Lecture 8: Resolution theorem-proving

(1)

Comp24412 Symbolic AI

Lecture 8: Resolution theorem-proving

Ian Pratt-Hartmann

Room KB2.38: email: [email protected]

2014–15

(2)

• In the previous Lecture, we met SATCHMO, a first-order theorem-prover implemented in a single page of Prolog.

• It has nice termination properties, but it not practically very useful.

• This lecture introduces a more serious approach to constructing proofs in first-order logic: resolution.

• Resolution is the basis of many widely-used theorem-provers today.

(3)

• We would like to be able to determine logical relationships such as validity

• For example,

∀x(artist(x) → ∃y (beekeeper(y ) ∧ admire(x, y )))

∀x(beekeeper(y ) → ∀y (artist(y ) ∧ admire(y , x) → despise(x, y )))

∀x(artist(x) → ∃y (beekeeper(y ) ∧ despises(y , x) ∧ admire(x, y )))

(4)

• Here is a more complex example

∀x∀y ∀z(p(x, y ) ∧ p(y , z) → p(x, z))

∀x∀y ∀z(q(x, y ) ∧ q(y , z) → q(x, z))

∀x∀y (p(x, y ) → p(y , x))

∀x∀y (p(x, y ) ∨ q(x, y ))

∀x∀yp(x, y ) ∨ ∀x∀yq(x, y )

(5)

• As with Satchmo, we first convert toclause form.

• Aliteral is an atomic formula or an atomic formula prefixed by

¬.

• Thus, p, q(a), ¬r (x , y ) are all literals; p ∨ p, q(x ) ∧ p and

¬¬r (x, y ) are not literals.

• A clauseis a literal or a collection of literals all joined by ∨.

• Thus ¬p, p ∨ q(x ), ¬p(a) ∨ ¬r (a, b) ∨ r are all clauses;

p ∧ q(a), p → p and ¬¬r (x , y ) are not clauses.

(6)

• The following result is easily established:

• Let ϕ be a quantifier-free formula. Then there exist clauses C₁, . . . , C_n such that

|= ϕ ↔ (C₁∧ C₂∧ . . . ∧ C_n).

• In other words, any quantifier-free formula is logically equivalent to the conjunction of some collection of clauses.

(7)

• A formula in the predicate calculus in which all the quantifiers are at the front is said to be inprenex form.

• The formulas

∀x(man(x) → mortal(x))

∀x∃y loves(x, y ) are in prenex form

• The formulas

∀x(boy(x) → ∃y (girl(y ) ∧ loves(x, y )))

∀x((boy(x) ∧ ∃y (girl(y ) ∧ loves(x, y ))) → happy(x)) are not.

(8)

• The following result can easily be established:

• Let ϕ be a formula. Then there exists a prenex form formula ψ such that

|= ϕ ↔ ψ.

• In other words, any formula is logically equivalent to a formula with all the quantifiers at the front.

(9)

• Example

∀x(boy(x) →∃y(girl(y ) ∧ loves(x , y ))) is equivalent to

∀x∃y(boy(x ) → (girl(y ) ∧ loves(x , y )))

• Example

∀x((boy(x) ∧ ∃y(girl(y ) ∧ loves(x , y ))) → happy(x )) is equivalent to

∀x∀y((boy(x ) ∧ girl(y ) ∧ loves(x , y )) → happy(x ))

(10)

• Given a prenex formula, existential quantifiers can be eliminated.

• Consider

(∃x )(man(x ) ∧ philosopher(x ))

• This is equisatisfiablewith

man(a) ∧ philosopher(a).

(This name must be new: i.e. not occurring in any other formulas.)

• Similarly

(∃y )(∀x )(loves(x , y )) is equisatisfiable with

(∀x )(loves(x , b)).

(11)

• Things are different however, for:

(∀x )(∃y )(loves(x , y )).

This formula is equisatisfiable with:

(∀x )(loves(x , f (x )))

where f is a function (called aSkolem function).

• Note the use of a function (rather than a constant): the formula says that everyone loves someone or other – but not necessarily the same person.

• We need a function whenever we want to eliminate an existential quantifier which is to the right of some universal quantifiers.

(12)

• These new constants and functions are called Skolem constantsand Skolem functions, respectively.

• The process of replacing existentially quantified variables with Skolem constants and functions is called Skolemization.

• Notice that Skolemization does not produce logically equivalent formulas; but it does produce equisatisfiable formulas.

(13)

• Suppose then, we put a formula ϕ in prenex form, and Skolemize.

• The result will be

∀x₁. . . ∀xnχ where χ is quantifier-free.

• But the universal quantifiers convey no information any more, so we might as well write:

χ

• Now we replace χ (the quantifier-free rump) by an equivalent collection of clauses.

• The result is said to be theclause form for ϕ.

(14)

• So we can massage any formula of the predicate calculus into a collection of clause form expressions (containing variables) of the form

L1∨ . . . ∨ L_N where the Li are literals

• Example:

∀x(boy(x) → ∃y (girl(y ) ∧ love(x, y ))) has, as its clausal form,

¬boy(x) ∨ girl(f (x)) ¬boy(x) ∨ love(x, f (x))

• Since we can equisatisfiably convert to clausal forms, we need only inference procedures which work on clausal forms.

(15)

• First, we look at the ground case.

• The rulemodus ponens:

p → q p

q

• Re-writing:

¬p ∨ q p

q

• Generalizing to longer clauses:

¬p ∨ C p ∨ D

C ∨ D

(16)

• This gives us a way to make inferences.

• Consider the formulas

boy(john) → (girl(mary) → loves(john, mary)), boy(john), girl(mary), ¬loves(john, mary).

• It should be obvious that these formulas are not (simultaneously satisfiable)

• In clausal form:

¬boy(john) ∨ ¬girl(mary) ∨ loves(john, mary), boy(john), girl(mary), ¬loves(john, mary).

(17)

• We can apply resolution repeatedly as follows:

¬loves(j, m)

boy(j)

girl(m) ¬boy(j) ∨ ¬girl(m) ∨ loves(j, m)

¬boy(j) ∨ loves(j, m) loves(j, m)

⊥.

• Since the empty clause ⊥ has been derived, it is evident (even to a computer) that the formulas we started with are not satisfiable.

(18)

• This gives us a way to test validity.

• For suppose we want to test whether ϕ1, . . . ϕn|= ψ.

• We convert

ϕ1, . . . ϕn, ¬ψ

to clausal form and try to use resolution and factoring to obtain a contradiction. If we succeed, the sequent is valid.

• In particular, we have shown the validity of the argument with premises

boy(john) → (girl(mary) → loves(john, mary)) boy(john) girl(mary)

and conclusionloves(john, mary).

(19)

• Next, we look at the non-groundcase.

• To apply resolution to non-ground clauses, we need the concept of unification.

• If A and A⁰ are atoms, we say that A and A⁰ are unifiable if there is a substitution σ of terms for variables such that Aσ = A⁰σ.

(20)

• Example 1:

man(x ) man(socrates) unify, under the substitution σ : x 7→ socrates.

• Example 2:

q(x , f (x )) q(u, v )

unify, under the substitution σ : u 7→ x ; v 7→ f (x ).

• Example 3:

p(x ) q(u)

do not unify.

• Example 4:

p(x , f (x )) p(f (y ), y ) do not unify either.

(21)

• We noted that

q(x , f (x )) q(u, v ) unify, under the substitution

σ : u 7→ x ; v 7→ f (x ) making the common term

q(x , f (x )).

• But they also unify under the substitution, say, σ : u 7→ a; v 7→ f (a); x 7→ a, making the common term

q(a, f (a)).

(22)

• If A and A⁰ are unifiable, then there is a substitution θ such that Aθ = A⁰θ and, for any substitution σ such that

Aσ = A⁰σ, we have σ = θρ for some substitution ρ.

• We say that θ is a most general unifier (m.g.u).

• The m.g.u. is unique up to renaming of variables.

• It can be efficiently computed . . .

(23)

• Now back to the analogue of modus ponens for non-ground clauses:

¬p(x) ∨ q(x) p(a)

q(a) .

• Generalizing again:

¬A ∨ C A⁰∨ D

(C ∨ D)θ ,

where A and A⁰ areunifiable atoms withm.g.u. θ

• This rule of inference is known as theresolution rule

(24)

• It is obvious that we can remove repeated literals from clauses:

p ∨ p ∨ p ∨ C p ∨ C ,

• Generalizing to the non-ground case gives thefactoring rule:

C (C θ)⁰,

with θ the m.g.u of some literals in C and ⁰ denoting deletion of repeated literals

(25)

• Again, we can use resolution and factoring to construct proofs:

¬lv(john, mary)

boy(john)

girl(mary) ¬boy(x) ∨ ¬girl(y ) ∨ lv(x, y )

¬boy(x) ∨ lv(x, mary) lv(john, mary)

⊥.

• If clause C is derivable from a set C of clauses by means of resolution and factoring, we write C ` C .

(26)

• As before, this gives us a way to make inferences.

• Consider the formulas

∀x(boy(x) → ∀y (girl(y ) → loves(x, y ))) boy(john)

girl(mary) ¬loves(john, mary).

• In clausal form:

¬boy(x) ∨ ¬girl(y ) ∨ loves(x, y ) boy(john)

• But these clauses imply ⊥.

• Therefore, the original set of formulas is unsatisfiable.

(27)

• As before: to test whether

ϕ₁, . . . ϕ_n|= ψ,

• convert

ϕ₁, . . . ϕ_n, ¬ψ

to clausal form and try to use resolution and factoring to obtain a contradiction.

• If you succeed, the sequent is valid

(28)

• In particular, in showing that the set of formulas

is unsatisfiable, we have shown that the argument

girl(mary) loves(john, mary) is valid.

(29)

• In fact, the following result can be proved:

Theorem

Let C be a set of clauses. Then C ` ⊥ if and only if the universal closure of C is unsatisfiable.

• This means that resolution and factoring are all the rules we need to determine entailments

• We say that the proof system in question is (sound and) complete.

• To think about: Why does this not give us an algorithm for testing validity?

(30)

• Let A be the collection of all atoms (over some signature).

• Let ≺ be a partial order on A.

• We say that ≺ is an A-ordering if, for all substitutions θ, A ≺ A⁰ implies Aθ ≺ A⁰θ.

• We extend any A-ordering ≺ to literals by ignoring negations.

(31)

• Example of A-ordering:

A ≺² A⁰ iff A involves a unary predicate and A⁰ involves a binary predicate.

• Example of A-ordering:

A ≺^d A⁰ iff d (A) < d (A⁰) and

d (x , A) < d (x , A⁰) for all variables x in A, where d (A) is the depth of A and d (x , A) is the term depth of x in A.

(32)

• Given an A-ordering ≺, ordered resolution is the same as resolution, namely

¬A ∨ C A⁰∨ D

(C ∨ D)θ ,

subject to the added restriction that, for every literal L in C , A 6≺ L, and for every literal L⁰ in D, A⁰6≺ L⁰.

• Similarly with ordered factoring.

• If clause C is derivable from a set C of clauses by means of

≺-ordered resolution and factoring, we write C `_≺C .

(33)

• Amazingly, ordering restrictions never compromise completeness

Theorem

Let C be a set of clauses and ≺ an A-ordering. Then C `≺⊥ if and only if C is unsatisfiable.

(34)

• Let us say that the1-variable fragment is the set of clauses C satisfying either of the following conditions:

• every literal of C is ground;

• there is a variable x such that, for every literal L of C , Vars(L) = {x }.

• The clauses

p(a) ∨ q(b, f (a)), p(x ) ∨ ¬r (h(x , g (x )), x ) are in the 1-variable fragment, but

p(a) ∨ q(x , f (a)), p(x ) ∨ ¬r (y , x ) are not in the 1-variable fragment.

• Note that resolution and factoring preserve membership in the 1-variable fragment.

(35)

• In general, resolution increases the functional depth of clauses:

p(x ) ∨ q(fx ) ¬r (gx⁰) ∨ s(x⁰) ∨ ¬q(x⁰) p(x ) ∨ ¬r (gfx ) ∨ s(fx )

• However, ≺_d-ordered resolution and factoring do not:

p(x ) ∨ q(fx ) ¬r (x⁰) ∨ s(x⁰) ∨ ¬q(x⁰) p(x ) ∨ ¬r (fx ) ∨ s(fx )

(36)

• Suppose we take a set C of clauses in the 1-variable fragment, and keep applying resolution and factoring to them.

• Obviously, only clauses featuring the signature of C will be generated.

• Because the functional depth is not increased, the total number of clauses that can be generated from C by

≺_d-ordered resolution and factoring is bounded by an exponential function of the size of the signature.

• Hence, the satisfiability of clauses in the 1-variable fragment (with a fixed depth-bound) can be algorithmically decided in exponential time.