1. CS.462 Artificial Intelligence SOMCHAI THANGSATHITYANGKUL Lecture 05 : Knowledge Base & First Order Logic

2. 2 Knowledge base A knowledge base KB is a set of sentences. Example KB: JerryGivingLecture  (TodayIsTuesday  TodayIsThursday)  JerryGivingLecture It is equivalent to a single long sentence: the conjunction of all sentences (JerryGivingLecture  (TodayIsTuesday  TodayIsThursday))   JerryGivingLecture

3. 3 Entailment Entailment is the relation of a sentence logically follows from other sentences.  |=   |=  if and only if, in every interpretation in which  is true,  is also true Deduction theorem:  |=  if and only if    is valid (always true)

4. 4 Natural Deduction Proof is a sequence of sentences First ones are premises (KB) Then, you can write down on line j the result of applying an inference rule to previous lines When  is on a line, you know KB  If inference rules are sound, then KB                          Modu s ponen s And- introduct ion And- eliminat ion Modu s tolens

5. 5 Natural deduction example StepFormulaDerivation 1 P  Q Given 2 PRPR 3 (Q  R)  S Given Prove S

6. 6 Natural deduction example KB: 1. JerryGivingLecture  (TodayIsTuesday  TodayIsThursday) 2.  JerryGivingLecture Prove:  TodayIsTuesday

7. 7 StepFormulaDerivation 1 JerryGivingLecture  (TodayIsTuesday  TodayIsThursday) Given 2  JerryGivingLecture Given 3 JerryGivingLecture  (TodayIsTuesday  TodayIsThursday) Biconditional elimination to 1. 4 (TodayIsTuesday  TodayIsThursday)  JerryGivingLecture Biconditional elimination to 1. 5  JerryGivingLecture   (TodayIsTuesday  TodayIsThursday) Contrapositive to 4.

8. 8 Propositional Resolution

9. 9 Propositional Resolution Example

10. 10 Resolution tree KB : (A  C  D)  (A  D  E)  (A   C) Prove : (D  E) Negated conclusion :  (D  E) Convert KB in the CNF, So we have KB: 1.(  A  C  D) 2.(A  D  E) 3.(  A   C) 4.  D 5.  E

11. 11 Resolution tree

12. 12 Try this (P → Q) → Q, (P → P) → R, (R → S) → ¬(S → Q) Prove R

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