we have $\gamma (\eta [^d/_x]) = \gamma (\eta )$. Intuitively, if $\gamma$ represents a formula possibly containing free variables, x cannot be among them. Conversely, $x \in sv(\gamma )$ if there is a function $\eta$ and elements $b, c \in D$ such that $\gamma (\eta [^b/_x]) \ne \gamma (\eta [^c/_x])$, intuitively meaning that x does occur free in $\gamma$.

1.3 A.3 Conjugate Modal Algebras

Algebras that employ more than one modal operator are said to be multimodal. We focus here on a particular kind of such algebras, called conjugate algebras.

Definition 20

A conjugate algebra $\mathcal {D}$ is a 8-tuple $\langle A, \vee , 0, \wedge , 1, \lnot , \lozenge _1, \lozenge _2 \rangle$ such that both 7-tuples $\langle A, \vee , 0, \wedge , 1, \lnot , \lozenge _1 \rangle$ and $\langle A, \vee , 0, \wedge , 1, \lnot , \lozenge _2 \rangle$ are modal algebras and moreover it holds $a \le \square _1 \lozenge _2 a \wedge \square _2 \lozenge _1 a$.

A conjugate algebra is complete if both the underlying modal algebras are so.

What is noteworthy is a well-known characterisation via just the $\lozenge$ operators.

Lemma 3

$\mathcal {D}$ is a conjugate algebra iff it holds $\lozenge _1 a \wedge b = 0 \Leftrightarrow a \wedge \lozenge _2 b = 0$.

Remark 13

The lemma is stated by using the more standard notion of the axiom on $\lozenge$. An alternative, friendlier version is $\lozenge _1 a \le b \Leftrightarrow a \le \square _2 b$. The proof of the equivalence between the two axioms is straightforward, and it exploits the following law holding in Boolean algebras, namely $a \wedge b = 0$ iff $a \le \lnot b$.