Heyting Mereology as a Framework for Spatial Reasoning

Abstract

In this paper it is shown that Heyting and Co-Heyting mereological systems provide a convenient conceptual framework for spatial reasoning, in which spatial concepts such as connectedness, interior parts, (exterior) contact, and boundary can be defined in a natural and intuitively appealing way. This fact refutes the wide-spread contention that mereology cannot deal with the more advanced aspects of spatial reasoning and therefore has to be enhanced by further non-mereological concepts to overcome its congenital limitations. The allegedly unmereological concept of boundary is treated in detail and shown to be essentially affected by mereological considerations. More precisely, the concept of boundary turns out to be realizable in a variety of different mereologically grounded versions. In particular, every part K of a Heyting algebra H gives rise to a well-behaved K-relative boundary operator.

Appendix (Proofs of Theorems (4.7) and (4.9))

Since there does not seem available in the literature a point-free proof of Stone’s equation (4.9) I have provided one by myself. It is elementary, but a bit involved in that it uses most of the results on Heyting algebras collected in Sect. 2.

Theorem 4.7

Let C be a Co-Heyting algebra, and a, b ∈ C. Then the following holds:

$$ {\text{bd}}({\text{a}} \wedge {\text{b}}) \vee {\text{bd}}({\text{a}} \vee {\text{b}}) = {\text{bd}}\left( {\text{a}} \right) \vee {\text{bd}}\left( {\text{b}} \right). $$

Proof

The direction ≤ is proved separately for bd(a ∧ b) and bd(a ∨ b). By definition one has

$$ \begin{aligned} {\text{bd}}({\text{a}} \wedge {\text{b}}) &= ({\text{a}} \wedge {\text{b}}) \wedge ({\text{a}} \wedge{\text{b}})^{\text{+}} \\ & = ({\text{a}} \wedge {\text{b}})\wedge ({\text{a}}^{\text{+}} \vee {\text{b}}^{\text{+}} ) \\& = ({\text{a}} \wedge {\text{b}}) \wedge{\text{a}}^{\text{+}} ) \vee ({\text{b}}^{\text{+}} \wedge{\text{b}}) \wedge {\text{a}}^{\text{+}} ) \\ & =({\text{bd}}\left( {\text{a}} \right) \wedge {\text{b}}) \vee({\text{bd}}\left( {\text{b}} \right) \wedge {\text{a}}^{\text{+}})) \le {\text{bd}}\left( {\text{a}} \right) \vee {\text{bd}}\left({\text{b}} \right). \\\end{aligned} $$

Now consider bd(a ∨ b). By definition of the boundary we obtain

$$ \begin{aligned} {\text{bd}}({\text{a}} \vee {\text{b}}) & =({\text{a}} \vee {\text{b}}) \wedge ({\text{a}} \vee{\text{b}})^{\text{+}} \\ & \le ({\text{a}} \vee {\text{b}})\wedge ({\text{a}}^{\text{+}} \wedge {\text{b}}^{\text{+}} ) \\& = ({\text{a}} \vee {\text{b}}) \wedge ({\text{a}}^{\text{+}}\wedge {\text{b}}^{\text{+}} ) \\ & = ({\text{a}} \wedge{\text{a}}^{\text{+}} \wedge {\text{b}}) \vee ({\text{b}} \wedge{\text{b}}^{\text{+}} \wedge {\text{a}}^{\text{+}} ) \\ & \le{\text{bd}}\left( {\text{a}} \right) \vee {\text{bd}}\left({\text{b}} \right). \\ \end{aligned} $$

Now let us prove the other half of (4.7), namely, bd(a) ∨ bd(b) ≤ bd(a ∧ b) ∨ bd(a ∨ b). Instead of proving it directly, for convenience we prove the dual proposition for Heyting algebras:

$$ {\text{cbd}}\left( {\text{a}} \right) \wedge {\text{cbd}}\left( {\text{b}} \right) \ge {\text{cbd}}({\text{a}} \vee {\text{b}}) \wedge {\text{cbd}}({\text{a}} \wedge {\text{b}}). $$

Spelt out in detail this amounts to

$$ (({\text{a}} \wedge {\text{b}}) \vee ({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} ) \le ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ). $$

Proving this proposition for Heyting algebras saves us a lot of asterisks, since for the proof we have to use the technical results of Sect. 2 formulated for Heyting algebras. So we get:

$$ \begin{gathered} (({\text{a}} \wedge {\text{b}}) \vee ({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} ) \le ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ) \hfill \\ \Leftrightarrow [(({\text{a}} \wedge {\text{b}}) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} )] \vee [({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} ) \, ] \, \le \, ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ). \hfill \\ \end{gathered} $$

This holds if and only if

$$ [(({\text{a}} \wedge {\text{b}}) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} )] \, \le \, ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ) $$

and

$$ [({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} )] \le ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ). $$

The inequality (8.1) is true, since clearly $({\text{a}} \wedge {\text{b}}) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}} \vee {\text{b}})^{*} ) \le ({\text{a}} \wedge {\text{b}}) \le ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} )$. So it remains to prove (8.2). Due to the de Morgan law (2.4)(6) valid for Heyting algebras this is equivalent to

$$ [({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \le ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} ). $$

Due to (2. 4)(3) is equivalent to

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \Rightarrow ({\text{a}} \vee {\text{a}}^{*} ) \wedge ({\text{b}} \vee {\text{b}}^{*} )] = 1. $$

By (2.4)(2) is equivalent to

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \Rightarrow ({\text{a}} \vee {\text{a}}^{*} )] = 1 $$

(8.1a)

and

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}}) \vee ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \Rightarrow (({\text{b}} \vee {\text{b}}^{*} )] = 1. $$

(8.1b)

Both equations are symmetric in a and b. Hence it is sufficient to prove one, say (8.5a). By applying the distributivity law, we obtain that (8.5b) is equivalent to

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}})] \vee [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \Rightarrow ({\text{a}} \vee {\text{a}}^{*} )] = 1. $$

(8.2)

Applying once again (2.4)(1) this is equivalent

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}})] \vee [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \le ({\text{a}} \vee {\text{a}}^{*} )]. $$

(8.3)

But clearly $[[( 1 {\text{a}} \wedge {\text{b}})^{*} ) \wedge ({\text{a}}^{*} \wedge {\text{b}}^{*} ))] \le ({\text{a}} \vee {\text{a}}^{*} )]$. Hence there is only left to prove

$$ [[({\text{a}} \wedge {\text{b}})^{*} ) \wedge (({\text{a}} \vee {\text{b}})] \le ({\text{a}} \vee {\text{a}}^{*} )] $$

(8.4)

Applying distributivity to the left side we obtain that (8.8) is equivalent to

$$ [({\text{a}} \wedge {\text{b}})^{*} ) \wedge {\text{a}}) \vee ({\text{a}} \wedge {\text{b}})^{*} ) \wedge {\text{b}})] \le ({\text{a}} \vee {\text{a}}^{*} ). $$

Obviously $[({\text{a}} \wedge {\text{b}})^{*} ) \wedge {\text{a}}) \le ({\text{a}} \vee {\text{a}}^{*} )$, and so we are left with $({\text{a}} \wedge {\text{b}})^{*} ) \wedge {\text{b}})] \le ({\text{a}} \vee {\text{a}}^{*} )$. Clearly this is implied by proving $({\text{a}} \wedge {\text{b}})^{*} ) \wedge {\text{b}})] \le {\text{a}}^{*}$. By the definition of ⇒ this holds if and only if $({\text{a}} \wedge {\text{b}})^{*} \le ({\text{b}} \Rightarrow {\text{a}}^{*} )$. Taking into account that ${\text{a}}^{*}$ is just (a ⇒ 0) one has to prove that (a ∧ b) ⇒ 0 ≤ (b ⇒ (a ⇒ 0), since (b ⇒ (a ⇒ 0) = (a ∧ b) ⇒ 0. But this true by (Borceux 3, Proposition 1.2.15, p. 13). This clinches the proof. ♦

Corrollary

Let (X, OX) be a topological space, and a, b ∈ OX. Then

$$ {\text{bd}}({\text{a}} \cap {\text{b}}) \cup {\text{bd}}({\text{a}} \cup {\text{b}}) = {\text{bd}}\left( {\text{a}} \right) \cup {\text{bd}}\left( {\text{b}} \right). $$

Proof

Apply (4.7) to the closed sets Ca and Cb and observe that bd(a) = bd(Ca) and bd(b) = bd(Cb).♦

Theorem 4.9

(A.H. Stone) Let (X, OX) be a topological space and and a, b ∈ PX. Then the following formula holds:

$$ {\text{bd}}({\text{a}} \cap {\text{b}}) \cup ({\text{bd}}({\text{a}} \cup {\text{b}})) \cup ({\text{bd}}\left( {\text{a}} \right) \cap {\text{bd}}\left( {\text{b}} \right)) = {\text{bd}}\left( {\text{a}} \right) \cup {\text{bd}}\left( {\text{b}} \right). $$

Proof

We first prove the inequality ⊆ . This is done separately for each of the three components of the left side. Trivially, one has (bd(a) ∩ bd(b)) ⊆ bd(a) ∪ bd(b). Hence we have to deal only with the first two factors of the right side. By definition of the boundary operator, for bd(a ∩ b) one obtains:

$$ \begin{aligned} {\text{bd}}({\text{a}} \cap {\text{b}}) & = {\text{cl}}({\text{a}} \cap {\text{b}}) \cap ({\text{cl}}({\mathbf{C}}({\text{a}} \cap {\text{b}})) \subseteq ({\text{cl(a)}} \cap {\text{cl(b)}}) \cap ({\text{cl(}}{\mathbf{C}}{\text{a)}} \cup {\text{cl(}}{\mathbf{C}}{\text{b)}}) \\ & = ({\text{cl(a)}} \cap {\text{cl(b)}} \cap {\text{cl(}}{\mathbf{C}}{\text{a)}}) \cup ({\text{cl(a)}}) \cap {\text{cl(b)}} \cap ({\text{cl(}}{\mathbf{C}}{\text{b)}}) \\ & = ({\text{bd(a)}} \cap {\text{cl(b)}}) \cup ({\text{bd(b)}} \cap ({\text{cl(a)}}) \\ & \subseteq ({\text{bd(a)}} \cup {\text{bd(b)}}). \\ \end{aligned} $$

To prove the inclusion of bd(a ∪ b) one argues in a similar manner:

$$ \begin{aligned} {\text{bd}}({\text{a}} \cup {\text{b}}) & = {\text{cl}}({\text{a}} \cup {\text{b}}) \cap {\text{cl}}({\mathbf{C}}({\text{a}} \cup {\text{b}})) \\ & = ({\text{cl}}\left( {\text{a}} \right) \cup {\text{cl}}\left( {\text{b}} \right)) \cap {\text{cl}}({\mathbf{C}}{\text{a}} \cap {\mathbf{C}}{\text{b}})) \\ & \subseteq ({\text{cl}}\left( {\text{a}} \right) \cup {\text{cl}}\left( {\text{b}} \right)) \cap \left( {{\text{cl}}\left( {{\mathbf{C}}{\text{a}}} \right)} \right) \cap {\text{cl}}\left( {{\mathbf{C}}{\text{b}}} \right)) \\ & = ({\text{bd}}\left( {\text{a}} \right) \cap {\text{cl}}\left( {{\mathbf{C}}{\text{b}}} \right)) \cup ({\text{bd}}\left( {\text{b}} \right) \cap \left( {{\text{cl}}\left( {{\mathbf{C}}{\text{a}}} \right)} \right) \\ & \subseteq {\text{bd}}\left( {\text{a}} \right) \cup {\text{bd}}\left( {\text{b}} \right). \\ \end{aligned} $$

This proves the first half of the equation. To prove the opposite direction, first note that the equation is symmetric in a and b, hence it is sufficient to prove

$$ {\text{bd(a)}} \subseteq {\text{bd}}({\text{a}} \cap {\text{b}}) \cup ({\text{bd}}({\text{a}} \cap {\text{b}})) \cup ({\text{bd(a)}} \cap {\text{bd(b)}}). $$

Assume α ∈ bd(a). Then α ∈ int(b) or α ∈ bd(b) or α ∈ int(Cb) (cf. Christenson and Voxman (1997), p. 29) This trichotomy is strict and exhaustive, i.e. int(b) ∪ bd(b) ∪ int(Cb) = X and its factors are disjoint.

Assume α ∈ int(b). Then there is an open neighborhood U(α) ⊆ b. By definition of a boundary point every open neighborhood V(α) intersects nontrivially both with a and Ca. We are going to show that in this case α ∈ bd(a ∩ b).

$$ \begin{aligned} {\text{bd}}({\text{a}} \cap {\text{b}}) & = {\text{cl}}(({\text{a}} \cap {\text{b}}) \cap {\text{cl}}({\mathbf{C(}}{\text{a}} \cap {\text{b}})). \\ & = {\text{cl}}(({\text{a}} \cap {\text{b}}) \cap {\text{cl}}({\mathbf{C}}{\text{a}} \cup {\mathbf{C}}{\text{b}}). \\ & = ({\text{cl}}(({\text{a}} \cap {\text{b}}) \cap {\text{cl}}\left( {{\mathbf{C}}{\text{a}}} \right)) \cup ({\text{cl}}(({\text{a}} \cap {\text{b}}) \cap {\text{cl}}\left( {{\mathbf{C}}{\text{b}}} \right)). \\ \end{aligned} $$

Since α ∈ bd(a) and bd(a) = bd(Ca) clearly α ∈ cl(Ca). Hence it is sufficient to show that α ∈ cl((a ∩ b). By definition of cl this is the case if and only if every open neighborhood V(α) of α has a non-empty intersection with (a ∩ b). Since α ∈ int(b) one may assume V(α) ⊆ b. Since V(α) ∩ a ≠ Ø since α ∈ bd(a) we obtain V(α) ∩ a ∩ b ≠ Ø, i.e. α is an element of cl(a ∩ b).

Now assume α ∈ int(Cb). We are going to show that this entails α ∈ bd(a ∪ b). By definition of bd we get

$$ \begin{aligned} {\text{bd}}({\text{a}} \cup {\text{b}}) & = {\text{cl}}({\text{a}} \cup {\text{b}}) \cap {\text{cl}}({\mathbf{C}}({\text{a}} \cup {\text{b}})) \\ & = ({\text{cl}}\left( {\text{a}} \right) \cup {\text{cl}}\left( {\text{b}} \right)) \cap {\text{cl}}({\mathbf{C}}{\text{a}} \cap {\mathbf{C}}{\text{b}})) \\ & = \, ({\text{cl}}\left( {\text{a}} \right) \cap {\text{cl}}({\mathbf{C}}{\text{a}} \cap {\mathbf{C}}{\text{b}})) \cup \left( {{\text{cl}}\left( {\text{b}} \right)} \right) \cap {\text{cl}}({\mathbf{C}}{\text{a}} \cap {\mathbf{C}}{\text{b}})). \\ \end{aligned} $$

Since α ∈ bd(a) clearly α ∈ cl(a). Hence in order to show that α ∈ bd(a ∪ b) it is sufficient to prove that α ∈ cl(Ca ∩ Cb)). This is true if and only if every open neighborhood V(α) of α has a non-empty intersection with Ca ∩ Cb. Since α ∈ int(Cb) we can assume V(α) ⊆ b. By assumption α ∈ bd(a) and therefore V(α) ⊆ b. V(α) ∩ a = V(α) ∩ a ∩ b ≠ Ø. Hence in this case one has α ∈ bd(a ∪ b) and therefore

$$ {\text{bd(a)}} \subseteq {\text{bd}}({\text{a}} \cap {\text{b}}) \cup ({\text{bd}}({\text{a}} \cap {\text{b}})) \cup ({\text{bd(a)}} \cap {\text{bd(b)}}). $$

The analogous inclusion for bd(b) is proved in exactly the same way. This clinches the proof of Stone’s theorem. ♦

Stone’s theorem holds for every (Boolean) closure algebra (B, ≤, c) not only for those that arise from topological spaces (X, OX). Recall that a closure algebra (B, ≤, c) is a Boolean algebra (B, ≤) endowed with an operator c on B that satisfies the algebraic version of the four Kuratowski axioms (cf. Kuratowski and Mostowski 1976, §8, 27, McKinsey and Tarski 1944, 145/146). Clearly, every topological space (X, OX) gives rise to a closure algebra (PX, ⊆, cl) and vice versa. According to Theorem (2.7) of McKinsey and Tarski (1944, 150) every closure algebra (B, ≤, c) is isomorphic to a subalgebra of a closure algebra of (PX, ⊆, ≤) of a topological space (X, OX). Thus, by embedding of (B, ≤, c) into the closure algebra (PX, ⊆, ≤) of a topological space one obtains that Stone’s equation also holds for the abstract closure algebra (B, ≤, c).