KNOWLEDGE REASONING & INFERENCE KNOWLEDGE REASONING & INFERENCE 1 Dr. Abbas Fadhil M. A. AL-Juboori Computer Science Dept. – Kerbala University

perceive object interpretation action REASONING

 Definition  A process of applying knowledge to arrive at solution  Requires the ability to infer conclusions from the available facts  To reason is to think clearly and logically, to draw reasonable inference or conclusion from known or assumed facts  It works through interaction of rules and data

Process of working with knowledge, facts and problem solving strategies to draw conclusions

Reasoning mechanisms that are used to relate between facts and rules to derive new conclusion or facts

 5 types:  Deductive  Inductive  Abductive  Analogical  Common-Sense

 Deductive reasoning  A process in which general premises are used to obtain specific inference.  E.g. 1: (1). rainy_day  wet_grass (2). wet_grass rainy_day  E.g. 2: (1). mother(X)  female(X) (2). mother(lily) female(lily)

 Inductive Reasoning  Human use to arrive new conclusion from a limited set of facts by the process of generalization.  E.g: P1: Monkeys in the National Zoo eats bananas P2: Monkeys in Taiping Zoo eats bananas All monkeys eat bananas

 Abductive reasoning  A form of deduction that allows for plausible inference, meaning that, the conclusion might follow from available information, but it might be wrong.  Given B is true, if A  B is true, then the conclusion A is true might be deduced, but it can be false.  E.g.: P1: rainy_day  wet_grass P2: wet_grass rainy_day? … or maybe someone watered the grass.

 Analogical Reasoning  Mapping of a mental model (developed from experience) to situation or objects.  E.g.: Q: “What are the working hours of engineers in the company” ▪ Mental model : Engineers are white-collar employees White collar employees work from 8am-5p Working hours for the engineers is from 8am-5pm.

 Common-Sense Reasoning  Based on heuristic knowledge gained through experience.  Relies more on good judgment than on exact logic.  E.g.: ▪ A loose fan usually causes strange noises.  Valuable for quick solutions.

CConcept 1 : Pattern matching AA process of matching between symbols, and between predicates, to determine whether they are identical and therefore, match to each other. CConcept 2 : Instantiation AA process of substituting or replacing a variable with another variable or a constant. OOnly applicable when dealing with FOPL statements. CConcept 3 : Unification AA process of matching identical predicates and instantiating the variables (if any) with another variables or constants. CCombine pattern matching with instantiation. OOnly applicable when dealing with FOPL statements.

 Two symbols/predicates are matched if they are identical  E.g.  tweety  tweety (match)  tweety  tweaty (doesn’t match)  fly  flies (doesn’t match)  man  men (doesn’t match)  teach(X, math)  teach(she, math) ▪ identical predicate name/functor ▪ identical # of arguments Pattern 1 Pattern 2 Pattern 4 Mismatch because unequal with Pattern 1 Pattern 3

 A variable can be instantiated with  Another variable (normally represent object) ▪ E.g.: X/Y; X/Z  Constant ▪ E.g.: X/math; Y/adam  E.g.: man(X), man(Y)..... X/Y or Y/X man(X), man(adam)..... X/adam teach(X, math), teach(joe,Y)..... X/joe; Y/math

IInstantiation involving rules TThe same variable in the same rule must be instantiated with the same value (can be either another variable or constant). EE.g.: R1 : have(X, wings)  flies(X)  isa(X, Y) flies(tweety) X/tweety R1 : have(tweety,wings)  flies(tweety)  isa(tweety,Y)

TTwo predicates unify when they pass pattern matching and instantiation test. EE.g.: 1.isa(X,mammal) and isa(whale,mammal )..... X/whale; mammal  mammal 2.isa(X,mammal) and isa(Y,Z)..... X/Y or Y/X; Z/mammal 3.isa(X,Y) and isa(whale,mammal)..... X/whale; Y/mammal 4.isa(X,Y) and isa(whale,fish)..... X/whale; Y/fish 5.isa(X,Y) and isa(P,Q)..... X/P or P/X; Y/Q or Q/Y 6.isa(X,Mammal) and isa(Whale,mammal)..... X/Whale or Whale/X; Mammal/mammal 7.isa(X,Y) and isa(whale,X)..... X/whale; Y/whale

 Two predicates cannot unify if they failed pattern matching and/or instantiation test.  E.g.: 1.isa(X,mammal) and isa(whale,fish )..... mammal  fish 2.isa(X,mammal) and is(Y,Z)..... isa  is 3.isa(X,Y) and isa(whale,mammal,animal)..... different # of arguments 4.isa(X,X) and isa(whale,mammal)..... whale  mammal 5.isa(X,mammal) and isa(whale,X)..... X/whale; X/mammal

 4 techniques:  Besides, there are also 2 control strategies:  Forward chaining  Backward chaining … will be discussed further in ES topic. Modus Ponens Modus Tollens Hypothetical Syllogism Resolution

 Definition: Rule of logic that asserts IF A and (A  B) are known to be true, then one can infer that B is true.  E.g. 1.It is sunny day (A) 2.If it is sunny (A), then we will go to the beach (B) 3.We will go to the beach (B) or (in PL) 1.A 2.A  B 3. B

Basic form: ABAB A B ((A  B)  A)  B..... a tautology

Example with rules with >1 propositions: A  B  C A B C ((((A  B)  C)  A)  B)  C

DDefinition: Rule of logic that asserts IF (a  b), and b is known to be not true, then one can infer that a is not true. BBasic form: A  B  B  A ((A  B)   B)   A

 Definition: Rule of logic that asserts IF (a  b) and (b  c) are true, (a  c) is also true.  Antecedent of the 1 st rule (b) is a premise of the 2 nd rule. The 3 rd rule concluded consists of:  the antecedent of the first rule (a) and  the consequent of the second rule (c).

 Basic form: ABAB BCBC A  C ((A  B)  (B  C))  (A  C)

 Definition  An inference strategy that is used to determine the truth of an insertion (new fact)  Structure:  (A v B)  (  B v C) = (A v C) * since B   B can be resolved.

 New insertion should contradict with original insertion.  A contradiction is simply two axioms that are logically contradictory (Eg: B and  B). ▪ If B is true, saying that B is false is definitely contradict.  Known as “proof by refutation”  Resolvent ▪ New expression from the resolution of existing axioms.

 Algorithm 1. Assume  P is TRUE 2.Show that the axioms and  P lead to a contradiction 3.Conclude that  P is FALSE since it leads to a contradiction. 4.Conclude that P is TRUE since  P is FALSE This process continues until a contradiction is produced.

 Steps: 1.Negate the goal (proof by refutation) 2.Transform rules into clausal forms 3.Produce a resolvent from contradict axioms P PP  A  BA  B  A  B  B  C  A  C resolvent

 Steps (c0ntinued): 4.Repeat Step 3 until P is obtained 5.Cancel both P and  P to procude [] to establish the goal. P PP [] ::::

 Equivalence rules 1. P  (Q  R)  (P  Q)  R 2. P  Q  Q  P 3. P  Q  Q  P 4.  (P  Q)  P   Q De Morgan 5.  (P  Q)  P   Q De Morgan 6. P  Q  Q   P 7.  P  P Double negation elimination 8. P  Q  P  Q Implication elimination 9. P  Q  (P  Q)  (Q  P) 10. P  Q  (P  Q)  (  P   Q) 11. P   P  False

1. IF temperature > 100 (A) THEN patient has high temperature (B). 2. IF patient has high temperature (B) THEN advise two aspirin (C). 3. temperature > 100 (A). STATE 1 1.A  B 2.B  C 3.A

STATE 2 1.A  B 2.B  C 3.A 4.B STATE 3 1.A  B 2.B  C 3.A 4.B 5.C Perform a resolution to prove C. Step 1: Proof by refutation... obtain  C (by negating the goal C)

Step 2: Convert rules into clausal forms 1.  A  B 2.  B  C 3.A..... The existing axiom 4.  C.... The new insertion, contradict with the goal Step 3: Produce resolvents  A  B  B  C  A  C Rules are converted using equivalence rules P  Q   P  Q 1.  A  B 2.  B  C 3.A 4.  C 5.  A  C

Step 4: Repeat Step 3 to produce another resolvents A  A  C C C is obtained. 1.  A  B 2.  B  C 3.A 4.  C 5.  A  C 6.C

Step 5: Cancel C and  C to establish []. C  C [] [] is established. 1.  A  B 2.  B  C 3.A 4.  C 5.  A  C 6.C 7.[]

A complete resolution process is shown as below:  A  B  B  C C A  A  C CC []

E.g.: All cats are animal. By nature, animal will die. Using resolution, prove that if Chico is a cat, then Chico will die. 1.  X cat(X)  animal(X). 2.  Y animal(Y)  will_die(Y). 3.cat(chico). 1.  X cat(X)  animal(X). 2.  Y animal(Y)  will_die(Y). 3.cat(chico).

 X cat(X)  animal(X) cat(chico) animal(chico)  Y animal(Y)  will_die(Y) will_die(chico) 1.  X cat(X)  animal(X). 2.  Y animal(Y)  will_die(Y). 3.cat(chico). 4.animal(chico). 5.  will_die(chico). 1.  X cat(X)  animal(X). 2.  Y animal(Y)  will_die(Y). 3.cat(chico). 4.animal(chico). 5.  will_die(chico). Step 1

Step 2 1.  X cat(X)  animal(X). 2.  Y animal(Y)  will_die(Y). 3.cat(chico). 4.animal(chico). 5.  will_die(chico). 1.  cat(X)  animal(X). 2.  animal(Y)  will_die(Y). 3.cat(chico). 4.animal(chico). 5.  will_die(chico). 1.  cat(X)  animal(X). 2.  animal(Y)  will_die(Y). 3.cat(chico). 4.animal(chico). 5.  will_die(chico). P  Q   P  Q

Proven! Contradiction between the goal will_die(ciko) and ¬wil_die(ciko). X/Y (note: Y/X is also true) Y/chico  cat(X)  animal(X)  animal(Y)  will_die(Y)  cat(Y)  will_die(Y) cat(chico)  will_die(chico) will_die(chico)

Q6: Use Modus Ponens to determine whether a cup supports a book or a book supports a cup. 1.ontopof(X,Y)  supports(X,Y). 2.isabove(X,Y)  (istouching(X,Y)  istouching(Y,X))  ontopof(X,Y). 3.isabove(cup,book). 4.istouching(cup,book).

ontopof(X,Y)  supports(Y,X) isabove(X,Y)  (istouching(X,Y)  istouching(Y,X))  ontopof(X,Y) isabove(cup,book) X/cup; Y/book isabove(cup, book)  (istouching(cup, book)  istouching(book, cup))  ontopof(cup, book) istouching(cup,book) ontopof(cup, book) X/cup; Y/book supports(book,cup) The answer for Q6

Q7: Use Resolution to prove hates(marcus,caesar). 1.man(marcus). 2.pompeians(marcus). 3.  X pompeians(X)  romans(X). 4.ruler(caesar). 5.  X romans(X)  loyalto(X,caesar)  hates(X,caesar). 6.  X loyalto(X,Y). 7.  X  Y man(X)  ruler(Y)  trytoassassinate(X,Y)   loyalto(X,Y). 8.trytoassassinate(marcus,caesar).

Q7: Use Resolution to prove hates(marcus,caesar). 1.man(marcus). 2.pompeians(marcus). 3.  X pompeians(X)  romans(X). 4.ruler(caesar). 5.  X romans(X)  loyalto(X,caesar)  hates(X,caesar). 6.  X loyalto(X,Y). 7.  X  Y man(X)  ruler(Y)  trytoassassinate(X,Y)   loyalto(X,Y). 8.trytoassassinate(marcus,caesar).

X/marcus (3)  pompeians(X)  romans(X) (2) pompeians(marcus) romans(marcus) (5)  romans(X)  loyalto(X,caesar)  hates(X,caesar). (7)  man(X)  ruler(Y)   trytoassassinate(X,Y)   loyalto(X,Y) loyalto(marcus,caesar)  hates(marcus,caesar) X/marcus; Y/caesar  man(marcus)   ruler(caesar)   trytoassassinate(marcus,caesar)  hates(marcus,caesar) hates(marcus,caesar) (4) ruler(caesar) (1) man(marcus)  ruler(caesar)   trytoassassinate(marcus,caesar)  hates(marcus,caesar)  trytoassassinate(marcus,caesar)  hates(marcus,caesar) (8)  trytoassassinate(marcus,caesar) (9)  hates(marcus,caesar) []