CS621 : Artificial Intelligence

1. CS621 : Artificial Intelligence Pushpak BhattacharyyaCSE Dept., IIT Bombay Lecture 12- Completeness Proof; Self References and Paradoxes 16th August, 2010

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

3. An example to illustrate the completeness proof

4. Completeness Proof • Statement If V(A) = T for all V, then |--A i.e. A is a theorem. In the example A is p(p V q) We need prove p(p V q) is a theorem given the tautology.

5. Lemma: If P, Q├ A and P, ~Q├ A then we show, P ├ A Proof: To prove this lemma we need a theorem as follows. Theorem:

6. Proof of Theorem i.e. ├ i.e. ├ i.e. ├ i.e. ├ i.e. ├

7. Proof of Lemma

8. An example to illustrate the completeness proof

9. Running the completeness proof For every row of the truth table set up a proof: • p, ~q |- p(p V q) ---(1) • p, q |- p(p V q) ---(2) • ~p, q |- p(p V q) ---(3) • ~p, ~q |- p(p V q) ---(4) • p |- p(p V q) ---(5) using (1) and (2) • ~ p |- p(p V q) ---(6) using (3) and (4) • |- p(p V q) ---(7) using (5) and (6) Hence p(p V q) is a theorem

10. Completeness Proof

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

12. We should 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

13. Completeness of Propositional Calculus Statement If V(A) = T for all V, then |--A i.e. A is a theorem. Lemma: If A consists of propositions P1, P2, …, Pn then P’1, P’2, …, P’n |-- A’, where A’ = A if V(A) = true = ~A otherwise Similarly for each P’i

14. Proof for Lemma Proof by induction on the number of ‘→’ symbols in A Basis: Number of ‘→’ symbols is zero. A is ℱ or P. This is true as, |-- (A → A) i.e. A → A is a theorem. Hypothesis: Let the lemma be true for number of ‘→’ symbols ≤ n. Induction: Let A which is B → C,contain n+1‘→’

15. Proof of Lemma (contd.) • Induction: By hypothesis, P’1, P’2, …, P’n |-- B’ P’1, P’2, …, P’n |-- C’ If we show that B’, C’ |-- A’ (A is B → C), then the proof is complete. For this we have to show: • B, C |-- B → C True as B, C, B |-- C • B, ~C |-- ~(B → C) True since B, ~C, B → C |-- ℱ • ~B, C |-- B → C True since ~B, C, B |-- C • ~B, ~C |-- B → C True since ~B, ~C, B, C → ℱ|-- ℱ • Hence the lemma is proved.

16. Proof of Theorem A is a tautology. There are 2n models corresponding to P1, P2, …, Pn propositions. Consider, P1, P2, …, Pn |-- A and P1, P2, …, ~Pn |-- A P1, P2, …, Pn-1 |-- Pn→ A and P1, P2, …, Pn-1 |-- ~Pn→ A RHS can be written as: |-- ((Pn → A) → ((~Pn → A) → A)) |-- (~Pn → A) → A |-- A Thus dropping the propositions progressively we show |-- A

17. Self Reference and Paradoxes

18. Paradox -1 “This statement is false” The truth of this cannot be decided

19. Paradox -2 (Russell Paradox or Barber Paradox) “In a city, a barber B shaves all and only those who do not shave themselves” Question: Does the barber shave himself? Cannot be answered

20. Paradox -3 (Richardian Paradox) Order the statements about properties of number in same order. E.g., “A prime no. is one that is divisible by itself of 1.” “A square no. is one that is product of 2 identical numbers.” . .

21. Paradox -3 (Richardian Paradox) Definition: A number is called Richardian if it does not have the property that it indexes. For example, in the above arrangement 2 is Richardian because it is not a square no.

22. Paradox -3 (Richardian Paradox) Now, suppose in this arrangement M is the number for the definition of Richardian M : “A no. is called Richardian … “ Question : is M Richardian? Cannot be answered.

23. Self Reference: source of paradoxes All these paradoxes came because of Self reference Confusion between what is inside a system and what is outside