Inference in first-order logic

Slides:



Advertisements
Similar presentations
Inference in First-Order Logic
Advertisements

Artificial Intelligence 8. The Resolution Method
Some Prolog Prolog is a logic programming language
Resolution Proof System for First Order Logic
Inference in first-order logic
First-Order Logic.
Logic: The Big Picture Propositional logic: atomic statements are facts –Inference via resolution is sound and complete (though likely computationally.
Inference Rules Universal Instantiation Existential Generalization
Knowledge & Reasoning Logical Reasoning: to have a computer automatically perform deduction or prove theorems Knowledge Representations: modern ways of.
Standard Logical Equivalences
Resolution.
ITCS 3153 Artificial Intelligence Lecture 15 First-Order Logic Chapter 9 Lecture 15 First-Order Logic Chapter 9.
Inference in first-order logic Chapter 9. Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward.
Automated Deduction resolution (Otter) backward-chaining (Prolog)
First Order Logic Resolution
Artificial Intelligence Inference in first-order logic Fall 2008 professor: Luigi Ceccaroni.
We have seen that we can use Generalized Modus Ponens (GMP) combined with search to see if a fact is entailed from a Knowledge Base. Unfortunately, there.
Methods of Proof Chapter 7, second half.. Proof methods Proof methods divide into (roughly) two kinds: Application of inference rules: Legitimate (sound)
For Friday No reading Homework: –Chapter 9, exercise 4 (This is VERY short – do it while you’re running your tests) Make sure you keep variables and constants.
13 Automated Reasoning 13.0 Introduction to Weak Methods in Theorem Proving 13.1 The General Problem Solver and Difference Tables 13.2 Resolution.
Methods of Proof Chapter 7, Part II. Proof methods Proof methods divide into (roughly) two kinds: Application of inference rules: Legitimate (sound) generation.
Logic.
Outline Recap Knowledge Representation I Textbook: Chapters 6, 7, 9 and 10.
Proof methods Proof methods divide into (roughly) two kinds: –Application of inference rules Legitimate (sound) generation of new sentences from old Proof.
Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9 Spring 2004.
Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9.
1 Automated Reasoning Introduction to Weak Methods in Theorem Proving 13.1The General Problem Solver and Difference Tables 13.2Resolution Theorem.
Inference and Resolution for Problem Solving
Inference in first-order logic Chapter 9. Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward.
Methods of Proof Chapter 7, second half.
Knoweldge Representation & Reasoning
Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9 Fall 2004.
Artificial Intelligence
Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9 Spring 2005.
INFERENCE IN FIRST-ORDER LOGIC IES 503 ARTIFICIAL INTELLIGENCE İPEK SÜĞÜT.
Inference in first-order logic I CIS 391 – Introduction to Artificial Intelligence AIMA Chapter (through p. 278) Chapter 9.5 (through p. 300)
Inference in First-Order logic Department of Computer Science & Engineering Indian Institute of Technology Kharagpur.
Inference in first-order logic Chapter 9. Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward.
CS 416 Artificial Intelligence Lecture 12 First-Order Logic Chapter 9 Lecture 12 First-Order Logic Chapter 9.
Logical Agents Logic Propositional Logic Summary
First-Order Logic and Plans Reading: C. 11 (Plans)
An Introduction to Artificial Intelligence – CE Chapter 7- Logical Agents Ramin Halavati
Unification Algorithm Input: a finite set Σ of simple expressions Output: a mgu for Σ (if Σ is unifiable) 1. Set k = 0 and  0 = . 2. If Σ  k is a singleton,
CS Introduction to AI Tutorial 8 Resolution Tutorial 8 Resolution.
Logical Agents Chapter 7. Knowledge bases Knowledge base (KB): set of sentences in a formal language Inference: deriving new sentences from the KB. E.g.:
Automated Reasoning Early AI explored how to automated several reasoning tasks – these were solved by what we might call weak problem solving methods as.
1 Inference in First-Order Logic CS 271: Fall 2009.
1 Inference in First Order Logic CS 171/271 (Chapter 9) Some text and images in these slides were drawn from Russel & Norvig’s published material.
An Introduction to Artificial Intelligence – CE Chapter 9 - Inference in first-order logic Ramin Halavati In which we define.
1/19 Inference in first-order logic Chapter 9- Part2 Modified by Vali Derhami.
© Copyright 2008 STI INNSBRUCK Intelligent Systems Propositional Logic.
Inference in first-order logic
Inference in First Order Logic. Outline Reducing first order inference to propositional inference Unification Generalized Modus Ponens Forward and backward.
First-Order Logic Reading: C. 8 and C. 9 Pente specifications handed back at end of class.
1 Propositional Logic Limits The expressive power of propositional logic is limited. The assumption is that everything can be expressed by simple facts.
Logical Agents Chapter 7. Outline Knowledge-based agents Propositional (Boolean) logic Equivalence, validity, satisfiability Inference rules and theorem.
Proof Methods for Propositional Logic CIS 391 – Intro to Artificial Intelligence.
Inference in First-Order Logic Chapter 9. Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward.
Logical Agents. Outline Knowledge-based agents Logic in general - models and entailment Propositional (Boolean) logic Equivalence, validity, satisfiability.
EA C461 Artificial Intelligence
Presented By S.Yamuna AP/CSE
Inference in first-order logic
Inference in First-Order Logic
Inference in First-Order Logic
Inference in first-order logic part 1
Artificial Intelligence
Inference in first-order logic
Inference in first-order logic part 1
Methods of Proof Chapter 7, second half.
Presentation transcript:

Inference in first-order logic

Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward chaining Backward chaining Resolution

Universal instantiation (UI) Every instantiation of a universally quantified sentence is entailed by it: v α Subst({v/g}, α) for any variable v and ground term g(without any variable) E.g., x King(x)  Greedy(x)  Evil(x) yields: King(John)  Greedy(John)  Evil(John) King(Richard)  Greedy(Richard)  Evil(Richard) King(Father(John))  Greedy(Father(John))  Evil(Father(John)) .

Existential instantiation (EI) For any sentence α, variable v, and constant symbol k that does not appear elsewhere in the knowledge base: v α Subst({v/k}, α) E.g., x Crown(x)  OnHead(x,John) yields: Crown(C1)  OnHead(C1,John) provided C1 is a new constant symbol, called a Skolem constant

Reduction to propositional inference Suppose the KB contains just the following: x King(x)  Greedy(x)  Evil(x) King(John) Greedy(John) Brother(Richard,John) Instantiating the universal sentence in all possible ways, we have: King(John)  Greedy(John)  Evil(John) King(Richard)  Greedy(Richard)  Evil(Richard) The new KB is propositionalized: proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc.

Reduction contd. Every FOL KB can be propositionalized so as to preserve entailment A ground sentence is entailed by new KB iff entailed by original KB Idea: propositionalize KB and query, apply resolution, return result Problem: with function symbols, there are infinitely many ground terms, e.g., Father(Father(Father(John)))

Reduction contd. Theorem: Herbrand (1930). If a sentence α is entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB Idea: For n = 0 to ∞ do create a propositional KB by instantiating with depth-$n$ terms see if α is entailed by this KB Problem: works if α is entailed, loops if α is not entailed Theorem: Turing (1936), Church (1936) Entailment for FOL is semidecidable algorithms exist that say yes to every entailed sentence no algorithm exists that also says no to every nonentailed sentence.

Problems with propositionalization Propositionalization seems to generate lots of irrelevant sentences. Example from: x King(x)  Greedy(x)  Evil(x) King(John) y Greedy(y) Brother(Richard,John) it seems obvious that Evil(John), but propositionalization produces lots of facts such as Greedy(Richard) that are irrelevant With p k-ary predicates and n constants, there are p·nk instantiations.

Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) Knows(John,x) Knows(y,OJ) Knows(John,x) Knows(y,Mother(y)) Knows(John,x) Knows(x,OJ) Standardizing apart eliminates overlap of variables, e.g., Knows(John,z27) Knows(z17,OJ) {x/Jane} {x/OJ, y/John} {x/Mother(John),y/John} No substitution possible yet.

Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unification finds substitutions that make different logical expressions look identical UNIFY takes two sentences and returns a unifier for them, if one exists UNIFY(p,q) =  where SUBST(,p) = SUBST (,q) Basically, find a  that makes the two clauses look alike

Unification Examples UNIFY(Knows(John,x), Knows(John,Jane)) = {x/Jane} UNIFY(Knows(John,x), Knows(y,Bill)) = {x/Bill, y/John} UNIFY(Knows(John,x), Knows(y,Mother(y))= {y/John, x/Mother(John) UNIFY(Knows(John,x), Knows(x,Elizabeth)) = fail Last example fails because x would have to be both John and Elizabeth We can avoid this problem by standardizing: The two statements now read UNIFY(Knows(John,x), Knows(z,Elizabeth)) This is solvable: UNIFY(Knows(John,x), Knows(z,Elizabeth)) = {x/Elizabeth,z/John}

Unification To unify Knows(John,x) and Knows(y,z) Can use θ = {y/John, x/z } Or θ = {y/John, x/John, z/John} The first unifier is more general than the second. There is a single most general unifier (MGU) that is unique up to renaming of variables. MGU = { y/John, x/z }

Unification Unification algorithm: Recursively explore the two expressions side by side Build up a unifier along the way Fail if two corresponding points do not match

The unification algorithm

The unification algorithm

Simple Example Brother(x,John)Father(Henry,y)Mother(z,John) Brother(Richard,x)Father(y,Richard)Mother(Eleanore,x)

Generalized Modus Ponens (GMP) p1', p2', … , pn', ( p1  p2  …  pn q) qθ p1' is King(John) p1 is King(x) p2' is Greedy(y) p2 is Greedy(x) θ is {x/John,y/John} q is Evil(x) q θ is Evil(John) GMP used with KB of definite clauses (exactly one positive literal) All variables assumed universally quantified where pi'θ = pi θ for all i

Soundness of GMP Need to show that p1', …, pn', (p1  …  pn  q) ╞ qθ provided that pi'θ = piθ for all I Lemma: For any sentence p, we have p ╞ pθ by UI (p1  …  pn  q) ╞ (p1  …  pn  q)θ = (p1θ  …  pnθ  qθ) p1', \; …, \;pn' ╞ p1'  …  pn' ╞ p1'θ  …  pn'θ From 1 and 2, qθ follows by ordinary Modus Ponens

Storage and Retrieval Use TELL and ASK to interact with Inference Engine Implemented with STORE and FETCH STORE(s) stores sentence s FETCH(q) returns all unifiers that the query q unifies with Example: q = Knows(John,x) KB is: Knows(John,Jane), Knows(y,Bill), Knows(y,Mother(y)) Result is 1={x/Jane}, 2=, 3= {John/y,x/Mother(y)}

Storage and Retrieval First approach: Create a long list of all propositions in Knowledge Base Attempt unification with all propositions in KB Works, but is inefficient Need to restrict unification attempts to sentences that have some chance of unifying Index facts in KB Predicate Indexing Index predicates: All “Knows” sentences in one bucket All “Loves” sentences in another Use Subsumption Lattice (see below)

Storing and Retrieval Subsumption Lattice Child is obtained from parent through a single substitution Lattice contains all possible queries that can be unified with it. Works well for small lattices Predicate with n arguments has a 2n lattice Structure of lattice depends on whether the base contains repeated variables Knows(x,y) Knows(x,John) Knows(x,x) Knows(John,x) Knows(John,John)

Forward Chaining Forward Chaining Idea: Start with atomic sentences in the KB Apply Modus Ponens Add new atomic sentences until no further inferences can be made Works well for a KB consisting of Situation  Response clauses when processing newly arrived data

Forward Chaining First Order Definite Clauses Disjunctions of literals of which exactly one is positive: Example: King(x)  Greedy(x)  Evil(x) King(John) Greedy(y) First Order Definite Clauses can include variables Variables are assumed to be universally quantified Greedy(y) means y Greedy(y) Not every KB can be converted into first definite clauses

Example knowledge base The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal

Example knowledge base contd. ... it is a crime for an American to sell weapons to hostile nations: American(x)  Weapon(y)  Sells(x,y,z)  Hostile(z)  Criminal(x) Nono … has some missiles, i.e., x Owns(Nono,x)  Missile(x): Owns(Nono,M1) and Missile(M1) … all of its missiles were sold to it by Colonel West Missile(x)  Owns(Nono,x)  Sells(West,x,Nono) Missiles are weapons: Missile(x)  Weapon(x) An enemy of America counts as "hostile“: Enemy(x,America)  Hostile(x) West, who is American … American(West) The country Nono, an enemy of America … Enemy(Nono,America)

Forward chaining algorithm

Forward chaining proof

Forward chaining proof

Forward chaining proof

Properties of forward chaining Sound and complete for first-order definite clauses Datalog = first-order definite clauses + no functions FC terminates for Datalog in finite number of iterations May not terminate in general if α is not entailed This is unavoidable: entailment with definite clauses is semidecidable

Efficiency of forward chaining Incremental forward chaining: no need to match a rule on iteration k if a premise wasn't added on iteration k-1  match each rule whose premise contains a newly added positive literal Matching itself can be expensive: Database indexing allows O(1) retrieval of known facts e.g., query Missile(x) retrieves Missile(M1) Forward chaining is widely used in deductive databases

Hard matching example Diff(wa,nt)  Diff(wa,sa)  Diff(nt,q)  Diff(nt,sa)  Diff(q,nsw)  Diff(q,sa)  Diff(nsw,v)  Diff(nsw,sa)  Diff(v,sa)  Colorable() Diff(Red,Blue) Diff (Red,Green) Diff(Green,Red) Diff(Green,Blue) Diff(Blue,Red) Diff(Blue,Green) Colorable() is inferred iff the CSP has a solution CSPs include 3SAT as a special case, hence matching is NP-hard

Backward Chaining Improves on Forward Chaining by not making irrelevant conclusions Alternatives to backward chaining: restrict forward chaining to a relevant set of forward rules Rewrite rules so that only relevant variable bindings are made: Use a magic set Example: Rewrite rule: Magic(x)American(x) Weapon(x) Sells(x,y,z) Hostile(z)Criminal(x) Add fact: Magic(West)

Backward Chaining Idea: Given a query, find all substitutions that satisfy the query. Algorithm: Work on lists of goals, starting with original query Algo finds every clause in the KB that unifies with the positive literal (head) and adds remainder (body) to list of goals

Backward chaining algorithm SUBST(COMPOSE(θ1, θ2), p) = SUBST(θ2, SUBST(θ1, p))

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Properties of backward chaining Depth-first recursive proof search: space is linear in size of proof Incomplete due to infinite loops fix by checking current goal against every goal on stack Inefficient due to repeated subgoals (both success and failure) fix using caching of previous results (extra space) Widely used for logic programming

Logic programming: Prolog Algorithm = Logic + Control Basis: backward chaining with Horn clauses + bells & whistles Widely used in Europe, Japan (basis of 5th Generation project) Program = set of clauses: head :- literal1, … literaln. criminal(X) :- american(X), weapon(Y), sells(X,Y,Z), hostile(Z). Depth-first, left-to-right backward chaining Built-in predicates for arithmetic etc., e.g., X is Y*Z+3 Built-in predicates that have side effects (e.g., input and output predicates, assert/retract predicates) Closed-world assumption ("negation as failure") e.g., given alive(X) :- not dead(X). alive(joe) succeeds if dead(joe) fails

Prolog Appending two lists to produce a third: append([],Y,Y). append([X|L],Y,[X|Z]) :- append(L,Y,Z). query: append(A,B,[1,2]) ? answers: A=[] B=[1,2] A=[1] B=[2] A=[1,2] B=[]

Prolog Has problems with repeated states and infinite paths Example: Path finding in graphs path(X,Z) :- link(X,Z) path(X,Z) :- path(X,Y),link(Y,Z) path(a,c) A B C link(a,c) link(Y,c) fail path(a,Y) { Y/b} link(a,b) {Y/b }

Prolog Has problems with repeated states and infinite paths Example: Path finding in graphs path(X,Z) :- path(X,Y),link(Y,Z) path(X,Z) :- link(X,Z) path(a,c) A B C path(a,Y) link(Y,b) fail path(a,Y’) link(Y’,Y) path(a,Y) link(Y,b)

Resolution Resolution for propositional logic is a complete inference procedure Existence of complete proof procedures in Mathematics would entail: All conjectures can be established mechanically All mathematics can be established as the logical consequence of a set of fundamental axioms Gődel 1930: Completeness Theorem for first order logic Any entailed sentence has a finite proof No algorithm given until J.A. Robinson’s resolution algorithm in 1965 Gődel 1931: Incompleteness Theorem: Any logical system with induction is necessarily incomplete There are sentences that are entailed, but not proof can be given

Resolution First order logic requires sentences in CNF Conjunctive Normal Form: Each clause is a disjunction of literals, but literals can contain variables, which are assumed to be universally quantified Example: Convert x American(x)  Weapon(y)  Sells(x,y,z)  Hostile(z)  Criminal(x) American(x)   Weapon(y)  Sells(x,y,z)   Hostile(z) Criminal(x)

Conversion to CNF Everyone who loves all animals is loved by someone: x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)] 1. Eliminate biconditionals and implications x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)] 2. Move  inwards: x p ≡ x p,  x p ≡ x p x [y (Animal(y)  Loves(x,y))]  [y Loves(y,x)] x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)] x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)]

Conversion to CNF Standardize variables: each quantifier should use a different one x [y Animal(y)  Loves(x,y)]  [z Loves(z,x)] Skolemize: a more general form of existential instantiation. Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables: x [Animal(F(x))  Loves(x,F(x))]  Loves(G(x),x) Drop universal quantifiers: [Animal(F(x))  Loves(x,F(x))]  Loves(G(x),x) Distribute  over  : [Animal(F(x))  Loves(G(x),x)]  [Loves(x,F(x))  Loves(G(x),x)]

Resolution Inference Rule Full first-order version: l1  ···  lk, m1  ···  mn Subst(θ ,l1  ···  li-1  li+1  ···  lk  m1  ···  mj-1  mj+1  ···  mn)θ where Unify(li, mj) = θ. The two clauses are assumed to be standardized apart so that they share no variables. For example, Rich(x)  Unhappy(x) Rich(Ken) Unhappy(Ken) with θ = {x/Ken} Apply resolution steps to CNF(KB  α); complete for FOL

Resolution Show KB ⊢ α by showing that KB   α is unsatisfyable

Resolution Example Everyone who loves all animals is loved by someone Anyone who kills an animal is loved by no one. Jack loves all animals. Either Jack or Curiosity killed the cat, who is named Tuna All dogs kill a cats Rintintin is a dog Question: Did Curiosity kill the cat?

Resolution Example Everyone who loves all animals is loved by someone. Formulate in FOL x [y [Animal(y)  Loves(x,y)]]  [z Loves(z,x)] Remove Implications x [[y Animal(y)  Loves(x,y)]]  [z Loves(z,x)] x [[y  Animal(y)  Loves(x,y)]]  [z Loves(z,x)] Move negation inward x [y   Animal(y)   Loves(x,y)]  [z Loves(z,x)] x [y Animal(y)   Loves(x,y)]  [z Loves(z,x)] Skolemize x [ Animal(F(x))   Loves(x,F(x))]  [Loves(G(x),x)] N.B.: Argument of Skolemization function are all universally qualified variables Drop universal quantifier [ Animal(F(x))   Loves(x,F(x))]  [Loves(G(x),x)] Use distributive law (and get two clauses) Animal(F(x))  Loves(G(x),x);  Loves(x,F(x))  Loves(G(x),x)

Resolution Example Anyone who kills an animal is loved by no one. Transfer to FOL x [y (Animal(y)  Kills(x,y)] (z Loves(z,x) Remove Implications x [y (Animal(y)  Kills(x,y)]  (z Loves(z,x) Move negations inwards x [ y Animal(y)   Kills(x,y)]  (z  Loves(z,x)) Remove quantifiers Animal(y)   Kills(x,y)   Loves(z,x)

Resolution Example Jack loves all animals. FOL form x [Animal(x)  Loves(Jack, x)] Remove implications x [Animal(x)  Loves(Jack, x)] Remove quantifier Animal(x)  Loves(Jack, x)

Resolution Example Either Jack or Curiosity killed the cat, who is named Tuna. FOL form Kills(Jack,Tuna)  Kills(Curiosity,Tuna); Cat(Tuna)

Resolution Example All cats are animals FOL form Remove implications x [Cat(x)  Animal(x) Remove implications x [Cat(x)  Animal(x)] Remove quantifier Cat(x)  Animal(x)

Resolution Example All dogs kill a cats FOL form Remove implications x [Dog(x)  y[Cat(y)  Kills(x,y)]] Remove implications x [Dog(x)  y[Cat(y)  Kills(x,y)]] Skolemize x [Dog(x)  [Cat(H(x))  Kills(x,H(x))]] Drop universal quantifiers Dog(x)  [Cat(H(x))  Kills(x,H(x))] Distribute (and obtain two clauses) Dog(x)  Cat(H(x); Dog(x)  Kills(x,H(x))]

Resolution Example Rintintin is a dog FOL form Dog(Rintintin)

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin)

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Question: Did Curiosity kill the cat? Kills(Curiosity,Tuna)]

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna)] Cat(Tuna) , Cat(x)  Animal(x) Unify(Cat(Tuna), Cat(x)) = { x/Tuna } First line thus resolves to: Animal(Tuna)

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)  Kills(Curiosity,Tuna), Kills(Curiosity,Tuna) Resolves to: Kills(Jack,Tuna)

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Animal(y)   Kills(x,y)   Loves(z,x), Animal(Tuna) Unify(Animal(Tuna), Animal(y)) = {y/Tuna} Resolves to:  Kills(x,Tuna)   Loves(z,x),

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)  Kills(x,Tuna)   Loves(z,x),  Loves(x,F(x))  Loves(G(x),x), Animal(z)  Loves(Jack, z) Unify( Loves(x,F(x)) , Loves(Jack, z)) = { x / Jack, z / F(x)} Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack)  Animal(F(Jack))

Resolution Example Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)  Kills(x,Tuna)   Loves(z,x) Loves(G(Jack),Jack)  Animal(F(Jack)) Animal(F(x))  Loves(G(x),x), Loves(G(Jack),Jack)  Animal(F(Jack)) Unify(Animal(F(x)) , Animal(F(Jack)))= { x / Jack} Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack)

Resolution Example  Kills(x,Tuna)   Loves(z,x), Loves(G(Jack),Jack) Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)  Kills(x,Tuna)   Loves(z,x) Loves(G(Jack),Jack)  Animal(F(Jack)) Loves(G(Jack),Jack)  Kills(x,Tuna)   Loves(z,x), Loves(G(Jack),Jack) Unify( Loves(z,x), Loves(G(Jack),Jack) ) = { x / Jack, z / G(Jack)} Resolvent clause is obtained by substituting the unification rule  Loves(G(Jack),Jack)

Resolution Example  Loves(G(Jack),Jack), Loves(G(Jack),Jack) Animal(F(x))  Loves(G(x),x)  Loves(x,F(x))  Loves(G(x),x) Animal(y)   Kills(x,y)   Loves(z,x) Animal(x)  Loves(Jack, x) Kills(Jack,Tuna)  Kills(Curiosity,Tuna) Cat(Tuna) Cat(x)  Animal(x) Dog(x)  Cat(H(x) Dog(x)  Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)  Kills(x,Tuna)   Loves(z,x) Loves(G(Jack),Jack)  Animal(F(Jack)) Loves(G(Jack),Jack)  Loves(G(Jack),Jack)  Loves(G(Jack),Jack), Loves(G(Jack),Jack) Resolvent clause is empty. Proof succeeded

Resolution Resolution is refutation-complete Theorem provers If a set of sentences is unsatisfiable, then resolution will be able to produce a contradiction Theorem provers Use control in order to be more efficient Focus of most research effort Separate control from knowledge base Example: Otter (Organized Technique for Theorem proving and Effective Research) A set of clauses known as the SoS - Set of Support The important facts about a problem Search if focused on resolving SoS with another axiom A set of usable axioms Background knowledge about problem field Rewrites / demodulators Rules to transform expressions into a canonical form Set of parameters or clauses that defines the control strategy to allow user to control search and filtering functions to eliminate useless subgoals

Theorem Prover Successes First formal proof of Gődel’s Incompleteness Theorem (1986) Robbins algebra (a simple set of axioms) is Boolean algebra (1996) Software verification: Remote agent spacecraft control program (2000)

Resolution proof: definite clauses