# Table Puzzle and Crypto-Arithmetic Program

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CS6364 Last Name: _____________________ First Name: __________________________ NetID (UTD email): __________________ Part1 Part2Par3Documentationand DemoTotal 100%2
CS6364 Last Name: _____________________ First Name: __________________________ NetID (UTD email): __________________ PART 1. You should show your work for the following problems, here in this document (.doc file). [You may use word Symbol to complete this part, or use "=" or "<=>" for , -> for , \/ or V for, /\ or ^ for , "exists" for , "all" for , and ~ (or "not") for .]#1. (1) The knowledge base KB has one Boolean formula: (x1 x2) (x2 x3)Find an equivalent knowledge base in Conjunctive Normal Form (CNF).Your Answer here:(since P -> Q = ~ P v Q and using demorgans law we can get the other one.) x1 x2 = ~ x1 v x2 or x1 ^ ~ x2 x2 x3 = ~ x2 v x3 or x2 ^ ~ x3 (since P< -> Q = (P -> Q) ^ (Q -> P) and using above derivations the knowledge base can be written as follows:)[(x1 /\ ~x2)\/( ~x2x3)] /\ [(x2 /\ ~ x3)\/( ~x1x2)](2) A logical knowledge base has the following Boolean formulas:XX YY Z(X Z) WUse inference rules to prove W XAnswer: Given, X is true.Y is also true as X -> Y is true.Z is also true as Y -> Z is true.And therefore by and induction W is true in (X ^ Z) -> W as X and Y are true. (3) KB is: P QP R(a) Prove or disprove using resolution-refutation that Q R(b) Prove or disprove using resolution-refutation that Q RSol:- We know that A -> B = ~A V B.3
CS6364 Last Name: _____________________ First Name: __________________________ NetID (UTD email): __________________ Therefore, P->Q = ~P v Q ------(1)~P->R = P v Q -------(2)From 1 and 2 ~P v Q P v Q Q v R ~ Q ^ ~ R (refuting by first ~ Q ~ R)NilTherefore ~ Q ~ R is true.~ Q -> R(4) You need to convert FOL formula into clause-form to do the proof by resolution-refutation.KB is: x y F(x, y). Prove using resolution-refutation that x y F(y, x).Ans:CNF FORM:x y F(x, y) = F(x, y).x y F(y, x) = F(y, x).Given that goal is x y F(y, x).Therefore ~ goal = ~ F(y,x). replace y with ‘a’ and replace x with ‘b’ in goal.P1: ~ F (b, a) {a| y, b| x}.refutation-resolution:we get ~ F (b, a) F(x, y) nilHence Proved.4

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