CS621: Artificial Intelligence
This lecture by Pushpak Bhattacharyya from IIT Bombay delves into the fundamental concepts of soundness, completeness, and consistency in propositional logic. It covers the definitions and proofs of soundness and completeness, illustrating their significance in the context of semantic and syntactic worlds. The discussion includes truth tables, tautologies, and essential theorems that demonstrate the relationship between provability and validity. This comprehensive exploration provides valuable insights for anyone studying artificial intelligence and formal logic.
CS621: Artificial Intelligence
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Presentation Transcript
CS621: Artificial Intelligence Pushpak BhattacharyyaCSE Dept., IIT Bombay Lecture 11- Soundness and Completeness; proof of soundness; start of proof of completeness 12th august, 2010
Soundness, Completeness &Consistency Soundness Semantic World ---------- Valuation, Tautology Syntactic World ---------- Theorems, Proofs Completeness * *
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’
V(F) = F V(AB) is defined through what is called the truth table V(A) V(B) V(AB) T F F T T T F F T F T T
Tautology An expression ‘E’ is a tautology if V(E) = T for all valuations of constituent propositions Each ‘valuation’ is called a ‘model’.
Soundness • Provability Validity • Completeness • Validity Provability
Soundness:Correctness of the System • Proved entities are indeed valid • Completeness:Power of the System • Valid things are indeed provable
Consistency The System should not be able to prove both P and ~P, i.e., should not be able to derive F
Examine the relation between Soundness & Consistency Soundness Consistency
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
To see that (FP) is a tautology two models V(P) = T V(P) = F V(FP) = T for both
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
Problem (P Q)(P Q) Semantic Proof A B P Q P Q P Q AB T F F T T T T T T T F F F F T F T F T T
To show syntactically (P Q) (P Q) i.e. [(P (Q F )) F ] [(P F ) Q]
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
QF; hypothesis (QF)(P(QF)); A1 P(QF); MP F; MP Thus we have a proof of the line we started with
Soundness Proof Hilbert Formalization of Propositional Calculus is sound. “Whatever is provable is valid”
Statement Given A1, A2, … ,An |- B V(B) is ‘T’ for all Vs for which V(Ai) = T
Proof Case 1 B is an axiom V(B) = T by actual observation Statement is correct
Case 2 B is one of Ais if V(Ai) = T, so is V(B) statement is correct
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
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
Soundness:Correctness of the System • Proved entities are indeed true/valid • Completeness:Power of the System • True things are indeed provable
Tautology An expression ‘E’ is a tautology if V(E) = T for all valuations of constituent propositions Each ‘valuation’ is called a ‘model’.
Necessary results Statement: (pq)((~pq)q) Proof: If we can show that (pq), (~pq) |- q Or, (pq), (~pq), qF |- F Then we are done.
Proof continued 1. (pq) H1 2. (~pq) H2 3. qF H3 4. (~pq) (~qp) theorem of contraposition 5. ~qp MP, 2, 4 6. P MP, 3,5 7. q MP, 6, 1 8. F MP,7,3 QED
How to prove contraposition To show (pq)(~q~p) Proof: pq, ~q, p |- F Very obvious!
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)
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
p, q |- p (p V q) i.e. p, q, p, ~p |- q same as 1
~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
Why all this? If we have shown p, q |- A and p, ~q |- A then we can show that p |- A
p |- (q A) also p |- (~q A) But (q A) ((~q A) A) is a theorem by MP twice p |- A
General Statement of the completeness proof If V(A) = T for all models then |- A
Elaborating, If P1, P2, …, Pn are constituent propositions of A and if V(A) = T for every model V(Pi) = T/F then |- A
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
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
Lemma If row has P1’, P2’, …, Pn’, A’ Then P1’, P2’, …, Pn’ |- A’ A very critical result linking syntax with semantics