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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 1 / 82 Outline

1 Propositional Logic Conjunctive Normal Form Propositional Resolution

2 First Order Logic Syntax Semantics Resolution for FOL PROLOG

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 2 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 3 / 82 Semantics: principle of compositionality

The meaning of a complex expression is determined by the meanings of 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))

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 4 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 5 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 6 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 6 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 7 / 82 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)

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 8 / 82 Resolution

John Alan Robinson, A Machine-Oriented Logic Based on the Resolution Principle J. ACM. 12 (1): 23-41, 1965.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 9 / 82 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 original set of clauses must have been unsatisﬁable.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 10 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 11 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 11 / 82 Propositional Resolution Rule

Propositional Resolution Rule:

C t {p} D t {¬p} C ∪ D where C and D are clauses and t denotes disjoint union.

C ∪ D is called the resolvent of C t {p} and D t {¬p}:

C ∪ D = Res(C t {p}, D t {¬p})

Resolution on clause forms:

{C1,..., C t {p}, D t {¬p},..., Cn} {C1,..., C ∪ D,..., Cn} Resolution on clause forms preserves satisﬁability.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 12 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 13 / 82 Proof by resolution

In order to show that φ1, φ2, . . . , φn |= ψ,

we show that φ1 ∧ φ2 ∧ ... ∧ φn ∧ ¬ψ is unsatisﬁable.

1 First, φ1 ∧ φ2 ∧ ... ∧ φn ∧ ¬ψ is converted into CNF. 2 Then, the resolution rule is applied to the resulting clauses. 3 The process continues until one of two things happens:

I two clauses resolve to yield the empty clause , in which case φ1, φ2, . . . , φn |= ψ. I there are no new clauses that can be added, in which case φ1, φ2, . . . , φn 6|= ψ;

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 14 / 82 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 6∈ S do S := S ∪ {C} If ∈ S then return unsatisﬁable else (i.e. if 6∈ S) return satisﬁable.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 15 / 82 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) Resolution Logic Short Course 2017 16 / 82 Example II We show that p 6|= 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: 1 {p} a clause of S 2 {¬p, ¬q} a clause of S 3 {¬q} resolvent of clauses 1 and 2 S is satisfiable, since there are no new resolvents that can be added and 6∈ S. Meghdad Ghari (IPM) Resolution Logic Short Course 2017 17 / 82 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: ¨H ¨ H ¨¨ HH ¨¨ HH p¯ p @ @ @ @ p¯q¯ pq¯ ¡A ¡ A ¡ A ¡ A p¯qr¯ ¯r

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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 18 / 82 Exercise

By the resolution algorithm show that: 1 ¬(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. 2 p → q 6|= ¬(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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 19 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 20 / 82 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, owns, comes between, . . . Functions: father of, best friend, one more than, beginning of . . .

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 21 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 22 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 23 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 24 / 82 First Order Theories

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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 25 / 82 Axioms of FOL

Axioms: 1 A ﬁnite complete set of axioms of propositional logic, 2 ∀xA → A{t/x}, where t is free for x in A, 3 ∀x(A → B) → (A → ∀xB), where x 6∈ FV (A), 4 A{t/x} → ∃xA, where t is free for x in A, 5 ∀x(A → B) → (∃xA → B), where x 6∈ FV (B), 6 ∀x(x = x),

7 t1 = t2 ∧ A{t1/x} → A{t2/x}. Rules: ` A → B ` A ` A MP Gen ` B ` ∀xA

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 26 / 82 Graph Theory

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

Proper axioms: 1 ∀x¬R(x, x), 2 ∀x∀y(R(x, y) → R(y, x)).

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 27 / 82 Group Theory

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

Proper axioms: 1 ∀x(e ∗ x = x ∗ e = x), 2 ∀x∀y∀z(x ∗ (y ∗ z) = (x ∗ y) ∗ z), 3 ∀x∃y(x ∗ y = y ∗ x = e).

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 28 / 82 Set Theory

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

Proper axioms: 1 ∀z(z ∈ x ↔ z ∈ y) → x = y, 2 ∀x(ϕ(x) → x ∈ y) → ∃w∀x(x ∈ w ↔ ϕ(x)), 3 ∃w∀z(z ∈ w ↔ ∃y ∈ x(z ∈ y)), 4 ∃w∀x(x ∈ w ↔ x ⊆ y), 5 ∃w∀x(x ∈ w ↔ ∃y ∈ z(x = F(y)), where F is a unary operation, 6 ∃x(∃y ∈ x∀z(z 6∈ y) ∧ ∀y ∈ x∃z ∈ x∀w(w ∈ z ↔ w ∈ y ∨ w = y)), 7 ∃y(y ∈ x) → ∃y ∈ x∀z ∈ y(z 6∈ x).

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 29 / 82 Peano Arithmetic PA

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

Proper axioms: 1 ∀x(s(x) 6= 0¯), 2 ∀x(x 6= 0¯ → (∃y)s(y) = x), 3 ∀x(x ⊕ 0¯ = x), 4 ∀x∀y(x ⊕ s(y) = s(x ⊕ y)), 5 ∀x(x 0¯ = 0¯), 6 ∀x∀y(x s(y) = (x y) ⊕ x). 7 ∀y[ϕ(0¯, y) ∧ ∀x(ϕ(x, y) → ϕ(s(x), y)) → ∀xϕ(x, y)].

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 30 / 82 Semantics of FOL

semantics Syntactic expressions of language 99K 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) Resolution Logic Short Course 2017 31 / 82 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. n n n I I(P ) ⊆ D , for P ∈ R. n n n I I(f ): D → D, for f ∈ F. I I(c) ∈ D, for c ∈ C.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 32 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 33 / 82 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.

0 s : V → N s : V → N s(x) = 6 s0(x) = 7 φ(x) is true under s φ(x) is false under s0

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 34 / 82 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.

0 s : V → N s : V → N s(x) = 6 s0(x) = 7 φ(x) is true under s φ(x) is false under s0

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 34 / 82 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]

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 35 / 82 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 I,s I,s I,s [f (t1,..., tn)] := I(f )(t1 ,..., tn ) for f ∈ F

Example Let L = {0¯, s, ⊕, } be the language of PA, N = (N, I) be the standard model of PA, and s0 is an assignment s.t. s0(x) = 3.

0 [s(0¯) ⊕ x]I,s = 4

0 [s(s(0¯)) s(x)]I,s = 8

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 36 / 82 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: I,s I,s M |=s P(t1,..., tn) iff (t1 ,..., tn ) ∈ I(P). M |=s ¬φ iff M 6|=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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 37 / 82 Example

Let φ(x) ≡ ∃y(x = y ⊕ y)

Let N = (N, I) be the standard model of PA.

0 s : V → N s : V → N s(x) = 6 s0(x) = 7

N |=s φ(x) N 6|=s0 φ(x)

In fact,

N |=s φ(x) iff s(x) is an even number.

Exercise Show that 6|= ∀y∃x p(x, y) → ∃x∀y p(x, y).

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 38 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 39 / 82 Completeness

Theorem FOL is sound and complete:

` φ iff |= φ

There exists a proof for φ iff φ is true in all models.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 40 / 82 Completeness

Theorem FOL is sound and complete:

` φ iff |= φ

There exists a proof for φ iff φ is true in all models.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 40 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 41 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 42 / 82 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

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 43 / 82 Prenex Normal Form (PNF)

Theorem For every formula A there is an equivalent formula A0 (prenex normal form of A) with the same free variables in which all quantiﬁers appear at the begining.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 44 / 82 ` ¬∀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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 45 / 82 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 ∀∗∃∗, ∀∗∃∀∗, ∀∗∃∃∀∗.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 46 / 82 Skolemization

Theorem For every sentence A in a given language L there is an equisatisﬁable universal formula A0 (Skolemization of A) in an expanded language L0 gotten by the addition of new function symbols.

Lemma

For any sentence A = ∀x1 ... ∀xn∃yB of a language L, A and 0 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))

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 47 / 82 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))] L0 = 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)

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 48 / 82 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 L0. ~ ~ 0 Then A is valid iff there are terms t1,..., tn of L such that ~ ~ ¬C(t1) ∨ ... ∨ ¬C(tn) is a tautology of propositional logic.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 49 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 50 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 50 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 50 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 50 / 82 Meghdad Ghari (IPM) Resolution Logic Short Course 2017 51 / 82 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.

Meghdad Ghari (IPM) Resolution Logic Short Course 2017 52 / 82 Uniﬁcation Algorithm