Methods of AI. Practice Session 7 Kai-Uwe Kühnberger Dec. 13 th, 2002

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Methods of AI Practice Session 7 Kai-Uwe Kühnberger Dec. 13 th, 2002

Today! As usual: Organizational issues! Midterm! Further plan! Proof Methods! Fourth part: Cases in direct proof! Some invalid proof techniques! Logic! Calculus vs. Model! Boolean algebras! Completeness for propositional logic! Assignment 3 and new assignments

Organization! Midterm! You did a very good job! Average:! German grade: 1.96! ECTS grade: B! I added some credits concerning two problems! Problem with arbitrary heuristics concerning A*! Problems with the difference between optimal solution and optimal opponent

Statistics German grade 1.0 1.3 1.7 2.0 2.3 2.7 3.0 3.3 3.7 4.0 5.0 ECTS grade A A B B C C C D E E F # of students 3 6 20 6 4 4 2 4 0 0 0

Statistics Midterm ECTS grade distribution 30 25 20 15 10 5 0 A B C Grade D E F R1

Statistics Midterm German grade distribution Number of students 25 20 15 10 5 0 1.0 1.3 1.7 2.0 2.3 2.7 3.0 3.3 3.7 4.0 5.0 German grade

Midterm! Congratulations! People passing the midterm with grade A! Karin Pietruska (100%)! Sven Lauer! Tobias Lang! Sebastian Bitzer! Kalia Dogbevi! Johannes Knabe! Till Meyer! Sepideh Sadaghiani! Frank Sanders

Organization! Further plan! Tuesday: 5 th assignment (will be online later)! Deadline: Friday, 10 th of January, 2003! Tuesday: 2 nd programming assignment online! Deadline: Friday, 10 th of January, 2003! Notice! Deadline 4 th assignment! Postponed to Monday, 16 th of December, 2002

Organization! Submission of solutions for assignments! We should use a common subject! Relevant information! Course name! Number of assignment! Family names of teams members! Example! MAI_Assignment 3_Mueller/Richter

Proof Methods: Part 4 Cases in direct proofs and equivalences between statements

Cases in direct proofs! Sometimes it is necessary to split a direct proof into different cases! Each case must be separately proven in order to establish the claim! Make sure that no case is left over! If you miss a case your proof is not valid

Example of a proof by cases! To show that the subset relation is a partial order relation on (X) for a given set X consider cases! Show reflexivity! a (X): a a! Show antisymmetry! a (X) b (X): (a b b a a = b)! Show transitivity! a (X) b (X) c (X): a b b c a b

Equivalences! The proof of an equivalence between two statements is essentially a characterization! A particular fact is characterized by the equivalent statement! An equivalence s t consist of two directions! : The left-to-right direction! : The right-to-left direction

Examples of equivalences! Completeness!! F iff " F! Claim: Let <D, > be partially ordered set. Then: <D, > is a complete lattice iff inf(x) exists in D for every subset X D! Def.: A partial ordered set <D, > is a complete lattice iff sup(x) and inf(x) exists for all X D! sup(x) is called the supremum of X (= the least upper bound) and inf(x) is called the infimum of X (the greatest lower bound)

Some invalid proof techniques! Proof by intimidation! Trivial! Proof by eminent authority! I saw the nobel prize winner downstairs and she said the statement is probably true! Proof by obfuscation! A long plotless sequence of true and/or meaningless syntactically related statements! Proof by the reference of inaccessible literature! The author cites a corollary of a theorem to be found in a privately circulated memoir of the Maldive Islands Philological Society, 1883.

Logic Calculi and Models in Propositional Logic

Remarks! In this session we only examine propositional logic! But a little bit more extended! Not only! ϕ but also! ϕ (deduction from hypotheses)! Similarities to the resolution facts of the lecture Resolution Calculus Compactness theorem The empty clause can be derived Refutation completeness Propositional Logic Maximal consistent sets Consistent sets Completeness

Calculus! A calculus is the syntactic side of a logical system! Exercise 3 (Assignment 4) specifies a calculus for propositional logic! It consists of axioms and rules! Axioms are supposed to be provable! Rules are the machinery to deduce new theorems! Calculi can look quite different! Many axioms but low number of rules (Exercise 3)! Many rules but low number of axioms (Calculus of natural deduction / Logic I)

Model! A model is an algebraic structure <U,I> where U is the universe and I is an interpretation function! A model is the semantic side of a logical system! For propositional logic:! U can be identified with the atomic propositions {p,q,r,...}! I maps atomic propositions to truth values! For predicate logic:! U is a set of objects of our domain of discourse {Paul, White House, Fermat s theorem,...}! I maps constants into U (I(c) = Paul) means that c refers to Paul

Properties! Important notations!! denotes provability in a calculus!! ϕ denotes that ϕ is provable from the axioms alone! Example: In the calculus of Exercise 3 all axioms are provable and every formula that can be derived by Modus Ponens is provable!! ϕ denotes that from one can prove ϕ! " denotes truth in a model! For propositional logic every formula that is a tautology according to the truth tables is true in every model! " ϕ denotes that ϕ is a tautology! " ϕ denotes that ϕ follows from

Important Features! Important concepts related to models! The interpretation function I is defined on {p,q,r,...}! Extension to complex propositional formulas is possible (recursion principle)! Call the extension I +! Deduction from hypothesis! " ϕ iff ψ : (I + (ψ) = T I + (ϕ) = T)! In order to be able to prove a completeness result we need Boolean algebras! Deduction theorem! {ϕ}! ψ, then! ϕ ψ

Boolean Algebras! Definition: A Boolean algebra is a tuple <D,,,!,"> such that! <D, > is a distributive lattice!! is the top and " the bottom element! is a complement operation! You show in the Exercise 4 of the 4th assignment! Such a structure can be represented alternatively by associativity, commutativity etc.

Boolean Algebras! Here is an example of a Boolean algebra! < (X),,,,X>! The intuition why Boolean algebras have something to do with propositional logic! Take the bottom element " as contradiction! Take the top element! as tautology! The other elements represent equivalence classes of formuals! Semantic equivalence: ϕ ψiff! ϕ ψ! If ϕ ψthen ϕ and ψ are in the same equivalence class [ϕ]

Boolean Algebras! What is the implication?! Intuitively, the implication in the calculus should correspond to the relation in the Boolean algebra! ϕ ψiff [ϕ] [ψ]! In predicate logic this Boolean algebra of equivalence classes of formulas is called Lindenbaum algebra! Consistency of a set of formulas! is consistent if! Falsum! Compare satisfiability of formulas! Maximal consistent set of formuals! Meaning: A set is consistent and it cannot be made larger without becoming inconsistent! Algebraic counterpart: ultrafilter

Relation between Boolean Algebras and logic! Maximal consistent set of formuals continued! In the lecture the compactness theorem was mentioned! This is the counterpart to maximal consistent sets! Completeness! Assume a calculus of propositional logic is given! An example is Exercise 3 of the 4 th assignment! Then:! ϕ iff " ϕ! The left side means: ϕ can be proven in the calculus! The right side means: ϕ is true in our model! Therefore: if a propositional formula can be proven (in the calculus), it is true. If a propositional formula is true it can be proven (in the calculus)

Idea of the proof of the completeness result! Steps to prove the completeness result for propositional logic!! ϕ " ϕ! Exercise 3: Everything that can be proven is a tautology! This can be extended to! ϕ! " ϕ! ϕ! (i) Show the contrapositive:! ϕ " ϕ! (ii) Assume! ϕ, show that { ϕ} is consistent! (iv) Show that { ϕ} corresponds to a filter F in the Lindenbaum algebra L! (v) Show that F can be made maximal consistent (ultrafilter)! (vi) Truth lemma: if X is maximal consistent and I(p) = T iff p X, then I + (ψ) = T iff ψ X! (vii) Show that { ϕ} is satisfiable

Assignment 5 Deadline: January 10 th 2003