Logical Representations and Resolution Boolean Logic Conjunctive normal

Logical Representations and Resolution

Boolean Logic • Conjunctive normal form • Resolution

Conjunctive Normal Form A literal is a variable or a negated variable. A clause is either a single literal or the disjunction of two or more literals. P, P P, and P Q R S are clauses. (R S ) and P Q are not clauses. A wff is in conjunctive normal form iff it is either a single clause or the conjunction of two or more clauses. (P Q R S) ( P R) is in cnf (P Q R S) ( P R) is not in cnf

Conjunctive Normal Form Theorem: Every wff can be converted to an equivalent wff in conjunctive normal form. Proof: By construction: conjunctive. Boolean(w: wff) = 1. Eliminate and from w, using the fact that P Q is equivalent to P Q. 2. Reduce the scope of each to a single term, using the facts: • Double negation: ( P) = P • de. Morgan’s laws: • (P Q) ( P Q) 3. Convert w to a conjunction of clauses using the fact that both and are associative and the fact that and distribute over each other.

Conversion to Conjunctive Normal Form Let w be the wff : P (R Q). Then w can be converted to conjunctive normal form as follows: Step 1 produces: P (R Q). Step 2 produces: P ( R Q). Step 3 produces: ( P R) ( P Q).

Resolution Suppose we know: winter summer winter summer nil winter summer winter cold summer cold

Resolution From the premises: Conclude: (P Q) and (R Q), (P R). Resolution is sound: If (P Q) and (R Q) are True, then: • • If Q is True, R must be True. If Q True, P must be True. Since either Q or Q must be True, P R must be True.

Resolution – The Algorithm To prove ST from A: 1. Convert A to a list L of clauses in conjunctive normal form. 2. Construct ST and add it to L. 3. Resolve.

Complementary Literals A pair of complementary literals is a pair of literals that are not mutually satisfiable. So two literals are complementary iff one is positive, one is negative, and they contain the same propositional symbol. For example, Q and Q. Choose two parent clauses that contain a pair of complementary literals. For example: C 1 = R 1 R 2 … Rj Q C 2 = S 1 S 2 … Sk Q. Resolve C 1 and C 2 to derive: R 1 R 2 … Rj S 1 S 2 … Sk

Resolution – Generating nil Consider: Q Q nil The empty clause is unsatisfiable since it contains no literals that can be made True. So if it is ever generated, the resolution procedure halts and reports that, since adding ST to A has led to a contradiction, ST is a theorem given A.

Resolution – The Algorithm resolve-Boolean(A, ST) = 1. Construct L, the list of of clauses from A. 2. Negate ST, convert the result to conjunctive normal form, and add the resulting clauses to L. 3. Until either nil is generated or no progress is being made do: 3. 1 Choose two parent clauses. 3. 2 Resolve the parent clauses together. 3. 3 If the resolvent is not nil and is not in L, add it to L. 4. If nil was generated, a contradiction has been found. Return success. ST must be true. 5. If nil was not generated and there was nothing left to do, return failure.

Resolution – An Example Prove R given: Given Axioms: P (P Q) R (S T) Q T Add: R Clauses: P P Q R S Q T R

Resolution – An Example P P Q R S Q T R P Q R R P Q T Q P Q T T nil

Resolution – Only Select One Pair to Resolve P Q P Q R (1) (2) Prove R: P Q R R P Q P Q ? nil But is R entailed by the two facts we have been given?

Facts in English (1) Marcus was a man. (2) Marcus was a Pompeian. (3) All Pompeians were Romans. (4) Caesar was a ruler. (5) All Romans were either loyal to Caesar or hated him. (6) Everyone is loyal to someone. (7) People only try to assassinate rulers they are not loyal to. (8) Marcus tried to assassinate Caesar.

An Example - Facts in FOL (1) Marcus was a man(Marcus) (2) Marcus was a Pompeian(Marcus) (3) All Pompeians were Romans. x Pompeian(x) Roman(x) (4) Caesar was a ruler(Caesar) (5) All Romans were either loyal to Caesar or hated him. x Roman(x) loyalto(x, Caesar) hate(x, Caesar) (6) Everyone is loyal to someone. x y loyalto(x, y) (7) People only try to assassinate rulers they are not loyal to. x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y) (8) Marcus tried to assassinate Caesar. tryassassinate(Marcus, Caesar)

Question Answering (1) Marcus was a man(Marcus) (2) Marcus was a Pompeian(Marcus) (3) All Pompeians were Romans. x Pompeian(x) Roman(x) (4) Caesar was a ruler(Caesar) (5) All Romans were either loyal to Caesar or hated him. x Roman(x) loyalto(x, Caesar) hate(x, Caesar) (6) Everyone is loyal to someone. x y loyalto(x, y) (7) People only try to assassinate rulers they are not loyal to. x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y) (8) Marcus tried to assassinate Caesar. tryassassinate(Marcus, Caesar) Was Marcus loyal to Caesar?

Reasoning Backward (1) man(Marcus) (2) Pompeian(Marcus) (3) x Pompeian(x) Roman(x) (4) ruler(Caesar) (5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar) (6) x y loyalto(x, y) (7) x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y) (8) tryassassinate(Marcus, Caesar) loyalto(Marcus, Caesar) (Marcus/x) (Caesar/y) person(Marcus) ruler(Caesar) tryassassinate(Marcus, Caesar)

Reasoning Backward (1) man(Marcus) (2) Pompeian(Marcus) (3) x Pompeian(x) Roman(x) (4) ruler(Caesar) (5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar) (6) x y loyalto(x, y) (7) x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y) (8) tryassassinate(Marcus, Caesar) (9) x man(x) person(x) loyalto(Marcus, Caesar) (Marcus/x) (Caesar/y) person(Marcus) (Marcus/x) man(Marcus) ruler(Caesar) tryassassinate(Marcus, Caesar)

Functions and Predicates (1) Marcus was a man(Marcus) (2) Marcus was a Pompeian(Marcus) (3) Marcus was born in 40 A. D. born(Marcus, 40) (4) All men are mortal. x man(x) mortal(x) (5, 6) All. Pompeians died when the volcano erupted in 79 A. D. erupted(volcano, 79) x Pompeian(x) died(x, 79) (7) No mortal lives longer than 150 years. x t 1 t 2 mortal(x) born(x, t 1) gt(t 2 -t 1, 150) dead(x, t 2) (8) It is now 2004. now = 2004 Is Marcus alive?

Functions and Predicates (1) man(Marcus) (2) Pompeian(Marcus) (3) born(Marcus, 40) (4) x man(x) mortal(x) (5) erupted(volcano, 79) (6) x Pompeian(x) died(x, 79) (7) x t 1 t 2 mortal(x) born(x, t 1) gt(t 2 -t 1, 150) dead(x, t 2) (8) now = 2007 alive(Marcus, now)

Functions and Predicates - Filling in the Blanks (1) man(Marcus) (2) Pompeian(Marcus) (3) born(Marcus, 40) (4) x man(x) mortal(x) (5) erupted(volcano, 79) (6) x Pompeian(x) died(x, 79) (7) x t 2 mortal(x) born(x, t 1) gt(t 2 -t 1, 150) dead(x, t 2) (8) now = 2007 (9 a) x t alive(x, t) dead(x, t) (9 b) x t dead(x, t) alive(x, t) (10) x t 2 died(x, t 1) gt(t 2, t 1) dead(x, t 2) alive(Marcus, now)

Showing that Marcus is Not Alive (10)(Marcus/x)(now/t 2) died(Marcus, t 1) alive(Marcus, now) (9 a) (Marcus/x) (now/t) dead(Marcus, now) (7) (Marcus/x)(now/t 2) gt(now, t 1) mortal(Marcus) born(Marcus, t 1) gt(now-t 1, 150) (5) (Marcus/x) (4) (Marcus/x) (79/t 1) Pompeian(Marcus) gt(now, 79) man(Marcus) (2) subst T (3) (40/t 1) born(Marcus, 40) gt(now-40, 150) (1) gt(2004, 79) eval T subst T T gt(2007 -40, 150) eval T

A Harder One Given: x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))] Roman(Isaac) hate(Isaac, Caesar) hate(Paulus, Marcus) thinkcrazy(Isaac, Paulus) Prove: know(Isaac, Marcus)

Prenex Normal Form A sentence in first-order logic is in prenex normal form iff it is of the form: <quantifier list> <matrix>, where <quantifier list> is a list of quantified variables and <matrix> is quantifier-free. x ( y ((P(x) Q(y)) z (R(x, y, z))) is not in pnf. x y z (P(x) Q(y)) R(x, y, z) is in pnf. Its matrix is: (P(x) Q(y)) R(x, y, z).

Clause Form A sentence is in clause form iff: • • • It has been converted to prenex normal form, Its quantifier list contains only universal quantifiers, Its quantifier list is no longer explicitly represented, It is in conjunctive normal form, and There are no variable names that appear in more than one clause. Begin with: x (P(x) Q(x)) y (S(y)) In prenex normal form: In clause form: x y (P(x) Q(x)) S(y) ( P(x) Q(x)) S(y)

Clause Form Simplifies the Process x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))] becomes: Roman(x) know(x, Marcus) hate(x, Caesar) hate(y, z) thinkcrazy(x, y)

Conversion to Clause Form - Step 1 1. Eliminate , using the fact that a b is equivalent to a b x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))] x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))]

Conversion to Clause Form - Step 2 2. Reduce the scope of each to a single term, using: • ( p) = p • de. Morgan’s laws • x P(x) x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))] x [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y z hate(y, z) thinkcrazy(x, y))]

Conversion to Clause Form - Step 3 3. Standardize variables so that each quantifier binds a unique variable. x P(x) x Q(x) x P(x) y Q(y)

Conversion to Clause Form - Step 4 4. Move all quantifiers to the left without changing their relative order. x [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y z hate(y, z) thinkcrazy(x, y))] x y z [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))] At this point, we have prenex normal form.

Conversion to Clause Form - Step 5 5. Eliminate existential quantifiers through the use of Skolem functions and constants. x Roman(x) Roman(S 1) x z father-of(x, z) x father-of(x, S 2(x))

Conversion to Clause Form - Step 6 6. Drop the prefix since all remaining quantifiers are universal. x y z [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))]

Conversion to Clause Form - Step 7 7. Convert the matrix into a conjunction of disjuncts by using: • Associative properties of and . [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))] Roman(x) know(x, Marcus) hate(x, Caesar) hate(y, z) thinkcrazy(x, y) • Distributivity of and . (P(x) Q(x)) T(x) (P(x) T(x)) (Q(x)) T(x))

Conversion to Clause Form - Step 8 8. Create a separate clause for each conjunct. (P(x) T(x)) (Q(x)) T(x)) (P(x) T(x)) (Q(x)) T(x))

Conversion to Clause Form - Step 9 9. Standardize apart the variables. (P(x) T(x)) (Q(x)) T(x)) (P(x) T(x)) (Q(y)) T(y))

Resolution in FOL To find a contradiction, we must show that the KB, augmented with P, is unsatisfiable. Herbrand’s theorem tells us: 1. To show that a set of clauses S is unsatisfiable, it is necessary to consider only interpretations over a particular set called the Herbrand universe of S, which is the set of all ground terms constructable from the following: 1. The function symbols in S, if any. 2. The constant symbols in S, if any. If none, then the constant symbol A. 2. A set of clauses S is unsatisfiable iff a finite subset of ground instances (in which all bound variables have had a value substituted for them) of S is unsatisfiable. Resolution is an algorithm that finds contradictions without enumerating most of the Herbrand universe.

Unification In propositional logic, it is easy to identify complementary literals such as P and P. But in FOL, what should we do about: x, y (hate(x, y)) hate(Marcus, Caesar) Or: hate(Marcus, Caesar) hate(Paulus, Caesar) Or: x, y (P(x, x)) P(Marcus, Caesar)

Before We Start Before we consider a set of formulas and begin to reason with them, we need to standardize the variables apart: Consider: x, y Lived-In(x, y) x, y Mother-Of(x, y) Although we have used x and y in both of these formulas, the variables in these formulas are logically distinct since they occur inside the scope of quantifiers that are independent (i. e. , neither lies inside the scope of the other). So an equivalent way to write these facts is: x 1, y 1 Lived-In(x 1, y 1) x 2, y 2 Mother-Of(x 2, y 2)

Unification We need an algorithm that does two things given a sentence S and a term T: • Determine whether T matches some term T´ in S, and • Return a substitution that can be applied to the remainder of S. Example: S: x, y In(x, y) Climate(y, moderate) Climate(x, moderate) T: In(Italy, Europe) The process we will call unification will determine that T matches the first term of S and it will return the substitution Italy/x, Europe/y. When we apply that to the remainder of S, we have: Climate(Europe, moderate) Climate(Italy, moderate)

The Unification Algorithm unify-for-resolution(lit 1, lit 2) = If either lit 1 or lit 2 is a variable or a constant then: Case (checking the conditions in order and executing only the first one that matches): lit 1 and lit 2 are identical: return nil. /* Succeed with no substitution. lit 1 is a variable that occurs in lit 2: return Fail. /* These two cases implement the lit 2 is a variable that occurs in lit 1: return Fail. occur check. lit 1 is a variable: return (lit 2/lit 1). lit 2 is a variable: return (lit 1/lit 2). otherwise: return Fail. /* No match. If the initial predicate or function symbols of lit 1 and lit 2 are not the same, return Fail. If lit 1 and lit 2 do not have the same number of arguments, return Fail. substitution-list = nil. For i = 1 to the number of arguments of lit 1 do Let S be the result of unify-for-resolution on the ith argument of lit 1 and of lit 2. If S contains Fail, return Fail. If S is not equal to nil then: Apply S to the remainder of both lit 1 and lit 2. Append S to substitution-list. Return substitution-list.

The Occur Check Suppose we are trying to unify: P(x, x) P(g(x), g(x)) What happens if we skip the occur check? x and g(x) unify and yield the substitution g(x)/x. Now we must apply that substitution : g(x) g(g(x)) g(g(g(x))) But if we’d standardized apart the variables before we started, this wouldn’t have been a problem.

The Occur Check is Necessary Even with Standardizing Apart Suppose we are trying to unify: P(x, f(x), x) P(f(a), a, a) x and f(a) unify and yield the substitution f(a)/x. Applying it: f(f(a)), f(a) a, a Without the occur check, f(f(a)) and a will unify, yielding the substitution f(f(a))/a. To apply it, we must replace all a’s in the remaining terms. But that process never terminates.

Most General Unifiers (MGUs) Suppose we are trying to prove x, y (A(x, y) B(x, y)) Using the fact z (A(John, z)) Unifying A(x, y) with A(John, z) yields John/x, z/y. Applying it: z (B(John, z)) But we could also have matched with the substitution John/x, John/y, and been left trying to prove B(John, John), which would be harder. The algorithm we have presented always returns the most general unifier. The MGU is unique up to variable name substitution.

Unification Examples Inputs [1] [2] [3] [4] [5] [6] [7] Roman(x), Roman(Paulus) Roman(x), Ancient(Paulus) Roman(father-of(Marcus)), Roman(x) Roman(father-of(Marcus)), Roman(Flavius), Roman(x), Roman(y), Roman(father-of(x)), Roman(x) Likes(x, y), Likes(Flavius, Marcus) Substitution Result Succeed Paulus/x Fail Succeed father-of(Marcus)/x Fail Succeed x/y Fail (fails occur check) Succeed Flavius/x, Marcus/y

FOL Resolution – An Example Prove: Mortal(Marcus) given: x (Man(x) Mortal(x)) Man(Marcus) Man(x) Mortal(x) Man(Marcus) Mortal(Marcus) Add: Man(x) Mortal(Marcus) Marcus/x Man(Marcus) nil

FOL Resolution – the Algorithm resolve-FOL(A, ST) = Construct L, the list of clauses from A. Rename all variables in ST so that they do not conflict with any variables in L. Negate ST, convert the result to clause form, and add the resulting clauses to L. Until either nil is generated or no progress is being made do: Choose from L two (parent) clauses that contain a pair CL of complementary literals. Resolve the parent clauses together to produce a resolvent: Initially, let the resolvent be the disjunction of all of the literals in both parent clauses except for the two literals in CL. Apply to all of the literals in the resolvent the substitution that was constructed when the literals in CL were unified. Rename all of the variables in the resolvent so that they do not conflict with any of the variables in L. If the resolvent is not nil and is not already in L, add it to L. If nil was generated, a contradiction has been found. Return success. ST must be true. If nil was not generated and there was nothing left to do, return failure. ST may or may not be true. But no proof of ST has been found.

Heuristics to Aid the Resolution Procedure • Only resolve pairs of clauses that contain complementary literals. P(x) Q(x) • Eliminate certain clauses as soon as they are generated: • Tautologies • Clauses that are subsumed by other clauses. • Set of support strategy: whenever possible resolve with a clause that is part of the statement we are trying to prove. • Unit preference strategy: whenever possible, resolve with clauses that have a single literal.

Subsumption Clause A subsumes clause B iff B must be true in any interpretation in which A is true. (A is more general than B. ) True P Q R T P Q P R T Q P Q W W

Subsumption in FOL Clause A subsumes clause B iff B must be true in any interpretation in which A is true. (A is more general than B. ) True x (P(x)) P(Marcus) x(P(x))

Using Resolution Painter(Leonardo) Composer(Palistrina) Country(Italy) Lived-In(Leonardo, Italy) In(Italy, Europe) Climate(Europe, moderate) Painter(Rubens) Lived-In(Rubens, Europe) x, y (In(x, y) Climate(y, moderate) Climate(x, moderate)) Prove that Leonardo was a painter who lived in a moderate climate. Prove: Negate it: y (Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate)) ( y (Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate)))

Converting to Clause From Painter(Leonardo) Composer(Palistrina) Country(Italy) Lived-In(Leonardo, Italy) In(Italy, Europe) Climate(Europe, moderate) Painter(Rubens) Lived-In(Rubens, Europe) x, y (In(x, y) Climate(y, moderate) Climate(x, moderate)) In(x, y) Climate(y, moderate) Climate(x, moderate) ( y (Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate))) y ( (Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate))) Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate)

Using Resolution 1. Painter(Leonardo) 2. Composer(Palistrina) 3. Country(Italy) 4. Lived-In(Leonardo, Italy) 5. In(Italy, Europe) 6. Climate(Europe, moderate) 7. Painter(Rubens) 8. Lived-In(Rubens, Europe) 9. Painter(Sargent) 10. In(x, y) Climate(y, moderate) Climate(x, moderate) Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate) Lived-In(Leonardo, Italy) Painter(Leonardo) Leonardo/x Lived-In(Leonardo, y 1) Climate(y 1, moderate) Italy/y 1 Climate(Italy, moderate)

Using Resolution 1. Painter(Leonardo) 2. Composer(Palistrina) 3. Country(Italy) 4. Lived-In(Leonardo, Italy) 5. In(Italy, Europe) 6. Climate(Europe, moderate) 7. Painter(Rubens) 8. Lived-In(Rubens, Europe) 9. Painter(Sargent) 10. In(x, y) Climate(y, moderate) Climate(x, moderate) Painter(Leonardo) Lived-In(Leonardo, y) Climate(y, moderate) Lived-In(Leonardo, Italy) Painter(Leonardo) Leonardo/x Lived-In(Leonardo, y 1) Climate(y 1, moderate) Italy/y 1 In(x, y) Climate(y, moderate) Climate(x, moderate) Climate(Italy, moderate) In(Italy, Europe) Italy/x In(Italy, y 2) Climate(y 2, moderate) Climate(Europe, moderate) Europe/y 2 Climate(Europe, moderate) nil

Answering Questions A Question: Name a painter who lived in a moderate climate. ? Painter(x) Lived-In(x, y) Climate(y, moderate) Construct an existence statement to be proved: x, y (Painter(x) Lived-In(x, y) Climate(y, moderate)) Negate it and resolve, tracking the bindings: ( x, y (Painter(x) Lived-In(x, y) Climate(y, moderate))) x, y ( (Painter(x) Lived-In(x, y) Climate(y, moderate))) Painter(x) Lived-In(x, y) Climate(y, moderate)

Tracking the Bindings Painter(x) Lived-In(x, y) Climate(y, moderate) Painter(Leonardo) Leonardo/x Painter(Leonardo) Lived-In(Leonardo, y 1) Climate(y 1, moderate) Lived-In(Leonardo, Italy) Italy/y 1 Painter(Leonardo) Climate(Italy, moderate) In(x, y) Climate(y, moderate) Climate(x, moderate) In(Italy, Europe) Climate(Europe, moderate) Italy/x Painter(Leonardo) In(Italy, y 2) Climate(y 2, moderate) Europe/y 2 Painter(Leonardo) Climate(Europe, moderate) Painter(Leonardo)

A Reminder About Standardizing Variables Apart Prove father(Chris, Bill) given: father(x, y) woman(x) {father(x, y) woman(x)} mother(x, y) woman(x) {mother(x, y) woman(x)} mother(Chris, Mary) Add: father(Chris, Bill) father(x, y) woman(x) (1) (2) (3) mother(x, y) woman(x) * father(x, y) mother(x, y) mother(Chris, Mary) Chris/x, Mary/y father(Chris, Mary)

Back to Marcus (1) man(Marcus) (2) Pompeian(Marcus) (3) x Pompeian(x) Roman(x) (4) ruler(Caesar) (5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar) (6) x y loyalto(x, y) (7) x y man(x) ruler(y) tryassassinate(x, y) loyalto(x, y) (8) tryassassinate(Marcus, Caesar) (1) man(Marcus) (2) Pompeian(Marcus) (3) Pompeian(x 1) Roman(x 1) (4) ruler(Caesar) (5) Roman(x 2) loyalto(x 2, Caesar) hate(x 2, Caesar) (6) loyalto(x 3, S 1(x 3)) (7) man(x 4) ruler(y 1) tryassassinate(x 4, y 1) loyalto(x 4, y 1) (8) tryassassinate(Marcus, Caesar)

Proving Marcus Not Loyal to Caesar loyalto(M, C) man(x 4) ruler(y 1) tryassassinate(x 4, y 1) loyalto(x 4, y 1) (M/x 4)(C/y 1) man(M) ruler(C) tryassassinate(M, C) nil ruler(C) tryassassinate(M, C) man(M)

Does Isaac Know Marcus? Given: 1. x [Roman(x) know(x, Marcus)] [hate(x, Caesar) ( y ( z hate(y, z)) thinkcrazy(x, y))] Roman(x) know(x, Marcus) hate(x, Caesar) hate(y, z) thinkcrazy(x, y) 2. Roman(Isaac) 3. hate(Isaac, Caesar) 4. hate(Paulus, Marcus) 5. thinkcrazy(Isaac, Paulus) Prove: know(Isaac, Marcus) * know(Isaac, Marcus)

Does Marcus Hate Caesar? Try to prove hate(M, C): Hate(M, C)

Try to Prove that Marcus Does Not Hate Caesar Try to prove hate(M, C): Hate(M, C)

Try (Again) to Prove that Marcus Does Not Hate Caesar (1) man(Marcus) (2) Pompeian(Marcus) (3) Pompeian(x 1) Roman(x 1) (4) ruler(Caesar) (5) Roman(x 2) loyalto(x 2, Caesar) hate(x 2, Caesar) (6) loyalto(x 3, S 1(x 3)) (7) man(x 4) ruler(y 1) tryassassinate(x 4, y 1) loyalto(x 4, y 1) (8) tryassassinate(Marcus, Caesar) (9) persecute(x 5, y 2) hate(y 2, x 5) persecute(x, y) hate(y, x) (10) hate(x 6, y 3) persecute(y 3, x 6) hate(x, y) persecute(y, x) Now try to prove hate(M, C): Hate(M, C)

A Winnie Example "And if anyone knows anything about anything, " said Bear to himself, "It's Owl who knows something about something, " he said, "Or my name is not Winnie-the-Pooh, " he said. "Which it is, " he added. "So there you are. "

When Does Resolution Work? Resolution is sound and refutation-complete. If there is a contradiction, resolution will eventually find it if resolvents are chosen systematically. But: • It may take exponential time. • Satisfiability is semi-decidable: if a set of clauses is satisfiable, resolution will never find a conflict and it may notice that it is failing to do so.

Satisfiability is Semi Decidable Suppose we want to know whether P(x) is true. There are three possibilities: (1) P(x) is entailed by KB and thus: P(x) KB is unsatisfiable. (2) P(x) is entailed by KB and thus: P(x) KB is unsatisfiable. (3)

Horn Clauses A Horn Clause is a clause in which at most one literal is positive. A(x) B(x) C(x) A(x) B(x) C(x) becomes a Horn clause Deciding entailment with Horn clauses can be done in time that is linear in the size of the KB by using either backward or forward chaining.