Checking the Size of Circumscribed Formulae

The circumscription of a propositional formula T may not be representable in polynomial space, unless the polynomial hierarchy collapses. This depends on the specific formula T , as some can be circumscribed in little space and others cannot. The problem considered in this article is whether this happens for a given formula or not. In particular, the complexity of deciding whether CIRC(T ) is equivalent to a formula of size bounded by k is studied. This theoretical question is relevant as circumscription has applications in temporal logics, diagnosis, default logic and belief revision. Keywords—Circumscription; computational complexity; belief revision.


I. INTRODUCTION
The circumscriptive reasoning mechanism requires a set of variables to be minimized [1], [2], that is, set to the logical value false whenever possible.Similarly to the closed world assumption [3], it formalizes the assumption that lack of information on certain conditions can be considered evidence that they do not hold.Applications include temporal domains [4], [5], diagnosis [6], induction [7] and belief revision [8].Contrary to the basic closed world assumption, circumscription takes into account all possible ways variables can be set to false; for example, x ∨ y is consistent with either ¬x and ¬y but not both, leading to the two possible cases (x∨y)∧¬x and (x∨y)∧¬y.These may be up to 2 n , if the number of variables is n: a trivial representation of the circumscribed formula may be exponential.However, it may be equivalent to a smaller formula.
Expressing propositional circumscription as a formula of size bounded by a polynomial has been proved not possible in general [9], unless the polynomial hierarchy collapses [10], a condition generally deemed unlikely.As a result, the problem of whether propositional circumscription can be represented in space bounded by some number k has not an obvious answer: it is possible in some cases but not in others.The problem considered in this article is whether this is possible; in particular, the complexity of this problem is studied.This is similar to the problem of minimizing propositional formulae: given a formula F , is there an equivalent formula of size bounded by k [11]?For circumscription, the question is whether the circumscription of a formula is equivalent to some formula of size bounded by k.For example, the circumscription of x ∨ y accounts for both (x ∨ y) ∧ ¬x and (x ∨ y) ∧ ¬y to be possible; therefore, the result is the formula ((x∨y)∧¬x)∨((x∨y)∧¬y).However, this formula is equivalent to (x ∧ ¬y) ∨ (¬x ∧ y).By the standard metric of formulae where size is defined as the number of variable occurrences, this formula has size 4. Therefore, the circumscription of x ∨ y is equivalent to a formula of size bounded by k = 4, but not for example k = 1 as no formula of a single variable is equivalent to (x ∧ ¬y) ∨ (¬x ∧ y).The answer is not this easy when the formula is more complex than x ∨ y.Indeed, it will be proved that the problem is hard for the complexity class Π p 2 , that is, harder than problems such as propositional satisfiability, vertex cover and Hamiltonian cycle [10].
The question of the size of the representation has an implementation impact.Indeed, verifying which conditions hold under the circumscription assumption amounts to CIRC(T ) |= C, where T represents the current information and C the condition to check, and this is an hard problem [12], [13], [14].However, if CIRC(T ) can be represented by a formula F of bounded size, the problem can be solved by first finding F and then solving the easier (coNP) problem F |= C. Once F is determined, any number of other conditions C 1 , C 2 , . . .can then be checked against F at the same cost.
Since circumscription is also used as the target of translation of several belief revision operators, the question concerns the dynamic of logic.Indeed, changing a formula to accommodate for new information is generally expected to produce a result of bounded size.
The article is organized as follows: the next section contains the formal definition of circumscription and the notations used in this article, plus two preliminary lemmas; in the section afterwards, the complexity of the problem of whether the circumscription of a formula can be represented in size bounded by some number is studied; the final section comments the practical implications of this analysis and its open problems.

II. PRELIMINARY RESULTS
Propositional formulae are denoted by the capital letters T and F , and are always assumed to be in Negation Normal Form (NNF).Sets of variables are denoted by X, Y and Z. Notation X ¬ indicates the set {¬x | x ∈ X}.The shorthand x ≡ y indicates (x ∧ ¬y) ∨ (¬x ∧ y).
Models are denoted by ω X , where the suffix X indicates the set of variables: ω X is a truth evaluation of the variables X, ω Y is a truth evaluation of the variables Y , etc. Models are identified by the sets of variables they assign to true; this allows to write ω X ⊆ ω X to mean that ω X assigns true to all variables ω X assigns true, but not necessarily the converse.The model assigning true to all variables X is denoted ω + X , the one assigning false to all ω − X .The following notation is used to denote a formula that represents a single model: If F is a formula over variables X ∪ Y and ω X a truth evaluation over X, the notation F | ω X indicates the formula obtained by replacing each variable X in F with its truth value according to ω X .
In this article, circumscription is defined over propositional logic, and restricted to the case where all variables are minimized.This gives rise to the following definition.
Definition 1: Given a formula T over variables X, its circumscription CIRC(T ) is defined as follows, where Some formulae T have small circumscription.For example, T = X has a circumscription equal to itself, since S = ∅ is the only subset of X ¬ satisfying the definition.Some other formulae have larger circumscription, such as T = X; indeed, for this formula S = X ¬ \{x} satisfies the definition for every x ∈ X.Some formulae do not even have polynomialsize equivalent representations of their circumscription [9].
Circumscription is simple to compute on formulae that imply either x, ¬x, or x ≡ x for some variables x and x : Property 1: The following equivalences hold: )) These are well-known properties.The third equivalence allows evaluating CIRC(T ) separately for x true and x false, if T does not contain x .
The size of formulae is defined by the following metrics.

Definition 2:
The size of a formula F , denoted ||F ||, is the number of variable occurrences in F .
For example, the size of (a ∧ ¬b) ∨ c ∨ ¬(¬a) is four, since the variable a occurs twice in it and b and c once each.According to this definition, the size of a formula and of its NNF form obtained by applying the De Morgan rules coincide.A bound on the size of a formula derives from its models.
Lemma 1: If a NNF formula F has a model that satisfies a literal l but not the modified model where the value of l is inverted, then F contains l.
Proof: Let F be a formula and ω X its model satisfying l.Let us assume, on the converse, that F does not mention the literal l.Since F is in NNF, no part of it is turned to false by changing the value of l from true to false.As a result, the model ω X obtained by changing the value of l in ω X satisfies F , contradicting the assumption of the lemma.
As a consequence, if a formula is satisfied by a model where x is true but not by the same model where x is false, and vice versa, then any formula equivalent to it contains both x and ¬x.Therefore, if a formula contains x ≡ y, either conjoined with a satisfiable formula not containing x and y or disjoined with a non-valid formula not containing x and y, then it must contain at least two literal occurrences for x and two for y.The following lemma shows a sufficient condition for the presence of a literal in a formula.
Lemma 2: Let F be a formula over X ∪ Y .For any truth evaluation ω X , no formula equivalent to F is smaller than the smallest formula equivalent to F | ω X .
Proof: Let T be a formula equivalent to F .Equivalence is preserved when replacing a variable with a truth value in both formulae.As a result, F | ω X ≡ T | ω X .Furthermore, such a replacement does not increase the number of literal occurrences in T , since it only replace some variables with either true or false.As a result, the size of T | ω X is less than or equal to the size of T .Since T | ω X is a formula equivalent to F | ω X , it is at least as large as the smallest formula equivalent to F | ω X .Since T is larger or has the same size, the claim is proved.
This lemma is useful when formulae contain parts that are satisfiable only for a specific truth evaluation of some variables X.Such formulae are built to the aim of generating a (relatively) large subformula whenever a condition is met.

III. THE SIZE OF CIRCUMSCRIPTIVE FORMULAE
In this section, we analyze the problem of deciding whether the circumscription of a formula can be represented by a formula of size bounded by an integer k, in unary notation.The unary notation is used to avoid exponentially-sized formulae to be taken into account.Equivalently, the problem could be reformulated as: is there any formula that is equivalent to CIRC(T ) and has size less or equal than another formula G? In this proof the following notations are used, where X and X are sets of variables in one-to-one correspondence and each x corresponds to x = c(x): Formula T and number k are as follows.
The reduction works as follows: X ≡ X allows expressing CIRC(T ) in terms of the disjunction of CIRC(T | ω X ) for all possible ω X ; if ∀X∃Y.F is true, all these formulae CIRC(T | ω X ) can be expressed in the same way, so that a single formula equivalent to CIRC(T ) exists with size bounded by k; otherwise, for the evaluation ω X that makes F false CIRC(T | ω X ) alone has size greater than k.
The first step employs the third equivalence of Property 1, when applied to every x ∈ X and its respective x ∈ X , since T contains X ≡ X : The second step of the proof is to analyze CIRC(T | ω X ) for an evaluation ω X .Formula T | ω X can be rewritten as follows.
In this last formula, ω X is the evaluation of X setting each variable in X to the opposite value of the corresponding variable in X.This formula does not contain any variable in X.Therefore, CIRC(T | ω X ) is defined by taking into account only the other variables: X , Y , Y and Z.Since X has a fixed value, it holds: The first subformula of circumscription  This proves that every model of the first subformula contains a model of the second, if F | ω X is satisfiable.If this is the case, the first subformula is irrelevant to circumscription.Otherwise, the second subformula is unsatisfiable.
The rest of the proof depends on whether F is satisfiable for every

IV. CONCLUSIONS
The problem of checking whether the circumscription of a formula can be represented by a formula of size bounded by k turned out to be Π p 2 -hard in Σ p 3 .These two classes are at the second and third level of the polynomial hierarchy, respectively.As a result, the problem cannot be solved by a propositional satisfiability solver.It can, however, be translated into a QBF and then passed as input to one of the existing QBF solvers [15].An open question is how much complexity decreases if the formulae are in Horn form, and in particular if some additional restriction makes the problem tractable.If k is in binary representation rather than unary, the question is whether CIRC(T ) can be represented by a formula that may be exponential, but still bounded by k.The necessity of considering such large formulae is likely to make this problem harder than with k in unary notation: polynomial space may not be sufficient to solve it.Indeed, assuming k in unary notation amounts to requiring the equivalent formula to have size comparable to that of the input data.This is equivalent to ask whether CIRC(T ) is equivalent to a formula of the same size of another formula G, for example.Allowing k to be stored in binary form with n bit allows the bound be as large as 2 n − 1.As a result, even formulae of exponential size are allowed as representations of CIRC(T ).What complicates the analysis is that the usual guess-and-check algorithm for finding such a formula does not work in polynomial space, as this may not be enough for even storing the formula.A cycle of the minimal models of T is still feasible, but this may not allow determining the size of a formula satisfied exactly by all of them, unless such a formula is explicitly produced.
which is also equivalent to (Z ≡ Z ) ∧ Y ∧ Y by applying the second and third equivalence of Property 1.For every z ∈ Z, this formula has a model that makes z true, but changing only the evaluation of z results in a model not satisfying this formula.The same applies to all variables in Z and Z and their negation, and to all variables in Y and Y .By Lemma 1, every formula equivalent to this one has size greater than or equal to 4|Z| + 2|Y | = 4m + 2n = 4(3n + ||F || + 1) + 2n = 12n + 4||F || + 4 + 2n = 16n + 4||F || + 4 > k.By Lemma 2, every formula equivalent to CIRC(T ) has size greater than or equal to this amount.www.ijarai.thesai.org

Theorem 1 :
The problem of deciding whether CIRC(T ) is equivalent to a formula F with ||F || ≤ k, where k is a number in unary notation, is in Σ p 3 .Proof: The problem can be reformulated as follows: check whether there exists a formula F that is equivalent to CIRC(T ) and ||F || ≤ k.The problem F |= CIRC(T ) is in coNP, since it amounts to check whether ω ⊂ ω for every ω |= T and ω |= F .Since coNP is a subclass of Π p in unary notation such that ∀X∃Y.F is valid if and only if CIRC(T ) is equivalent to a formula of size ≤ k.Let us assume, without loss of generality, that |X| = |Y | = n.The reduction introduces a set of new variables X in in one-on-one correspondence with X.It also introduces a set of new variables Y in correspondence with Y and a set of new variables Z of cardinality m = 3n + ||F || + 1.
The problem can be proved hard for the class Π p 2 .Theorem 2: The problem of deciding whether CIRC(T ) is equivalent to a formula T with ||T || ≤ k is Π p 2 -hard.Proof: Let F be a formula over variables X ∪ Y .The proof shows how to build in polynomial time a formula T and www.ijarai.thesai.org a number k which ω + Y and ω + Y set all variables in Y and Y to true.This model contains a model of the second subformula if F | ω X is satisfiable.Indeed, let ω Y be the model that satisfies F | ω X .This model is contained in ω + Y .The model ω Y that assigns y ∈ Y to true if and only if the corresponding y ∈ Y is false in ω and is contained in ω + Y .A model of the second subformula is therefore ω Y ∪ω Y ∪ω − Z ∪ω − Z , where ω − Z ∪ω − Z set all variables to false and are therefore contained in ω Z ∪ ω Z .