Logic seminar 4 Herbrand’s theorem 07.11.2005. Slobodan Petrović.

Logic seminar 4

Herbrand’s theorem

07.11.2005.

Slobodan Petrović


• A general decision procedure to verify the validity (inconsistency) of a formula:– Tried by Leibnitz, Peano, Hilbert, …– Proved inexistent by Church and Turing (1936)

• Church and Turing independently showed that there is no general decision procedure to check the validity of formulas of the first order logic.


• There are, however, proof procedures which can verify that a formula is valid if it is valid indeed.

• For invalid formulas, these procedures never terminate in general.

• This is the best that one can expect from a proof procedure.


• By definition, a valid formula is a formula that is true under all interpretations.

• Herbrand (1930) developed an algorithm to find an interpretation (if it exists) that can falsify a given formula.

• If the given formula is indeed valid, no such interpretation can exist and Herbrand’s algorithm halts after a finite number of trials.


• Herbrand’s procedure is a refutation procedure.

• Instead of proving a formula valid, it proves that the negation of the formula is inconsistent.

• There is no loss of generality in using refutation procedures.

• The refutation procedures are applied to a “standard” form of a formula.


• The standard form:– A formula of the first order logic is transformed

into prenex normal form:• The matrix contains no quantifiers

• The prefix is a sequence of quantifiers.

– The matrix is transformed into a conjunctive normal form.

– The existential quantifiers in the prefix are eliminated by using Skolem functions.


• Skolem standard form (standard form)– Let a formula F be already in a prenex normal

form (Q1x1)…(Qnxn)M, where M is in a conjunctive normal form.

– Let Qr be an existential quantifier in the prefix (Q1x1)…(Qnxn), 1≤r≤n.


• Skolem standard form (cont.)– If no universal quantifier appears before Qr, we do

the following:• choose a new constant c different from other constants

occurring in M;

• replace all xr appearing in M by c;

• delete (Qrxr) from the prefix.


• Skolem standard form (cont.)– If Qs1

,…,Qsm are all the universal quantifiers

appearing before Qr, 1≤s1≤s2≤…≤sm≤r, we do the following:

• choose a new m-place function symbol f different from other function symbols occurring in M;

• replace all xr appearing in M by f(xs1,xs2

,…,xsm);

• delete (Qrxr) from the prefix.


• Skolem standard form (cont.)– After this process is applied to all the existential

quantifiers in the prefix, a Skolem standard form of the formula F is obtained.

– The constants and functions used to replace the existential variables are called Skolem functions.

– This process is called Skolemization.


• Example– Obtain a standard form of the formula

• (x)(y)(z)(u)(v)(w)P(x,y,z,u,v,w)

– (x) is preceded by no universal quantifiers, (u) is preceded by (y) and (z), and (w) by (y), (z) and (v).

– Therefore, we replace the existential variable x by a constant a, u by a two-place function f(y,z), and w by a three-place function g(y,z,v).

– The standard form of the formula is• (y)(z)(v)P(a,y,z,f(y,z),v,g(y,z,v)).

Herbrand’s theorem• Example

– Obtain a standard form of the formula• (x)(y)(z)((~P(x,y)Q(x,z))R(x,y,z)).

– First, the matrix is transformed into a conjunctive normal form:

• (x)(y)(z)((~P(x,y)R(x,y,z))(Q(x,z)R(x,y,z))). – Then, since (y) and (z) are both preceded by

(x), the existential variables y and z are replaced, respectively by one-place functions f(x) and g(x).

– The standard form of the formula is• (x)((~P(x,f(x))R(x,f(x),g(x)))(Q(x,g(x))R(x,f(x),g(x)))).

Herbrand’s theorem• Definition

– A clause is a disjunction of literals.

• When it is convenient, it is possible to regard a set of literals as synonymous with a clause.

• Example– PQ~R={P,Q,~R}

• A clause consisting of r literals is called an r-literal clause.

• A one-literal clause is called a unit clause.

Herbrand’s theorem• When a clause contains no literal, it is called

the empty clause.

• Since the empty clause has no literal that can be satisfied by an interpretation, the empty clause is always false.

• The empty clause is denoted by □.

• Example – the clauses are– ~P(x,f(x))R(x,f(x),g(x))

– Q(x,g(x))R(x,f(x),g(x)).

Herbrand’s theorem• A set S of clauses is regarded as a conjunction

of all clauses in S.

• Every variable in S is considered governed by a universal quantifier.

• Thus, a standard form can be represented by a set of clauses.


– The standard form of the formula• (x)((~P(x,f(x))R(x,f(x),g(x)))(Q(x,g(x))R(x,f(x),g(x))))

– can be represented by a set of clauses• {~P(x,f(x))R(x,f(x),g(x)), Q(x,g(x))R(x,f(x),g(x))}.

• Theorem– Let S be a set of clauses that represents a standard

form of a formula F. – Then F is inconsistent if and only if S is

inconsistent.

Herbrand’s theorem• Let S be a standard form of a formula F.

• If F is inconsistent, then F=S.

• If F is not inconsistent, F is not equivalent to S in general.

• Example– F=(x)P(x),– S=P(a).– S is a standard form of F.

Herbrand’s theorem• Example (cont.)

– The interpretation I:• Domain: D={1,2}

• Assignment for P: P(1)=F, P(2)=T.

– F is true in I, but S is false in I. That is, FS.

• A formula may have more than one standard form.• When a formula F is transformed into a standard

formula S, existential quantifiers should be replaced by the simplest possible Skolem functions – with the least number of arguments.

Herbrand’s theorem• Existential quantifiers should be moved to the

left as far as possible.

• If F=F1…Fn, it is possible to separately obtain a set Si of clauses, where Si represents a standard form of Fi, i=1,…,n.

• Let S=S1…Sn.

• Then F is inconsistent if and only if S is inconsistent.


– The theorem: If xx=e for all x in group G, where is a binary operator and e is the identity in G, then G is commutative.

– We first symbolize the theorem and some other axioms of group theory.

– Then we represent the negation of the theorem as a set of clauses.


– The axioms of the group:• A1: x,yG implies that xyG (closure property)

• A2: x,y,zG implies that x(yz)=(xy)z (associativity property)

• A3: xe=ex=x for all xG (identity property)

• A4: For every xG there exists an element x-1G such that xx-1=x-1x=e (inverse property).


– Let P(x,y,z) stand for xy=z and i(x) for x-1.– Then, the axioms of the group can be represented

by:• A1’: (x)(y)(z)P(x,y,z)

• A2’: (x)(y)(z)(u)(v)(w)(P(x,y,u)P(y,z,v)P(u,z,w)P(x,v,w))

(x)(y)(z)(u)(v)(w)(P(x,y,u)P(y,z,v)P(x,v,w)P(y,z,w))

• A3’: (x)P(x,e,x)(x)P(e,x,x)

• A4’: (x)P(x,i(x),e)(x)P(i(x),x,e).


– The conclusion of the theorem:• B: If xx=e for all xG, then G is commutative, i.e.,

uv=vu for all u,vG.

– B formalized:• B’: (x)P(x,x,e)((u)(v)(w)(P(u,v,w)P(v,u,w))).

– The entire theorem is represented by the formula:• F=A1’…A4’B’.

• Thus, ~F=A1’A2’A3’A4’~B’.


– To obtain a set S of clauses for ~F, we first obtain a set Si of clauses for each axiom Ai’, i=1, 2, 3, 4:

• S1: {P(x,y,f(x,y))}

• S2: {~P(x,y,u)~P(y,z,v)~P(u,z,w)P(x,v,w),

~P(x,y,u)~P(y,z,v)~P(x,v,w)P(u,z,w)}

• S3: {P(x,e,x),P(e,x,x)}

• S4: {P(x,i(x),e),P(i(x),x,e)}.


• ~B’=~((x)P(x,x,e)((u)(v)(w)(P(u,v,w)P(v,u,w))))

=~(~(x)P(x,x,e)((u)(v)(w)(~P(u,v,w)P(v,u,w))))

=(x)P(x,x,e)~((u)(v)(w)(~P(u,v,w)P(v,u,w)))

=(x)P(x,x,e)(u)(v)(w)(P(u,v,w)~P(v,u,w)).

– A set of clauses for ~B’:• T: {P(x,x,e),P(a,b,c),~P(b,a,c)}.


– The set S=S1S2S3S4T:• (1) P(x,y,f(x,y))• (2) ~P(x,y,u)~P(y,z,v)~P(u,z,w)P(x,v,w)• (3) ~P(x,y,u)~P(y,z,v)~P(x,v,w)P(u,z,w)• (4) P(x,e,x)• (5) P(e,x,x)• (6) P(x,i(x),e)• (7) P(i(x),x,e)• (8) P(x,x,e)• (9) P(a,b,c)• (10) ~P(b,a,c).


• F is valid if and only if S is inconsistent.

• Refutation procedures are used to prove theorems.

• The input to a refutation procedure is always a set of clauses.

• Instead of “inconsistent” (“consistent”) “unsatisfiable” (“satisfiable”) is used, respectively for sets of clauses.


• The Herbrand universe of a set of clauses– A set of clauses S is unsatisfiable if and only if it is

false under all interpretation over all domains.– It is impossible to consider all interpretations over all

domains.– It would be desirable to have one special domain H

such that S is unsatisfiable if and only if S is false under all the interpretations over this domain.

– There exists such a domain – Herbrand universe of S.


• The Herbrand universe of a set of clauses– Let H0 be the set of constants appearing in S.

– If no constant appears in S, then H0 is to consist of a single constant, H0={a}.

– For i=0,1,2,… let Hi+1 be the union of Hi and the set of all terms of the form fn(t1,…,tn) for all n-place functions fn occurring in S, where tj, j=1,…,n, are members of the set Hi.

– Hi is called the i-level constant set of S.


• The Herbrand universe of a set of clauses– Example 1: S={P(a),P(x)P(f(x))}

• H0={a}

• H1={a,f(a)}

• H2={a,f(a),f(f(a))}

• .

• .

• H={a,f(a),f(f(a)),f(f(f(a))),…}


• The Herbrand universe of a set of clauses– Example 2: S={P(x)Q(x),R(z),T(y)W(y)}

• There is no constant in S, so we let H0={a}

• There is no function symbol in S, hence H=H0=H1=…={a}

– Example 3: S={P(f(x),a,g(y),b)}• H0={a,b}

• H1={a,b,f(a),f(b),g(a),g(b)}

• H2={a,b,f(a),f(b),g(a),g(b),f(f(a)),f(f(b)),f(g(a)),f(g(b)),g(f(a)),g(f(b)),g(g(a)),g(g(b))}

• …


• The Herbrand universe of a set of clauses– Expression – a term, a set of terms, an atom, a set of

atoms, a literal, a clause, or a set of clauses.– When no variable appears in an expression, it is

called a ground expression.– It is possible to use a ground term, a ground atom, a

ground literal, and a ground clause – this means that no variable occurs in respective expressions.

– A subexpression of an expression E is an expression that occurs in E.


• The Herbrand universe of a set of clauses– Let S be a set of clauses. The set of ground atoms of

the form Pn(t1,…,tn) for all n-place predicates Pn occurring in S, where t1,…,tn are elements of the Herbrand universe of S, is called the atom set, or the Herbrand base of S.

– A ground instance of a clause C of the set S of clauses is a clause obtained by replacing variables in C by members of the Herbrand universe of S.


• The Herbrand universe of a set of clauses– Example

• S={P(x),Q(f(y))R(y)}

• C=P(x) is a clause in S

• H={a,f(a),f(f(a)),…} is the Herbrand universe of S.

• P(a) and P(f(f(a))) are ground instances of C.


• The Herbrand universe of a set of clauses– Let S be a set of clauses.– An interpretation over the Herbrand universe of S is

an assignment of constants, function symbols and predicate symbols occurring in S.


• The Herbrand universe of a set of clauses– Let S be a set of clauses, H the Herbrand universe of

S and I an interpretation of S over H. I is an H-interpretation of S if:

• I maps all constants in S to themselves.

• Let f be an n-place function symbol and h1,…,hn be elements of H. In I, f is assigned a function that maps (h1,…,hn) (an element in Hn) to f(h1,…,hn) (an element in H).


• The Herbrand universe of a set of clauses– There is no restriction on the assignment to each

n-place predicate symbol in S.

– Let A={A1,A2,…,An,…} be the atom set of S.

– An H-interpretation I can be conveniently represented by a set I={m1,m2,…,mn,…} in which mj is either Aj or ~Aj for j=1,2,…

– If mj is Aj then Aj is assigned “true”, otherwise Aj is assigned “false”.


• The Herbrand universe of a set of clauses– Example: S={P(x)Q(x),R(f(y))}

• The Herbrand universe of S is H={a,f(a),f(f(a)),…}.

• Predicate symbols: P, Q, R.

• The atom set of S:– A={P(a),Q(a),R(a),P(f(a)),Q(f(a)),R(f(a)),…}.

• Some H-interpretations for S:– I1={P(a),Q(a),R(a),P(f(a)),Q(f(a)),R(f(a)),…}

– I2={~P(a),~Q(a),~R(a),~P(f(a)),~Q(f(a)),~R(f(a)),…}

– I3={P(a),Q(a),~R(a),P(f(a)),Q(f(a)),~R(f(a)),…}


• The Herbrand universe of a set of clauses– An interpretation of a set S of clauses does not

necessarily have to be defined over the Herbrand universe of S.

– Thus an interpretation may not be an H-interpretation.

– Example:• S={P(x),Q(y,f(y,a))}

• D={1,2}


• The Herbrand universe of a set of clauses– Example (cont.) – an interpretation of S:

a f(1,1) f(1,2) f(2,1) f(2,2)

2 1 2 2 1

P(1) P(2) Q(1,1) Q(1,2) Q(2,1) Q(2,2)

T F F T F T


• The Herbrand universe of a set of clauses– Example (cont.) – we can define an H-interpretation I*

corresponding to I.

– First we find the atom set of S• A={P(a),Q(a,a),P(f(a,a)),Q(a,f(a,a)),Q(f(a,a),a),Q(f(a,a),f(a,a)),…}

– Next we evaluate each member of A by using the given table• P(a)=P(2)=F

• Q(a,a)=Q(2,2)=T

• P(f(a,a))=P(f(2,2))=P(1)=T

• Q(a,f(a,a))=Q(2,f(2,2))=Q(2,1)=F


• The Herbrand universe of a set of clauses– Example (cont.)

• Q(f(a,a),a)=Q(f(2,2),2)=Q(1,2)=T

• Q(f(a,a),f(a,a))=Q(f(2,2),f(2,2))=Q(1,1)=T

– H-interpretation I* corresponding to I

– I*={~P(a),Q(a,a),P(f(a,a)),~Q(a,f(a,a)),Q(f(a,a),a),

~Q(f(a,a),f(a,a)),…}.


• The Herbrand universe of a set of clauses– If there is no constant in S, the element a used to initiate the

Herbrand universe of S can be mapped into any element of the domain D.

– If there is more than one element in D, then there is more than one H-interpretation corresponding to I.

– Example: S={P(x),Q(y,f(y,z))}, D={1,2}


• The Herbrand universe of a set of clauses– Example (cont.):

f(1,1) f(1,2) f(2,1) f(2,2)

1 2 2 1

P(1) P(2) Q(1,1) Q(1,2) Q(2,1) Q(2,2)

T F F T F T


• The Herbrand universe of a set of clauses– Example (cont.):

– Then the two H-interpretations corresponding to I are:• I*={~P(a),Q(a,a),P(f(a,a)),~Q(a,f(a,a)),Q(f(a,a),a),

~Q(f(a,a),f(a,a)),…} if a=2,

• I*={P(a),~Q(a,a),P(f(a,a)),~Q(a,f(a,a)),~Q(f(a,a),a),

~Q(f(a,a),f(a,a)),…} if a=1.


• The Herbrand universe of a set of clauses– Definition

• Given an interpretation I over a domain D, an H-interpretation I* corresponding to I is an H-interpretation that satisfies the condition:

• Let h1,…,hn be elements of H (the Herbrand universe of S). Let every hi be mapped to some di in D. If P(d1,…,dn) is assigned T (F) by I, then P(h1,…,hn) is also assigned T(F) in I*.

– Lemma:• If an interpretation I over some domain D satisfies a set S of

clauses, then any H-interpretation I* corresponding to I also satisfies S.


• The Herbrand universe of a set of clauses– Theorem

• A set S of clauses is unsatisfiable if and only if S is false under all the H-interpretations of S.

– We need consider only H-interpretations for checking whether or not a set of clauses is unsatisfiable.

– Thus, whenever the term “interpretation” is used, a H-interpretation is meant.


• The Herbrand universe of a set of clauses– Let denote an empty set. Then:

• A ground instance C’ of a clause C is satisfied by an interpretation I if and only if there is a ground literal L’ in C’ such that L’ is also in I, i.e. C’I.

• A clause C is satisfied by an interpretation I if and only if every ground instance of C is satisfied by I.

• A clause C is falsified by an interpretation I if and only if there is at least one ground instance C’ of C such that C’ is not satisfied by I.

• A set S of clauses is unsatisfiable if and only if for every interpretation I there is at least one ground instance C’ of some clause C in S such that C’ is not satisfied by I.


• The Herbrand universe of a set of clauses– Example: Consider the clause C=P(x)Q(f(x)). Let I1, I2,

and I3 be defined as follows:

• I1={P(a),Q(a),P(f(a)),Q(f(a)),P(f(f(a))),Q(f(f(a))),…}



– C is satisfied by I1 and I2, but falsified by I3.

– Example: S={P(x),P(a)}. The only two H-interpretations are: I1={P(a)}, I2={P(a)}.

– S is falsified by both H-interpretations and therefore is unsatisfiable.


• Semantic trees– Introduced by Robinson, 1968.

– It can be shown that finding a proof for a set of clauses is equivalent to generating a semantic tree.

– Definition• If A is an atom, then the two literals A and A are said to be each

other’s complement.

• The set {A,A} is called a complementary pair.

– A clause is a tautology if it contains a complementary pair.


• Semantic trees– Definition

• Given a set S of clauses, let A be the atom set of S.

• A semantic tree for S is a (downward) tree T, where each link is attached with a finite set of atoms or negations of atoms from A in such a way that:

– For each node N, there are only finitely many immediate links L1,…,Ln from N. Let Qi be the conjunction of all the literals in the set attached to Li, i=1,…,n. Then Q1Q2…Qn is a valid propositional formula.

– For each node N, let I(N) be the union of all the sets attached to the links of the branch of T down to and including N. Then I(N) does not contain any complementary pair.


• Semantic trees– Definition

• Let A={A1,A2,…,Ak,…} be the atom set of a set S of clauses.

• A semantic tree for S is said to be complete if and only if for every tip node N of the semantic tree, i.e. a node that has no links sprouting from it, I(N) contains either Ai or Ai for i=1,2,…

– Example:• Let A={P,Q,R} be the atom set of a set S of clauses.

• Two examples of complete semantic trees for S:


• Semantic trees– Example (cont.)


• Semantic trees– Example:

• S={P(x),P(a)}

• The atom set of S is {P(a)}

• A complete semantic tree for S:

– Example:• S={P(x),Q(f(x))}

• The atom set of S is {P(a),Q(a),P(f(a)),Q(f(a)),P(f(f(a))),Q(f(f(a))),…}

• A semantic tree for S:


• Semantic trees– Example (cont.)

Herbrand’s theorem• Semantic trees

– For each node N in a semantic tree for S, I(N) is a subset of some interpretation for S.

– I(N) is called partial interpretation for S.

– When the atom set A of a set S of clauses is infinite, any complete semantic tree for S is infinite.

– A complete semantic tree for S corresponds to an exhaustive survey of all possible interpretations for S.

– If S is unsatisfiable, then S fails to be true in each of the interpretations.

– We may stop expanding nodes from a node N if I(N) falsifies S.


– Definition – A node N is a failure node if I(N) falsifies some ground instance of a clause in S, but I(N’) does not falsify any ground instance of a clause in S for every ancestor node N’ of N.

– Definition – A semantic tree T is said to be closed if and only if every branch of T terminates at a failure node.

– A node N of a closed semantic tree is called an inference node if all the immediate descendant nodes of N are failure nodes.


– Example:• Let S={P,QR,PQ,PR}.

• The atom set of S is A={P,Q,R}.

• A complete semantic tree for S:


– Example (cont.):• A closed semantic tree for S:


– Example:• Consider S={P(x),P(x)Q(f(x)),Q(f(a))}

• The atom set of S: A={P(a),Q(a),P(f(a)),Q(f(a)),…}

• A closed semantic tree for S:

Herbrand’s theorem• To test whether a set S of clauses is unsatisfiable, we

need consider only interpretations over the Herbrand universe of S.

• If S is false under all interpretations over the Herbrand universe of S, then we can conclude that S is unsatisfiable.

• Since there are usually many, possibly an infinite number, of these interpretations, we should organize them in a systematic way – by using a semantic tree.

Herbrand’s theorem• Herbrand’s theorem – Version 1 (a classical one).

– A set S of clauses is unsatisfiable if and only if there is a finite unsatisfiable set S’ of ground instances of clauses of S.

• Herbrand’s theorem – Version 2 (with semantic trees)– A set S of clauses is unsatisfiable if and only if

corresponding to every complete semantic tree of S, there is a finite closed semantic tree.


• Example:– Let S={P(x),P(f(a))}.– S is unsatisfiable.– Hence, by Herbrand’s theorem, there is a finite

unsatisfiable set S’ of ground instances of clauses in S.

– One of these sets is S’={P(f(a)),P(f(a))}


• Example:– Let S={P(x)Q(f(x),x),P(g(b)),Q(y,z)}– S is unsatisfiable.– One of the unsatisfiable sets of ground instances of

clauses in S:• S’={P(g(b))Q(f(g(b)),g(b)),P(g(b)),Q(f(g(b)),g(b))}


• Example:– Let S consist of the following clauses:

• S={P(x,y,u)P(y,z,v)P(x,v,w)P(u,z,w),

P(x,y,u)P(y,z,v)P(u,z,w)P(x,v,w),

P(g(x,y),x,y),P(x,h(x,y),y),P(x,y,f(x,y)),

P(k(x),x,k(x))}

– S is unsatisfiable.– It is not easy to find by hand a finite unsatisfiable

set S’ of ground instances of clauses in S.


• Example (cont.):– One way to find such a set S’ is to generate a

closed semantic tree T’ for S.– Then the set S’ of all the ground instances falsified

at all the failure nodes of T’ is such a desired set.– Each ground clause in S’ is a ground instance of

some clause in S.– S’ is unsatisfiable.


• Example (cont.):– The set S’:

S’={P(a,h(a,a),a),P(k(h(a,a)),h(a,a),k(h(a,a))),

P(g(a,k(h(a,a))),a,k(h(a,a))),

P(g(a,k(h(a,a))),a,k(h(a,a)))P(a,h(a,a),a)

P(g(a,k(h(a,a))),a,k(h(a,a)))

P(k(h(a,a)),h(a,a),k(h(a,a)))}


• Implementations of Herbrand’s theorem– The classical version of the Herbrand’s theorem

suggests a refutation procedure.– Given an unsatisfiable set of clauses to prove, if

there is a mechanical procedure that can successively generate sets S1’,…Sn’,… of ground instances of clauses in S and can successively test S1’,S2’,… for unsatisfiability, then it is guaranteed by Herbrand’s theorem that this procedure can detect a finite N such that Sn’ is unsatisfiable.


• Gilmore’s implementation– In 1960, Gilmore wrote a computer program that

successively generated S0’,S1’,… , where Si’ is the set of all the ground instances obtained by replacing variables in S by the constants in the i-level constant set Hi of S.

– Since each Si’ is a conjunction of ground clauses, it is possible to use any method of the propositional logic to check its unsatisfiability.


• Gilmore’s implementation (cont.)– Gilmore used the multiplication method.

– As each Si’ is produced, it is multiplied out into a disjunctive normal form.

– Any conjunction in the disjunctive normal form containing a complementary pair is removed.

– Should some Si’ be empty, then Si’ is unsatisfiable and a proof is found.


• Gilmore’s implementation (cont.)– Example: S={P(x),P(a)}.

• H0={a}

• S0’=P(a)P(a)=□.

• Thus S is proved to be unsatisfiable.


• Gilmore’s implementation (cont.)– Example: S={P(a),P(x)Q(f(x)),Q(f(a))}.

• H0={a}

• S0’=P(a)(P(a)Q(f(a)))Q(f(a))

=(P(a)P(a)Q(f(a)))(P(a)Q(f(a))Q(f(a)))

=□□=□.

• Thus S is proved to be unsatisfiable.


• Gilmore’s implementation (cont.)– The multiplication method used by Gilmore is

inefficient.– Even for a small set of ten two-literal ground

clauses, there are 210 conjunctions.

Herbrand’s theorem• The method of Davis and Putnam

– Let S be a set of ground clauses.– The method of Davis and Putnam consists of four

rules:• Tautology:

– Delete all the ground clauses from S that are tautologies. The remaining set S’ is unsatisfiable if and only if S is.

• One-literal rule: – If there is a unit ground clause L in S, obtain S’ from S by

deleting those ground clauses in S containing L. If S’ is empty, S is satisfiable. Otherwise, obtain a set S’’ from S’ by deleting L from S’. S’’ is unsatisfiable if and only if S is. If L is a unit ground clause, then the clause becomes □ when L is deleted from it.


• The method of Davis and Putnam– The method of Davis and Putnam rules (cont.):

• Pure-literal rule:– A literal L in a ground clause of S is said to be pure in S if and

only if the literal L does not appear in any ground clause in S. If a literal L is pure in S, delete all the ground clauses containing L. The remaining set S’ is unsatisfiable if and only if S is.


• The method of Davis and Putnam– The method of Davis and Putnam rules (cont.):

• Splitting rule:– If the set S can be put into the form:

(A1L)…(AmL)(B1L)…(BnL)R,

where Ai, Bi, and R are free of L and L, then obtain the sets S1=A1…AmR and S2=B1…BnR. S is unsatisfiable if and only if (S1S2) is unsatisfiable, i.e. both S1 and S2 are unsatisfiable.


• The method of Davis and Putnam– Example: Show that

S=(PQR)(PQ)PRU is unsatisfiable.(PQR)(PQ)PRU(QR)(Q)RU (One-literal rule on P)

RRU (One-literal rule on Q)

□U (One-literal rule on R)

– Since the last formula contains the empty clause □, S is unsatisfiable.


• The method of Davis and Putnam– Example: Show that S=(PQ)Q(PQR) is

satisfiable.(PQ)Q(PQR)

P(PR) (One-literal rule on Q)

R (One-literal rule on P)

■ (One-literal rule on R)

– Thus, S is satisfiable.



S=(PQ)(PQ)(QR)(QR) is satisfiable.(PQ)(PQ)(QR)(QR)

(Q(QR)(QR))

(Q(QR)(QR)) (Splitting rule on P)

RR (One-literal rule on Q and Q)

■■ (One-literal rule on R)




S=(PQ)(PQ)(RQ)(RQ) is satisfiable.(PQ)(PQ)(RQ)(RQ)

(RQ)(RQ) (Pure-literal rule on P)

■ (Pure-literal rule on R)



• The method of Davis and Putnam– The method of Davis and Putnam is more efficient

than the Gilmore’s multiplication method.– This method can be applied to any formula in the

propositional logic.– The procedure is:

• Transform the given propositional formula into a conjunctive normal form.

• Apply the four rules on the conjunctive normal form.

Logic seminar 4 Herbrand’s theorem 07.11.2005. Slobodan Petrović.

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