Where is Wumpus Propositional logic (cont…) Reasoning where is wumpus 鬼はどこですか？ Propositional logic (cont…) 　　　命題論理 Reasoning where is wumpus 　鬼がいる場所を推理する

limination introduction negation resolution complex atomic conjunction disjunction time-dependent

elimination 削除 introduction 導入 negation 否定 resolution 解決 atomic 原子の referent 指示物 conjunction 連結 disjunction 分離 dependent 従属関係の

命題論理: 意味論 論理積 A∧B AかつB 論理和 A∨B AまたはB 否定 ￢A Aでない 含意 A⇒B AならばBを意味する 同等 A⇔B (AならばB)かつ(BならばA) S1 S2 S1 S2 S1 is true, then S2 is true. S1 is false, then S2 is either true or false S1 is true, then S2 is true. S1 is false, then S2 is false S1 S2 S1  S2 white  false white  false

then B is either true or false This relation between sentences is called entailment. A |= B This relation between sentences is called implication. A  B AならばB」(A→B)は、Aが真ならばBが真のときだけ真、Aが偽ならばBの真偽にかかわらず真となります。 A is true, then B is true. A is false, then B is either true or false

課題：真偽値の計算 p = T, q = F, r = Tのとき (p  q)  r p = T, q = F, r = Fのとき p = F, q = F, r = Tのとき (p  q)  r p = F, q = F, r = Fのとき (p  q)  r 　これは論理式の意味（真偽値）の計算例である．

Seven inference rules for propositional Logic Modus Ponens And-Elimination And-Introduction Or-Introduction Double-Negation Elimination Unit Resolution Logic connectives:    ,  i 1  2 … n 1  2 … n 1, 2, …, n 1  2  …  n i        ,   (α　または　β　,　not β　) →　α　   ,     (α　または　β　,　not β　または　γ ) →　 α　または　γ　である   

The knowledge base p p A p Percept sentences: there is no smell in the square [1,1]  S1,1 there is no breeze in the square [1,1]  B1,1 there is no smell in the square [2,1]  S2,1 there is breeze in the square [2,1]  　B2,1 there is smell in the square [1,2]  S1,2 there is no breeze in the square [1,2]  B1,2 s s b b p w g p A p

The knowledge base p p A p knowledge sentences: w g If a square has no smell, then neither the square nor any of its adjacent squares can house a wumpus. R1: S1,1  W1,1  W1,2  W2,1 R2: S2,1  W1,1  W2,1  W2,2  W3,1 If there is smell in [1,2], then there must be a wumpus in [1,2] or in one or more of the neighboring squares. R3: S1,2  W1,3  W1,2  W2,2  W1,1 If a square has no breeze, then neither the square nor any of its adjacent squares can have a pit. R4: B1,1  P1,1  P1,2  P2,1 R5: B1,2  P1,1  P1,2 P2,2  P1,3 If there is breeze in [2,1], then there must be a pit in [2,1] or in one or more of the neighboring squares. R6: B2,1  P3,1  P2,1  P2,2  P1,1 s s b b p w g p A p

Inferring knowledge using propositional logic Concerning with the 6 squares, [1,1], [2,1], [1,2], [3,1], [2,2], [1,3], there are 12 symbols, S1,1, S2,1, S1,2, B1,1, B2,1, B1,2, W1,1, W1,2, W2,1, W2,2, W3,1, W1,3 The process of finding a wumpus in [1,3] as follows: 1. Apply R1 to S1,1, we obtain W1,1  W1,2  W2,1 2. Apply And-Elimination, we obtain W1,1 W1,2 W2,1 3. Apply R2 and And-Elimination to S2,1, we obtain W1,1 　W2,2 W2,1 W3,1 4. Apply R3 and the unit resolution to S1,2, we obtain ( is W1,3W1,2 W2,2 and  is W1,1 ) W1,3  W1,2  W2,2 5. Apply the unit resolution again, we obtain ( is W1,3 W1,2 and  is W2,2 ) W1,3  W1,2 6. Apply the unit resolution again, we obtain ( is W1,3 and  is W1,2 ) W1,3 Here is the answer: the wumpus is in [1,3], that is, W1,3 is true. i 1  2 … n    ,   R4: S1,2  W1,3  W1,2  W2,2  W1,1 s s b b p w g p A p

Problem with propositional logic Too many propositions  too many rules to define a competent agent The world is changing, propositions are changing with time.  do not know how many time-dependent propositions we will need have to go back and rewrite time-dependent version of each rule. The problem with proposition logic is that it only has one representational device: the proposition!!! The solutions to the problem is to introduce other logic first-order logic That can represent objects and relations between objects in addition to propositions.