25.12.2015general resolution method1 general resolution method in fol lesson 8

21.04.2023 General resolution Method 1

General Resolution method in FOL

Lesson 8

General resolution Method 221.04.2023

Typical problems to solve

Proving logical validity of a formula: Formula F is logically valid, denoted |= F, iff F is true

in all interpretations, i.e., every interpretation structure is its model

Proving the validity of an argument: An argument is valid, denoted P1, …, Pn |= Q, iff the

formula Q is true in all the models of the set of assumptions P1, …, Pn

What follows from the assumptions? P1, …, Pn |= ?


Typical problems, tasks

Their solution by examining the infinite set of models is difficult; it is not possible to automatize semantic proofs

Therefore we look for syntactic (mechanic) methods of proving One of them is the method of semantic tableau based on the

transformation of a formula into its disjunctive normal form (disadvantage: we often need to perform many transformations using the distributive laws)

Now we are going to learn a very effective method that is used also for automatic theorem proving and programming in logic: the resolution method for FOL (generalization of the resolution method for propositional logic)

Rezolution method makes use of the conjunctive normal form of a formula

21.04.202321.04.2023 General resolution MethodGeneral resolution Method 44

ResoluResolution methodtion method It is applicable to a formula in the It is applicable to a formula in the

special kind of the special kind of the conjunctive normal conjunctive normal form: form: Skolem cSkolem clausal formlausal form

It makes use of the It makes use of the indirect proofindirect proof It serves as a theoretical foundation for It serves as a theoretical foundation for

automatic theorem provingautomatic theorem proving and and logic logic programmingprogramming; ;

In particular for the In particular for the programming programming language language PROLOGPROLOG


Resolution methodResolution methodIn order to generalize the resolution method of In order to generalize the resolution method of

propositional logic, there are propositional logic, there are two problemstwo problems::1)1) Transformation into the Skolem clausal form: Transformation into the Skolem clausal form:

xx11……xxnn [ [CC11 … … CCmm]]where where CCii are clauses, i.e., disjunctions of literals, are clauses, i.e., disjunctions of literals, e.g.: P(e.g.: P(xx) ) Q(f(Q(f(x,yx,y))))

2)2) In order to apply the resolution rule on a pair of a In order to apply the resolution rule on a pair of a positive and negative literals, the literals have to be positive and negative literals, the literals have to be identical; we have to identical; we have to unifyunify them. them. For instance: For instance: PP((xx) ) QQ(f((f(xx)),y,y))

PP((gg((yy)) )) QQ((x,zx,z))

The literals The literals PP((xx), ), PP((gg((yy)) might have been )) might have been eliminated, if they were identical; we have to eliminated, if they were identical; we have to unify unify them by substituting them by substituting gg((yy) for ) for x.x.


Transformation into the clausal form Transformation into the clausal form (Skolem)(Skolem)

1.1. Existential generalization (if there are free varaibles, Existential generalization (if there are free varaibles, then bind them by then bind them by ( (consistencyconsistency preserving) preserving)

2.2. Elimination of dispensable quantifiersElimination of dispensable quantifiers3.3. Elimination of the connectives Elimination of the connectives , , 4.4. Moving negation inside (applying de Morgan laws)Moving negation inside (applying de Morgan laws)5.5. Renaming variablesRenaming variables6.6. Moving quantifiers to the rightMoving quantifiers to the right7.7. Elimination of existential quantifiersElimination of existential quantifiers

(Skolemization – (Skolemization – consistencyconsistency preserving) preserving)8.8. Moving general quantifiers to the leftMoving general quantifiers to the left9.9. Applying distributive lawsApplying distributive laws


Resolution method – example. |= [x y P(x,y) y z Q(y,z)] x y z [P(x,y) Q(y,z)]

1. Negating the formula and variables renaming:

[x y’ P(x,y’) y z Q(y,z)] u v w [P(u,v) Q(v,w)]

2. Elimination of existential quantifiers. MIND !

[x P(x, f(x)) y Q(y, g(y))] v w [P(a,v) Q(v,w)]

3. Moving the quantifiers to the left:

x y v w [P(x, f(x)) Q(y, g(y)) [P(a,v) Q(v,w)]]

4. The formula is obviously not satisfiable; how to prove it?

5. Write down the clauses and perform the unification of literals, i.e., substitute appropriate terms for variables, so that to be able to apply the resolution rule:


Resolution method – example 1. P(x, f(x)) 2. Q(y, g(y))3. P(a,v) Q(v,w) Now we have to apply resolution rule. In order to resolve

clauses 1. and 3., we have to substitute the term a for the variable x and the term f(a) for variable v.

4. Q(f(a),w) resolving 1., 3., substitution: x/a, v/f(a)5. resolving 2., 4., substitution: y/f(a),

w/g(f(a))

QED – the negated formula is a contradiction, which means that the original formula is logically valid

Note: When unifying the literals, it is necessary to substitute for all the occurrences of a variable!


Skolem clausal formLiteral: atomic formula or the negation of an atomic formula.

Examples: P(x, g(y)), Q(z)

Clause: disjunction of literals, e.g.: P(x, g(y)) Q(z)

Skolem clausal form: x1…xn [C1 … Cm], a closed formula, where Ci are clauses, disjunctions of literals, e.g.: P(x) Q(f(x,y))

Skolemization (elimination of existential quantifiers):

x y1...yn A(x, y1,...,yn) y1...yn A(c, y1,...,yn) where c is a new, in the language not used constant

y1...yn x A(x, y1,...,yn) y1...yn A(f(y1,...,yn), y1,...,yn) where f is a new, in the language not used functional

symbol


Skolemization is consistency preservingy1...yn x A(x, y1,...,yn) y1...yn A(f(y1,...,yn), y1,...,yn)

B BS

Proof: Let the structure I be a model of the formula BS (Skolemised). Then for any n-tuple of individuals i1, i2,…,in it holds that (n+1)-tuple of individuals i1, i2,…,in, fU (i1, i2,…,in) AU, where fU is the function assigned to the symbol f (in the interpretation I) and AU is the relation defined by the formula A in the interpretation I. Then the interpretation structure I is also a model of the formula B =

y1y2…yn x A(x, y1, y2,…,yn). Hence every model of the formula BS is a model of the formula B, but not

vice versa, ie BS |= B. The set of models MBS of the formula BS is a subset of the set of models

MB of the formula B: MBS MB. If MBS (i.e., BS is satisfiable) then MB , i.e., B is satisfiable as well. If MB = , i.e. B is not satisfiable, then MBS = , i.e., BS is not satisfiable,

which is sufficient for the indirect proof.


Thoralf Albert SkolemBorn: 23 May 1887 in Sandsvaer, Buskerud, Norway

Died: 23 March 1963 in Oslo, Norway

Skolem was remarkably productive publishing around 180 papers on topics such as Diophantine equations, mathematical logic, group theory, lattice theory and set theory


Transformation into the Skolem clausal formA AS:1. Existential generalization (consistency preserving)2. Elimination of dispensable quantifiers (that do not bind any variables –

equivalent transformation) 3. Elimination of , (applying the PL laws – equivalent transformation)4. Moving negation inside (applying de Morgan laws – equivalent

transformation) 5. Renaming variables (equivalent transformation)6. Moving quantifiers to the right (equivalent transformation):

Qx (A @ B(x)) A @ Qx B(x), Qx (A(x) @ B) Qx A(x) @ B, where @ is a conjunction or disjunction connective, Q is a quantifier (general or existential)

7. Elimination of existential quantifiers(Skolemize the subformulas Qx B(x), Qx A(x) obtain in the step 6)

8. Moving general quantifiers to the left (equivalent transformation)9. Applying distributive laws (equivalent transformation)


Transforming A AS (Skolem) The resulting formula AS is not equivalent to the original

formula A (nor does it follow from A), but it holds that if A has a model then AS has a model.

It also holds that AS |= A, but the two formulas are not equivalent.

The non-equivalent steps are: Existential generalisation Skolemization (elimination of existential quantifiers)

This is the reason why the resolution method is used as an indirect proof.

It can be used for a direct proof only if the assumptions do not contain any existential quantifier


Transformation: A AS

Always follow the above Skolem algorithm (9 steps). Some of them can be omitted, but do not transform into the conjunctive normal form first, and then Skolemize (as you can sometimes read). In other words, do not omit the step 6.

Example: the importance of the step 6 (quantifiers to the right – moved to the subformulas containing the quantified variables).

We will illustrate an erroneous practice of omitting the step 6 by (not) proving that|= x A(x) x A(x):

1. negating the formula: x A(x) x A(x), 2. Renaming variables: x A(x) y A(y), 3. Moving quantifiers to the left: 4. x [A(x) y A(y)] x y [A(x) A(y)], and 5. Skolemize: x [A(x) A(f(x))]. But the terms x and f(x) cannot be unified! Thus we did not prove the above

tautology, though the resolution method is a complete proof method. Each logically valid formula is provable:

|= A | A The problem consists in the fact that we Skolemized after moving quantifiers to the

left. We have to do it before, when the quantifiers are to the right – at the subformulas that they actually quantify.


Example: A AS

x {P(x) z {y [Q(x,y) P(f(x1))] y [Q(x,y) P(x)]}}

1.,2. Existential generalisation and elimination of z: x1 x {P(x) {y [Q(x,y) P(f(x1))] y [Q(x,y) P(x)]}}

3.,4. Renaming y, elimination of : x1 x {P(x) {y [Q(x,y) P(f(x1))] z [Q(x,z) P(x)]}}

5.,6. Negation moved inside, quantifiers to the right: x1 x {P(x) {[y Q(x,y) P(f(x1))] [z Q(x,z) P(x)]}}

7. Elimination of the existential quantifiers: x {P(x) {[Q(x,g(x)) P(f(a))] [z Q(x,z) P(x)]}}

8. Quantifier z moved to the left: x z {P(x) {[Q(x,g(x)) P(f(a))] [Q(x,z) P(x)]}}

9. Distributive law: x z {[P(x) Q(x,g(x))] [P(x) P(f(a))] [P(x) Q(x,z) P(x)]}

10. simplifying: x {[P(x) Q(x,g(x))] [P(x) P(f(a))]}


The way of proving by resolution Proof of the logical validity of a formula A:

1. Negate the formula A2. Transform A into the clausal Skolem form (A)S

3. Step-by-step apply the resolution rule (via unifying the literals) and try to prove that the formula (A)S is not satisfiable, hence also the formula A does not have a model, which means that A is logically valid.

Proof of the validity of an argument P1,…,Pn | Z

1. Negate Z2. Transform the assumptions and the negated conclusion into the

Skolem clausal form3. Step-by-step apply the resolution rule (via unifying the literals) and

try to prove the inconsistency of the set {P1,…,Pn, Z}


Unification of literals Example: x y z v [P(x, f(x)) Q(y, h(y)) (P(a, z) Q(z, v))] How to prove that the formula does not have a model?

1. P(x, f(x))2. Q(y, h(y))3. P(a, z) Q(z, v) Now if we want to resolve the respective clauses; but there are no identical terms in the respective pairs of literals. However, all the variables are bound by the general quantifier. Hence we can apply the law of specialization, and thus substitute terms for variables in order to find the witness of inconsistency:

In order to unify literals of 1 and 3, we use the substitution x/a, z/f(a).


Unification of literalsAfter substituting we obtain the following clauses:1’. P(a, f(a))2. Q(y, h(y))3‘. P(a, f(a)) Q(f(a), v)

Now resolving 1‘ and 3‘ we obtain:4. Q(f(a), v)

Now in order to resolve 2. and 4., we again use a substitution:y/f(a), v/h(f(a))

and get2‘. Q(f(a), h(f(a)) )4’. Q(f(a), h(f(a)) )5.

By resolving 2’, 4’ we obtain an empty clause that does not have a model. Hence the original formula A does not have a model, it is a contradiction.


Unification of literalsUp to now we sought particular substitutions in order to unify

literals ad hoc, intuitively looking for opposite “similar” literals

We need to find an algorithm of unification. There are many such algorithms, we will examine two of them:a) Herbrand procedureb) Robinson unification algorithmAd (a) A formula F is not satisfiable iff it is not true in any

interpretation structure, over any universe of discourse. It might be useful to find just one universe such that F is a contradiction iff F is not satisfiable over any interpretation structure over this universe.

Is there any such universe? Yes, it was discovered by Herbrand, and it is called Herbrand’s universe.


Jacques HerbrandBorn: 12 Feb 1908 in Paris, France

Died: 27 July 1931 in La Bérarde, Isère, France

École Normale Supérieure at the age of 17 (!)

His doctoral thesis was approved in April 1929

On a holiday in the Alps he died in a mountaineering accident at the age of 23


Herbrand procedureHerbrand universe HF: consists of all the possible terms made

up from the constants and functional symbols occurring in the formula; if there is no constant in the formula, then use any constant.

Example: For the formula A = x [P(a) Q(b) P(f(x))]

the HA = {a, b, f(a), f(b), f(f(a)), f(f(b)), …} For the formula B = x y P( (f(x), y, g(x,y) )

the HB = {a, f(a), g(a,a), f(f(a)), g(a,f(a)), g(f(a),a), …} Basic instances of a clause: all the variables are substituted for by the elements of Herbrand universe

Theorem (Herbrand): A formula A in the Skolem clausal form is not satisfiable iff there is a finite conjunction of its basic instances that is not satisfiable.


Herbrand procedure

Example: 1. P(x, f(x))2. Q(y, h(y))3. P(a, z) Q(z, v)

Here the HA = {a, f(a), h(a), f(f(a)), f(h(a)), h(f(a)), h(h(a)), …}. Now we create basic instances, substituting elements of (the ordered) Herbrand universe for variables:

Substitution 1: {x/a, y/a, z/a, v/a} P(a, f(a)) Q(a, h(a)) [P(a, a) Q(a, a)]Substitution 2: {x/a, y/a, z/a, v/f(a)} P(a, f(a)) Q(a, h(a)) [P(a, a) Q(a, f(a))] etc., till we find Substitution n: {x/a, y/f(a), z/f(a), v/h(f(a))} P(a,f(a)) Q(f(a),h(f(a))) [P(a,f(a)) Q(f(a),h(f(a)))]

Contradiction


Herbrand Procedure

Herbrand procedure is a method that partially decides whether a given formula F is a contradiction; if F is a contradiction then Herbrand procedure answers YES after a finite number of steps, otherwise it may not answer at all.

(One of the corollaries of Gödel’s theorem; there is no procedure that would decide it totally.)

Problem: Space and time complexity of the algorithm; the number of basic instances we need to generate till we find the witness of contradiction is often to large to be computationally tractable.


John Alan Robinson

In his 1965 article "A Machine-Oriented Logic Based on the Resolution Principle" John Alan Robinson laid an important foundation for a whole branch of automated deduction systems. The original Prolog is essentially a refined automated theorem prover based on Robinson's resolution


Robinson’s algorithm (1965) Let A be a formula containing variables xi, i=1,2,...,n. Let

us denote ={x1/t1, x2/t2,...,xn/tn} the simultaneous substitution of terms ti for all the

occurrences of variables xi for i=1,2,...,n. Then let A be the formula that arises from A by performing the substitution .

Unification (unification substitution) of formulas A, B is the substitution such that

A = B. The general unification of formulas A, B is the

unification such that any other unification of formulas A, B it holds that = (where , it is not an empty substitution); in other words, the general unification leaves as much as possible variables free.


Robinson’s algorithm (1965)It is the algorithm of finding the general unification: Assume that A = P(t1, t2,...,tn), B = P(s1,s2,...,sn),

where (for the sake of simplicity) at least one of the terms ti, si is a variable.

1. For i = 1,2,...,n do:a) If ti = si then i = (the empty substitution).

b) If ti si then check whether the terms ti, si are such that one of them is a variable, say x, and the other term, say r, does not contain this variable x.

c) If Yes, then i = {x/r}.d) If No, then finish the cycle: the formulas A, B are not unifiable.

2. If the cycle were not finished by the exception ad (d), then A, B are unifiable, and the general unification = 12...n (the composed substitution).


Robinson’s algorithm

Example: A = Px, f(x), u, B = Py, z, g(x,y) 1 = {x/y},

A1 = P(y, f(y), u, B1 = Py, z, g(y,y) 2 = {z/f(y)},

A12 = P(y, f(y), u, B12 = Py, f(y), g(y,y) 3 = {u/g(y,y)},

A123 = P(y, f(y), g(y,y), B123 = Py, f(y), g(y,y).

The composed substitution =123 is the general unification of the formulas A, B: {x/y, z/f(y), u/g(y,y)}


General resolution rule

A L1, B L2

A B

where is the general unification of L1, L2: L1 = L2


Example: prove the logical validity of

xy [{P(x,y) Q(x, f(g(x)))} {R(x) x Q(x, f(g(x)))} x R(x)] x P(x, g(x))

1. Negate the formula:

xy [{P(x,y) Q(x, f(g(x)))} {R(x) x Q(x, f(g(x)))} x R(x)] x P(x, g(x))

2. Eliminate existential quantifiers, and rename the variables:

xy [{P(x,y) Q(x, f(g(x)))} {R(x) Q(a, f(g(a)))} R(b)] z P(z, g(z))

3. Write down the clauses (after eliminating the implication):


Examplexy [{P(x,y) Q(x, f(g(x)))}

{R(x) Q(a, f(g(a)))} R(b)] z P(z, g(z))

1. P(x,y) Q(x, f(g(x)))2. R(x) Q(a, f(g(a)))3. R(b)4. P(z, g(z))5. Q(x, f(g(x))) 1, 4: z/x, y/g(x)6. Q(a, f(g(a))) 2, 3: x/b 7. 5, 6: x/a

Note: Parent clauses to resolve are chosen in such a way that leaves as much as possible variables free (for further substitutions)

25.12.2015general resolution method1 general resolution method in fol lesson 8

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