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CS344 : Introduction to Artificial Intelligence Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 15, 16, 17- Completeness Proof; Self References and Paradoxes Week of 9/2/09 (with a quiz on 9 th )

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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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Running the completeness proof For every row of the truth table set up a proof: 1. p, ~q |- p (p V q) 2. p, q |- p (p V q) 3. ~p, q |- p (p V q) 4. ~p, ~q |- p (p V q)

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Completeness Proof

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

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If we can show P 1 ’, P 2 ’, …, P n ’ |- A’ For every row where P i ’ = P i if V(P i ) = T = ~Pi 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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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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Answer to Quiz-1 (9/2/09)

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Question: proof of MR theorem “Prove that if monotone restriction is satisfied, the path to a node taken up for expansion is optimal.” i.e., for a node n being expanded, g(n) = g*(n) Arpit’s Maheswari’s contribution in the proof is acknowledged

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Proof Lemma: If monotone restriction is satisfied, f values of nodes expanded is non decreasing. i.e., if n q is expanded after n p, f(n p ) ≤ f(n q ) Proof of Lemma: Case 1 : n q is child of n p g(n q ) + h(n q ) = g(n p ) +c(n p, n q ) + h(n q ) but, c(n p, n q ) + h(n q ) ≥ h(n p ) hence, f(n q ) ≥ g(n p ) + h(n p ) = f(n p ) npnp g nqnq h(n p ) h(n q )

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Proof of Lemma Case 2 : n p and n q are both in open list, n q expanded after n p by definition of A*, f(n p ) ≤ f(n q ) Case 3 : n q did not exist on open list when n p was expanded. But there was another node n r which led to n q later i.e., f(n p ) ≤ f(n r )... ≤ f(n q ) Thus, the lemma is proved.

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Actual Proof of the Theorem Proof by induction on the length of the path to an expanded node n. n q is in OL. Base Case: k = 0 g(s) = g*(s) statement is true Hypothesis: Suppose the statement is true for length = k. i.e., g(n p ) = g(n k ) = g*(n p ) s n n1n1 nqnq npnp k All in closed list

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Actual Proof of the Theorem Induction: To examine g(n) = g(n k+1 ) By lemma, f(n) ≤ f(n q ) i.e., g(n) + h(n) ≤ g(n q ) + h(n q ) c(n p, n) + g(n p ) + h(n) ≤ g(n q ) + h(n q ) Add c(n q, n) to both sides, c(n p, n) + g(n p ) +h(n) + c(n q, n) ≤ g(n q ) + c(n q, n) +h(n q ) But, h(n) + c(n q, n) ≥ h(n q ) hence, c(n p, n) + g(n p ) ≤ g(n q ) + c(n q, n) Thus, g(n) = g*(n) nqnq n g h(n q ) h(n) g(n) through n p g(n) through n q

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Assignment-2 Create a system to prove syntactically the theoremhood of any propositional calculus expression. This will be an automatic theorem prover for propositional calculus. The input to the system will be any propositional calculus expression and the output a yes/no answer depending on whether the expression is a theorem or not.

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