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Resolution for First Order Logic Meghdad Ghari Institute for Research in Fundamental Sciences (IPM) School of Mathematics-Isfahan Branch Logic Group http://math.ipm.ac.ir/Isfahan/Logic-Group.htm Logic Short Course II, Computational Predicate

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Resolution for First Order Logic

Meghdad Ghari

Institute for Research in Fundamental Sciences (IPM) School of Mathematics-Isfahan Branch Logic Group http://math.ipm.ac.ir/Isfahan/Logic-Group.htm

Logic Short Course II, Computational Predicate Logic March 9, 2017

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Outline

1

Propositional Logic Conjunctive Normal Form Propositional Resolution

2

First Order Logic Syntax Semantics Resolution for FOL PROLOG

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Language

Atom = {p1, p2, p3, . . .}, A (finite or infinite) set of atomic propositions (or propositional variables). L = Atom ∪ {(, ), ¬, ∧, ∨, →} Formulas of propositional logic are defined by the following grammar in Backus Naur form (BNF): φ ::= p | (¬φ) | (φ ∧ φ) | (φ ∨ φ) | (φ → φ), p ∈ Atom

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Semantics: principle of compositionality

The meaning of a complex expression is determined by the meanings

f its constituent expressions and the rules used to combine them.

Valuations: v : Atom → {T, F} Valuations can be extended to the set of all formulas as follows: v(¬φ) = 1 − v(φ) v(φ ∧ ψ) = min(v(φ), v(ψ)) v(φ ∨ ψ) = max(v(φ), v(ψ)) v(φ → ψ) = min(1, 1 − v(φ) + v(ψ))

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Decideability

Definition

A logical system is decidable if there is an effective method (or algorithm) for determining whether arbitrary formulas are theorems of the logical system. Truth tables provide a decision procedure for propositional logic.

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Validity and satisfiability

Theorem

Let φ be a formula of propositional logic. Then φ is satisfiable iff ¬φ is not valid.

Theorem

Given formulas φ1, φ2, . . . , φn and ψ of propositional logic, φ1, φ2, . . . , φn | = ψ iff | = φ1 ∧ φ2 ∧ . . . ∧ φn → ψ

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Validity and satisfiability

Theorem

Let φ be a formula of propositional logic. Then φ is satisfiable iff ¬φ is not valid.

Theorem

Given formulas φ1, φ2, . . . , φn and ψ of propositional logic, φ1, φ2, . . . , φn | = ψ iff | = φ1 ∧ φ2 ∧ . . . ∧ φn → ψ

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Conjunctive Normal Form (CNF)

L ::= p | ¬p, p ∈ Atom Literal D ::= L | L ∨ D Clause C ::= D | D ∧ C Conjunctive normal form

Example (Set representation)

CNF

Clausal form

(¬q ∨ p ∨ r) ∧ (¬p ∨ r) ∧ q ∧ (r ∨ ¬p) {{¬q, p, r}, {¬p, r}, {q}}

Facts

The empty clsuse is always false and unsatisfiable. The empty clause form ∅ is valid. A clause form that has is a contradiction, and hence unsatisfiable.

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Converting formulas to CNF

Theorem

Every formula in propositional logic can be transformed into an equivalent formula in CNF Step 1. Elimination of → and ↔: φ → ψ ≡ ¬φ ∨ ψ φ ↔ ψ ≡ (¬φ ∨ ψ) ∧ (φ ∨ ¬ψ) Step 2. Converting to negation normal forms (NNF): ¬¬φ ≡ φ ¬(φ ∧ ψ) ≡ ¬φ ∨ ¬ψ ¬(φ ∨ ψ) ≡ ¬φ ∧ ¬ψ Step 3. Distributivity rules: φ ∧ (ψ ∨ σ) ≡ (φ ∧ ψ) ∨ (φ ∧ σ) φ ∨ (ψ ∧ σ) ≡ (φ ∨ ψ) ∧ (φ ∨ σ)

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Resolution

John Alan Robinson, A Machine-Oriented Logic Based on the Resolution Principle

J. ACM. 12 (1): 23-41, 1965.

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Propositional Resolution

Resolution is a refutation procedure used to check if a formula in clausal form is unsatisfiable. The resolution procedure consists of a sequence of applications of the resolution rule to a set of clauses. The rule maintains satisfiability: if a set of clauses is satisfiable, so is the set of clauses produced by an application of the rule. Therefore, if the (unsatisfiable) empty clause is ever obtained, the

riginal set of clauses must have been unsatisfiable.

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Propositional Resolution Rule

Resolution: ¬φ ∨ ψ φ ∨ χ ψ ∨ χ Resolution is sound (preserves validity). Modus Ponens: φ → ψ φ ψ ¬φ ∨ ψ φ ∨ ⊤ ψ ∨ χ Modus Ponens is a special case of resolution.

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Propositional Resolution Rule

Resolution: ¬φ ∨ ψ φ ∨ χ ψ ∨ χ Resolution is sound (preserves validity). Modus Ponens: φ → ψ φ ψ ¬φ ∨ ψ φ ∨ ⊤ ψ ∨ χ Modus Ponens is a special case of resolution.

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Propositional Resolution Rule

Propositional Resolution Rule: C ⊔ {p} D ⊔ {¬p} C ∪ D where C and D are clauses and ⊔ denotes disjoint union. C ∪ D is called the resolvent of C ⊔ {p} and D ⊔ {¬p}: C ∪ D = Res(C ⊔ {p}, D ⊔ {¬p}) Resolution on clause forms: {C1, . . . , C ⊔ {p}, D ⊔ {¬p}, . . . , Cn} {C1, . . . , C ∪ D, . . . , Cn} Resolution on clause forms preserves satisfiability.

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Proof by resolution

Definition

A resolution refutation of S is a resolution proof of empty clause from S.

Theorem

S is unsatisfiable iff there is a resolution refutation of S.

Checking validity

φ is valid, if ¬φ is unsatisfiable or there is a resolution refutation of ¬φ.

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Proof by resolution

In order to show that φ1, φ2, . . . , φn | = ψ, we show that φ1 ∧ φ2 ∧ . . . ∧ φn ∧ ¬ψ is unsatisfiable.

First, φ1 ∧ φ2 ∧ . . . ∧ φn ∧ ¬ψ is converted into CNF .

Then, the resolution rule is applied to the resulting clauses.

The process continues until one of two things happens:

◮ two clauses resolve to yield the empty clause , in which case

φ1, φ2, . . . , φn | = ψ.

◮ there are no new clauses that can be added, in which case

φ1, φ2, . . . , φn | = ψ;

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Resolution algorithm

/* Input: A set of clauses S / / Output: S is satisfiable or unsatisfiable */ while there are clauses C1, C2 ∈ S and C := Res(C1, C2) such that C ∈ S do S := S ∪ {C} If ∈ S then return unsatisfiable else (i.e. if ∈ S) return satisfiable.

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Example I

We prove that | = ¬(p ∨ q) → (¬p ∧ ¬q). To this end, we show that ¬[¬(p ∨ q) → (¬p ∧ ¬q)] (1) is unsatisfiable. The CNF of (1) is: ¬p ∧ ¬q ∧ (p ∨ q), and in clausal form: {{¬p}, {¬q}, {p, q}}. The following is a resolution refutation of it: {¬p} {p, q} {q} {¬q}

Meghdad Ghari (IPM)

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Example II

We show that p | = p ∧ q. To this end, we show that p ∧ ¬(p ∧ q) (2) is satisfiable. The CNF of (2) is: p ∧ (¬p ∨ ¬q), and in clausal form: S = {{p}, {¬p, ¬q}}. Using resolution algorithm we update S as follows:

{p} a clause of S

{¬p, ¬q} a clause of S

{¬q} resolvent of clauses 1 and 2 S is satisfiable, since there are no new resolvents that can be added and ∈ S.

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Resolution refutation as a tree

In order to prove that (p → (q → r)) → ((p → q) → (p → r)) is valid, we show that its negation is unsatisfiable. The following is a resolution refutation tree of its negation:

p p ¯ p¯ q ¯ pq ¯ p¯ qr ¯ r

✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍ ❍

❅ ❅ ❅ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆

¯ p denotes ¬p. ¯ pq denotes the clause {¬p, q}.

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Exercise

By the resolution algorithm show that:

¬(p → ¬q) | = p. Counterexample in natural language: it is not the case that if it is raining then the ground is not wet | = it is raining.

p → q | = ¬(p → ¬q). Counterexample in natural language: if it is raining then the ground is wet | = it is not the case that if it is raining then the ground is not wet.

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Ontological commitment

Ontological commitment: what the logic assumes about the nature of reality: propositional logic assumes that there are facts that either hold or do not hold in the world. First-order logic assumes that the world consists of objects with certain relations among them that do or do not hold.

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Language

Objects: people, houses, numbers, theories, Ronald McDonald, colors, baseball games, wars, centuries . . . Relations: these can be unary relations or properties such as red, round, prime, . . ., or more general n-ary relations such as brother of, bigger than, inside, part of, has color, occurred after,

wns, comes between, . . .

Functions: father of, best friend, one more than, beginning of . . .

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Language

Variables: V = {x1, x2, x3, . . .} Logical operators: L = {⊥, ¬, ∨, ∧, →, ∀x, ∃x, =}x∈V for x ∈ V First order language: L(R, F, C) = L ∪ R ∪ F ∪ C R is a finite or countable set of relation symbols (or predicate symbols) of any arity. F is a finite or countable set of function symbols of any arity. C is a finite or countable set of constant symbols.

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Language

Terms (Tm): t ::= x | c | f(t, t, . . . , t), where x ∈ V, c ∈ C, f ∈ F. Formulas (Fml): φ ::= P(t, t, . . . , t) | t = t | (¬φ) | (φ∧φ) | (φ∨φ) | (φ → φ) | (∀x)φ | ∃xφ, where P ∈ R, t ∈ Tm, x ∈ V

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Examples

Natural language Logical formula John walks W(j) John sees Mary S(j, m) John gives Mary the book G(j, m, b) He walks W(x) John sees her S(j, x) Someone walks ∃xW(x) Some boy walks ∃x(B(x) ∧ W(x)) Everyone walks ∀xW(x) Every girl sees Mary ∀x(G(x) → S(x, m)) Someone creates everythings ∃x∀yC(x, y) Natural language processing

Computers translate

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First Order Theories

First order logic is a framework to formalize mathematical theories. It is a universal language for talking about structures.

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Axioms of FOL

Axioms:

A finite complete set of axioms of propositional logic,

∀xA → A{t/x}, where t is free for x in A,

∀x(A → B) → (A → ∀xB), where x ∈ FV(A),

A{t/x} → ∃xA, where t is free for x in A,

∀x(A → B) → (∃xA → B), where x ∈ FV(B),

∀x(x = x),

t1 = t2 ∧ A{t1/x} → A{t2/x}. Rules: ⊢ A → B ⊢ A ⊢ B MP ⊢ A ⊢ ∀xAGen

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Graph Theory

L = {R}, R is a predicate symbol with two arguments. Proper axioms:

∀x¬R(x, x),

∀x∀y(R(x, y) → R(y, x)).

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Group Theory

L = {∗, e}, ∗ is a binary function symbol and e is a constant symbol. Proper axioms:

∀x(e ∗ x = x ∗ e = x),

∀x∀y∀z(x ∗ (y ∗ z) = (x ∗ y) ∗ z),

∀x∃y(x ∗ y = y ∗ x = e).

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Set Theory

L = {∈}, ∈ is a predicate symbol with two arguments. Proper axioms:

∀z(z ∈ x ↔ z ∈ y) → x = y,

∀x(ϕ(x) → x ∈ y) → ∃w∀x(x ∈ w ↔ ϕ(x)),

∃w∀z(z ∈ w ↔ ∃y ∈ x(z ∈ y)),

∃w∀x(x ∈ w ↔ x ⊆ y),

∃w∀x(x ∈ w ↔ ∃y ∈ z(x = F(y)), where F is a unary operation,

∃x(∃y ∈ x∀z(z ∈ y) ∧ ∀y ∈ x∃z ∈ x∀w(w ∈ z ↔ w ∈ y ∨ w = y)),

∃y(y ∈ x) → ∃y ∈ x∀z ∈ y(z ∈ x).

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Peano Arithmetic PA

L = {¯ 0, s, ⊕, ⊙}, ¯ 0 is a constant symbol, s is a unary function symbol and ⊕, ⊙ are binary function symbol. Proper axioms:

∀x(s(x) = ¯ 0),

∀x(x = ¯ 0 → (∃y)s(y) = x),

∀x(x ⊕ ¯ 0 = x),

∀x∀y(x ⊕ s(y) = s(x ⊕ y)),

∀x(x ⊙ ¯ 0 = ¯ 0),

∀x∀y(x ⊙ s(y) = (x ⊙ y) ⊕ x).

∀y[ϕ(¯ 0, y) ∧ ∀x(ϕ(x, y) → ϕ(s(x), y)) → ∀xϕ(x, y)].

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Semantics of FOL

Syntactic expressions of language

semantics

Real situations

Example

M(x) ⇋ x is a man B(y) ⇋ y is a bicycle R(x, y) ⇋ x rides y ∃x∃y(M(x) ∧ B(y) ∧ R(x, y))

Meghdad Ghari (IPM)

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Models

Definition

A model for the first-order language L(R, F, C) is a pair M = (D, I) such that Domain: D is a non-empty set of individual objects. Interpretation: I is a function mapping relation symbols, function symbols, and constant symbols to relations on D, functions on D, and objects in D.

◮ I(Pn) ⊆ Dn, for Pn ∈ R. ◮ I(f n) : Dn → D, for f n ∈ F. ◮ I(c) ∈ D, for c ∈ C. Meghdad Ghari (IPM) Resolution Logic Short Course 2017 32 / 82

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Example

Let L = {¯ 0, s, ⊕, ⊙} be the language of PA. The standard model N = (N, I) of PA is defined as follows: I(¯ 0) = 0. I(s) = suc, where suc : N → N is suc(n) = n + 1. I(⊕) = +, where + : N × N → N is +(n, m) = n + m. I(⊙) = ·, where · : N × N → N is ·(n, m) = n · m.

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Variable assignment

Is φ(x) true in natural numbers? φ(x) ≡ ∃y(x = y ⊕ y) Variable assignment in a model M = (D, I): s : V → D

Example

Let N = (N, I) be the standard model of PA. s : V → N s′ : V → N s(x) = 6 s′(x) = 7 φ(x) is true under s φ(x) is false under s′

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Variable assignment

Is φ(x) true in natural numbers? φ(x) ≡ ∃y(x = y ⊕ y) Variable assignment in a model M = (D, I): s : V → D

Example

Let N = (N, I) be the standard model of PA. s : V → N s′ : V → N s(x) = 6 s′(x) = 7 φ(x) is true under s φ(x) is false under s′

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A variable assignment models a memory state of a computer, where the variables are “addresses" that store “current values". A computer typically works by successively replacing values in addresses by new values. s[x := a] is the variable assignment that is completely like σ except that x now gets the value a (where σ might have assigned a different value).

Example

Let D = {1, 2, 3} and V = {x, y, z}. s s[x := 2]

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Values of terms

Given a model M = (D, I) and variable assignment s in M, we define the value (or interpretation) of terms as follows: xI,s := s(x), for x ∈ V cI,s := I(c), for c ∈ C [f(t1, . . . , tn)]I,s := I(f)(tI,s

1 , . . . , tI,s n )

for f ∈ F

Example

Let L = {¯ 0, s, ⊕, ⊙} be the language of PA, N = (N, I) be the standard model of PA, and s′ is an assignment s.t. s′(x) = 3. [s(¯ 0) ⊕ x]I,s′ = 4 [s(s(¯ 0)) ⊙ s(x)]I,s′ = 8

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Tarski’s theory of truth

Notation: M | =s φ φ is ture in M under assignment s

Definition

Let M = (D, I) be a model and s be an assignment in M: M | =s P(t1, . . . , tn) iff (tI,s

1 , . . . , tI,s n ) ∈ I(P).

M | =s ¬φ iff M | =s φ. M | =s φ ∧ ψ iff M | =s φ and M | =s ψ. M | =s φ → ψ iff M | =s φ implies M | =s ψ. M | =s ∀xφ iff M | =s[x:=a] φ for all a ∈ D. M | =s ∃xφ iff M | =s[x:=a] φ for some a ∈ D.

Definition

φ is (logically) valid if M | =s φ for every model M and every assignment s in M. Notation: | = φ.

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Example

Let φ(x) ≡ ∃y(x = y ⊕ y) Let N = (N, I) be the standard model of PA. s : V → N s′ : V → N s(x) = 6 s′(x) = 7 N | =s φ(x) N | =s′ φ(x) In fact, N | =s φ(x) iff s(x) is an even number.

Exercise

Show that | = ∀y∃x p(x, y) → ∃x∀y p(x, y).

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Semantics

Valuations: v : Formulas → {T, F} v(¬φ) = 1 − v(φ) v(φ ∧ ψ) = min(v(φ), v(ψ)) v(φ ∨ ψ) = max(v(φ), v(ψ)) v(φ → ψ) = min(1, 1 − v(φ) + v(ψ)) v(∀xφ) = min{v(φ[¯ a/x]) | a ∈ D} v(∃xφ) = max{v(φ[¯ a/x]) | a ∈ D} ∀ is a generalization of ∧, and ∃ is a generalization of ∨. Semantics of FOL does not provide a decision procedure for FOL.

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Completeness

Theorem

FOL is sound and complete: ⊢ φ iff | = φ There exists a proof for φ iff φ is true in all models.

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Completeness

Theorem

FOL is sound and complete: ⊢ φ iff | = φ There exists a proof for φ iff φ is true in all models.

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Undecidability

Theorem

Validity in first order logic is undecidable.

Proof.

The proof that there is no decision procedure for validity describes an algorithm that takes an arbitrary Turing machine T and generates a formula ST in first-order logic, such that | = ST iff T halts. If there were a decision procedure for validity, this construction would give us an decision procedure for the halting problem.

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Syllogism

All A are B ∀x(A(x) → B(x)) No A are B ¬∃x(A(x) ∧ B(x)) or ∀x(A(x) → ¬B(x)) Some A are B ∃x(A(x) ∧ B(x)) Not all A are B ¬∀x(A(x) → B(x)) or ∃x(A(x) ∧ ¬B(x)) The language of Syllogism is a small fragment of predicate logic: monadic predicate logic, which has unary predicates only.

Theorem

Validity in monadic predicate logic is decidable. Johan van Benthem, Hans van Ditmarsch, Jan van Eijck, Jan Jaspars, Logic in Action New Edition, November 23, 2016

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CNF in Predicate Logic

In propositional logic, every formula is equivalent to one in CNF , but this is not true in first-order logic. However, a formula in first-order logic can be transformed into an equisatisfiable formula in CNF . Arbitrary formulas in predicate logic ↓ Formulas in prenex normal form (PNF) ↓ Skolemization ↓ Formulas in clausal form or CNF

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Prenex Normal Form (PNF)

Theorem

For every formula A there is an equivalent formula A′ (prenex normal form of A) with the same free variables in which all quantifiers appear at the begining.

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⊢ ¬∀xA ↔ ∃x¬A. ⊢ ¬∃xA ↔ ∀x¬A. ⊢ (∀xA ∗ B) ↔ ∀y(A[y/x] ∗ B). ⊢ (A ∗ ∀xB) ↔ ∀y(A ∗ B[y/x]). ⊢ (∃xA ∗ B) ↔ ∃y(A[y/x] ∗ B). ⊢ (A ∗ ∃xB) ↔ ∃y(A ∗ B[y/x]). ⊢ (∀xA → B) ↔ ∃y(A[y/x] → B). ⊢ (A → ∀xB) ↔ ∀y(A → B[y/x]). ⊢ (∃xA → B) ↔ ∀y(A[y/x] → B). ⊢ (A → ∃xB) ↔ ∃y(A → B[y/x]). where ∗ ∈ {∧, ∨}, and y does not occur in A, B.

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A decidable fragment

Theorem

There are decision procedures for the validity of PNF formulas without constant and function symbols whose prefixes are of one of the forms (where m, n ≥ 0): ∀x1 . . . ∀xn∃y1 . . . ∃ym ∀x1 . . . ∀xn∃y∀z1 . . . ∀zm ∀x1 . . . ∀xn∃y1∃y2∀z1 . . . ∀zm These classes are conveniently abbreviated ∀∗∃∗, ∀∗∃∀∗, ∀∗∃∃∀∗.

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Skolemization

Theorem

For every sentence A in a given language L there is an equisatisfiable universal formula A′ (Skolemization of A) in an expanded language L′ gotten by the addition of new function symbols.

Lemma

For any sentence A = ∀x1 . . . ∀xn∃yB of a language L, A and A′ = ∀x1 . . . ∀xnB[f(x1, . . . , xn)/y] are equisatisfiable, where f is a function symbol not in L.

Example

∃x∀yP(x, y)

∀yP(c, y)

∀x∃yP(x, y)

∀xP(x, f(x))

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Example

∀x∀y[∃z(P(x, z) ∧ P(y, z)) → ∃uQ(x, y, u)] ↓ ∀x∀y∀w[P(x, w) ∧ P(y, w) → ∃uQ(x, y, u)] ↓ ∀x∀y∀w∃v[P(x, w) ∧ P(y, w) → Q(x, y, v)] ↓ ∀x∀y∀w[P(x, w) ∧ P(y, w) → Q(x, y, f(x, y, w))] L′ = L ∪ {f} ↓ P(x, w) ∧ P(y, w) → Q(x, y, f(x, y, w)) ↓ ¬P(x, w) ∨ ¬P(y, w) ∨ Q(x, y, f(x, y, w)) Clausal form (CNF)

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Herbrand Theorem

Reduction of first-order logic to propositional logic.

Theorem

Let φ = (∃y1, . . . , yn)F(y1, . . . , yn) and F is quantifier-free. Then φ is valid if and only if there exists a finite sequence of terms tij, possibly in an expansion of the language, with 1 ≤ i ≤ k and 1 ≤ j ≤ n, such that F(t11, . . . , t1n) ∨ . . . ∨ F(tk1, . . . , tkn) is a tautology of propositional logic.

Theorem

Let A be a sentence in PNF in a language L, B a prenex equivalent of ¬A, and C( x) a quantifier-free Skolemization of B in the expanded languge L′. Then A is valid iff there are terms t1, . . . , tn of L′ such that ¬C( t1) ∨ . . . ∨ ¬C( tn) is a tautology of propositional logic.

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Substitution and unifier

Definition

An expression is any term or literal. A substitution is a mapping θ : V → Tm, denoted by {t1/x1, . . . , tn/xn}. The result of applying θ on expression E is denoted by Eθ. A set of expressions S = {E1, . . . , En} is unifiable if there is a substitution θ, called a unifier for S, such that E1θ = . . . = Enθ. A unifier θ for S is a most general unifier (mgu) for S if for every unifier σ for S there is a substitution λ such that σ = θλ.

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Substitution and unifier

Definition

An expression is any term or literal. A substitution is a mapping θ : V → Tm, denoted by {t1/x1, . . . , tn/xn}. The result of applying θ on expression E is denoted by Eθ. A set of expressions S = {E1, . . . , En} is unifiable if there is a substitution θ, called a unifier for S, such that E1θ = . . . = Enθ. A unifier θ for S is a most general unifier (mgu) for S if for every unifier σ for S there is a substitution λ such that σ = θλ.

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Substitution and unifier

Definition

An expression is any term or literal. A substitution is a mapping θ : V → Tm, denoted by {t1/x1, . . . , tn/xn}. The result of applying θ on expression E is denoted by Eθ. A set of expressions S = {E1, . . . , En} is unifiable if there is a substitution θ, called a unifier for S, such that E1θ = . . . = Enθ. A unifier θ for S is a most general unifier (mgu) for S if for every unifier σ for S there is a substitution λ such that σ = θλ.

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Substitution and unifier

Definition

An expression is any term or literal. A substitution is a mapping θ : V → Tm, denoted by {t1/x1, . . . , tn/xn}. The result of applying θ on expression E is denoted by Eθ. A set of expressions S = {E1, . . . , En} is unifiable if there is a substitution θ, called a unifier for S, such that E1θ = . . . = Enθ. A unifier θ for S is a most general unifier (mgu) for S if for every unifier σ for S there is a substitution λ such that σ = θλ.

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Disagreement set

Definition

For a finite nonempty set of expressions S, the disagreement set D(S) is the set of subexpressions of S that begin at the leftmost position at which elements of S differ.

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Example

We find a mgu for S = {P(x, g(b)), P(h(y, a), z)} by the unification algorithm.

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Example (Step 1)

σ := ǫ S = {P(x, g(b)), P(h(y, a), z)}

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Example (Step 2)

D(S) = {x, h(y, a)} σ := ǫ{h(y, a)/x} = {h(y, a)/x} S := Sσ = {P(h(y, a), g(b)), P(h(y, a), z)}

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Example (Step 3)

D(S) = {g(b), z} σ := {h(y, a)/x}{g(b)/z} = {h(y, a)/x, g(b)/z} S := Sσ = {P(h(y, a), g(b)), P(h(y, a), g(b))} = {P(h(y, a), g(b))}

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Example

Thus the mgu for S = {P(x, g(b)), P(h(y, a), z)} is σ = {h(y, a)/x, g(b)/z}.

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Unification Algorithm

/Input: A set S = ∅ of literals / /* Output: A most general unifier for S or report not unifiable */ σ := ǫ (the empty substitution) while S is not a singleton do choose a disagreement pair x ∈ V, t ∈ Tm in D(S) s.t. x ∈ t if there is no such disagreement pair then return “non-unifiable" else σ := σ{t/x} and S := Sσ return σ

Exercise

Apply the unification algorithm to the following sets:

{P(a, y, f(y)), P(z, z, u)}, where a ∈ C, y, z, u ∈ V, f ∈ F.

{P(g(x), y), P(y, y), P(y, f(u))}, where x, y, u ∈ V, f ∈ F.

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Resolution Rule

Let C ⊔ {P t1, . . . , P tn} and D ⊔ {¬P s1, . . . , ¬P sm} be clauses which by renaming of variables have no variables in common. Resolution Rule: C ⊔ {P t1, . . . , P tn} D ⊔ {¬P s1, . . . , ¬P sm} Cσ ∪ Dσ where σ is an mgu for {P t1, . . . , P tn, P s1, . . . , P sm}. Cσ ∪ Dσ is the resolvent of the parent clauses.

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Definition

A resolution proof of C from a given formula S is a finite sequence C1, C2, . . . , Cn such that each Ci is either a clause in S or obtained by renamig variables of a clause in S or is inferred from earlier member of the sequence by the resolution rule and Cn = C. In this case we write S ⊢R C. A resolution refutation of S is a resolution proof of empty clause from S.

Theorem

S is unsatisfiable iff there is a resolution refutation of S.

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Resolution algorithm

/* Input: A set of clauses S / / Output: S is satisfiable or unsatisfiable */ S0 := S Loop do Choose a new pair of clauses C1, C2 ∈ Si C := Res(C1, C2) If C = then return unsatisfiable else Si+1 := Si ∪ {C} If Si+1 := Si for all pairs of clauses then return satisfiable. While an unsatisfiable set of clauses will eventually produce under a suitable systematic execution of the procedure, the existence of infinite models means that the resolution procedure on a satisfiable set of clauses may never terminate, so resolution is not a decision procedure for first order logic.

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Example

We want to show that A1, A2 ⊢R A3, A1 := ∀x∀y∀z[P(x, y) ∧ P(y, z) → P(x, z)] (transitivity) A2 := ∀x∀y[P(x, y) → P(y, x)] (symmetry) A3 := ∀x∀y∀z[P(x, y) ∧ P(x, z) → P(y, z)] (Euclideanness) Note that ¬A3 := ∃x∃y∃z[P(x, y) ∧ P(x, z) ∧ ¬P(y, z)] The clausal form of A1, A2, ¬A3 are {{¬P(x, y), ¬P(y, z), P(x, z)}} {{¬P(u, v), P(v, u)}} {{P(a, b)}, {P(a, c)}, {¬P(b, c)}}

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Example

We want to show that A1, A2 ⊢R A3, A1 := ∀x∀y∀z[P(x, y) ∧ P(y, z) → P(x, z)] (transitivity) A2 := ∀x∀y[P(x, y) → P(y, x)] (symmetry) A3 := ∀x∀y∀z[P(x, y) ∧ P(x, z) → P(y, z)] (Euclideanness) Note that ¬A3 := ∃x∃y∃z[P(x, y) ∧ P(x, z) ∧ ¬P(y, z)] The clausal form of A1, A2, ¬A3 are {{¬P(x, y), ¬P(y, z), P(x, z)}} {{¬P(u, v), P(v, u)}} {{P(a, b)}, {P(a, c)}, {¬P(b, c)}}

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Example

We want to show that A1, A2 ⊢R A3, A1 := ∀x∀y∀z[P(x, y) ∧ P(y, z) → P(x, z)] (transitivity) A2 := ∀x∀y[P(x, y) → P(y, x)] (symmetry) A3 := ∀x∀y∀z[P(x, y) ∧ P(x, z) → P(y, z)] (Euclideanness) Note that ¬A3 := ∃x∃y∃z[P(x, y) ∧ P(x, z) ∧ ¬P(y, z)] The clausal form of A1, A2, ¬A3 are {{¬P(x, y), ¬P(y, z), P(x, z)}} {{¬P(u, v), P(v, u)}} {{P(a, b)}, {P(a, c)}, {¬P(b, c)}}

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Example

Resolution proof:

{¬P(x, y), ¬P(y, z), P(x, z)}

{¬P(u, v), P(v, u)}

{P(a, b)}

{P(a, c)}

{¬P(b, c)}

{P(b, a)} Res(2,3), {a/u, b/v}

{¬P(a, z), P(b, z)} Res(1,6), {b/x, a/y}

{¬P(a, c)} Res(5,7), {c/z}

Res(4,8)

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Exercise

Using resolution prove that father(y, x) ∧ male(x) → son(x, y), father(jim, tim), male(tim) ⊢ son(tim, jim) I.e. show that the clause S = {son(tim, jim)} follows from the following clauses

C1 = {son(x, y), ¬father(y, x), ¬male(x)}.

C2 = {father(jim, tim)}.

C3 = {male(tim)}.

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Example 1 (Prover 9)

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Example 2 (Prover 9)

⊢ ∃x∀yp(x, y) → ∀y∃xp(x, y)

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Example 3 (Prover 9)

⊢ ∀y∃xp(x, y) → ∃x∀yp(x, y)

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Example 4 (Prover 9)

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Other forms of resolution

Ground resolution Positive resolution Hyperresolution Semantic resolution Set-of-support resolution Unit resolution Input resolution Linear resolution SLD resolution ... Samuel R. Buss, An Introduction to Proof Theory, In Handbook of Proof Theory, Elsevier 1998

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