Embedding Proof Problems into Query-answering Problems and

Problem Solving by Equivalent Transformation

Kiyoshi Akama

and Ekawit Nantajeewarawat

Information Initiative Center, Hokkaido University, Hokkaido, Japan

Computer Science, Sirindhorn International Institute of Technology, Thammasat University, Pathumthani, Thailand

Keywords:

Question-answering Problems, Proof Problems, Equivalent Transformation, Solving Logical Problems.

Abstract:

A proof problem is a “yes/no” problem concerning with checking whether one logical formula is a logical

consequence of another logical formula, while a query-answering problem (QA problem) is an “all-answers

ﬁnding” problem concerning with ﬁnding all ground instances of a query atomic formula that are logical

consequences of a given logical formula. In order to establish a precise relation between these two problem

classes, the concept of an embedding mapping is introduced. When one problem class can be embedded into

another problem class at low computational cost, the former class can be regarded as a subclass of the latter

class and, consequently, problems in the former class can be solved through a method for solving problems in

the latter one. Construction of low-cost embedding mappings from proof problems to QA problems is demon-

strated. By such embedding, proof problems can be solved using a procedure for solving QA problems. A

procedure for solving QA problems based on the equivalent transformation principle is presented. Application

of the procedure to the two problem classes is illustrated.

1 INTRODUCTION

Given a ﬁrst-order formula K, representing back-

ground knowledge, and an atomic formula (atom) a,

representing a query, a query-answering problem (QA

problem) is to ﬁnd the set of all ground instances of a

that are logical consequences of K. Characteristically,

it is an “all-answers ﬁnding” problem, i.e., all ground

instances of the query atom satisfying the require-

ment must be found. A proof problem, by contrast,

is a “yes/no” problem; it is concerned with checking

whether or not one given logical formula is a logical

consequence of another given logical formula.

Historically, works on logic-based automated rea-

soning have been centered around proof problems

(Chang and Lee, 1973; Gallier, 1986; Fitting, 1996;

Newborn, 2000). Methods for solving proof prob-

lems were developed, e.g., tableau-based methods

(Beth, 1955) and resolution-based methods (Robin-

son, 1965), and they have been subsequently adapted

to address other classes of logical problems, including

some speciﬁc subclasses of QA problems, e.g., QA

problems on deﬁnite clauses (Lloyd, 1987). As op-

posed to such a proof-centered approach, we present

in this paper a direct approach towards solving QA

problems on the basis of the equivalent transforma-

tion (ET) principle. We show that proof problems can

naturally be considered as QA problems of a special

form; therefore, a method for solving QA problems

also lends itself to solve proof problems in a straight-

forward way.

In order to clearly understand the relation between

proof problems and QA problems, we introduce the

notion of an embedding mapping from one problem

class to another problem class. Using an embedding

mapping, we demonstrate that proof problems can

be formulated as a subclass of QA problems. We

propose a framework for solving QA problems by

ET. A given input QA problem on ﬁrst-order logic is

converted into an equivalent QA problem on an ex-

tended clause space, called the ECLS

space, through

meaning-preserving Skolemization (Akama and Nan-

tajeewarawat, 2011). The obtained QA problem is

then successively transformed on the ECLS

space by

application of ET rules until the answer to the original

problem can be readily obtained. With an embedding

mapping from proof problems to QA problems, this

framework can be used for solving proof problems.

To begin with, Section 2 formalizes QA problems

and proof problems. Section 3 deﬁnes an embedding

mapping and shows how to embed proof problems

into QA problems. Section 4 introduces extended

253

Akama K. and Nantajeewarawat E..

Embedding Proof Problems into Query-answering Problems and Problem Solving by Equivalent Transformation.

DOI: 10.5220/0004546202530260

In Proceedings of the International Conference on Knowledge Engineering and Ontology Development (KEOD-2013), pages 253-260

ISBN: 978-989-8565-81-5

 2013 SCITEPRESS (Science and Technology Publications, Lda.)

clauses, the extended space ECLS

and QA problems

on this space. Section 5 presents our ET-based proce-

dure for solving QA problems. Section 6 deﬁnes un-

folding transformation on the ECLS

space and pro-

vides some other ET rules on this space. Section 7

illustrates application of our framework. Section 8

concludes the paper.

2 QA PROBLEMS AND PROOF

PROBLEMS

2.1 Interpretations and Models

In this paper, an atom occurring in a ﬁrst-order for-

mula can be either a usual atom or a constraint atom.

The semantics of ﬁrst-order formulas based on a log-

ical structure given in (Akama and Nantajeewarawat,

2012) is used. The set of all ground usual atoms, de-

noted by G, is taken as the interpretation domain. An

interpretation is a subset of G . A ground usual atom g

is true with respect to an interpretation I iff g belongs

to I. Unlike ground usual atoms, the truth values of

ground constraint atoms are predetermined indepen-

dently of interpretations. A model of a ﬁrst-order for-

mula E is an interpretation that satisﬁes E. The set

of all models of a ﬁrst-order formula E is denoted by

Models(E). Given ﬁrst-order formulas E

and E

, E

is a logical consequence of E

iff every model of E

is a model of E

2.2 QA Problems

A query-answering problem (QA problem) is a pair

hK, ai, where K is a ﬁrst-order formula, representing

background knowledge, and a is a usual atom, repre-

senting a query. The answer to a QA problem hK, ai,

denoted by answer

(hK, ai), is deﬁned as the set of

all ground instances of a that are logical consequences

of K. Using Models(K), the answer to a QA problem

hK, ai can be equivalently deﬁned as

answer

(hK, ai) = (

Models(K)) ∩ rep(a),

where rep(a) denotes the set of all ground instances

of a. Accordingly, a QA problem can also be seen

as a model-intersection problem. When no confu-

sion is caused, answer

(hK, ai) is often written as

answer

(K, a).

2.3 Proof Problems

A proof problem is a pair hE

, E

i, where E

and E

are ﬁrst-order formulas, and the answer to this prob-

lem, denoted by answer

(hE

, E

i), is deﬁned by

answer

(hE

, E

i) =







“yes” if E

is a logical

consequence of E

“no” otherwise.

It is well known that a proof problem hE

, E

i can be

converted into the problem of determining whether

∧ ¬E

is unsatisﬁable (Chang and Lee, 1973),

i.e., whether E

∧ ¬E

has no model. As a result,

answer

(hE

, E

i) can be equivalently deﬁned by

answer

(hE

, E

i) =







“yes” if Models(E

∧ ¬E

)

is the empty set,

“no” otherwise.

When no confusion is caused, answer

(hE

, E

i) is

often written as answer

, E

3 EMBEDDING PROOF

PROBLEMS INTO QA

PROBLEMS

3.1 Embedding Mappings

The notion of a class of problems and that of an em-

bedding mapping are formalized below.

Deﬁnition 1. A class C of problems is a triple

hPROB, ANS, answeri, where

1. PROB and ANS are sets,

2. answer is a mapping from PROB to ANS.

The sets PROB and ANS are called the problem space

and the answer space, respectively, of C. Their el-

ements are called problems and (possible) answers,

respectively, in C. Given a problem prb ∈ PROB,

answer(prb) is the answer to prb in C.

Deﬁnition 2. Let C

= hPROB

, ANS

, answer

i and

= hPROB

, ANS

, answer

i be classes of prob-

lems. An embedding mapping from C

to C

is a pair

hπ, αi, where π is an injective mapping from PROB

to PROB

and α is a partial mapping from ANS

ANS

such that for any prb ∈ PROB

, answer

(prb) =

α(answer

(π(prb))).

Let C

and C

be classes of problems. Suppose

that (i) there exists an embedding mapping hπ, αi

from C

to C

, (ii) there exists a procedure P for solv-

ing problems in C

, and (iii) there also exist a proce-

dure P

for realizing π and a procedure P

for real-

izing α. Then a procedure for solving problems in

can be obtained by making the composition of the

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procedures P

, P and P

. C

is regarded as a subclass

of C

iff there exists an embedding mapping hπ, αi

from C

to C

such that π and α can be realized at

low computational cost.

3.2 Embedding Proof Problems into QA

Problems

Next, we show how to embed proof problems into QA

problems. Assume that:

• C

= hPROB

, ANS

, answer

i is the class of

QA problems deﬁned by Section 2.2, i.e., PROB

is the set of all QA problems, ANS

is the power

set of G, and answer

: PROB

→ ANS

is given

by Section 2.2.

• C

= hPROB

, ANS

, answer

i is the class of

proof problems deﬁned by Section 2.3, i.e.,

PROB

is the set of all proof problems, ANS

{“yes”,“no”}, and answer

: PROB

→ ANS

given by Section 2.3.

In order to construct an embedding mapping from

to C

, we want to construct from any ar-

bitrary given proof problem hE

, E

i a QA prob-

lem hK, yesi such that answer

, E

) = “yes” iff

answer

(K, yes) = {yes}, where yes is a 0-ary pred-

icate symbol and the atom yes occurs in neither E

nor E

. The following approaches can be taken for

constructing such a formula K:

• Construct K such that every model of K contains

yes iff answer

, E

) = “yes”.

• Construct K such that K has no model iff

answer

, E

) = “yes”.

We refer to the ﬁrst approach as positive construction,

and the second one as negative construction. They are

given below.

3.2.1 Embedding using Positive Construction

Positive construction of an embedding mapping from

to C

can be obtained by Proposition 1.

Proposition 1. Let E

and E

be ﬁrst-order formulas.

Assume that:

1. yes is a 0-ary predicate symbol and yes occurs in

neither E

nor E

2. prb

is the proof problem hE

, E

3. prb

is the QA problem hyes ↔ (E

→ E

), yesi.

Then answer

(prb

) = “yes” iff answer

(prb

) =

{yes}.

Proposition 1 determines an embedding mapping

hπ

, α

i from C

to C

as follows:

• For any proof problem hE

, E

(hE

, E

i) = hyes ↔ (E

→ E

), yesi.

• α

({yes}) = “yes” and α

(

0) = “no”.

3.2.2 Embedding using Negative Construction

The next proposition illuminates negative construc-

tion of an embedding mapping from C

to C

Proposition 2. Let E

and E

be ﬁrst-order formulas.

Assume that:

1. yes is a 0-ary predicate symbol and yes occurs in

neither E

nor E

2. prb

is the proof problem hE

, E

3. prb

is the QA problem hE

∧ ¬E

, yesi.

Then answer

(prb

) = “yes” iff answer

(prb

) =

{yes}.

Proposition 2 determines an embedding mapping

hπ

, α

i from C

to C

as follows:

• For any proof problem hE

, E

(hE

, E

i) = hE

∧ ¬E

, yesi.

• α

({yes}) = “yes” and α

(

0) = “no”.

4 QA PROBLEMS ON AN

EXTENDED SPACE

To solve a QA problem hK, ai on ﬁrst-order logic,

the ﬁrst-order formula K is usually converted into a

conjunctive normal form. The conversion involves

removal of existential quantiﬁcations by Skolemiza-

tion, i.e., by replacement of an existentially quantiﬁed

variable with a Skolem term determined by a relevant

part of a formula prenex. The classical Skolemiza-

tion, however, does not preserve the logical mean-

ing of a formula—the formula resulting from Sko-

lemization is not necessarily equivalent to the orig-

inal one (Chang and Lee, 1973). In (Akama and

Nantajeewarawat, 2011), a theory for extending the

space of ﬁrst-order formulas was developed and how

meaning-preservingSkolemization can be achievedin

the obtained extended space was shown. A procedure

for converting ﬁrst-order formulas into extended con-

junctive normal forms in an extended clause space,

called the ECLS

space, was also presented.

The basic idea of meaning-preserving Skolemiza-

tion is to use existentially quantiﬁed function vari-

ables instead of usual Skolem functions. Func-

tion variables, extended clauses, extended conjunctive

normal forms and QA problems on ECLS

are intro-

duced below.

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4.1 Function Constants, Function

Variables and func-Atoms

A usual function symbol, say f, in ﬁrst-order logic

denotes an unevaluated function; it is used for con-

structing from existing terms, say t

, . . . ,t

, a syntac-

tically new term, e.g., f(t

, . . . ,t

), possibly recur-

sively, without evaluating the new term f(t

, . . . ,t

A different class of functions is used in the extended

space. A function in this class is an actual mathe-

matical function, say h, on ground terms; when it

takes ground terms, say t

, . . . ,t

, as input, h(t

, . . . ,t

)

is evaluated for determining an output ground term.

We called a function in this class a function constant.

Variables of a new type, called function variables, are

introduced; each of them can be instantiated into a

function constant or a function variable, but not into a

usual term.

In order to clearly separate function constants

and function variables from usual function symbols

and usual terms, a new built-in predicate symbol

func is introduced. Given any n-ary function con-

stant or n-ary function variable

f, an expression

func(

f, t

, . . . ,t

n+1

), where the t

are usual terms,

is considered as an atom of a new type, called a func-

atom. When

f is a function constant and the t

are

all ground, the truth value of this atom is evaluated as

follows: it is true iff

f(t

, . . . ,t

) = t

n+1

4.2 Extended Clauses

An extended clause C is a closed formula of the form

∀v

, . . . , ∀v

: (a

∨ ··· ∨ a

∨ ¬b

∨ ··· ∨ ¬b

∨ ¬f

∨ ··· ∨ ¬f

where v

, . . . , v

are usual variables, each of a

, . . . ,

, b

, . . . , b

is a usual atom or a constraint atom,

and f

, . . . , f

are func-atoms. It is often written sim-

ply as (a

, . . . , a

← b

, . . . , b

, f

, . . . , f

). The sets

, . . . , a

} and {b

, . . . , b

, f

, . . . , f

} are called the

left-hand side and the right-hand side, respectively, of

the extended clause C, denoted by lhs(C) and rhs(C),

respectively. When n = 0, C is called a negative ex-

tended clause. When n = 1, C is called an extended

deﬁnite clause, the only atom in lhs(C) is called the

head of C, denoted by head(C), and the set rhs(C) is

also called the body of C, denoted by body(C). When

n > 1, C is called a multi-head extended clause. All

usual variables in an extended clause are universally

quantiﬁed and their scope is restricted to the clause

itself. When no confusion is caused, an extended

clause, a negative extended clause, an extended deﬁ-

nite clause and a multi-head extended clause will also

be called a clause, a negative clause, a deﬁnite clause

and a multi-head clause, respectively.

An extended normal form called existentially

quantiﬁed conjunctive normal form (ECNF) is a for-

mula of the form ∃v

, . . . , ∃v

: (C

∧ ··· ∧ C

where v

, . . . , v

are function variables and C

, . . . ,

are extended clauses. It is often identiﬁed with

the set {C

, . . . ,C

}, with implicit existential quantiﬁ-

cations of function variables and implicit clause con-

junction. Function variables in such a clause set are

all existentially quantiﬁed and their scope covers en-

tirely all clauses in the set.

4.3 QA Problems on ECLS

The set of all ECNFs is referred to as the extended

clause space (ECLS

). By the above identiﬁcation of

an ECNF with a clause set, we often regard an ele-

ment of ECLS

as a set of (extended) clauses. With

occurrences of function variables, clauses contained

in a clause set in the ECLS

space are connected

through shared function variables. By instantiating

all function variables in such a clause set into func-

tion constants, clauses in the obtained set are totally

separated.

A QA problem hCs, ai such that Cs is a clause

set in ECLS

and a is a usual atom is called a QA

problem on ECLS

. Given a QA problem hK, ai

on ﬁrst-order logic, the ﬁrst-order formula K can be

converted equivalently by meaning-preserving Skol-

emization, using the conversion procedure given in

(Akama and Nantajeewarawat, 2011), into a clause

set Cs in the ECLS

space. The obtained clause set

Cs may be further transformed equivalently in this

space for problem simpliﬁcation, by using unfolding

and other transformation rules.

5 SOLVING QA PROBLEMS

Using the notation introduced in Sections 5.1 and 5.2,

our ET-based procedure is presented in Section 5.3.

5.1 Inclusion of Query Information

The following notation is used. A set A of usual atoms

is said to be closed iff for any a ∈ A and any substi-

tution θ for usual variables, aθ belongs to A. Assume

that (i) A is the set of all usual atoms, (ii) A

and A

are disjoint closed subsets of A, and (iii) φ is a bijec-

tion from A

to A

such that for any a ∈ A

and any

substitution θ for usual variables, φ(aθ) = φ(a)θ. For

any i, j ∈ {1, 2}, an extended clause C is said to be

from A

to A

iff all usual atoms in rhs(C) belong to

and all those in lhs(C) belong to A

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Let hK, ai be a QA problem such that K is a ﬁrst-

order formula in which all usual atoms belong to A

and a ∈ A

. As will be detailed in Section 5.3, to solve

this problem using ET, K is transformed by meaning-

preserving transformation into a set Cs of extended

clauses from A

to A

and a singleton set Q consist-

ing only of the clause (φ(a) ← a) from A

to A

constructed from the query atom a. The resulting

QA problem hCs∪Q, φ(a)i is then successively trans-

formed using ET rules.

5.2 Triples for Transformation

In order to make a clear separation between a set of

extended clauses from A

to A

and a set of those

from A

to A

in a transformation process of QA

problems, the following notation is introduced: Given

a set Cs of extended clauses from A

to A

, a set Q

of extended clauses from A

to A

and an atom b in

, let the triple hCs, Q, bi denote the QA problem

hCs ∪ Q, bi. A QA problem hCs, Q, bi can be trans-

formed by changing Cs, by changing Q, or by chang-

ing both Cs and Q.

Deﬁnition 3. A transformation of a QA problem

hCs, Q, bi into a QA problem hCs

′

, Q

′

, bi is equiva-

lent transformation (ET) iff answer

(Cs ∪ Q, b) =

answer

(Cs

′

∪ Q

′

, b).

5.3 A Procedure for Solving QA

Problems by ET

Let A

be a closed set of usual atoms. Assume that a

QA problem hK, ai is given, where K is a ﬁrst-order

formula in which all usual atoms belong to A

and

a ∈ A

. To solve the QA problem hK, ai using ET,

perform the following steps:

1. Transform K by meaning-preserving Skolemiza-

tion into a clause set Cs in the ECLS

space.

2. Determine (i) a closed set A

of usual atoms such

that A

and A

are disjoint and (ii) a bijection φ

from A

to A

such that for any a ∈ A

and any

substitution θ for usual variables, φ(aθ) = φ(a)θ.

3. Successively transform the QA problem hCs,

{(φ(a) ← a)}, φ(a)i in the ECLS

space using un-

folding and other ET rules (see Section 6).

4. Assume that the transformation yields a QA prob-

lem hCs

′

, Q, φ(a)i. Then:

(a) If Models(Cs

′

) =

0, then output rep(a) as the

answer.

(b) If Models(Cs

′

) 6=

0 and Q is a set of unit clauses

such that the head of each clause in Q is an in-

stance of φ(a), then output as the answer the set

−1

(

[

C∈Q

rep(head(C))).

It is shown in (Akama and Nantajeewarawat, 2013)

that the obtained answer is always correct.

The set A

and the bijection φ satisfying the re-

quirement of Step 2 can be determined as follows:

First, introduce a new predicate symbol for each pred-

icate symbol occurring in A

. Next, let A

be the

atom set obtained from A

by replacing the predicate

of each atom in A

with the new predicate introduced

for it. Finally, for each atom a ∈ A

, let φ(a) be the

atom obtained from a by such predicate replacement.

6 ET RULES ON ECLS

Next, ET rules for unfolding and deﬁnite-clause re-

moval are presented, along with some other ET rules.

6.1 Unfolding Operation on ECLS

Assume that (i) Cs is a set of extended clauses, (ii) D

is a set of extended deﬁnite clauses, and (iii) occ is

an occurrence of an atom b in the right-hand side of a

clause C in Cs. By unfolding Cs using D at occ, Cs is

transformed into

(Cs− {C}) ∪ (

[

{resolvent(C,C

′

, b) | C

′

∈ D}),

where for each C

′

∈ D, resolvent(C,C

′

, b) is deﬁned

as follows, assuming that ρ is a renaming substitution

for usual variables such that C and C

′

ρ have no usual

variable in common:

• If b and head(C

′

ρ) are not uniﬁable, then

resolvent(C,C

′

, b) =

• If they are uniﬁable, then resolvent(C,C

′

, b) =

′′

}, where C

′′

is the clause obtained from C and

′

ρ as follows, assuming that θ is the most general

uniﬁer of b and head(C

′

ρ):

– lhs(C

′′

) = lhs(Cθ)

– rhs(C

′′

) = (rhs(Cθ) − {bθ}) ∪ body(C

′

ρθ)

The resulting clause set is denoted by UNFOLD(Cs,

D, occ).

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6.2 ET by Unfolding and Deﬁnite-clause

Removal

Let Atoms(p) denote the set of all atoms having a

predicate p. ET rules on ECLS

for unfolding and

for deﬁnite-clause removal are described below.

6.2.1 ET by Unfolding

Let hCs, ai be a QA problem on ECLS

. Assume that:

1. q is the predicate of the query atom a.

2. p is a predicate such that p 6= q.

3. D is a set of extended deﬁnite clauses in Cs that

satisﬁes the following conditions:

(a) For any C ∈ D, head(C) ∈ Atoms(p).

(b) For anyC

′

∈ Cs− D, lhs(C

′

) ∩ Atoms(p) =

4. occ is an occurrence of an atom in Atoms(p) in the

right-hand side of an extended clause in Cs− D.

Then hCs, ai can be equivalently transformed into the

QA problem hUNFOLD(Cs, D, occ), ai.

6.2.2 ET by Deﬁnite-clause Removal

Let hCs, ai be a QA problem on ECLS

. Assume that:

1. q is the predicate of the query atom a.

2. p is a predicate such that p 6= q.

3. D is a set of extended deﬁnite clauses in Cs that

satisﬁes the following conditions:

(a) For any C ∈ D, head(C) ∈ Atoms(p).

(b) For anyC

′

∈ Cs− D, lhs(C

′

) ∩ Atoms(p) =

4. For any C

′

∈ Cs− D, rhs(C

′

) ∩ Atoms(p) =

Then hCs, ai can be equivalently transformed into the

QA problem hCs − D, ai.

6.3 Some other ET Rules on ECLS

Next, ET rules for merging func-atoms having the

same call pattern, for removing isolated func-atoms,

and for removing subsumed clauses are presented.

They are used in examples in Section 7.

6.3.1 Merging func-Atoms with the Same

Invocation Pattern

Let hCs, ai be a QA problem on ECLS

. Suppose that

C ∈ Cs and rhs(C) contains func-atoms f

and f

that

differ only in their last arguments. Then:

1. If the last arguments of f

and f

are uniﬁable,

with their most general uniﬁer being θ, and C

′

an extended clause such that

• lhs(C

′

) = lhs(Cθ), and

• rhs(C

′

) = (rhs(C) − {f

})θ,

then hCs, ai can be equivalently transformed into

the QA problem h(Cs− {C}) ∪ {C

′

}, ai.

2. If their last arguments are not uniﬁable, then

hCs, ai can be equivalently transformed into the

QA problem hCs− {C}, ai.

6.3.2 Elimination of Isolated func-Atoms

A func-atom func(h, t

, . . . ,t

, v), where v is a usual

variable, is said to be isolated in an extended clause C

iff there is only one occurrence of v in C.

Now let hCs, ai be a QA problem on ECLS

. Sup-

pose that:

1. C ∈ Cs such that C contains a func-atom that is

isolated in C.

2. C

′

is the extended clause obtained from C by re-

moving all func-atoms that are isolated in C.

Then hCs, ai can be equivalently transformed into the

QA problem h(Cs− {C}) ∪ {C

′

}, ai.

6.3.3 Elimination of Subsumed Clauses

An extended clause C

is said to subsume an ex-

tended clause C

iff there exists a substitution θ

for usual variables such that lhs(C

)θ ⊆ lhs(C

) and

rhs(C

)θ ⊆ rhs(C

A subsumed clause can be removed as follows:

Let hCs, ai be a QA problem on ECLS

. If Cs con-

tains extended clauses C

and C

such that C

sub-

sumes C

, then hCs, ai can be equivalently trans-

formed into the QA problem hCs− {C

}, ai.

7 EXAMPLES

Example 1 demonstrates how the procedure in Sec-

tion 5.3 solves a QA problem using the ET rules in

Section 6. Example 2 shows how to apply the proce-

dure to solve a proof problem based on the embedding

mapping in Section 3.2.2.

Example 1. Consider the “Tax-cut” problem dis-

cussed in (Motik et al., 2005). This problem is to

ﬁnd all persons who can have discounted tax, with the

knowledge that (i) any person who has two children

or more can get discounted tax, (ii) men and women

are not the same, (iii) a person’s mother is always a

woman, (iv) Peter has a child named Paul, (v) Paul is

a man, and (vi) Peter has a child, who is someone’s

mother. This background knowledge is represented

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in ﬁrst-order logic as the formulas F

–F

below, as-

suming that hc, ns, tc, mn, wm and mo stand, respec-

tively, for hasChild, notSame, TaxCut, Man, Woman

and motherOf:

: ∀x: ((∃y

∃y

(hc(x, y

) ∧ hc(x, y

) ∧ ns(y

, y

))) → tc(x))

: ∀x∀y: ((mn(x) ∧ wm(y)) → ns(x, y))

: ∀x: ((∃y : mo(x, y)) → wm(x))

: hc(Peter, Paul)

: mn(Paul)

: ∃x: (hc(Peter, x) ∧ (∃y: mo(x, y)))

Accordingly, the “Tax-cut” problem is formulated as

the QA problem hK, tc(x)i, where K is the conjunc-

tion of F

–F

. Using the meaning-preserving Skol-

emization procedure given in (Akama and Nantajee-

warawat, 2011), the ﬁrst-order formula K is trans-

formed into a clause set Cs consisting of the following

extended clauses:

: tc(x) ← hc(x, y

), hc(x, y

), ns(y

, y

)

: ns(x, y) ← mn(x), wm(y)

: wm(x) ← mo(x, y)

: hc(Peter, Paul) ←

: mn(Paul) ←

: hc(Peter, x) ← func(h

, x)

: mo(x, y) ← func(h

, x), func(h

, y)

The clauses C

and C

together represent the ﬁrst-

order formula F

, where h

and h

are 0-ary function

variables.

Assume that all usual atoms occurring in Cs be-

long to A

, ans is a newly introduced unary predicate

symbol, all ans-atoms belong to A

, and for any term

t, φ(tc(t)) = ans(t). Let

= (ans(x) ← tc(x)).

To solve the QA problem hK, tc(x)i, the QA problem

hCs, {C

}, ans(x)i is successively transformed by ap-

plying the ET rules in Section 6 as follows:

1. By unfolding C

at tc(x) using {C

}, C

is re-

placed with:

: ans(x) ← hc(x, y

), hc(x, y

), ns(y

, y

)

2. By unfoldingC

at the last body atom using {C

is replaced with:

: ans(x) ← hc(x, y

), hc(x, y

), mn(y

), wm(y

)

3. By unfolding C

at the third body atom using

}, C

is replaced with:

: ans(x) ← hc(x, Paul), hc(x, y

), wm(y

)

4. By unfolding C

at the last body atom using

}, C

is replaced with:

: ans(x) ← hc(x, Paul), hc(x, y

), mo(y

, z)

5. By unfolding C

at the last body atom using

}, C

is replaced with:

: ans(x) ← hc(x, Paul), hc(x, y

), func(h

, y

func(h

, z)

6. By removing an isolated func-atom, C

is re-

placed with:

: ans(x) ← hc(x, Paul), hc(x, y

), func(h

, y

)

7. By unfolding C

at the ﬁrst body atom using

}, C

is replaced with:

: ans(Peter) ← hc(Peter, y

), func(h

, y

)

: ans(Peter) ← func(h

, Paul), hc(Peter, y

func(h

, y

)

8. By merging func-atoms with the same invocation

pattern, C

is replaced with:

: ans(Peter) ← func(h

, Paul), hc(Peter, Paul)

9. Since C

is subsumed by C

, C

is removed.

10. By unfolding C

at the ﬁrst body atom using

}, C

is replaced with:

: ans(Peter) ← func(h

, Paul)

: ans(Peter) ← func(h

, y

), func(h

, y

)

11. By deﬁnite-clause removal,C

–C

are removed.

12. By merging func-atoms with the same invocation

pattern, C

is replaced with:

: ans(Peter) ← func(h

, y

)

13. By removing an isolated func-atom, C

is re-

placed with:

: ans(Peter) ←

14. Since C

is subsumed by C

, C

is removed.

The resulting QA problem is h

0, {C

}, ans(x)i. Since

Models(

0) 6=

0 and C

is a unit clause whose head is

an instance of φ(tc(x)), the answer to the “Tax-cut”

problem hK, tc(x)i is determined by

−1

(

{rep(head(C

))}) = φ

−1

({ans(Peter)})

= {tc(Peter)},

i.e., Peter is the only one who gets discounted tax.

Example 2. Refer to the description of the “Tax-cut”

problem, the ﬁrst-order formulas F

–F

, the clauses

–C

and the clause set Cs = {C

, . . . ,C

} in Exam-

ple 1. From the background knowledge of the “Tax-

cut” problem, suppose that we want to prove the ex-

istence of someone who gets discounted tax. This

problem is formulated as the proof problem hE

, E

where E

is the conjunction of F

–F

and E

is the

ﬁrst-order formula ∃x : tc(x).

Using Proposition 2, this proof problem is con-

verted into the QA problem hE

∧¬E

, yesi. Using the

procedure in Section 5.3, this QA problem is solved

as follows:

EmbeddingProofProblemsintoQuery-answeringProblemsandProblemSolvingbyEquivalentTransformation

259

• Convert E

∧ ¬E

by meaning-preserving Skol-

emization, resulting in the clause set Cs ∪ {C

′

where C

′

is the negative clause (← tc(x)).

• Transform the QA problem

hCs∪ {C

′

}, {(φ(yes) ← yes)}, φ(yes)i

using ET rules. By following the transformation

Steps 1–14 in Example 1 except that the initial

target clause is C

′

instead of C

, the clauses C

′

–

′

are successively produced, where for each i ∈

{8, . . . , 20},

– lhs(C

′

) =

0, and

– rhs(C

′

) = rhs(C

and C

–C

are removed. As a result, Cs ∪ {C

′

}

is transformed into {C

′

}, where C

′

= (←), and

the QA problem

h{C

′

}, {(φ(yes) ← yes)}, φ(yes)i

is obtained.

• SinceC

′

is the empty clause, the clause set {C

′

}

has no model, i.e., Models({C

′

}) =

0. So the

procedure outputs rep(yes) = {yes} as the answer

to the QA problem hE

∧ ¬E

, yesi.

It follows from Proposition 2 that the answer to the

proof problem hE

, E

i is “yes”, i.e., there exists

someone who gets discounted tax.

8 CONCLUSIONS

Previous approaches to solving QA problems are

proof-centered. They were developed for speciﬁc

subclasses of QA problems; for example, answer-

ing queries in logic programming and deductive

databases can be regarded as solving QA problems on

deﬁnite clauses and those on a restricted form of def-

inite clauses, respectively. There has been no general

solution method for QA problems on full ﬁrst-order

formulas.

QA problems on full ﬁrst-order logic are consid-

ered in this paper. We introduced the concept of em-

bedding and proposed how to embed proof problems

into QA problems. This embedding leads to a uni-

ﬁed approach to dealing with proof problems and QA

problems, allowing one to use a method for solving

QA problems to solve proof problems. It enables

a QA-problem-centered approach to solving logical

problems.

Equivalent transformation (ET) is one of the most

fundamental principles of computation, and it pro-

vides a simple and general basis for veriﬁcation of

computation correctness. We proposed a framework

for solving QA problems by ET. All computation

steps in this framework are ET steps, including trans-

formation of a ﬁrst-order formula into an equivalent

formula in the extended clause space ECLS

and

transformation of extended clauses on ECLS

. To the

best of our knowledge, this is the only framework for

dealing with the full class of QA problems on ﬁrst-

order formulas.

Since many kinds of ET rules can be employed,

the proposed ET-based framework opens up a wide

range of possibilities for computation paths to be

taken. As a result, the framework enables develop-

ment of a large variety of methods for solving logical

problems. The range of possible computation meth-

ods can also be further extended by using computa-

tion spaces other than ECLS

. Proof by resolution

can be seen as one speciﬁc example of these possible

methods. As demonstrated in (Akama and Nantajee-

warawat, 2012), it can be realized by using two kinds

of ET rules, i.e., resolution and factoring ET rules, on

a computation space that differs slightly from ECLS

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