1 Introduction
Minkowski spacetime refers to the flat spacetime of special relativity, where in any inertial coordinates , light travels in straight lines at a angle to the time axis. To view this as a Kripke frame, there are at least four natural accessibility relations to chose from: reflexive and irreflexive versions of ‘can reach with a lightspeedorslower signal’ and ‘can reach with a slowerthanlight signal’. See below for the formal definitions in the twodimensional case.
For the basic modal language, Goldblatt found that, regardless of the number of spatial dimensions, both reflexive choices produce the logic (reflexivity, transitivity, confluence) [5]. Shapirovsky and Shehtman proved the irreflexive slowerthanlight logic is —transitivity, seriality, confluence, twodensity (see Section 2)—again regardless of dimension [15]. Both and are complete [13]. In contrast, Shapirovsky has shown that with irreflexive exactly lightspeed accessibility, validity is undecidable [14].
For the basic temporal language, the problems of axiomatising and determining the complexity of the validities of these frames had all been open for decades—Shehtman recommended an investigation of the twodimensional case (one time and one space dimension) in the concluding remarks of [16]. In the twodimensional case, starting with coordinates , we may rotate the axes through 45 to get coordinates , where . With these coordinates the reflexive lightspeedorslower relation , and the irreflexive slowerthanlightspeed relation , are given by
(1) 
where on the right are the usual irreflexive/reflexive orderings of the reals, and are obtained from by deleting/adding the identity, respectively.
Recently, Hirsch and Reynolds managed to show that for this twodimensional case, with either reflexive or irreflexive lightspeedorslower accessibility, the validity problem is complete [7]. However, they were unable to obtain decidability/complexity results for slowerthanlight accessibility. Indeed decidability appears in item (2) in the list of open problems at the end of their paper. In this paper we solve Hirsch and Reynolds’ problem, eventually proving the following.
Theorem 6 On the frame consisting of twodimensional Minkowski spacetime equipped with the irreflexive slowerthanlight accessibility relation, the set of validities of the basic temporal language is complete. The same is true with reflexive slowerthanlight accessibility.
The proof of Theorem 6 follows that of [7] very closely. The only additional insight needed is that all pertinent information about the behaviour of a valuation on a lightline can be captured in a finite way—see Definition 2—, despite all of a lightline’s points being mutually inaccessible.
The proof of the main result is structured as follows.

Given a fixed formula whose satisfiability is to be determined, we define the maximal consistent sets of subformulas/negated subformulas of , where consistency is with respect to the class of all temporal frames.

We define surrectangles by recording a maximal consistent set at each point of a rectangle, together with some information about the maximal consistent sets holding near but beyond the boundaries of the rectangle.

Starting in Section 3, we define biboundaries, also based on the maximal consistent sets. Biboundaries have finite specifications and the intuition for them is as a record of the information contained near the boundary of a surrectangle, plus a little from its interior. Indeed we define the biboundary determined by a given surrectangle .

We define three operations on biboundaries: joins, limits, and shuffles, and we define the set of fabricated biboundaries to be those biboundaries formed by iterating these operations, starting from certain basic biboundaries. This is all computable, and the iterative procedure must terminate, for the biboundaries are finite in number.

We show that every fabricated biboundary is given by for some surrectangle , by describing a recursive construction of . (Section 4)

Conversely, we show that for every surrectangle , the biboundary is fabricated. (Section 5)

We deduce the decidability of the validity problem over our frame.

In Section 6, we give a procedure for deciding if a biboundary is fabricated using polynomial space, and also note the satisfiability task is hard. Consequently our result that validity is decidable is refined to validity being complete.
In [6], Halpern and Shoham introduced a family of modal logics in which the entities under discussion are intervals. These logics have subsequently come to be highly influential, and their axiomatisability, decidability, and complexity extensively studied [19, 2, 9, 4, 11, 10, 3]. In Section 7, we describe how the proof of the complexity of validity for twodimensional Minkowski spacetime can be adapted to prove completeness of the temporal logic of intervals where the accessibility relation is , or its reflexive closure.
2 Preliminaries and surrectangles
We often, but not exclusively, follow the terminology and notation of [7]. We take as primitive the propositional connectives and , and modal operators and ; the usual abbreviations apply. The semantics is the usual semantics on temporal frames—Kripke frames for which the accessibility relation for is the converse of that for . An example of a formula that is valid over twodimensional slowerthanlight frames but not over irreflexive lightspeedorslower frames is the twodensity formula
asserting that if precedes both and , then there exists that is between and and between and .
Throughout, is a fixed formula whose satisfiability is to be determined. The closure of is the set of all subformulas and negated subformulas of . A maximal consistent set is a subset of that is satisfiable in some temporal frame and is maximal with respect to the inclusion ordering, subject to the satisfiability constraint. We denote the set of maximal consistent sets by . It is well known that satisfiability in a temporal frame is decidable, indeed complete [17]. Hence the set of all maximal consistent sets of is a (total) computable function of . The relation on given by
is transitive, and so defines a preorder on reflexive elements. We call a equivalence class a cluster and write for the partial order on clusters induced by . The notation means but not equal. We extend these notations to compare a cluster and a (not necessarily reflexive) maximal consistent set as follows. Write if for all we have , and write if for all we have (. Similarly for and .
A formula of the form is a future defect of a maximal consistent set if . A future defect of a set of maximal consistent sets is a future defect of any member of unless we explicitly identify as a cluster. The formula is a future defect of a cluster if is contained in some , but, for all , we have . A future defect is passed up to a set of maximal consistent sets if either or belongs to some . A past defect is defined similarly.
Recall from (1) the ordering and the associated ‘can reach with a slowerthanlight signal’ Kripke frame . Subsets of inherit the same ordering. The notation denotes the set . (The set is drawn from should be clear.) We define the operations and by
(According to lightspeedorslower accessibility is join and is meet). We also define the partial order on by , , and . When this holds we say that is ‘northwest’ of .
The diagonal dual of a condition on a frame/model whose domain is a subset of is the condition that holds on the frame/model obtained by swapping the  and axes (which will not affect accessibility). The temporal dual is the result of reversing the axis and reversing the axis (which reverses accessibility), and also swapping and in formulas. When we say ‘all duals’ we mean the diagonal dual, the temporal dual, and the diagonal temporal dual.
Suppose we have a preorderpreserving map from a subset of to and that is defined on for some an open neighbourhood (in ) of . Then it is easy to see there is a unique cluster such that there exists a neighbourhood of such that only takes values in on . We denote this cluster by . The analogous value for we denote .
Let be two distinct points on a horizontal line segment in the domain of the preorderpreserving map . Then so we do not know the relation between and . However, it is easy to see that . Similarly for . If is constantly equal to on an open line segment and defined at the left end of then also equals there.^{3}^{3}3For line segments, ‘open’ and ‘closed’ have their usual meanings of ‘excludes end points’ and ‘includes end points’, respectively. Similarly for with right ends. All diagonal duals of statements in this paragraph (that is, statements for vertical lines) hold similarly.
The following concept will be used in definitions that follow, and informs the way we think of a bitrace (Definition 2) as specifying behaviour on a horizontal or vertical line segment. Let and be clusters. The interpolant of and is the set of all with such that all future defects of are passed up to , and all past defects of are passed down to .
A bitrace (of length ) is two sequences and of clusters, and one sequence , with the following constraints.

For each we have , and the interpolant of and is nonempty.

For each either or .

For each we have .

For each all future defects of are passed up to and all past defects passed down to .
Formally, we consider a bitrace to be the interleaving of its three sequences, with the advantage that we can use the notation to indicate a bitrace. We call the ’s the upper clusters, the ’s the lower clusters, and the initial clusters, and and the final clusters. Each pair is a cluster pair of the bitrace, and each a transition value. The top part of Figure 1 suggests how to visualise a bitrace. There are only finitely many bitraces (because of the second condition in their definition).
Let and be bitraces, and . Then is defined and equal to if this is a bitrace. It is also defined if the final clusters of equal the initial clusters of , and is in the interpolant of and , in which case it equals .
We now define surrectangles. Intuitively, the domain of a surrectangle is a rectangle plus infinitesimally more beyond any closed edges of the rectangle (a ‘surreal rectangle’), and a surrectangle records a valuation on this domain.
A rectangle is a product of two intervals of (with unbounded and singlepoint intervals both allowed); it is degenerate if either interval is a single point, otherwise it is nondegenerate. An edge of a rectangle is an edge of the closure of in and is not considered to include its end points; a closed edge of is an edge of contained in . An upper edge of is either a horizontal edge with maximal vertical component, or a vertical edge with maximal horizontal component, a lower edge is defined dually. A rectangle is open/closed if it is open/closed in . The notation , for points and , signifies the closed rectangle .
A surrectangle consists of the following data.

A preorderpreserving map , for some nondegenerate rectangle . We call the rectangle and the core map of the surrectangle.

For each closed horizontal upper edge of with endpoints and ,^{4}^{4}4Here, it could be that and/or ; this does not present any problems. a finite sequence of clusters and a sequence , with , defining to be . Similar finite sequences for any closed vertical and/or lower edges.^{5}^{5}5For lower edges, these supplementary clusters are the ones denoted .
And is required to satisfy the following constraints.

For each closed horizontal upper edge as above, is constant on each open line segment (let this constant cluster be ), and forms a bitrace, where is defined to be , for each . Also, all duals of this constraint. (Figure 2 suggests how to visualise a closed edge.)

For any point , if either

resolved internally: there is such that ,

passed upwards: has a boundary point either due north or due east of such that is passed up to .


The temporal dual of (4) holds.
In (2), we call the supplementary clusters of , we call the transition points of , and we call the transition values. Note that the clusters are determined by in (3), and for edges not contained in (for example if the rectangle is unbounded in the corresponding direction) supplementary clusters and transition points are not defined.
Let and be respectively the lowerleft and upperright corners of . Then and are necessarily defined. The height of the surrectangle is the maximum possible length of a chain of clusters (not necessarily in the image of ) from its lower cluster to its upper cluster . Surrectangles also inherit descriptions such as open/closed from their underlying rectangle.
3 Biboundaries
A biboundary is a partial map on . It must be defined on and , and be clustervalued there. If is defined on , or , it is bitracevalued there, and if defined on , or , it is valued there. It is defined on if and only if it is defined on and ; similarly for and , for and , and for and . The following conditions must also be satisfied.

If is defined, then and every future defect of is passed up to .

If is defined, then the initial upper cluster of equals .

If is defined, then it is less than or equal to the initial upper cluster of , and every future defect of is passed up to that cluster.

Every future defect of is passed up to the interpolant of the final clusters of , to the interpolant of the final clusters of , or to , with defined in the places appropriate to the case.
Since there are only finitely many maximal consistent sets, clusters, and bitraces, there are only finitely many biboundaries. A biboundary is closed if it is defined on , and (hence also on , and ).
Let be a surrectangle. The biboundary determined by is defined in the obvious way.
We now define three types of operations on biboundaries: joins, limits, and shuffles. See Figure 3 for visual representations of limits and shuffles.
A biboundary is the vertical join of biboundaries and , written , if

and are both defined and are equal,

either , and are all defined, , and , or and are all undefined; similarly for ,

agrees with on , and , and with on , and .
The diagonaldual concept is a horizontal join, written .^{6}^{6}6The and subscripts indicate the orientation of the shared edge.
A biboundary is the southeastern limit of a biboundary using biboundaries if

,

the lower cluster of is constantly ,

the upper cluster of is constantly ,

agrees with over ,

, if defined, is a bitrace where the upper cluster is constantly ,

, if defined, is a bitrace where the lower cluster is constantly .
A northwestern limit is defined dually.
If is a set of biboundaries and there are such that is the southeastern limit of using , then is a southeastern limit over . Northwestern limits are dual. We say that is a limit over if it is either a southeastern or northwestern limit over .
Let be a collection of closed biboundaries. The biboundary is a shuffle of if there is a nonempty set such that

if is defined, then all upper clusters of equal ,

every future defect of is passed up to some , or there is a such that is passed up to either , , the interpolant of some cluster pair of or , or some transition value of or ,

for all , we have , all future defects of are passed up to , and all upper clusters of equal ,

for all , we have , and all future defects of every are passed up to ,
Now we are ready to define the biboundaries that we proceed to show are precisely those obtained from surrectangles.
A ground fabricated biboundary is a biboundary such that

,

if is defined, then all lower clusters of equal ,

all duals of (2) hold.
A fabricated biboundary is either a ground fabricated biboundary, or a biboundary obtained recursively as the join, limit, or shuffle of fabricated biboundaries.
4 From fabricated biboundaries to surrectangles
In this section we show that every fabricated biboundary is the biboundary obtained from some surrectangle, by describing how to construct such a surrectangle from a given biboundary. We use the recursive structure of fabricated biboundaries as given by their definition.
When we say a function fills densely with we mean that for all , and for each the set is dense in . It is clear that if is an open subset of and is a cluster, then there exists that fills densely with (and this remains true when restrictions are placed on the behaviour of outside of ). Similarly when is an open line segment. Further, if for a biboundary we have a surrectangle satisfying , we may assume that for every closed edge of and each associated cluster pair between transition points , the core map of fills densely with the interpolant of and . Hence if surrectangles and have a common edge , on which we obtain the same bitrace and transition points from both surrectangles, then we may assume and agree on .
Let be a surrectangle with core map and let and be orderpreserving bijections . Then there is a surrectangle with core map , the same supplementary clusters and transition values as , and a transition point for every transition point of . Moreover, yields the same biboundary as .
The proof of Lemma 4 is routine, and omitted. The proofs of the following four lemmas may be found in the appendix. Figure 3 illustrates the proofs of the last two. [ground biboundaries] Let be a ground fabricated biboundary. Then there exists a surrectangle such that .
[joins] Let and be biboundaries such that the vertical join exists, and suppose there exist surrectangles and with and . Then there exists a surrectangle such that .
Similarly for horizontal joins.
[limits] Let be a southeastern limit of using , and suppose there are surrectangles such that , for . Then there exists a surrectangle such that . Similarly for northwestern limits.
[shuffles] Let be a shuffle of , and suppose that for all there is a surrectangle with . Then there is a surrectangle with .
By Lemma 4, the set of biboundaries that can be obtained from surrectangles contains the ground fabricated biboundaries. By Lemma 4, Lemma 4, and Lemma 3, it is closed under joins, limits, and shuffles. Hence contains all fabricated biboundaries. This proves, as promised, the following.
Let be a fabricated biboundary. Then there exists a surrectangle such that .
5 From surrectangles to fabricated biboundaries
In this section we show that every biboundary obtained from a surrectangle is a fabricated biboundary (Lemma 5). We do this by induction on the height of the surrectangle.
If a surrectangle has height , then the upper and lower clusters of are equal, and from that it is easy to see that all conditions for to be a ground fabricated biboundary are satisfied.
In the remainder of this section, we assume that for every surrectangle of height no greater than , the biboundary is fabricated, and aim to prove the statement for all surrectangles of height . First, note that by Lemma 4, we only need prove the result for surrectangles with bounded domains. So henceforth we take bounded domains as an assumption.
Let be a surrectangle with domain , core map , northwest corner , southeast corner , lower cluster , and upper cluster . Let be the subset of that maps to the lower cluster, and . Define to be the closure (in ) of the intersection of the boundary of with the interior of . Define similarly. Elementary topology shows that and are homeomorphic to closed (proper) line segments, meeting the boundary of only at their endpoints, and linearly ordered by ; see [7, Lemmas 2.10 and 2.11] for details. The notation signifies the surrectangle formed in the obvious way by restriction of to (assuming this is nondegenerate). It is straightforward to check that is indeed a surrectangle.
As before, proofs of lemmas are contained in the appendix.
If and are disjoint then is fabricated.
If and are not disjoint, define a binary relation over by letting for all nondegenerate rectangles the biboundary is fabricated—clearly reflexive and symmetric, and also transitive (use a join of four biboundaries and the induction hypothesis), so an equivalence relation. By using joins and the induction hypothesis if necessary, we may assume . We aim to show that .
Let be an infinite sequence converging to . If for all we have then . Let —a closed set, inheriting a linear order from , and a subset of . Define a binary relation over as the smallest equivalence relation such that

whenever is an immediate successor of ,

whenever is degenerate,

all equivalence classes are topologically closed (in , equivalently, in ).
implies . For any equivalence class , let be the extreme points with respect to (equal if is a singleton). As is closed, . Write for the closed rectangle (a singleton if and only if is a singleton). Since we know .
Either all elements of are equivalent, or there are uncountably many singleton equivalence classes of .
Let , where is the equivalence class of , and let where is the equivalence class of . Since is the most southeastern point in , and has no immediate successor in , there is no point of due south of , similarly no point of due east of . Dual conditions hold for . It follows that is constantly on the south and west edges of , and is constantly on the north and east edges.
Since and we know that and . The next lemma shows that . We omit the proof, which is to check each of the conditions of Definition 3. Suppose the upper cluster on the south and west edges of is constantly , and the lower cluster on the north and east edges of is constantly . Then is a shuffle of where ranges over equivalence classes. Using joins and the induction hypothesis, this proves . Hence we have obtained our goal.
Let be any surrectangle. The biboundary is fabricated.
A biboundary is of the form for some surrectangle if and only if is a fabricated biboundary.
Now we are in a position to prove our first main result.
It is decidable whether a formula of the basic temporal language is valid on the frame consisting of twodimensional Minkowski spacetime equipped with the irreflexive slowerthanlight accessibility relation. The same is true with reflexive slowerthanlight accessibility.
Decidability of validity is equivalent to decidability of satisfiability; we prove the latter.
We first show that satisfiability of on is equivalent to the existence of an open surrectangle having some point assigned a maximal consistent set containing . An open surrectangle consists only of its core map . Given a valuation on and a point at which holds, define by , and then will be a surrectangle on with . Conversely, given such an , for any propositional variable if appears in , define , and otherwise define arbitrarily. Then holds at under valuation .
Let and define , , and similarly, in the evident way. The existence of an open surrectangle having some point assigned a maximal consistent set containing is equivalent to the existence of four surrectangles with domains , , , and , agreeing on their shared edges and at , and with contained in the maximal consistent set assigned to . By Lemma 5, this is in turn equivalent to the existence of four fabricated biboundaries of the appropriate types that match up in the appropriate way.
Since satisfiability of is equivalent to the existence of a set of four fabricated biboundaries with properties that are easily checked, and the set of all fabricated biboundaries is finite and computable, the satisfiability problem is decidable. This completes the proof of the irreflexive case.
The reflexive case follows by the reduction given by recursively replacing subformulas of of the form with , and similarly for .
6 A PSPACE procedure for fabricated biboundaries
In this section, we refine the decidability results of Theorem 5 to show the validity problems are complete.
On the frame consisting of twodimensional Minkowski spacetime equipped with the irreflexive slowerthanlight accessibility relation, the set of validities of the basic temporal language is complete. The same is true with reflexive slowerthanlight accessibility.
As mentioned in the introduction, for the reflexive frame, the validities of the purely modal fragment of the basic temporal language form S4.2, and for the irreflexive frame, OI.2. These are both known to be complete [13], so the validity problems for the entire basic temporal language are hard.
We provide a nondeterministic polynomial space algorithm for satisfiability, for the irreflexive frame. Hence validity is in . By the Immerman–Szelepcsényi theorem [8, 18], , and by Savitch’s theorem [12], , giving the result. The reflexive case follows by the same reduction as before.
Throughout this section, let the length of be . By structural induction, the length of a formula bounds the number of its subformulas, so the cardinality of is linear in . Hence any maximal consistent set—a subset of —can be stored using a linear number of bits, and is at most exponential in . All pertinent information about any cluster can also be stored in a linear number of bits, for we only need record . The maximal length of a chain of distinct clusters or irreflexive members of is also linear in . Hence any biboundary can be stored using a quadratic number of bits, and the number of biboundaries is exponential in .
Having nondeterministically chosen a bit string representing some —a putative maximal consistent set—the conjunct over all formulas in is quadratic in , so its satisfiability can be determined in polynomial space [17]. Hence we can determine if a chosen bit string represents a maximal consistent set using polynomial space.
We can similarly nondeterministically ‘guess’ sets of the form for some cluster , using polynomial space. To do this, first guess a maximal consistent set and check it is reflexive; keep in memory. Then onebyone for each bit string representing some , check and that . If so, add all elements of to an ongoing collection of formulas, discarding after each iteration.
From the preceding discussion, it is clear we can also determine if a chosen string of bits represents a biboundary using polynomial space.
The algorithm for satisfiability of first nondeterministically chooses four bit strings and checks they represent four compatible biboundaries, with at the appropriate corners—performed using polynomial space. Then for each of these in turn, it is checked whether the biboundary is fabricated. The remainder of the proof is devoted to showing that this check can be performed using polynomial space. The procedure to do this is shown in Algorithm 1.
We assume that during execution of Algorithm 1, the formula is a global constant (and therefore is too). All choices are made nondeterministically. At a check the algorithm fails if the check fails, otherwise it proceeds. A tailcall uses tail recursion. The algorithm succeeds if it terminates without failure.
Clearly Algorithm 1 can only succeed for a biboundary if is fabricated. Let be the amount of space required by Algorithm 1 to succeed for any fabricated biboundary of height . We prove that (as a function of and ) by induction on . Hence Algorithm 1 requires space to check any biboundary.
If is a fabricated biboundary of height , then it is ground, so success can be achieved by taking the first branch, requiring space .
For the inductive step, we show that . The proof has a similar structure to the proof in the previous section. Let be a fabricated biboundary of height , and be a surrectangle such that . Define , , , and as in the previous section. Let be the space required to store any biboundary. Thus is a quadratic function of , independent of .
As always, proofs of lemmas are relegated to the appendix.
If and are disjoint, the space required is bounded by .
If and intersect only on the boundary of , the space required is bounded by .
As in the previous section, let . Additionally, let .
If the biboundary of is a shuffle of a subset of , the space required is bounded by .
If and are each the limit of elements of in the interior of , then the biboundary of is the shuffle of , using , where and .
Now suppose only one of , is a limit of elements of in the interior of . Say is such a limit, but is not. If has a direct predecessor , then is in the interior of . If there are points of (strictly) due north of , let the northmost one be . Then either is a limit, and using a join we can reduced to the case of the previous paragraph, or there is a point either an immediate predecessor or due west of . In the second case, is in the interior of . Similarly if there are points of due west of . In each case we obtain an in the interior of and the biboundary of can be checked in , by Lemma 6.
We are going to obtain the biboundary of as the join of four biboundaries. In the southeast corner is . If the biboundary of is of reduced height then we are done. Otherwise, the other three biboundaries we use are modifications of restrictions. For the northwest corner, take the biboundary of . Modify its south edge to the bitrace and its east edge to . This is still a biboundary; call it . For the northeast corner, take the biboundary of (which is ground). Modify its west edge to the bitrace . This is still a biboundary (and still ground). Similarly for the southwest corner.
The biboundary of is the join of these four biboundaries. It only remains to show is fabricated, and that this can be checked efficiently. As is of full height, its upper cluster is and lower . By appealing to the facts that is a surrectangle and is also a surrectangle, we can see that is the shuffle of , using , where , and . The set is a subset of , so by Lemma 6, the biboundary can be checked using no more than space. Hence can been checked within , as promised.
The case where neither nor are limits is similar.
7 Halpern–Shoham logic
In this section we explain how the procedure for the temporal validities of can be modified to procedures performing the same function for certain Halpern–Shoham logics of intervals on the real line.
In Halpern–Shoham logic, intervals are identified with pairs of points, with either (strict interval semantics) or (nonstrict semantics).^{7}^{7}7Thus there is no distinction between open, halfopen, and closed intervals, and unbounded intervals are not present. There are thirteen different atomic relations that may hold between two strict intervals. Following Allen [1], we call these , , , , , , , , , , , , and .^{8}^{8}8When pointintervals are present, pairs of intervals of the form for are considered by Halpern and Shoham to stand in the relation (and not in ). Similarly for , , and . Modalities that may be included in a Halpern–Shoham logic are any corresponding to a relation given by the union of some of these thirteen.
Let be the open halfplane . The frame is precisely the frame of strict intervals of with the relation . Hence the temporal logic of is the strict Halpern–Shoham logic of with two modalities corresponding to this relation and its converse. As in [5], it can be shown that an arbitrary finitely generated directed partial order is a morphic image of the reflexive closure of . Hence an arbitrary purely modal formula (not using the past modality) is in S4.2 if and only if it is valid for the reflexive closure of , and hence the validity of temporal formulas over this frame is PSPACEhard. The hardness of irreflexive itself, follows.
A surtriangle is similar to a surrectangle, but the domain of the core map is either , for some , or a similar set where either is replaced by , or is replaced by , or both. A finite sequence of upper supplementary clusters and transition points are defined along
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