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CS/MATH 1019 Discrete Math for Computer Science MATH 1019 Winter 2024 First Midterm Exam Questions with Verified Solutions | 100% Pass | Graded A+ |£11.91
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York University
CS/MATH 1019 Discrete Math for Computer Science
CS/MATH 1019 Exam
Course Title and Number: CS/MATH 1019 Discrete Math for CS
Exam Date: Midterm and Final Exam 2024- 2025
Instructor: [Insert Instructor’s Name]
Student Name: [Insert Student’s Name]
Student ID: [Insert Student ID]
Examination
180 minutes
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5. Please answer each question below and click Submit when you have
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CS/MATH 1019 Discrete Math for Computer
Science
MATH 1019 Winter 2024 First Midterm Exam
Questions with Verified Solutions | 100% Pass |
Graded A+ |
York University
Math 1019 Winter 2022
First Midterm Exam Solutions
Question 1. (12 points) Decide whether the proposition (¬q ' r) x (q x (r ( F)) is satisfiable.
Prove your answer. Hint: You don’t have to write the entire truth table.
Solution. If q is true and r is false then ¬q ' r is false and q x (r ( F) is also false,
therefore (¬q ' r) x (q x (r ( F)) is true. Therefore (¬q ' r) x (q x (r ( F)) is
satisfiable.
□
2
Question 2. (8+8=16 points)
(a) Use a sequence of logical equivalences to show that (p ' T) → (q ( r) is logically
equivalent to (¬q ' ¬r) → ¬p. (Do not use truth tables.)
(b) Prove that ∀x(P (x) ( Q(x)) is NOT logically equivalent to ∀xP (x) ( ∀xQ(x).
Solution. (a) (Using truth tables is a wrong solution.)
(p ' T) → (q ( r) ≡ p → (q ( r) logical equiv. p ' T ≡ p
≡ ¬(q ( r) → ¬p logical equiv. p → q ≡ ¬q → ¬p
≡ (¬q ' ¬r) → ¬p de Morgan law
(b) We need to find a domain and interpretations of P (x) and Q(x) such that the given proposi-
tions have different truth values. For example, let the domain be the set of integers, P (x) is the state-
ment “x is an even integer” and Q(x) is the statement “x is an odd integer”. Then ∀x(P (x) (
Q(x)) states that “every integer is even or odd” which is true. However, ∀xP (x) ( ∀xQ(x)
states that “every integer is even or every integer is odd” which is clearly false. Therefore
∀x(P (x) ( Q(x)) is not logically equivalent to ∀xP (x) ( ∀xQ(x).
□
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