Desklib - Online Library for Study Material with Solved Assignments, Essays, Dissertations
Added on -2019-09-13
Desklib is an online library for study material with solved assignments, essays, dissertations, and more. It offers a vast collection of study material for various subjects and courses. Students can submit their assignments with ease and get better grades. The library offers resources for CS6364 and includes a sample assignment with solutions for Part 1 and Part 2. It also includes Prolog programming problems and a Prover9 tutorial.
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CS6364 Last Name: _____________________ First Name: __________________________ NetID (UTD email): __________________ CS6364Last Name: __________________First Name: __________________NetID (UTD email): _______________ @utdallas.eduAssignment #3 Due: 11/14 Monday NoonSubmission: Create a folder (named assign3-abc123456 where abc123456 is your UTD NetID – email) and submit its zip file. The folder should contains this doc file (in .doc or .rft), with the answers for each parts.For Part1 and Part2, you provide all your answers in this document. Copy and Paste all the source program listing and its output for Part3 and Part4. You may use the font size of 8 or 9 for your program listing or output. Format the document, by default, with space=single-spaced, font-size=10, and page-number in the bottom-center of each page. Have a summary of your document to show the table of content, to help your TA/Grader tofollow your document.Please organize your document in a presentable manner (that is, easy-to-follow and professionally formatted, and not overwhelming [that is, to make font size be 14 or bigger, to make it double- or triple-spaced to have the number of pages doubled or tripled, etc.]).For Part2 and Part3, also submit all the files (source and output) in a zip file, including thisdocument. ** Do not make any of these problems or your answers posted or available in Internet or do not share with any others. All of the assignment should be done by yourself.Scoresheetpart130%#1.1#1.2#1.3#1.4#1.5#1.6#2 CSPpar2Prolog30%1234part3Prover930%12345671
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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