2 Solved Problems on Logical Equivalence Rules - Quiz 6 | CMSC 250, Quizzes of Discrete Structures and Graph Theory

Download 2 Solved Problems on Logical Equivalence Rules - Quiz 6 | CMSC 250 and more Quizzes Discrete Structures and Graph Theory in PDF only on Docsity! Name (printed): Student ID #: Section # (or TA’s: name and time) CMSC 250 Quiz #6 Wednesday, Oct. 8, 2003 Write all answers legibly in the space provided. The number of points possible for each question is indicated in square brackets – the total number of points on the quiz is 30, and you will have exactly 20 minutes to complete this quiz. You may not use calculators, textbooks or any other aids during this quiz. You may use any of the definitions given in the text and you may use the fact that every integer is either even or odd and the fact that no integer is both even and odd without proving those statements for this quiz. 1. [20 pnts.] Use the handout of the “Logical Equivalence Rules” and the “Rules of Inference” to prove each of the the following. Both are Valid Arguments - you need to prove each of the following without using a truth table. You may assume domain D is non empty and that the a,b,and c mentioned are members of the domain D. a. P1 ∀x ∈ D[P (x) ∧Q(x)] P2 ∀y ∈ D[R(y) →∼ Q(y)] P3 ∀z ∈ D[∼ P (z) ∨M(z)] P4 P (a) −−−−−−−−−−−−−− therefore ∃x ∈ D[M(x)∧ ∼ R(x)] Line # Logical Statement Name of Rule Line Numbers Used 1 2 3 4 5 6 7 8 9 10 11 12 TURN OVER b. P1 ∀x ∈ D Q(x) → S(x) P2 ∀y ∈ D ∼ (S(y) ∧ T (y)) P3 ∀z ∈ D M(z) ∨ T (z) P4 ∀w ∈ D ∼ (∼ T (w) ∧Q(w)) P5 ∀x ∈ D ∼ M(x) → S(x) therefore ∀k ∈ D M(k) Line # Logical Statement Name of Rule Line Numbers Used 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 TURN TO NEXT PAGE