# PIB Pertemuan 12 Resolution & Refutation (review) Short Quiz

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PIB Pertemuan 12 Resolution & Refutation (review) Short Quiz

Resolution & Refutation

General Steps Translate facts and rules into predicate logic
Tranform sentence into CNF (conjunctive normal form) / horn clause / well formed formula

Specific Steps To Produce CNF
Eliminate implications & bi-conditionals Move  inwards Standardize variables: each quantifier has different ones Skolemize: choose a “fact” to eliminate  Eliminate  Distribute ^ over v

Short Quiz (1) Buktikan bahwa “Marcus membenci Caesar” Facts Rules
Marcus adalah seorang manusia Marcus orang Pompei Marcus mencoba membunuh Caesar Caesar adalah seorang penguasa Rules Semua orang Pompei adalah orang Romawi Semua orang Romawi setia pada Caesar atau membenci Caesar Setiap orang setia pada minimal 1 orang Orang hanya mencoba membunuh penguasa yang kepadanya mereka tidak setia

Translation into Predicate Logic
Facts: man(marcus) pompeian(marcus) tryassassinate(marcus,caesar) ruler(caesar) Rules: x pompeian(x)  roman(x) x roman(x)  loyalto(x, caesar) v hate(x,caesar) x y loyalto(x, y) x y man(x) ^ ruler(y) ^ tryassassinate(x,y)  loyalto(x,y) Prove that: hate(marcus, caesar), use refutation.

Inference on “Marcus’s Case”
Make CNF from rules pompeian(x) v roman(x) roman(x) v loyalto(x,caesar) v hate(x,caesar) loyalto(x,y) (man(x) ^ ruler(y) ^ tryassassinate(x,y)) v loyalto(x,y) man(x) v ruler(y) v tryassassinate(x,y) v loyalto(x,y) Apply substitutions into variable, and try to change rules into facts.

Introduction to prolog
PROLOG = “Programmation en logique” (Marseille, 1972) Declarative programming language with procedural elements Used for problems of AI / knowledge-based (expert) systems Motivation: reconcile use of logic as declarative knowledge representation with procedural representation of knowledge Strengths: Logical descriptions of problems, instead of HOW to solve them  let computer work out the solution Well-suited for problems involving search Simple programs can be understood by reading the code Limitations Flow of control / procedural semantics

Facts <predicate_name>(arg1, arg2…).
Prolog-program = collection of clauses = meta programming Facts Rules Goals (queries), shaped liked facts Facts describe properties of objects and relations between objects; comparable to tables in a relational database student Name interest Name Subject Hans Hans Math Tina Tina Datalogi Lars Lars Physics Frida Frida Math Prolog notation: student(hans). student(tina). student(lars). student(frida). Prolog notation: interest(hans,math). interest(tina,datalogi). interest(lars,physics). interest(frida,math). Prolog notation (facts): <predicate_name>(arg1, arg2…).

Rules Simple IF-THEN statements
“Every reasonable student is interested in math.” interest(X,math) :- student(X). head body All specified conditions in the body (all clauses) must be true to make the predicate in the head true. Conjunctions (AND connected): mother(X,Person) :- parent(X,Person),sex(X,female). Disjunctions (OR connected): interest(X,prolog) :- interest(X,artificial_intelligence). interest(X,prolog) :- interest(X,logic).

Goals Goals are queries One ore more subgoals ?- student(thomas).
=> no Pattern matching: a fact matches a goal if Same predicate Same corresponding arguments. Goals can contain variables: ?- student(X). => X = hans ; => X = tina ; => X = lars ; => X = frida ; => no. Variables are instantiated( included in rules),but cannot be declared!

Prolog’s Inference mechanism
Leftmost-depth-first search for solutions Matching: either two terms are identical, or they become identical by variable substitution (resolution based on pred.logic) Processing of subgoals from left to right Backtracking (from goal try to substitute variables into facts or rules) 1: s(a). 2: s(b). 3: q(a). 4: p(X) :- s(X). 5: p(Y) :- q(Y). ?- p(Z). p(Z) 4 5 s(Z) q(Z) 1 2 3 Z=a Z=b Z=a

starting from a goal (theorem) prove the goal by searching for rules whose ”head” (action part) matches the goal Given are the following rules: 1 X 2 H 3 4 OR 5 F&C B&E facts prove AND AND F C B E AND B A C B

SWI Prolog http://www.swi-prolog.org (ver. 5.6.62)
Graphical User Interface (XPCE) Interface with Java, C++ Steps: Consult / Compile Goal search / Execute Debug > Graphical Debugger Trace: show you how the unification & substitution occurs

Exercise 1 Use prolog to prove the “Marcus’s Case” man(marcus).
pompeian(marcus). ruler(caesar). tryassassinate(marcus, caesar). roman(X) :- pompeian(X). hate(X, Y) :- man(X), ruler(Y), tryassassinate(X, Y). loyalto(X, Y) :- man(X), ruler(Y).

Exercise 2: Family Relation (Kinship)
George x Mum Spencer x Kydd Elizabeth x Philip Margaret Diana x Charles Anne x Mark Andrew x Sarah Edward William Harry Peter Zara Beatrice Eugenie

Exercise 2 (cont’d) Buat fakta berdasarkan pohon keluarga di atas
Buat aturan untuk aturan: saudaraLaki(X,Y)  X berjenis kelamin laki-laki, X dan Y adalah anak dari seseorang, X dan Y bukan orang yang sama saudaraPerempuan(X,Y)  X berjenis kelamin perempuan, X dan Y adalah anak dari seseorang, X dan Y bukan orang yang sama bersaudaraKandung(X,Y)  X dan Y memiliki ayah dan ibu yang sama, X dan Y bukan orang yang sama kakek(X,Y,Z)  X adalah ayah dari Y, Y adalah ayah dari Z nenek(X,Y,Z)  X adalah ibu dari Y, Y adalah ibu dari Z

Prolog’s List Simple data structure to process non-numeric relation
A list is a set of ordered sequential elements [2,2,1,1,4,4,5,6] [b,u,d,i] [budi, iwan, wati] How to use a list? Simple case: empty [] Ordered element: [head|tail] [2,2,1,1,4,4,5,6]  [ 2 | [2,1,1,4,4,5,6] ]

Rules on Prolog’s List [a,b,c] unifies with [Head|Tail] resulting in Head=a and Tail=[b,c] [a] unifies with [H|T] resulting in H=a and T=[] [a,b,c] unifies with [a|T] resulting in T=[b,c] [a,b,c] doesn't unify with [b|T] [] doesn't unify with [H|T] [] unifies with []. Two empty lists always can be unified.

Example “Member List” Predicate: member(X,List). X is a member of List, in case of: 1. X is a head, or 2. X is a member of the tail member(X,[X|Tail]). member(X,[Head|Tail]):- member(X,Tail).

Example “Reverse” reverse([],X,X). reverse([X|Y],Z,W) :- reverse(Y,[X|Z],W). reverse([1,2,3],[],A)                    reverse([2,3],[1],A)         reverse([3],[2,1],A)         reverse([],[3,2,1],A)               true  A = [3,2,1]

Please try this (1) [a,b,c,d]=[a,[b,c,d]]. [a,b,c,d]=[a|[b,c,d]].
[]=_. []=[_]. []=[_|[]]. What should Prolog give as result?

Please try this (2) Develop a dictionary to translate number from Indonesian to English and vice versa, example: Translate([satu, sembilan],Y).  Y = [one,nine] Translate(Z,[one]).  Z = [satu] Translate([],[]).

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