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CS621: Artificial Intelligence

CS621: Artificial Intelligence. Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 11- Soundness and Completeness; proof of soundness; start of proof of completeness 12 th august, 2010. Soundness, Completeness & Consistency. Soundness. Semantic World ---------- Valuation,

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CS621: Artificial Intelligence

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  1. CS621: Artificial Intelligence Pushpak BhattacharyyaCSE Dept., IIT Bombay Lecture 11- Soundness and Completeness; proof of soundness; start of proof of completeness 12th august, 2010

  2. Soundness, Completeness &Consistency Soundness Semantic World ---------- Valuation, Tautology Syntactic World ---------- Theorems, Proofs Completeness * *

  3. Introduce Semantics in Propositional logic Valuation Function V Definition of V V(F ) = F Where F is called ‘false’ and is one of the two symbols (T, F) Syntactic ‘false Semantic ‘false’

  4. V(F) = F V(AB) is defined through what is called the truth table V(A) V(B) V(AB) T F F T T T F F T F T T

  5. Tautology An expression ‘E’ is a tautology if V(E) = T for all valuations of constituent propositions Each ‘valuation’ is called a ‘model’.

  6. Soundness • Provability Validity • Completeness • Validity Provability

  7. Soundness:Correctness of the System • Proved entities are indeed valid • Completeness:Power of the System • Valid things are indeed provable

  8. Consistency The System should not be able to prove both P and ~P, i.e., should not be able to derive F

  9. Examine the relation between Soundness & Consistency Soundness Consistency

  10. If a System is inconsistent, i.e., can derive F , it can prove any expression to be a theorem. Because F P is a theorem

  11. To see that (FP) is a tautology two models V(P) = T V(P) = F V(FP) = T for both

  12. If a system is Sound & Complete, it does not matter how you “Prove” or “show the validity” Take the Syntactic Path or the Semantic Path

  13. Problem (P Q)(P Q) Semantic Proof A B P Q P Q P Q AB T F F T T T T T T T F F F F T F T F T T

  14. To show syntactically (P Q) (P Q) i.e. [(P (Q F )) F ] [(P F ) Q]

  15. If we can establish (P (Q F )) F , (P F ), Q F ⊢ F This is shown as Q F hypothesis (Q F ) (P (Q F)) A1

  16. QF; hypothesis (QF)(P(QF)); A1 P(QF); MP F; MP Thus we have a proof of the line we started with

  17. Soundness Proof Hilbert Formalization of Propositional Calculus is sound. “Whatever is provable is valid”

  18. Statement Given A1, A2, … ,An |- B V(B) is ‘T’ for all Vs for which V(Ai) = T

  19. Proof Case 1 B is an axiom V(B) = T by actual observation Statement is correct

  20. Case 2 B is one of Ais if V(Ai) = T, so is V(B) statement is correct

  21. Case 3 B is the result of MP on Ei & Ej Ej is Ei B Suppose V(B) = F Then either V(Ei) = F or V(Ej) = F . . . Ei . . . Ej . . . B

  22. i.e. Ei/Ej is result of MP of two expressions coming before them Thus we progressively deal with shorter and shorter proof body. Ultimately we hit an axiom/hypothesis. Hence V(B) = T Soundness proved

  23. Towards Completeness Proof

  24. Soundness:Correctness of the System • Proved entities are indeed true/valid • Completeness:Power of the System • True things are indeed provable

  25. Tautology An expression ‘E’ is a tautology if V(E) = T for all valuations of constituent propositions Each ‘valuation’ is called a ‘model’.

  26. Necessary results Statement: (pq)((~pq)q) Proof: If we can show that (pq), (~pq) |- q Or, (pq), (~pq), qF |- F Then we are done.

  27. Proof continued 1. (pq) H1 2. (~pq) H2 3. qF H3 4. (~pq) (~qp) theorem of contraposition 5. ~qp MP, 2, 4 6. P MP, 3,5 7. q MP, 6, 1 8. F MP,7,3 QED

  28. How to prove contraposition To show (pq)(~q~p) Proof: pq, ~q, p |- F Very obvious!

  29. An example to illustrate the completeness proof

  30. Running the completeness proof For every row of the truth table set up a proof: • p, ~q |- p(p V q) • p, q |- p(p V q) • ~p, q |- p(p V q) • ~p, ~q |- p(p V q)

  31. p, ~q |- p  (p V q) i.e. p, ~q, p |- p V q p, ~q, p, ~p |- q p, ~q, p, ~p |- F |- F  q |- q

  32. p, q |- p  (p V q) i.e. p, q, p, ~p |- q same as 1

  33. ~p, q |- p  (p V q) ~p, q, p, ~p |- q Same as 1, since F is derived 4. ~p, ~q |- p  (p V q) Same as 1, since F is derived

  34. Why all this? If we have shown p, q |- A and p, ~q |- A then we can show that p |- A

  35. p |- (q  A) also p |- (~q  A) But (q  A)  ((~q  A)  A) is a theorem by MP twice p |- A

  36. General Statement of the completeness proof If V(A) = T for all models then |- A

  37. Elaborating, If P1, P2, …, Pn are constituent propositions of A and if V(A) = T for every model V(Pi) = T/F then |- A

  38. We have a truth table with 2n rows P1 P2 P3 . . . Pn A F F F . . . F T F F F . . . T T . . . T T T . . . T T

  39. If we can show P1’, P2’, …, Pn’ |- A’ For every row where Pi’ = Pi if V(Pi) = T = ~Pi if V(Pi) = F And A’ = A if V(A) = T = ~A if V(A) = F

  40. Lemma If row has P1’, P2’, …, Pn’, A’ Then P1’, P2’, …, Pn’ |- A’ A very critical result linking syntax with semantics

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