lOMoARcPSD|46692070
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
Instructions:
1. Read each question carefully.
2. Answer all questions.
3. Use the provided answer sheet to mark your responses.
4. Ensure all answers are final before submitting the exam.
5. Please answer each question below and click Submit when you have
completed the Exam.
6. This test has a time limit, The test will save and submit automatically
when the time expires
7. This is Exam which will assess your knowledge on the course Learning
Resources.
Good Luck!
Downloaded by Jack mah
()
, lOMoARcPSD|46692070
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).
□
Downloaded by Jack mah
()
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
Instructions:
1. Read each question carefully.
2. Answer all questions.
3. Use the provided answer sheet to mark your responses.
4. Ensure all answers are final before submitting the exam.
5. Please answer each question below and click Submit when you have
completed the Exam.
6. This test has a time limit, The test will save and submit automatically
when the time expires
7. This is Exam which will assess your knowledge on the course Learning
Resources.
Good Luck!
Downloaded by Jack mah
()
, lOMoARcPSD|46692070
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).
□
Downloaded by Jack mah
()
