Name (printed):Student ID #:Section # (or TA’s:name and time)CMSC 250 Quiz #5 Wednesday, Feb. 25, 2004Write all answers legibly in the space provided. The number of points p os sible for each question is indicatedin square brackets – the total number of points on the quiz is 30, and you will have exactly 15 minutes tocomplete this quiz. You may not use calculators, textbooks or any other aids during this quiz.1. [10 pnts.] Use an Euler diagram to determine if each of the following represents a valid argument.Make sure to label the parts of the diagram. If the argument is invalid you must have the Eulerdiagram represent that fact. If the argument is valid, just draw one of the Euler diagrams whichshow a valid interpretation.Only cats are nice.Some cats are yellow.———therefore: Some yellow things are nice.Circle One: Valid InvalidSome people are frightened things.All frightened things act strange.———therefore: All people act strange.Circle One: Valid Invalid↓ TURN OVER ↓2. [6 pnts.] Translate the individual statements into symbolic notation in the first available column andthen write either “∀MP ” or “∀M T ” or “∀instantiation” to tell which rule was used to reach theconclusion shown or say that the argument is “not valid” by any of these in the second availablecolumn. You must use the Universe of all things as your domain for any quantified variables, youmay use “b” as a name (instantiation) to represent Bessy, and you may use without indication any“double negation” considerations. Predicates: C(x)= “x is a cow”, G(x) = “x eats grass”, T(x) =“x is a tiger”, and P(x) = “x is a good pet”.a) All cows eat grass.Bessy my pet eats grass.therefore Bessy is a cowb) No tigers eat grass.Bessy my pet eats grass.therefore Bessy is not a tiger.c) All grass eating animals make good pets.Bessy my pet eats grass.therefore Bessy is a good pet.3. [14 pnts.] Using only the rules provided on the handout of the “Logical Equivalence Rules” and the“Rules of Inference” to prove the following. It is a Valid Argument - you need to prove it withoutusing a truth table.P1 ∀x ∈ D, [P (x) ∧ Q(x)]P2 ∀y ∈ D, [R(y) →∼ Q(y)]P3 ∀z ∈ D, [∼ P (z) ∨ M (z)]P4 P (a) where a ∈ Dtherefore ∃x ∈ D, [M(x)∧ ∼ R (x)]Line#Logical Statement Name of Rule Line
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