An Alternative To The Iteration Operator Of Propositional Dynamic Logic Marcos Alexandre Castilho 1 IRIT - Universite Paul abatier and UFPR - Universidade Federal do Parana (Brazil) Andreas Herzig IRIT - Universite Paul abatier e-mail: fcastilho,herzigg@irit.fr February 19, 1996 Abstract In this paper we present an alternative to the iteration operator of Propositional Dynamic Logic (PDL) and a tableau system for the resulting logic. We show the correctness and completeness of the method. REPORT 96-05-R 1 This work was supported by CAPE/BRAZIL
1 Introduction In this work we present an alternative to the iteration operator () of Propositional Dynamic Logic (PDL) [Har84]. The way we do this is to replace this operator by the necessity operator (2) of modal logic 4 plus an interaction axiom of multimodal logic. The resulting logic (called A) is weaker than PDL, but is strong enough to reason about actions and plans. In section 2 we present the language, semantics and axiomatics of A. In section 3 we show how one can reason about actions and plans in A. In section 4 we present a tableau system for A, to provide an automatic way of deciding whether a formula is a theorem or not. ections 5 and 6 present, respectively, the correctness and completeness results for the tableau system. 2 The logic A In this section we present the logic A, the language, semantics and axiomatics. We also present the notation to be used in the rest of the paper. 2.1 Language The language of A is as follows: Let ACT 0 be a set of atomic actions (ACT 0 = f; ; : : :g) and let F OR 0 be a set of atomic formulas, or just atoms (F OR 0 = fp; Q; : : :g). We construct inductively the set ACT of actions and the set F OR of formulas in the following way: > 2 F OR;? 2 F OR; F OR 0 F OR if A; B 2 F OR then :A 2 F OR, A _ B 2 F OR and 2A 2 F OR. if A 2 F OR and 2 ACT then []A 2 F OR ACT 0 ACT ; 2 ACT 0 if ; 2 ACT then (; ) 2 ACT and ( [ ) 2 ACT if A 2 F OR then A? 2 ACT We use ^,!, $ as usual, hia to abbreviate :[]:A and 3A to abbreviate :2:A as in modal logic. We also use [; ]A for [][]A, [ [ ]A for ([]A ^ []A), [A?]B for (A! B), []A for A and [if A then else ]B for ((A! []B) ^ (:A! []B)) As in dynamic logic we read \do nothing", A? \continue if A holds", [ \do either or nondeterministically", (; ) \do followed by " and []A \every possible execution of action leads to a situation in which A is true". We read 2A \all plans leads to a situation in which A is true". A plan is a (possibly empty) sequence of actions. Finally, if a formula does not contain modal operators, we call it a classical formula. 1
2.2 emantics We dene standard models for our logic A as quadruples of the form = (W; (R ) 2ACT 0 ; R2; j=), where W is a set of worlds, each R and R2 W W is a binary relation (called an accessibility relation) and j= is the satisfaction relation with the following truth condition: j= w P i w 2 (P ), where : F OR 0?! 2 W j= w > not j= w? j= w :A i not j= w A j= w A ^ B i j= w A and j= w B j= w A! B i if j= w A then j= w B j= w A _ B i j= w A or j= w B j= w A $ B i j= w A if and only if j= w B j= w []A i 8w 0 2 W such that wr w 0, j= w 0 A. j= w 2A i 8w 0 2 W such that wr2w 0, j= w 0 A. We impose no restrictions on the (R ) family of relations, but we require R2 to be reexive and transitive. Moreover, we suppose R2 R ; 2 ACT 0. Validity and satisability are dened as usual. We denote j=a P to mean \P is valid in A". 2.3 Axiomatics The following are the axioms and rules of inference schemes for A, where A; B 2 F OR and 2 ACT : P L: All tautologies of classical propositional calculus K(): [](A! B)! ([]A! []B) K(2): 2(A! B)! (2A! 2B) T (2): 2A! A 4(2): 2A! 22A I(2; ): 2A! []A (M:P:): if A and A! B then infer B (RN2): if A then infer 2A We denote `A A to mean \A is a theorem of A" and theoremhood is dened as usual. Remark.: A may be seen as a multimodal logic made up from Propositional Dynamic Logic (PDL) [Har84] without the iteration operator (*), plus the well known modal logic 4 for the 2- operator plus the interaction axioms fi(2; ) : 2A! []Ag. In fact, A is an abbreviation for K() + 4(2) + I(2; ). 3 Reasoning about actions and plans in A In this section we show how the components of systems for reasoning about actions and plans may be written in the formalism of our logic A. We present usual tasks and classical problems of reasoning about actions and plans in terms of A, but we don't propose any solution for them. This is subject of future research and is presented now only as a motivation. 2
3.1 Representation Let ACT 0 stand for atomic actions and ACT stand for compound actions built with elements of ACT 0 and the operators \", \?", \;", \[", \if... then... else... ". For example, we may have atomic actions like \load", \shoot" and \wait", and more complicated compound actions like \load; wait; shoot" and \if dead then else load; shoot". Let KB be the knowledge base, the set of classical formulas representing the actual world. Let IC ACT be the set of Action Laws. They have the form 2(A! []B), where A is the precondition and B is the postcondition of the action. pecial action laws have the form 2(A! []A) and are called frame axioms. Finally, we have the set IC CLA of Integrity Constraints, whose elements have the form 2A, where A is classical. In planning systems we also have the set Goal, whose formulas have the form 3A, where A is classical. Example: KB : walking ^ loaded. IC CLA : 2(alive $ :dead), 2(loaded $ :empty), 2(walking! alive). IC ACT : hwaiti> hloadi> loaded! [shoot]dead, loaded! [shoot]empty, (frame axioms:) alive! [wait]alive, loaded! [wait]loaded walking! [load]walking, dead! [shoot]dead Goal : 3dead, 3alive 3.2 Reasoning tasks Once we have a representation for the world and for the actions laws, the essentials tasks we are interested includes: 1. Planning: We know the actual and the desired world, and we want to know about the existence of a sequence of actions that lead from the actual situation to the desired one. In other words we want to know whether `A KB ^ IC ACT ^ IC CLA! Goal is a theorem of A. We claim that as antecedents do not contain positive 3?operators, from the proof of a planning problem we can construct a plan. For example, we may have to know if there is a proof of the formula \:loaded ^ 2(loaded! [shoot]dead) ^ 2[load]loaded! 3dead". The proof of this theorem gives us a plan. 2. Prediction: We know the actual world (KB 1 ) and we want to point out the correct state of the world we may obtain (KB 2 ) if we apply in KB 1 a given sequence of actions. In the formalism we have, what we want to know is whether `A KB 1 ^ IC ACT ^ IC CLA! []KB 2 is a theorem of A. 3. Postdiction: We want to know the preceding state of the world (KB 1 ) having only the actual world (KB 2 ) and knowing that a given sequence of actions was executed. In other words, we want to know whether `A []KB 2 ^ IC ACT ^ IC CLA! KB 1 is a theorem of the logic. 3
3.3 Reasoning problems In order to realize tasks as those described in the previous section, three main problems arise. These problems were pointed out the rst time by McCarthy and Hayes [MH69]: 1. The frame problem is that in real world, it is impractical to write down all non-consequences of an action. In other words, after an action, one must say everything that do not change in the world. In terms of A, the frame problem is that the set of frame axioms is too big, and in fact it contain almost all formulas of IC ACT. E.g. we don't want to write the frame axioms alive! [wait]alive, loaded! [wait]loaded, etc... 2. The ramication problem is the dual of the frame problem, i.e., it is dicult to take into account all the (direct and indirect) consequences of an action, i.e., it is dicult to take into account the set IC CLA. E.g. after the execution of [shoot] in loaded! [shoot]dead, one must also conclude that :walking and :loaded. 3. The qualication problem is related with the diculty to write down the potentially immense number of preconditions of an action. E.g. the preconditions of the action of starting a car. Moreover, it may be computationally intractable to check all qualications for each action to be executed. We note that with our formalism it is possible to solve the ramication problem. In fact the solution is straightforward since we have the set IC CLA. We don't have solutions to the frame neither the qualication problem, but for the former we claim that it is possible to nd such a solution following the work of Gasquet and Herzig [GH95]. 4 A-Tableau rules In this section we present a tableau system for A. We will use a generalization of Fitting's signed formulas notation [Fit83]. Thus, if A is a formula we write: V 2 to mean T2A or F3A V to mean T []A or F hia 2 to mean T3A or F2A to mean T hia or F []A Moreover, the result of removing the modal operator of V 2, V, 2 or is denoted by a subscripted 0. For example, if V 2 = 2A then V 2 0 = A. We also consider the following denitions in order to simplify notation in the next section. Let be a set of formulas and 2 ACT. Then 2 = fv 2 : V 2 2 g = fv 0 : V 2 g When we want to evidence some formula A in a set of formulas we will write ; fag. In this case, we consider that A =2. The - and -rules 2 are the same as in [Fit83] and are not listed. The other rules are divided in two categories: those that do not correspond to a world change in the semantics (the V-rules) and those that do correspond to a world change in the semantics (the -rules, represented in the gures by an arrow). In the former, like the - and -rules, we simply add a formula to the set, and in the latter we erase all formulas and rewrite new ones in the new world. They are 3 : 2 and appear in boldface when they are used as rule names and not to denote dynamic operators. 3 This method does not have a V -rule. 4
V 2 -rule 2 -rule -rule from V 2 from ; 2 from ; # # infer V 2 0 infer 2 ; 2 0 infer 2 ; ; 0 Example: If we have the following formulas in a world w, where ; 2 ACT, A i 2 F OR: w 1. T3A 1 2. T3A 2 3. T2A 3 4. T2A 4 5. T []A 5 6. T []A 6 7. T hia 7 8. T hia 8 We can apply the A-tableau rules to obtain (in the same world): w 9. T A 3 3, V 2 -rule 10. T A 4 4, V 2 -rule or we can apply the A-tableau rules to obtain (in dierent worlds w 1 and w 2 ): w 1 w 2 1'. T2A3 1". T2A3 2'. T2A 4 1, 2 -rule or 3'. T A 1 2". T2A 4 3". T A 5 4". T A 7 7, -rule 5 Correctness In this section we will present the correctness proof of the A-tableau system. In order to do this we have to show two lemmas. Denition 1 If w is a world in a model = (W; (R ) 2ACT 0 ; R2; j=) and is a set of signed formulas, we write j= w when j= w A, 8A 2. Lemma 1 uppose = (W; (R ) 2ACT 0 ; R2; j=) is a standard model, is a set of signed formulas, w 2 W and w 0 2 W is any world such that wr w 0 (and hence wr2w 0, since R2 R ). Then: 1. j= w implies j= w 0 [ fv 2 0 : V 2 2 g. 2. j= w implies j= w 0 2. 3. j= w V 2 implies j= w V 2 0. Proof 1. If j= w then j= w A, 8A 2. If A is of the form V, then by denition, 8w 0 2, wr w 0, j= w 0 V 0. o j= w 0 fv 0 jv 2 g, i.e., j= w 0. And since R2 R ; 8 2 ACT we have wr2w 0. o for each V 2 2 we have by denition j= w 0 V2 0. Thus j= w 0 fv2 0 : V2 2 g. Hence j= w 0 [ fv 2 0 : V 2 2 g. 5
2. Let w 00 be any world in W such that w 0 R2w 00. ince wr2w 0 and R2 is transitive, wr2w 00. o, for each V 2 2, if j= w V2 it follows j= w 0 V2 0, i.e., 8w00 2 such that w 0 R2w 00 we have wr2w 00 and so j= w 00 V2 0, in other words, j= w 0 V2. Hence j= w 0 2. 3. ince j= w V2, we have by denition j= w 0 V2 0, 8w0 2 such that wr2w 0. But R2 is reexive, so in particular we have wr2w and hence j= w V 2 0. Denition 2 A tableau is A-satisable if some branch of it is A-satisable. A branch of a tableau is A-satisable if the set of formulas on it is A-satisable. A set of formulas is A-satisable if there is a world w in a standard model such that j= w. Lemma 2 uppose T is a A-satisable tableau. Then T 0, the tableau that results from a single A-tableau rule applied to T, is also A-satisable. Proof If T is a A-satisable tableau then some branch of it is A-satisable. If a branch other then one on which we apply the rule is A-satisable that branch is still present in T 0 and is still A-satisable. Hence T 0 is A-satisable. uppose is a A-satisable branch of T and the tableau rule is applied on. We analyze the consequences for each tableau rule: i) case -rule is applied ( = ; is replaced by 1 = ; ; 1 ; 2 ). Now is A-satisable, so in some A-model = (W; (R ) 2ACT 0 ; R2; j=) there is a world w such that j= w and j= w. By denition j= w 1 and j= w 2, and hence 1 is A-satisable. ii) case -rule is applied ( = ; is replaced by 1 = ; ; 1 and 2 = ; ; 2 ). Now is A-satisable, so in some A-model = (W; (R ) 2ACT 0 ; R2; j=) there is a world w such that j= w and j= w. Hence j= w 1 or j= w 2, and hence either 1 or 2 is A-satisable. iii) case V 2 -rule is applied ( = ; V 2 is replaced by 1 = ; V 2 ; V 2 0 ). Now is A-satisable, so in some A-model = (W; (R ) 2ACT 0 ; R2; j=) there is a world w such that j= w and j= w V 2. By lemma 1 we also have j= w V2 0. Hence 1 is A-satisable. iv) case 2 -rule is applied ( = ; 2 is replaced by 1 = 2 ; 2 0 ). Now is A-satisable, so in some A-model = (W; (R ) 2ACT 0 ; R2; j=) there is a world w such that j= w and j= w 2. But if j= w 2 then, by denition, for some world w 1 2 such that wr2w 1 we have j= w 1 2 0. Moreover, by lemma 1, if j= w then j= w 1 2. Hence 1 is A-satisable in w 1 2, so 1 is A-satisable. v) case -rule is applied ( = ; is replaced by 1 = 2 ; ; 0 ). Now is A-satisable, so in some A-model = (W; (R ) 2ACT 0 ; R2; j=) there is a world w such that j= w and j= w. But if j= w then for some world w 1 2 such that wr w 1 we have j= w 1 0 and by lemma 1, if j= w then j= w 1 2 and j= w 1 (recall R2 R ; 8 2 ACT, so if wr w 1 then wr2w 1 ). o again 1 is A-satisable in w 1 2. Hence 1 is A-satisable. There are no more cases. Hence this completes the proof. Theorem 1 (Correctness theorem for A) If A has an A-tableau proof then j=a A. Proof uppose there is a tableau proof starting by F A that arrives at a closed tableau. If ff Ag were A-satisable, we would start with an A-satisable tableau, and by lemma 2 every A-tableau we get will also be A-satisable. But this is a contradiction because a closed tableau is not A-satisable. Hence F A is not A-satisable, i.e., there is no A-model such that j= 6 w A,for some world w 2. In other words, for every A-model we have j= w A, that is, j=a A. 6 Completeness In this section we will prove the completeness of our tableau system. We will use the method proposed in [Fit83], i.e., we will rst show consistency properties for A (section 6.1), then we will show the model existency theorem for A (section 6.2), and nally (in section 6.3) we will present the completeness proof. 6
6.1 Consistency properties for A In this section we present the denition of A-consistency property and we show two important lemmas. Denition 3 [A-Consistency Property] Let IC be a collection of sets of signed formulas. IC is an A-consistency property if it meets the following conditions, for each 2 IC: 0. contains no atomic signed formula T A and its conjugate F A; contains neither F > nor T?. 1. 2 ) [ f 1 ; 2 g 2 IC 2. 2 ) [ f 1 g 2 IC or [ f 2 g 2 IC 3. 2 2 ) 2 [ f 2 0 g 2 IC 4. 2 ) 2 [ [ f 0 g 2 IC 5. V 2 2 ) [ fv 2 0 g 2 IC If 2 IC we say is IC-consistent else is IC-inconsistent. Denition 4 Let IC be an A-consistency property and let U be a set of signed formulas. We call IC U-compatible if, for each 2 IC and for each Z 2 U, [ fzg 2 IC. Lemma 3 Let IC be an A-consistency property. Let IC 0 contain all subsets of all members of IC. Then IC 0 is also an A-consistency property, extending IC and closed under subsets. Also, if IC is U-compatible, so is IC 0. Proof We check each item of the denition: 0. Members of IC 0 are subsets of members of IC, which do not contain with an atomic formula and its conjugate. o IC 0 either. For the same reason IC 0 does not contain with F > or T?. 1. uppose 2 IC 0 and 2. Then 9T 2 IC, T and, since IC is an A-consistency property, T [ f 1 ; 2 g 2 IC. But, by construction of IC 0, and the fact [ f 1 ; 2 g T [ f 1 ; 2 g, we have [ f 1 ; 2 g 2 IC 0. 2. uppose 2 IC 0 and 2. Then 9T 2 IC, T and, since IC is an A-consistency property, T [ f 1 g 2 IC or T [ f 2 g 2 IC. Again by construction of IC 0, and the fact [ f 1 g T [ f 1 g and [ f 2 g T [ f 2 g, we have [ f 1 g 2 IC 0 or [ f 2 g 2 IC 0. 3. uppose 2 IC 0 and 2 2. Then 9T 2 IC, T, and then 2 T 2 (recall denition of 2 in section 4). Now, since IC is an A-consistency property, T 2 [ f 2 0 g 2 IC. Again by construction of IC 0 and the fact that 2 [ f 2 0 g T 2 [ f 2 0 g, we have 2 [ f 2 0 g IC 0. 4. uppose 2 IC 0 and 2. Then 9T 2 IC, T and since IC is an A-consistency property, T 2 [ T [ f 2 0 g 2 IC. But if T then by the denitions of section 4, T and 2 T 2. ince 2 [ [ f 0 g T 2 [ T [ f 0 g and by the denition of IC 0, we conclude 2 [ [ f 0 g 2 IC 0. 5. uppose 2 IC 0 and V 2 2. Then 9T 2 IC, T and since IC is an A-consistency property, T [ fv 2 0 g 2 IC. ince T implies [ fv2 0 g T [ fv2 0 g and by the denition of IC0, we conclude [ fv 2 0 g 2 IC 0. Hence IC 0 is a A-consistency property. Obviously, IC 0 extends IC and is closed under subsentences. We show the U-compatibility: Let 2 IC 0 and Z 2 U. Now T 2 IC and hence T [ fzg 2 IC (which is U-compatible). It follows that [ fzg T [ fzg 2 IC 0. This concludes the proof that IC 0 is U-compatible. 7
Lemma 4 Let IC 0 be an A-consistency property closed under subsets. Let IC 00 consist of those sets all of whose nite subsets are in IC 0. Then IC 00 is also an A-consistency property, extending IC 0 and meeting the condition that 2 IC 00 if and only if all nite subsets of belong to IC 00. Also IC 00 is U compatible if IC 0 is. Proof Again, we check each item of the denition: 0. If 9 2 IC 00 with an atomic formula and its conjugate, then the nite set of this two formulas belongs to IC 0. This is a contradiction, since IC 0 is an A-consistency property. For the same reason, neither F > nor T? are in. 1. uppose 2 IC 00 and 2 and, by contradiction, [ f 1 ; 2 g =2 IC 00. Hence there is a nite set F 1 [ f 1 ; 2 g such that F 1 =2 IC 0. But 2 IC 00, then all its nite subsets are in IC 0, so F 1 cannot be a subset of alone, hence 1 or 2 or both are in F 1. Now let F1 1 be F 1? f 1 g, F1 2 be F 1? f 2 g and F1 12 be F 1? f 1 ; 2 g. Thus we have: F1 1 is a nite set with F 1 1 but F1 1 [ f 1g =2 IC 0 or F1 2 is a nite set with F 1 2 but F 1 2 [ f 2g =2 IC 0 or F1 1 2 is a nite set with F1 1 2 but F1 1 2 [ f 1 ; 2 g =2 IC 0 or all of them. Let F = F1 1 [ F1 2 [ F1 1 2 [ fg. then F is nite and F and since 2 IC 00, F 2 IC 0. But IC 0 is an A-consistency property, and 2 F, hence F [ f 1 g 2 IC 0, F [ f 2 g 2 IC 0, and F [ f 1 ; 2 g 2 IC 0. ince IC 0 is closed under subsets F 1 2 IC 0, which is a contradiction. o [ f 1 ; 2 g 2 IC 00. 2. uppose 2 IC 00 and 2 and, by contradiction, [ f 1 g =2 IC 00 and [ f 2 g =2 IC 00. ince [ f 1 g =2 IC 00 there is a nite subset F 1 [ f 1 g such that F 1 =2 IC 0. But 2 IC 00 then, since all nite subsets of are in IC 0, F 1 cannot be a subset of alone, hence 1 2 F 1. We write F1 1 for F 1? f 1 g. Thus F1 1 is a nite set and F1 1 but F1 1 [ f 1 g =2 IC 0. imilarly, F 2 [ f 2 g, F 2 nite and F 2 =2 IC 0. Then we write F2 1 for F 2? f 2 g and conclude F2 1 is nite, F2 1 but F2 1 [ f 2 g =2 IC 0. Let F = F1 1 [ F 2 1 [ fg. F is nite and F and 2 IC00, then F 2 IC 0 which is an A-consistency property. But 2 F, hence F [ f 1 g 2 IC 0 or F [ f 2 g 2 IC 0. But IC 0 is closed under subsets, hence F1 1 [ f 1 g 2 IC 0 or F2 1 [ f 2 g 2 IC 0. Thus we have a contradiction. Hence [ f 1 g 2 IC 00 or [ f 2 g 2 IC 00. 3. uppose 2 IC 00 and 2 2 and, by contradiction, 2 [ f 2 0 g =2 IC 00. Then there is some nite set F 1 2 [ f 2 0 g =2 IC 0. But F 1 cannot be a subset of 2 alone, since 2, and all nite subsets of are in IC 0. o 2 2 F 1. Let F 0 1 = F 1? f 2 0 g. We have F 0 1 nite, F 0 1 2, but F 0 1 [ f2 0 g =2 IC0. Let F = F 0 1 [ f2 g. F is nite, F and 2 IC 00. o F 2 IC 0. But IC 0 is an A-consistency property, so F 2 [ f 2 0 g 2 IC 0. By construction we have F 2 = F12 0, and since all elements of F 0 1 belong to 2 (again by construction), we have F 0 1 = F12 0. Hence F 2 = F1. 0 o, we have a contradiction. Hence 2 [ f 2 0 g 2 IC 00. 4. uppose 2 IC 00, 2 and, by contradiction, 2 [ [f 0 g =2 IC 00. Then there is some nite set F 2 [ [ f 0 g =2 IC 0. We consider two cases depending on where f 0 g is in F or not: i) f0 g =2 F. Then F 2 [, and F is nite. We construct F = f[]b : B 2 (F \ )g [ (F \ 2 ). o we have a nite F, and IC 00. o F 2 IC 0. But F is nite, so F 2 IC 00. ince f g 2 we have a nite F [ f g 2 IC 00, so F [ f g 2 IC 0, and since IC 0 is an A-consistency property, we have F [ f 0 g 2 IC0. Finally, since IC 0 is closed under subsets, we conclude F 2 IC 0, which is a contradiction. o we conclude 2 [ [ f 0 g 2 IC00. ii) f 0 g 2 F. Then let F 1 be F? f 0 g. We have F 1 nite, F 1 2 [, so just as in the previous case, we conclude F 1 [ f 0 g 2 IC0, which is a contradiction. 5. uppose 2 IC 00, V 2 2 and, by contradiction, [ fv 2 0 g =2 IC 00. Then there is some nite set F 1 2 [ fv 2 0 g =2 IC 0. But F 1 can not be a subset of alone, since all nite subsets of are in IC 0. o V 2 0 2 F 1. Let F 0 1 = F 1? fv 2 0 g. We have F 0 1 nite, F 0 1 but F 0 1 [ fv2 0 g =2 IC0. Now, let F = F 0 1 [ fv2 g. F is nite F and 2 IC 00, so F [ fv 2 0 g 2 IC0. IC 0 is closed under subsets, so it follows that F 0 1 [ fv 2 0 g 2 IC 0 and we have a contradiction. Hence, [ fv 2 0 g 2 IC 00. 8
ince IC 0 is closed under subsets, if 2 IC 0, all nite subsets of are in IC 0, hence 2 IC 00. Thus IC 00 extends IC 0. uppose F is nite. If F 2 IC 00 then all nite subsets of F must be in IC 0, in particular F 2 IC 0. Conversely, if F 2 IC 0, by the previous paragraph, F 2 IC 00. Hence IC 0 and IC 00 agree on nite subsets. Now, 2 IC 00 if and only if all nite subsets of are in IC 0 (by denition) if and only if all nite subsets of are in IC 00 (by the previous paragraph). For the U-compatibility, suppose 2 IC 00 and Z 2 U. 2 IC 00, so all nite subsets of belongs to IC 0. Let F nite and F. F 2 IC 0 and IC 0 is U-compatible, hence F [ fzg 2 IC 0. Hence all nite subsets of [ fzg belongs to IC 00 (by the previous paragraph). This concludes the proof. Denition 5 A collection IC of sets is said to be of nite character provided 2 IC i each nite subset of belongs to IC. Proposition 1 Any (U-compatible) A-consistency property may be extended to a (U-compatible) A-consistency property of nite character. Proof This follows immediately from lemmas 3 and 4. Proposition 2 Let IC be an A-consistency property of nite character. Each member of IC may be extended to a maximal member of IC. Proof The proof is identical as in [Fit83]. 6.2 Model existence theorem for A In this section we present the proof of the model existence theorem of A. This theorem will be used In the next section to show the completeness of A. Theorem 2 (Model existence theorem for A) Let IC be an A-consistency property. If is ICconsistent then is A-satisable. Proof Let IC be an A-consistency property, and suppose is IC-consistent ( 2 IC). We show is A-satisable. By proposition 1, IC IC 0, where IC 0 is also an A-consistency property and is of nite character. We use IC 0 to create an A-model. Let W be the set of all maximal members of IC 0 (there are some, by proposition 2). Let ACT 0 be a set of atomic actions. For v; w 2 W let vr w mean v w and vr2w mean v 2 w. We note the following: (1) v w ) v 2 w 2 and v w. (2) v 2 = v 22 and v 2 v. We claim (W; (R ) ;2ACT 0 ; R2) is a A-frame. In fact, there is no restriction on R and since v 2 v, R2 is reexive. Now suppose ur2v and vr2w, i.e., u 2 v and v 2 w. Thus, by the above observation, u 22 v 2. And since u 2 = u 22 we have u 2 v 2 w. Hence u 2 w, i.e., ur2w. o R2 is transitive. Hence (W; (R ) ;2ACT ; R2) is a A-frame. Now we dene an interpretation v in (W; (R ) 2ACT 0 ; R2) as follows: for each w 2 W and for each propositional variable P, set T if T P 2 w v(w; P ) = F otherwise The interpretation v determines a unique model = (W; (R ) 2ACT 0 ; R2; j=) which is an A-model. We claim: for each w 2 W and for each signed formula Z, Z 2 w implies j= w Z. We prove this by induction on the size of Z: Base step: If Z is of size 0, Z is T P or F P, where P is a variable. Then: T P 2 w ) v(w; P ) = T (denitions of v). o j= w P. F P 2 w ) T P =2 w (IC 0 is an A-consistency property so it does not contains an atomic formula and its conjugate). o v(w; P ) = F (denition of v). o 6j= w P. Induction Hypothesis (IH): uppose Z 2 w implies j= w Z is valid for all w 2 W and all signed formulas of size less or equal than n 0. 9
uppose now Z is of size greater than n > 0. We have several cases depending on where Z is of type ; ; 2 ; ; V 2 ; V : i) Case Z = : then 2 w implies w [ f 1 ; 2 g 2 IC 0 (since IC 0 is a A-consistency property). o 1 ; 2 2 w (since w is maximal). Hence j= w 1 and j= w 2 (IH). o j= w. ii) Case Z = : then 2 w implies (since IC 0 is a A-consistency property) w [ f 1 g 2 IC 0 or w [ f 2 g 2 IC 0. o 1 2 w or 2 2 w (since w is maximal). Hence j= w 1 or j= w 2 (IH). o j= w. iii) Case Z = 2 : then 2 2 w implies w 2 [ f 2 0 g 2 IC 0 (IC 0 is a A-consistency property). Now let w 0 be a maximal extension in IC 0 of w 2 [ f 2 0 g. Then w 0 2 W. ince 2 0 2 w 0 and 2 0 is of lower size than 2, by IH, j= w 0 2 0. ince w 2 w 0 we have wr2w 0. o j= w 2. iv) Case Z = : then 2 w implies w [w 2 [f0 g 2 IC 0. Now let w 0 be a maximal extension in IC 0 of w [ w 2 [ f0 g. Then w 0 2 W. ince 0 2 w 0 and 0 is of lower size than, by IH j= w 0 0. But w w 0, so wr w 0. Hence j= w. v) Case Z = V 2 : then V 2 2 w. Take any w 0 2 W such that wr2w 0, that is, w 2 w 0. We claim V 2 0 2 w0. In fact, if V 2 2 w then V 2 2 w 2, so V 2 2 w 0. ince V 2 2 w 0 we have w 0 [ fv 2 0 g 2 IC0 (and IC 0 is an A-consistency property). But w 0 is maximal in IC 0, so V 2 0 2 w 0. Now V 2 0 2 w 0, then j= w V2 0 0 by IH. o we have 8w 0 2 W such that wr2w 0 ; j= w V2 0 0. Hence, j= w V 2. vi) Case Z = V : then V 2 w. Take any w 0 2 W such that wr w 0, that is, w w 0. If V 2 w then by the denition of w we conclude V 0 2 w. o V 0 2 w 0. Hence by (IH), j= w 0 V 0. Thus we have 8w 0 2 W such that wr w 0 ; j= w 0 V 0. Hence j= w V. To nish the proof, recall we started with an A-consistency property IC and a set 2 IC. Then 2 IC 0. Extend to a maximal member w of IC 0 and by the above remarks w is A-satisable, hence the subset as well is. Theorem 3 (trong model existence theorem for A) Let U be a set of signed formulas and let IC be an A-consistency property that is U-compatible. If is IC-consistent, then is satisable at a world in some A-model = (W; (R ) 2ACT 0 ; R2; j=) in which the members of U hold at every possible world. Proof In addition to the previous proof, suppose IC is U-compatible. Then so is IC 0 (if w is a maximal member of IC 0 then U w, because Z 2 U then w [ fzg 2 IC 0 since IC 0 is U-compatible). But w is maximal so w [ fzg = w. Hence Z 2 w. Then j= w U for each w 2 W. This completes the proof. 6.3 Completeness theorem of A In this section we prove the completeness theorem for A. We will need to show a lemma. Denition 6 uppose fz 1 ; ; Z n g is a nite set of formulas. By an A-tableau for fz 1 ; ; Z n g we mean any tableau that begins: Z 1 Z 2. Z n and continues by applications of the ; ; V 2 ; 2 and - rules. Denition 7 A nite set of signed formulas is A-consistent if no A-tableau for closes. Lemma 5 Let IC be the collection of all A-consistent sets. IC is an A-consistency property. 10
Proof We check each item of the denition: 0. uppose 2 IC. If contains an atomic formula and its conjugate then is a closed tableau. o is not A-consistent, thus =2 IC, which is a contradiction. o does not contain a formula and its conjugate. For the same reason does contain neither F > nor T?. 1. uppose 2 IC, 2 and [ f 1 ; 2 g =2 IC. Then there is a closed tableau of the form: 1 2 where represents the tableau structure below the occurrence of 1 and 2. But since 2 we may begin an A-tableau proof by an application of an -rule, which results in: 1 2 Now we just copy the structure represented by and we get a closed tableau for. o is not A-consistent, thus =2 IC, which is a contradiction. Hence [ f 1 ; 2 g 2 IC. 2. uppose 2 IC, 2, and [ f 1 g =2 IC and [ f 2 g =2 IC. If [ f 1 g =2 IC then there is a closed tableau of the form: 1 1 where 1 represents the tableau structure below the occurrence of 1. And if [ f 2 g =2 IC then there is a closed tableau of the form: 2 2 where 2 represents the tableau structure below the occurrence of 2. But 2, so we may begin an A-tableau proof by a -rule, like this:. & 1 2 and we can copy the structures 1 and 2 below the respective occurrences of 1 and 2. In this way we obtain:. & 1 2 1 2 11
which is a closed tableau for. o is not A-consistent, thus =2 IC, which is a contradiction. Hence [ f 1 g 2 IC or [ f 2 g 2 IC. 3. uppose 2 IC, 2 2 and 2 [ f 2 0 g =2 IC. Then there is a closed tableau of the form: 2 2 0 where represents the tableau structure below the occurrence of 2 0. Now, since 2 2, we may begin an A-tableau proof by a 2 -rule in to obtain: 2 # 2 2 0 and if we copy the structure represented by we get a closed tableau for. Thus is not A-consistent and so =2 IC, which is a contradiction. Hence 2 [ f 2 0 g 2 IC. 4. uppose 2 IC, 2 and 2 [ [ f 0 g =2 IC. Then there is a closed tableau of the form: 2 0 where represents the tableau structure below the occurrence of 0. Now, since 2, we may begin an A-tableau proof by a -rule in to obtain: # 2 0 and if we copy the structure represented by we get a closed tableau for. Thus is not A-consistent and so =2 IC, which is a contradiction. Hence 2 [ [ f 2 0 g 2 IC. 5. uppose 2 IC, V 2 2 and [ fv 2 0 g =2 IC. Then there is a closed tableau of the form: V 2 0 where represents the tableau structure below the occurrence of V 2 0. Now, since V 2 2, we may begin an A-tableau proof by a V 2 -rule in to obtain: V 2 V 2 0 and if we copy the structure represented by we get a closed tableau for. Thus is not A-consistent and so =2 IC, which is a contradiction. Hence [ fv 2 0 g 2 IC. 12
This nishes the proof of the lemma. Theorem 4 (Completeness theorem for A) Let X be a formula. A-tableau proof. If j=a X then X has an Proof uppose A has no A-tableau proof. Then there is no closed tableau for ff Ag. Hence by lemma 5, ff Ag 2 IC. By the model existence theorem, ff Ag is A-satisable, hence A is not valid in all A-models. This is a contradiction. 7 Conclusion We have presented the denition of the logic A, which is a version of Propositional Dynamic Logic (PDL) that we claim suitable for reasoning about actions and plans. This logic is an alternative to PDL in the sense that it presents an alternative to the iteration operator. It is a formal system with \almost" the same expressive power of PDL, in the sense that we are able to speak about atomic and compound actions, as well as action laws, integrity constraints, goals and the existence of plans. We have presented a tableau system for this logic in order to allow automatic deductions of facts. We have showed correctness and completeness results. We have presented open problems which are subjetc of future research. References [Fit83] M. Fitting. Proof methods for modal and intuitionistic logics. Reidel Publishing Company, 1983. [GH95] O. Gasquet and A. Herzig. Reasoning about actions using dependence relations. Technical report, IRIT - Universite Paul abatier, 1995. [Har84] D. Harel. Dynamic logic. In D. Gabbay and F. Gunthner, editors, Handbook of Philosophical Logic, volume II, pages 497{604. D. Reidel Publishing Company, 1984. [MH69] J. McCarthy and P. Hayes. ome philosophical problems from the standpoint of articial intelligence. In B. Meltzer and D. Michie, editors, Machine Intelligence, volume 4, pages 463{502. Edinburgh University Press, 1969. 13