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

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

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An example to illustrate the completeness proof pqp (p V q) TFT TTT FTT FFT

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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.

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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:

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

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Proof of Lemma

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An example to illustrate the completeness proof pqp (p V q) TFT TTT FTT FFT

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

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Completeness Proof

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P 1 P 2 P 3... P n A F F F... F T F F F... T T. T T T... T T We have a truth table with 2 n rows

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We should show P 1 ’, P 2 ’, …, P n ’ |- A’ For every row where P i ’ = P i if V(P i ) = T = ~P i if V(P i ) = F And A’ = A if V(A) = T = ~A if V(A) = F

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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 P 1, P 2, …, P n then P’ 1, P’ 2, …, P’ n |-- A’, where A’ = A if V(A) = true = ~Aotherwise Similarly for each P’ i

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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 ‘→’

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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. Proof of Lemma (contd.)

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

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Self Reference and Paradoxes

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

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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

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

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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.

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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.

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Self Reference: source of paradoxes All these paradoxes came because of 1. Self reference 2. Confusion between what is inside a system and what is outside

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