The Last Paper on the Halpern-Shoham Interval Temporal Logic
TThe Last Paper on the Halpern–Shoham Interval Temporal Logic
Draft (October 4, 2018)Jerzy Marcinkowski, Jakub MichaliszynInstitute of Computer Science,University Of Wroclaw,ul. Joliot-Curie 15, 50-383 Wroclaw, Poland {jma,jmi}@cs.uni.wroc.pl
Abstract
The Halpern–Shoham logic is a modal logic of time in-tervals. Some effort has been put in last ten years to clas-sify fragments of this beautiful logic with respect to decid-ability of its satisfiability problem. We contribute to thiseffort by showing – what we believe is quite an unexpectedresult – that the logic of subintervals, the fragment of theHalpern–Shoham where only the operator “during”, or D ,is allowed, is undecidable over discrete structures. This issurprising as this logic is decidable over dense orders [14]and its reflexive variant is known to be decidable over dis-crete structures [13]. Our result subsumes a lot of previ-ous results for the discrete case, like the undecidability for ABE [10], BE [11], BD [12], ADB , A ¯ AD , and so on[2, 6]. In classical temporal logic structures are defined by as-signing properties (propositional variables) to points of time(which is an ordering, discrete or dense). However, not allphenomena can be well described by such logics. Some-times we need to talk about actions (processes) that takesome time and we would like to be able to say that one suchaction takes place, for example, during or after another.The Halpern–Shoham logic [10], which is the subject ofthis paper, is one of the modal logics of time intervals. Judg-ing by the number of papers published, and by the amountof work devoted to the research on it, this logic is probablythe most influential of time interval logics. But historicallyit was not the first one. Actually, the earliest papers aboutintervals in context of modal logic were written by philoso-phers, e.g., [9]. In computer science, the earliest attempts to formalize time intervals were process logic [17, 19] andinterval temporal logic [15]. Relations between intervalsin linear orders from an algebraic point of view were firststudied systematically by Allen [1].The Halpern–Shoham logic is a modal temporal logic,where the elements of a model are no longer — like in clas-sical temporal logics — points in time, but rather pairs ofpoints in time. Any such pair — call it [ p, q ] , where q is notearlier than p — can be viewed as a (closed) time interval,that is, the set of all time points between p and q . HS logicdoes not assume anything about order — it can be discreteor continuous, linear or branching, complete or not.Halpern and Shoham introduce six modal operators, act-ing on intervals. Their operators are “begins” B , “during” D , “ends” E , “meets” A , “later” L , “overlaps” O and thesix inverses of those operators: ¯ B, ¯ D, ¯ E, ¯ A, ¯ L , ¯ O . It is easyto see that the set of operators is redundant. The ,,more ex-pressive” of them, which are A, B and E can define D ( B and E suffice for that – a prefix of my suffix is my infix)and L (here A is enough –“later” means “meets an intervalthat meets”). The operator O can be expressed using E and ¯ B . In their paper, Halpern and Shoham show that (satisfia-bility of formulae of) their logic is undecidable. Their proofrequires logic with five operators ( B, E and A are explic-itly used in the formulae and, as we mentioned above, once B, E and A are allowed, D and L come for free) so theystate a question about decidable fragments of their logic.Considerable effort has been put since this time to settlethis question. First, it was shown [11] that the BE fragmentis undecidable. Recently, negative results were also givenfor the classes B ¯ E , ¯ BE , ¯ BE , A ¯ AD, ¯ AD ∗ ¯ B, ¯ AD ∗ B [2, 6],and BD [12]. Another elegant negative result was that O ¯ O is undecidable over discrete orders [3, 4].On the positive side, it was shown that some small frag-ments, like B ¯ B or E ¯ E , are decidable and easy to translate1 a r X i v : . [ c s . L O ] O c t nto standard, point-based modal logic [8]. The fragmentusing only A and ¯ A is a bit harder and its decidability wasonly recently shown [6, 7]. Obviously, this last result im-plies decidability of L ¯ L as L is expressible by A . Anotherfragment known to be decidable is AB ¯ B [16].The last interesting fragment of the Halpern and Shohamlogic of unknown status was the, apparently very simple,fragment with the single operator D (,,during”), which wecall here the logic of sub-intervals . Since D does not seemto have much expressive power (an example of a formulawould be ,,each morning I spend a while thinking of you”or ,,each nice period of my life contains an unpleasant frag-ment”) logic of sub-intervals was widely believed to be de-cidable. A number of decidability results concerning vari-ants of this logic has been published. For example, it wasshown in ([5, 14]) than satisfiability of formulae of logic ofsubintervals is decidable over dense structures. In [13]. de-cidability is proved for (slightly less expressive) ,,reflexive D ”. The results in [21] imply that D (as well as some richerfragments of the HS logic) is decidable if we allow models,in which not all the intervals defined by the ordering areelements of the Kripke structure.In this paper we show that satisfiability of formulae fromthe D fragment is undecidable over the class of finite order-ings as well as over the class of all discrete orderings. Ourresult subsumes the negative results for the discrete case for ABE [10], BE [11], BD [12] and ADB , A ¯ AD [2, 6]. Our contribution consists of the proofs of the followingtwo theorems:
Theorem 1.
The satisfiability problem for the formulae ofthe logic of subintervals, over models which are subordersof the order (cid:104) Z , ≤(cid:105) , is undecidable. Since truth value of a formula is defined with respect to amodel and an initial interval in this model (see Preliminar-ies), and since the only allowed operator is D , which meansthat the truth value of a formula in a given interval dependsonly on the labeling of this interval and its subintervals The-orem 1 can be restated as: The satisfiability problem for theformulae of the logic of subintervals, over finite models isundecidable , and it is this version that will be proved inSection 3 .
Theorem 2.
The satisfiability problem for the formulae ofthe logic of sub-intervals, over all discrete models, is unde-cidable.
Orderings.
As in [10], we say that a total order (cid:104) D , ≤(cid:105) is discrete if each element is either minimal (maximal) or has a unique predecessor (successor); in other words for all a, b ∈ D if a < b , then there exist points a (cid:48) , b (cid:48) such that a < a (cid:48) , b (cid:48) < b and there exists no c with a < c < a (cid:48) or b (cid:48) < c < b . Semantic of the D fragment of logic HS (logic of sub-intervals). Let (cid:104) D , ≤(cid:105) be a discrete ordered set .An interval over D is a pair [ a, b ] with a, b ∈ D and a ≤ b . A labeling is a function γ : I( D ) → P ( V ar ) , where I( D ) is a set of all intervals over D and V ar is a finite set ofvariables. A structure of the form M = (cid:104) I( D ) , γ (cid:105) is called a model .We say that an interval [ a, b ] is a leaf iff it has no sub-intervals (i.e. a = b ).The truth values of formulae are determined by the fol-lowing (natural) semantic rules:1. For all v ∈ V ar we have M , [ a, b ] | = v iff v ∈ γ ([ a, b ]) .2. M , [ a, b ] | = ¬ ϕ iff M , [ a, b ] (cid:54)| = ϕ .3. M , [ a, b ] | = ϕ ∧ ϕ iff M , [ a, b ] | = ϕ and M , [ a, b ] | = ϕ .4. M , [ a, b ] | = (cid:104) D (cid:105) ϕ iff there exists an interval [ a (cid:48) , b (cid:48) ] such that M , [ a (cid:48) , b (cid:48) ] | = ϕ , a ≤ a (cid:48) , b (cid:48) ≤ b , and [ a, b ] (cid:54) = [ a (cid:48) , b (cid:48) ] . In that case we say that [ a, b ] sees [ a (cid:48) , b (cid:48) ] .Boolean connectives ∨ , ⇒ , ⇔ are introduced in the stan-dard way. We abbreviate ¬(cid:104) D (cid:105)¬ ϕ by [ D ] ϕ and ϕ ∧ [ D ] ϕ by [ G ] ϕ .A formula ϕ is said to be satisfiable in a class of or-derings D if there exist a structure D ∈ D , a labeling γ ,and an interval [ a, b ] , called the initial interval , such that (cid:104) I( D ) , γ (cid:105) , [ a, b ] | = ϕ . A formula is satisfiable in a givenordering D if it is satisfiable in { D } .
3. Proof of Theorem 1
In Section 3 only consider finite orderings.
Our representation.
We imagine the Kripke structure ofintervals of a finite ordering as a directed acyclic graph,where intervals are vertices and each interval [ a, b ] withthe length greater that has two successors: [ a + 1 , b ] and [ a, b − . Each level of this representation contains intervalsof the same length (see Fig. 1). L A In this section we will, for a given two-counter finite au-tomaton (Minsky machine) A , define a regular language L A To keep the notation light, we will identify the order (cid:104) D , ≤(cid:105) with itsset D igure 1. Our representation of order (cid:104){ a, a +1 , . . . , b } , ≤(cid:105) . whose words will almost encode the computation of A (be-ginning from the empty counters).Let Q be the set of states of A , and let Q (cid:48) = { q (cid:48) : q ∈ Q } . Define B = { f, f l , f r , f (cid:48) , f (cid:48) l , f (cid:48) r , s, s l , s r , s (cid:48) , s (cid:48) l , s (cid:48) r } .The alphabet Σ of L A will consist of all the elements of Q ∪ Q (cid:48) (jointly called states ) and of all the subsets (pos-sibly empty) of B which consist of at most 4 elements:at most one of them from { f, f l , f r } , at most one from { f (cid:48) , f (cid:48) l , f (cid:48) r } , at most one from { s, s l , s r } , and at most onefrom { s (cid:48) , s (cid:48) l , s (cid:48) r } .Symbols of Σ containing f l or f (cid:48) l ( s l or s (cid:48) l ) will be called first (resp. second ) counters . Symbols of Σ containing f r or f (cid:48) r ( s r or s (cid:48) r ) will be called first (resp. second ) shadows (or shadows of the first/the second counter ).The language L A will consist of all the words w over Σ which satisfy all the following six conditions: • The first symbol of w is the beginning state q of A and the last symbol of w is either q or q (cid:48) , where q isone of the final states of A .By a configuration we will mean a maximal sub-word of w , whose first element is a state (called the state of thisconfiguration ) and which contains exactly one state (so that w is split into disjoint configurations). A configuration willbe called even if its state is from Q and odd if it is from Q (cid:48) . • Odd and even configurations alternate in w . • Each configuration, except of the last one (which onlyconsists of the state) contains exactly one first counterand exactly one second counter. If a configuration iseven, then its first counter contains f l and its second See Lemma 1 for an explanation what we mean by ”almost”. By a sub-word we mean a sequence of consecutive elements of a word,an infix. counter contains s l . If a configurations is odd, then itsfirst counter contains f (cid:48) l and its second counter con-tains s (cid:48) l . The first non-state symbol of the first config-uration is both a first counter and a second counter. • The are no shadows in the first and the last configu-ration. Each configuration, except of the first and thelast, contains exactly one first shadow and exactly onesecond shadow. If a configuration is even, then its firstshadow contains f (cid:48) r and its second shadow contains s (cid:48) r .If a configurations is odd, then its first shadow contains f r and its second shadow contains s r .It follows, from the conditions above, that if (in a wordfrom the language L A ) there is a counter containing f l ( f (cid:48) l , s l , s (cid:48) l ) then there is its shadow f r (resp. f (cid:48) r , s r , s (cid:48) r ) inthe subsequent configuration. Call a sub-word beginningwith first (second) counter and ending with its shadow afirst (resp. second ) shade . Notice, that the above conditionsimply in particular that each state (except of the first oneand last one) is in exactly one first shade and in exactly onesecond shade. • A non-state symbol of w contains f ( f (cid:48) , s, s (cid:48) ) if andonly if it is inside some shade beginning with f l (resp. f (cid:48) l , s l , s (cid:48) l )The last condition defining L A will depend on the in-structions of the automaton A . We say that a configurationhas first ( second ) counter equal zero if the first non-statesymbol of this configuration contains f l or f (cid:48) l (resp. s l or s (cid:48) l ). It is good to think, that the number of symbols beforethe first/second counter is the value of this counter in thegiven configuration. Notice that the first configuration of a w ∈ L A is indeed the initial configuration of A – its state is q and both its counters equal 0.Since the format of an instruction of A is: If in state q the first counterequals/does not equal 0 andthe second counterequals/does not equal 0then change the state to q anddecrease/increase/keep unchangedthe first counter anddecrease/increase/keep unchangedthe second counter. it is clear what we mean, saying that configuration C matches the assumption of the instruction I . • If C and C are subsequent configurations in w , and C matches the assumption of an instruction I , then:3 If I changes the state into q then the state of C is q . – If I orders the first (second) counter to remain un-changed, then the first (resp. second) counter in C coincides with the first (resp. second) shadowin C . – If I orders the first (second) counter to be de-creased, then the first (resp. second) counter in C is the direct predecessor of the first (resp. sec-ond) shadow in C . – If I orders the first (second) counter to be in-creased, then the first (resp. second) counter in C is the direct successor of the first (resp. sec-ond) shadow in C .This completes the definition of the language L A . It isclear, that it is regular. Our main tool will be the following: Lemma 1.
The following two conditions are equivalent:(i) Automaton A , started from the initial state q andempty counters, accepts.(ii) There exists a word w ∈ L A and a natural number n such that: • each configuration in w (except of the last one,consisting of a single symbol) has length n − • each shade in w has length n (this includes thetwo symbols in the two ends of a shade).Proof. For the ⇒ direction consider an accepting compu-tation of A and take n as any number greater than all thenumbers that appear on the two counters of A during thiscomputation. For the ⇐ direction notice that the distanceconstraint from (ii) imply, that the distance between a stateand the subsequent first (second) shadow equals the valueof the first (resp. second) counter in the previous config-uration. Together with the last of the six conditions defin-ing L A this implies that the subsequent configurations in w ∈ L A can indeed be seen as subsequent configurations inthe valid computation of A .Since the halting problem for two-counter automata isundecidable, the proof of Theorem 1 will be completedwhen we write, for a given automaton A , a formula Ψ ofthe language of the logic of sub-intervals which is satisfi-able (in a finite model) if and only if condition (ii) fromLemma 1 holds. Actually, what the formula Ψ is going tosay is, more or less, that the word written (with the use ofthe labeling function γ ) in the leaves of the model is a word w as described in Lemma 1 (ii).In the following subsections we are going to write for-mulae Φ orient , Φ L A , Φ cloud and Φ length , such that Φ orient ∧ Φ L A ∧ Φ cloud ∧ Φ length will be the formula Ψ we want. As we said, we want to write a formula saying that theword written in the leaves of the model is the w describedin Lemma 1 (ii).The first problem we need to overcome is the symmetryof D – the operator does not see a difference between pastand future, or between left and right, so how can we distin-guish between the beginning of w and its end? We deal withthis problem by introducing five variables L, R, s , s , s and writing a formula Φ orient which will be satisfied by aninterval [ a, b ] if [ a, a ] is the only interval that satisfies L and [ b, b ] is the only interval that satisfies R , or [ b, b ] is the onlyinterval that satisfies L and [ a, a ] is the only interval thatsatisfies R , and if all the following conditions hold: • any interval that satisfies L satisfies also s ; • each leaf is labeled either with s or with s or with s ; • each interval labeled with s or with s or with s is aleaf; • if c, d, e are three consecutive leaves of [ a, b ] and if s i holds in c , s j holds in d and s k holds in e then { i, j, k } = { , , } .If [ a, b ] | = Φ orient then the leaf of [ a, b ] where L holds(resp. where R holds) will be called the left (resp. the right)end of [ a, b ] .Let exactly one of ( X ) = (cid:87) x ∈ X ( x ∧ (cid:86) x (cid:48) ∈ X \{ x } ¬ x (cid:48) ) be a formula saying (which is not hard to guess) that exactlyone variable from the set X is true in the current interval. Φ orient is a conjunction of the following formulae.(i) [ D ](([ D ] ⊥ ⇒ exactly one of ( { s , s , s } ) ∧ ( s ∨ s ∨ s ⇒ [ D ] ⊥ ))) (ii) [ D ]( (cid:104) D (cid:105)(cid:104) D (cid:105)(cid:62) ⇒ (cid:104) D (cid:105) s ∧ (cid:104) D (cid:105) s ∧ (cid:104) D (cid:105) s ) (iii) [ D ]( L ⇒ s ) (iv) (cid:104) D (cid:105) R ∧ (cid:104) D (cid:105) L (v) [ D ]( L ⇒ ¬ R ) (vi) [ D ]([ D ][ D ] ⊥ ∧ (cid:104) D (cid:105) L ⇒ ¬(cid:104) D (cid:105) s ) (vii) (cid:87) i ∈{ , , } [ D ]([ D ][ D ] ⊥ ∧ (cid:104) D (cid:105) R ⇒ ¬(cid:104) D (cid:105) s i ) Formulae (i), (ii), and (iii) express the property definedby the conjunction of the four items above (notice, that [ D ] ⊥ means that the current interval is a leaf).Formula (iv) says that there exists an interval labeledwith R and an interval labeled with L .4ormula (v) states that intervals labeled with L are alsolabeled with s , and intervals labeled with R are labeledwith s , so they are leaves.Formula (vi) guarantees that no interval containing ex-actly 2 leaves, which is a super-interval of an interval la-beled with L , can contain a sub-interval labeled with s . Itimplies that an interval labeled with L can only have onesuper-interval containing exactly 2 leaves — if there weretwo, then their common super-interval containing 3 leaveswould not have a sub-interval labeled with s , what wouldcontradict (ii).Finally, formula (vii), finally, works like (vi) but for R .We have to use disjunction in this case since we do not knowwhich s i is satisfied in the interval labeled with R .In the rest of paper we restrict our attention to modelssatisfying formula Φ orient , and treat the leaf labeled with L as the leftmost element of the model. In this section we show how to make sure that consecu-tive leaves of the model, read from L to R , are labeled withvariables that represent a word of a given regular language. Lemma 2.
Let A = (cid:104) Σ , Q , q , F , δ (cid:105) , where q ∈ Q , F ⊆Q , δ ⊆ Q× Σ ×Q be a finite–state automaton (deterministicor not, it does not matter).There exists a formula ψ A of the D fragment of Halpern–Shoham logic over alphabet Q ∪ Σ that is satisfiable (withrespect to the valuation of the variables from Q ) if and onlyif the word, over the alphabet Σ written in the leaves of themodel, read from L to R , belongs to the language acceptedby A .Proof. It is enough to write a conjunction of the followingproperties.1. In every leaf, exactly one letter from Σ is satisfied (sothere is indeed a word written in the leaves).2. Each leaf is labeled with exactly one variable from Q .3. For each interval with the length , if this intervalcontains an interval labeled with s i , with a ∈ Σ and with q ∈ Q and another interval labeled with s ( i +1) mod , and with q (cid:48) ∈ Q , then (cid:104) q, a, q (cid:48) (cid:105) ∈ δ .4. Interval labeled with R is labeled with such q ∈ Q and a ∈ Σ that (cid:104) q, a, q (cid:48) (cid:105) ∈ δ for some q (cid:48) ∈ F .5. Interval labeled with L is labeled with q .Clearly, a model satisfies properties 1-5 if and only if itsleaves are labeled with an accepting run of A on the wordover Σ written in its leaves. The formulae of the D fragmentof Halpern–Shoham logic expressing properties 1-5 are nothard to write: 1. [ G ](([ D ] ⊥ ⇒ exactly one of (Σ)) ∧ ( (cid:87) Σ ⇒ [ D ] ⊥ )) [ G ](([ D ] ⊥ ⇒ exactly one of ( Q )) ∧ ( (cid:87) Q ⇒ [ D ] ⊥ )) [ G ]([ D ][ D ] ⊥ ∧ (cid:104) D (cid:105) s i ∧ (cid:104) D (cid:105) s i +1 mod ⇒ (cid:87) (cid:104) q,a,q (cid:48) (cid:105)∈ δ (cid:104) D (cid:105) ( s i ∧ q ∧ a ) ∧ (cid:104) D (cid:105) ( s i +1 mod ∧ q (cid:48) )) ,for each i ∈ { , , } [ G ]( R ⇒ (cid:87) (cid:104) q,a,q (cid:48) (cid:105)∈ δ,q (cid:48) ∈F ( q ∧ a )) [ G ]( L ⇒ q ) Now, let A be a finite automaton recognizing language L A from Section 3.1 and put Φ L A = ψ A . We still need to make sure, that there exists n such thateach configuration (except of the last one) has length n − and that each shade has the length exactly n . Let us startwith: Definition 1.
Let
M = (cid:104) I( D ) , γ (cid:105) be a model and p a vari-able. We call p a cloud if there exists k ∈ N such that p ∈ γ ([ a, b ]) if and only if the length of [ a, b ] is exactly k . So one can view a cloud as a set of all intervals of somefixed length. Notice, that if the current interval has length k then exactly k + 1 leaves are reachable from this segmentwith the operator D .We want to write a formula of the language D fragmentof Halpern-Shoham logic saying that p is a cloud. In orderto do that we use an additional variable e . The idea is thatan interval [ a, a + n ] satisfies e iff [ a + 1 , a + n + 1] doesnot. Figure 2. An example of a cloud.
Let Φ cloud be a conjunction of the following formulae.1. (cid:104) D (cid:105) p — there exists at least one point that satisfies p .2. [ D ]( p ⇒ [ D ] ¬ p ) — intervals labeled with p cannotcontain intervals labeled with p .3. [ G ](( (cid:104) D (cid:105) p ) ⇒ ( (cid:104) D (cid:105) ( p ∧ e )) ∧ ( (cid:104) D (cid:105) ( p ∧ ¬ e ))) — eachinterval that contains an interval labeled with p actuallycontains at least two such intervals — one labeled with e and one with ¬ e .5 emma 3. If M , [ a M , b M ] | = Φ cloud , where a M and b M areendpoints of M , then p is a cloud.Proof. We will prove that if an interval [ x, y ] is labeled with p , then also [ x + 1 , y + 1] is labeled with p . A symmetricproof shows that the same holds for [ x − , y − , so allthe intervals of length equal to m , where m is the length of [ x, y ] , are labeled with p .This will imply that no other intervals can be labeledwith p and p is indeed a cloud. This is because each suchinterval either has a length greater than m , and thus con-tains an interval of length m , and as such labeled with p , orhas a length smaller than m , and is contained in an intervallabeled by p , in both cases contradicting (ii).Consider an interval [ x, y ] labeled with p . Interval [ x, y + 1] contains an interval labeled with p , so it has tocontain two different intervals labeled with p – one labeledwith e and the other one with ¬ e . Suppose without loss ofgenerality that [ x, y ] is the one labeled with e , and let uscall the second one [ u, t ] . If t < y + 1 , then [ u, t ] is a sub-interval of [ x, y ] and is labeled with p , a contradiction. So t = y + 1 .Let us assume that u > x + 1 . The interval [ u − , y + 1] must contain two different intervals labeled with p . Oneof them is [ x, y + 1] , and it cannot contain another intervallabeled with p , so the other one must be a sub-interval of [ u − , y ] . But then it is a sub-interval of [ x, y ] (because u − > x + 1 − x ) which also is labeled with p — acontradiction. So u = x + 1 . Let us now concentrate on models which satisfy Φ orient ∧ Φ L A ∧ Φ cloud . Since Φ cloud is satisfied then p is a cloud.Let n − denote number of leaves contained in the intervalsthat form the cloud. Our goal is to write a formula Φ length that would guarantee the following properties:1. Configurations and shades are not too short. If you seetwo states (i.e. more than an entire configuration) or anentire shade, then you must see a lot, at least n leaves.So you must be high enough. Higher than the cloud.2. Configurations and shades are not too long. If you onlysee an interior of a configuration (i.e. you do not seestates) or an interior of some shade, then you do notsee much, at most n − leaves. So you must be underthe cloud.Once we do that, the formula Ψ = Φ orient ∧ Φ L A ∧ Φ cloud ∧ Φ length will be satisfiable if and only if there existsa word satisfying the conditions from Lemma 1 (ii) – it isstraightforward how to translate such a word into a modelof Ψ and vice versa. So put Φ length = Φ ,c length ∧ Φ ,s length ∧ Φ ,c length ∧ Φ ,s length where: Φ ,c length = [ G ]( (cid:86) q ∈ Q,q (cid:48) ∈ Q (cid:48) ( (cid:104) D (cid:105) q ∧ (cid:104) D (cid:105) q (cid:48) ) ⇒ (cid:104) D (cid:105) p )Φ ,c length = [ G ]( (cid:86) q ∈ Q [ D ] ¬ q ⇒ ¬ p ∧ [ D ] ¬ p ) Formulae for shades are a little bit more com-plex. Let F l ( F (cid:48) l , S l , S (cid:48) l , F, F (cid:48) , S, S (cid:48) , F r , F (cid:48) r , S r , S (cid:48) r resp.) be a set of symbols that contain f l ( f (cid:48) l , s l , s (cid:48) l , f, f (cid:48) , s, s (cid:48) , f r , f (cid:48) r , s r , s (cid:48) r resp.), and T = {(cid:104) F l , F, F r (cid:105) , (cid:104) F (cid:48) l , F (cid:48) , F (cid:48) r (cid:105) , (cid:104) S l , S, S r (cid:105) , (cid:104) S (cid:48) l , S (cid:48) , S (cid:48) r (cid:105)} . Φ ,s length =[ G ]( (cid:86) (cid:104) T l ,T,T r (cid:105)∈T ( (cid:104) D (cid:105) (cid:87) T l ∧ (cid:104) D (cid:105) (cid:87) T r ) ⇒ (cid:104) D (cid:105) p )Φ ,s length =[ G ]( (cid:94) (cid:104) T l ,T,T r (cid:105)∈T ( (cid:104) D (cid:105) (cid:95) T ∧ ¬(cid:104) D (cid:105) (cid:95) ( T l ∪ T r )) ⇒ ¬ p ∧ [ D ] ¬ p )
4. Proof of Theorem 2
Unfinished