Q1. Let p, q, r be the propositions: p: You have flu. q: You miss the final examination. r: You pass the course. Express each of the following propositions as an English sentence.
1) p → q 3) q → ï¿¢r 5) (p → ï¿¢r) ∨ (q → ï¿¢r)
2) ï¿¢q ↔ r 4) p ∨ q ∨ r 6) (p ∧ q) ∨ (ï¿¢q ∧ r)
Solution:If you have flu, you miss the final examination.
You pass the course if and only if you do not miss the final examination.
If you miss the final examination, you will not pass the course.
You either have the flu, miss the final examination or pass the course.
You have flu and you will miss the exam or you do not miss the exam you will pass the course.
Q2. State the converse, contrapositive and inverse of each of the conditional statement. Answer: InverseIf it does not snow tonight, then I will not stay at home.
ConverseIf I stay at home, then it snows tonight.
ContrapositiveIf I don’t stay at home then it won’t be snowing tonight.
InverseI don’t go to the beach whenever it is not a sunny summer day.
ConverseWhenever I go to the beach, it is a sunny summer day.
ContrapositiveIf it is not a sunny summer day then I will not go to the beach.
InverseIf I do not stay up late, it is necessary that I don’t sleep until noon.
ConverseIf I sleep until noon, then I stayed up late.
ContrapositiveIf I don’t sleep until noon, then I don’t need to stay up late.
InverseIf it does not snow today, then I will not ski tomorrow.
ConverseIf I ski tomorrow, it will snow today.
ContrapositiveIf I don’t ski tomorrow, it will not snow today.
InverseI don’t come to class if there is not going to be a quiz.
ConverseIf there is going to be a quiz then I will come to class.
ContrapositiveIf there is not going to be a quiz then will not come to the class.
InverseIf a positive integer is not prime, it has divisors other than 1 and itself.
ConverseIf a positive integer has no divisors other than 1 and itself, it is prime.
ContrapositiveIf a positive integer has divisors other than 1 and itself, it is not prime.
Q3. Is the assertion “This statement is false” a proposition? Motivate your answer.Answer. The Assertion “This statement is false” is not a proposition because it is an incomplete description which has no reference.
Q4. Use De Morgan’s Law to find the negation of each of the following statements.|
p |
q |
p → q |
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
q |
r |
q → r |
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
p |
r |
p → r |
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
p → q |
q → r |
(p → q) ∧ (q → r) |
|
1 |
1 |
1 |
|
1 |
1 |
1 |
|
0 |
0 |
0 |
|
1 |
1 |
1 |
|
(p → q) ∧ (q → r) |
p → r |
(p → q) ∧ (q → r) → (p → r) |
|
1 |
1 |
1 |
|
1 |
1 |
1 |
|
0 |
0 |
1 |
|
1 |
1 |
1 |
Thus (p → q) ∧ (q → r) → (p → r) is a Tautology.
Q6. Show that (p ∨ q) ∧ (ï¿¢p ∨ r) → (q ∨ r) is tautology. Solution.
|
p |
q |
p ∨ q |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
|
p |
r |
ï¿¢p |
ï¿¢p ∨ r |
|
0 |
0 |
1 |
1 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
0 |
0 |
|
1 |
1 |
0 |
1 |
|
p ∨ q |
ï¿¢p ∨ r |
(p ∨ q) ∧ (ï¿¢p ∨ r) |
|
0 |
1 |
0 |
|
1 |
1 |
1 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
q |
r |
q ∨ r |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
|
(p ∨ q) ∧ (ï¿¢p ∨ r) |
q ∨ r |
(p ∨ q) ∧ (ï¿¢p ∨ r) → (q ∨ r) |
|
0 |
0 |
1 |
|
1 |
1 |
1 |
|
0 |
1 |
1 |
|
1 |
1 |
1 |
Thus (p ∨ q) ∧ (ï¿¢p ∨ r) → (q ∨ r) is tautology.
Q7. Let N(x) be the statement “x has visited north”, where the domain consist of the students in UOH. Express each of these quantifications in English.a) ∃xN(x) b) ∀xN(x) c) ¬∃xN(x)
d) ∃x¬N(x) e)¬∀xN(x) f) ∀x¬N(x)
Answer.
Q8. Show that (using law) the following propositions are logically equivalent.
p ↔ q ≡ (p → q) ∧ (q → p)
≡ (¬p ∨ q) ∧ (¬q ∨ p)
≡ ((¬p ∨ q) ∧ ¬q) ∨ ((¬p ∨ q) ∧ p), Distributive law
≡ ((¬p ∧ ¬q) ∨ (q ∧ ¬q)) ∨ ((¬p ∧ p) ∨ (q ∧ p)), Distributive law (twice)
≡ ((¬p ∧ ¬q) ∨ F) ∨ (F ∨ (q ∧ p)), Negation law (twice)
≡ (¬p ∧ ¬q) ∨ (q ∧ p), Identity law (twice)
≡ (p ∧ q) ∨ (¬p ∧ ¬q), Commutative law (twice)
(p → q) ∧ (p → r) ≡ (¬p ∧ q) ∧ (¬p ∧ r) Logical equivalence using conditionals
≡ (¬p ∧ ¬p) ∧ (q ∧ r) Associative and Commutative Laws
≡ ¬p ∧ (q ∧ r) Idempotent law
≡ p → (q ∧ r) Logical equivalence using conditionals.
Assignment No 1 (Solved) _ 0.pdf
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