Class 12

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.10 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.10

Question 1.
Express the truth of each of the following statements by Venn diagrams:
(i) Some hardworking students are obedient.
Solution:
Let U : set of all students
S : set of all hardworking students
O : set of all obedient students.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ O ≠ φ

(ii) No circles are polygons.
Solution:
Let U : set of closed geometrical figures in the plane
P : set of all polygons
C : set of all circles.
Then the Venn diagram represents the truth of the given statement is as follows:

P ∩ C = φ

(iii) All teachers are scholars and scholars are teachers.
Solution:
Let U : set of all human beings
T : set of all teachers
S : set of all scholars.
Then the Venn diagram represents the truth of the given statement is as below:

(iv) If a quadrilateral is a rhombus, then it is a parallelogram.
Solution:
Let U : set of all quadrilaterals
R : set of all rhombus
P : set of all parallelograms.
Then the Venn diagram represents the truth of the given statement is as below:

R ⊂ P

Question 2.
Draw the Venn diagrams for the truth of the following statements:
(i) Some share brokers are chartered accountants.
Solution:
Let U : set of all human beings
S : set of all share brokers
C : set of all chartered accountants.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ C ≠ φ

(ii) No wicket-keeper is a bowler in a cricket team.
Solution:
Let U : set of all human beings
W : set of all wicket keepers
B : set of all bowlers.
Then the Venn diagram represents the truth of the given statement is as follows:

W ∩ B = φ

Question 3.
Represent the following statements by Venn diagrams:
(i) Some non-resident Indians are not rich.
Solution:
Let U : set of all human beings
N : set of all non-resident Indians
R : set of all rich people.
Then the Venn diagram represents the truth of the given statement is as below:

N – R ≠ φ

(ii) No circle is a rectangle.
Solution:
Let U : set of all geometrical figures
C : set of all circles
R : set of all rectangles
Then the Venn diagram represents the truth of the given statement is as below:

C ∩ R = φ

(iii) If n is a prime number and n ≠ 2, then it is odd.
Solution:
Let U : set of all real numbers
P : set of all prime numbers n, where n ≠ 2
O : set of all odd numbers.
Then the Venn diagram represents the truth of the given statement is as below:

P ⊂ O

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.9 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.9

Question 1.
Without using truth table, show that
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~(q ∨ p) …..(Conditional Law)
≡ [~p ∧ (~(q ∨ p)] ∨ [q ∧ (~q ∨ p] …..(Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] ……(Distributive Law)
≡ [(~p ∧ ~q) ∨ c] ∨ [c ∨ (q ∧ p)] …..(Complement Law)
≡ (~p ∧ ~q) ∨ (q ∧ p) ……(Identity Law)
≡ (~p ∧ ~q) ∨ (p ∧ q) ……(Commutative Law)
≡ (p ∧ q) ∨ (~p∧ q) ……(Commutative Law)
≡ RHS.

(ii) p ∧ [~p ∨ q) ∨ (~q)] ≡ p
Solution:
LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.

(iii) ~[(p ∧ q) → ~(q)] ≡ p ∧ q
Solution:
LHS = ~[(p ∧ q) → ~(~q)]
≡ (p ∧ q) ∧ ~(~q) ……(Negation of implication)
≡ (p ∧ q) ∧ q …..(Negation of negation)
≡ p ∧ (q ∧ q) …..(Associative Law)
≡ P ∧ q ……(Idempotent Law)
= RHS

(iv) ~r → ~(p ∧ q) ≡ [~(q → r)] → (~p)
Solution:
LHS = ~r → ~(p ∧ q)
≡ ~q → (~p ∨ ~q) ……(De Morgan’s Law)
≡ ~(~r) ∨ (~p ∨~q) …..(Conditional Law)
≡ r ∨ (~p ∨ ~q) …..(Involution Law)
≡ r ∨ ~q ∨ ~p …..(Commutative Law)
≡ (~q ∨ r) ∨ (~p) ……(Commutative Law)
≡ ~(q → r) ∨ (~p) ……(Conditional Law)
≡ ~(q → r) → (~p) ……(Conditional Law)
= RHS.

(v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Solution:
LHS = (p ∨ q) → r
≡ ~(p → q) ∨ r ……..(Conditional Law)
≡ (~p ∧ ~q) ∨ r ……….(De Morgan’s Law)
≡ (~p ∨ r) ∧ (~q ∨ r) ………..(Distributive Law)
≡ (p → r) ∧ (q → r) …….(Conditional Law)
= RHS.

Question 2.
Using the algebra of statement, prove that:
(i) [p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p] ≡ p
Solution:
LHS = [p ∧ (q ∨ r)] ∨ [ ~r ∧ ~q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p] ……(Associative Law)
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ……(Commutative Law)
≡ [p ∧ (q ∨ r)] ∨ [ ~ (q ∨ r) ∧ p] ……(De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] ……(Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~ (q ∨ r) ] …..(Distributive Law)
≡ p ∧ t …….(Complement Law)
≡ p ……(Identity Law)
= RHS.

(ii) (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q) ≡ p ∨ ~q
Solution:
LHS = (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~q) ∨ (~p ∧ ~q)] ……(Associative Law)
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~q ∧ ~p)] …..(CommutativeLaw)
≡ (p ∧ q) ∨ [ ~q ∧ (p ∨ ~ p)] …..(Distributive Law)
≡ (p ∧ q) ∨ (~q ∧ t) ……(Complement Law)
≡ (p ∧ q) ∨ (~q) …….(Identity Law)
≡ (p ∨ ~q) ∧ (q ∨ ~q) ……(Distributive Law)
≡ (p ∨ ~q) ∧ t …….(Complement Law)
≡ p ∨ ~q …..(Identity Law)
= RHS.

(iii) (p ∨ q) ∧ (~p ∨ ~q) ≡ (p ∧ ~q) ∨ (~p ∧ q)
Solution:
LHS = (p ∨ q) ∧ (~p ∨ ~q)
≡ [p ∧ (~p ∨ ~q)] ∨ [q ∧ (~p ∨ ~q)] ……(Distributive Law)
≡ [(p ∧ ~p) ∨ (p ∧ ~q)] ∨ [q ∧ ~p) ∨ (q ∧ ~q)] ……(Distributive Law)
≡ [c ∨ (p ∧ ~q)] ∨ [(q ∧ ~p) ∨ c] ……(Complement Law)
≡ (p ∧ ~q) ∨ (q ∧ ~p) ……..(Identity Law)
≡ (p ∧ ~q) ∨ (~p ∧ q) ………(Commutative Law)
= RHS.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.8

Question 1.
Write the negation of each of the following statements:
(i) All the stars are shining if it is night.
Solution:
The given statement can be written as:
If it is night, then all the stars are shining.
Let p : It is night.
q : All the stars are shining.
Then the symbolic form of the given statement is p → q
Since, ~(p → q) ≡ p ∧ ~q,
the negation of the given statement is ‘It is night and all the stars are not shining.’

(ii) ∀ n ∈ N, n + 1 > 0.
Solution:
The negation of the given statement is
‘∃ n ∈ N, such that n + 1 ≤ 0.’

(iii) ∃ n ∈ N, such that (n² + 2) is odd number.
Solution:
The negation of the given statement is
‘∀ n ∈ N, n² + 2 is not an odd number.’

(iv) Some continuous functions are differentiable.
Solution:
The negation of a given statement is ‘All continuous functions are not differentiable.’

Question 2.
Using the rules of negation, write the negations of the following:
(i) (p → r) ∧ q
Solution:
The negation of (p → r) ∧ q is
~[(p → r) ∧ q] ≡ ~(p → r) ∨ (~q) …..[Negation of conjunction]
≡ (p ∧ ~r) ∨ (~q) ……[Negation of implication]

(ii) ~(p ∨ q) → r
Solution:
The negation of ~(p ∨ q) → r is
~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ (~r) …..[Negation of implication]
≡ (~p ∧ ~q) ∧ (~r) ……[Negation of disjunction]

(iii) (~p ∧ q) ∧ (~q ∨ ~r)
Solution:
The negation of (~p ∧ q) ∧ (~q ∨ ~r) is
~[(~p ∧ q) ∧ (~q ∨ ~ r)] ≡ ~(~p ∧ q) ∨ ~(~q ∨ ~r) ……[Negation of conjunction]
≡ [~(~p) ∨ ~q] ∨ [~(~q) ∧ ~(~r)] … [Negation of conjunction and disjunction]
≡ (p ∨ ~q) ∨ (q ∧ r) …..[Negation of negation]

Question 3.
Write the converse, inverse, and contrapositive of the following statements:
(i) If it snows, then they do not drive the car.
Solution:
Let p : It snows.
q : They do not drive the car.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If they do not drive the car, then it snows.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If it does not snow, then they drive the car.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If they drive the car, then it does not snow.

(ii) If he studies, then he will go to college.
Solution:
Let p : He studies.
q : He will go to college.
Then two symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If he will go to college, then he studies.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If he does not study, then he will not go to college.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If he will not go to college, then he does not study.

Question 4.
With proper justification, state the negation of each of the following:
(i) (p → q) ∨ (p → r)
Solution:
The negation of (p → q) ∨ (p → r) is
~[(p → q) ∨ (p → r)] ≡ ~(p → q) ∧ ~(p → r) …..[Negation of disjunction]
≡ (p ∧ ~q) ∧ (p ∧ ~r) …[Negation of implication]

(ii) (p ↔ q) ∨ (~q → ~r)
Solution:
The negation of (p ↔ q) ∨ (~q → ~r) is
~[(p ↔ q) ∨ (~q → ~r)] ≡ ~(p ↔ q) ∧ ~(~q → ~r) …..[Negation of disjunction]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~q ∧ ~(~r)] ……[Negation of biconditional and implication]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~q ∧ r) ……[Negation of negation]

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~[(p → q) ∧ r] ≡ ~(p → q) ∨ (~r) …..[Negation of conjunction]
≡ (p ∧ ~q) ∨ (~r) …..[Negation of implication]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.7

Question 1.
Write the dual of each of the following:
(i) (p ∨ q) ∨ r
Solution:
(p ∧ q) ∧ r

(ii) ~(p ∨ q) ∧ [p ∨ ~(q ∧ ~r)]
Solution:
~(p ∧ q) ∨ [p ∧ ~(q ∨ ~r)]

(iii) p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
Solution:
p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

(iv) ~(p ∧ q) ≡ ~p ∨ ~q
Solution:
~(p ∨ q) ≡ ~p ∧ ~q

Question 2.
Write the dual statement of each of the following compound statements:
(i) 13 is prime number and India is a democratic country.
Solution:
13 is prime number or India is a democratic country.

(ii) Karina is very good or everybody likes her.
Solution:
Karina is very good and everybody likes her.

(iii) Radha and Sushmita can not read Urdu.
Solution:
Radha or Sushmita can not read Urdu.

(iv) A number is real number and the square of the number is non-negative.
Solution:
A number is real number or the square of the number is non-negative.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.6

Question 1.
Prepare the truth tables for the following statement patterns:
(i) p → (~p ∨ q)
Solution:
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (~p ∨ q) ∧ (~p ∨ ~q)
Solution:

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
Here are three statements and 4 connectives.
∴ there are 2 × 2 × 2 = 8 rows and 3 + 4 = 7 columns in the truth table.

(iv) (p ∧ q) ∨ ~r
Solution:

Question 2.
Examine, whether each of the following statement patterns is a tautology or a contradiction or a contingency:
(i) q ∨ [~(p ∧ q)]
Solution:

All the entries in the last column of the above truth table are T.
∴ q ∨ [~(p ∧ q)] is a tautology.

(ii) (~q ∧ p) ∧ (p ∧ ~p)
Solution:

All the entries in the last column of the above truth table are F.
∴ (~q ∧ p) ∧ (p ∧ ~p) is a contradiction.

(iii) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(iv) ~p → (p → ~q)
Solution:

All the entries in the last column of the truth table are T.
∴ p → (p → ~q) is a tautology.

Question 3.
Prove that each of the following statement pattern is a tautology:
(i) (p ∧ q) → q
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → q is a tautology.

(ii) (p → q) ↔ (~q → ~p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~q → ~p) is a tautology.

(iii) (~p ∧ ~q) → (p → q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∧ ~q) → (p → q) is a tautology.

(iv) (~p ∨ ~q) ↔ ~(p ∧ q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∨ ~q) ↔ ~(p ∧ q) is a tautology.

Question 4.
Prove that each of the following statement pattern is a contradiction:
(i) (p ∨ q) ∧ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

(ii) (p ∧ q) ∧ ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∧ ~p is a contradiction.

(iii) (p ∧ q) ∧ (~p ∨ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∧ q) ∧ (~p ∨ ~q) is a contradiction.

(iv) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

Question 5.
Show that each of the following statement pattern is a contingency:
(i) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(ii) (p → q) ↔ (~p ∧ q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ↔ (~p ∧ q) is a contingency.

(iii) p ∧ [(p → ~q) → q]
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ p ∧ [(p → ~q) → q] is a contingency.

(iv) (p → q) ∧ (p → r)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ∧ (p → r) is a contingency.

Question 6.
Using the truth table, verify:
(i) p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

(ii) p → (p → q) ≡ ~q → (p → q)
Solution:

The entries in columns 5 and 6 are identical.
∴ p → (p → q) ≡ ~q → (p → q)

(iii) ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q
Solution:

The entries in columns 5, 7 and 8 are identical.
∴ ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q.

(iv) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:

The entries in columns 3 and 7 are identical.
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 7.
Prove that the following pairs of statement patterns are equivalent:
(i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(ii) p ↔ q and (p → q) ∧ (q → p)
Solution:

The entries in columns 3 and 6 are identical.
∴ p ↔ q ≡ (p → q) ∧ (q → p)

(iii) p → q and ~q → ~p and ~p ∨ q
Solution:

The entries in columns 5, 6 and 7 are identical.
∴ p → q ≡ ~q → ~p ≡ ~p ∨ q.

(iv) ~(p ∧ q) and ~p ∨ ~q
Solution:

The entries in columns 6 and 7 are identical.
∴ ~(p ∧ q) ≡ ~p ∨ ~q.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.5

Question 1.
Use qualifiers to convert each of the following open sentences defined on N, into a true statement:
(i) x² + 3x – 10 = 0
Solution:
∃ x ∈ N, such that x² + 3x – 10 = 0 is a true statement
(x = 2 ∈ N satisfy x² + 3x – 10 = 0)

(ii) 3x – 4 < 9
Solution:
∃ x ∈ N, such that 3x – 4 < 9 is a true statement.
(x = 1, 2, 3, 4 ∈ N satisfy 3x – 4 < 9)

(iii) n² ≥ 1
Solution:
∀ n ∈ N, n² ≥ 1 is a true statement.
(All n ∈ N satisfy n² ≥ 1)

(iv) 2n – 1 = 5
Solution:
∃ x ∈ N, such that 2n – 1 = 5 is a true statement.
(n = 3 ∈ N satisfy 2n – 1 = 5)

(v) y + 4 > 6
Solution:
∃ y ∈ N, such that y + 4 > 6 is a true statement.
(y = 3, 4, 5, … ∈ N satisfy y + 4 > 6

(vi) 3y – 2 ≤ 9
Solution:
∃ y ∈ N, such that 2y ≤ 9 is a true statement.
(y = 1, 2, 3 ∈ N satisfy 3y – 2 ≤ 9).

Question 2.
If B = {2, 3, 5, 6, 7}, determine the truth value of each of the following:
(i) ∀ x ∈ B, x is a prime number.
Solution:
(i) x = 6 ∈ B does not satisfy x is a prime number.
So, the given statement is false, hence its truth value is F.

(ii) ∃ n ∈ B, such that n + 6 > 12.
Solution:
Clearly n = 7 ∈ B satisfies n + 6 > 12.
So, the given statement is true, hence its truth value is T.

(iii) ∃ n ∈ B, such that 2n + 2 < 4.
Solution:
No element n ∈ B satisfy 2n + 2 < 4.
So, the given statement is false, hence its truth value is F.

(iv) ∀ y ∈ B, y² is negative.
Solution:
No element y ∈ B satisfy y² is negative.
So, the given statement is false, hence its truth value is F.

(v) ∀ y ∈ B, (y – 5) ∈ N.
Solution:
y = 2 ∈ B, y = 3 ∈ B and y = 5 ∈ B do not satisfy (y – 5) ∈ N.
So, the given statement is false, hence its truth value is F.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.4

Question 1.
Write the following statements in symbolic form:
(i) If the triangle is equilateral, then it is equiangular.
Solution:
Let p : Triangle is equilateral.
q : It is equiangular.
Then the symbolic form of the given statement is p → q.

(ii) It is not true that ‘i’ is a real number.
Solution:
Let p : ‘i’ is a real number.
Then the symbolic form of the given statement is ~p.

(iii) Even though it is not cloudy, it is still raining.
Solution:
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is ~p ∧ q.

(iv) Milk is white if and only if the sky is not blue.
Solution:
Let p : Milk is white.
q : Sky is blue.
Then the symbolic form of the given statement is p ↔ (~q).

(v) Stock prices are high if and only if stocks are rising.
Solution:
Let p : Stock prices are high.
q : stocks are rising.
Then the symbolic form of the given statement is p ↔ q

(vi) If Kutub-Minar is in Delhi, then Taj Mahal is in Agra.
Solution:
Let p : Kutub-Minar is in Delhi.
q : Taj Mahal is in Agra.
Then the symbolic form of the given statement is p → q

Question 2.
Find the truth value of each of the following statements:
(i) It is not true that 3 – 7i is a real number.
Solution:
Let p : 3 – 7i be a real number.
Then the symbolic form of the given statement is ~p.
The truth value of p is F.
∴ the truth value of ~p is T. ….[~F ≡ T]

(ii) If a joint venture is a temporary partnership, then a discount on purchase is credited to the supplier.
Solution:
Let p : Joint venture is a temporary partnership.
q : Discount on purchases is credited to the supplier.
Then the symbolic form of the given statement is p → q.
The truth values of p and q are T and F respectively.
∴ the truth value of p → q is F. …..[T → F ≡ F]

(iii) Every accountant is free to apply his own accounting rules if and only if machinery is an asset.
Solution:
Let p : Every accountant is free to apply his own accounting rules.
q : Machinery is an asset.
Then the symbolic form of the given statement is p ↔ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ↔ q is F. ….[F ↔ T ≡ F]

(iv) Neither 27 is a prime number nor divisible by 4.
Solution:
Let p : 27 is a prime number.
q : 27 is divisible by 4.
Then the symbolic form of the given statement is ~p ∧ ~q.
The truth values of both p and q are F.
∴ the truth value of ~p ∧ ~q is T. …..[~F ∧ ~F ≡ T ∧ T ≡ T]

(v) 3 is a prime number and an odd number.
Solution:
Let p : 3 be a prime number.
q : 3 is an odd number.
Then the symbolic form of the given statement is p ∧ q
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

Question 3.
If p and q are true and r and s are false, find the true value of each of the following statements:
(i) p ∧ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∧ (q ∧ r) ≡ T ∧ (T ∧ F)
≡ T ∧ F
≡ F
Hence, the truth value of the given statement is false.

(ii) (p → q) ∨ (r ∧ s)
Solution:
(p → q) ∨ (r ∧ s) ≡ (T → T) ∨ (F ∧ F)
≡ T ∨ F
≡ T
Hence, the truth value of the given statement is true.

(iii) ~[(~p ∨ s) ∧ (~q ∧ r)]
Solution:
~[(~p ∨ s) ∧ (~q ∧ r)] ≡ ~[(~ T ∨ F) ∧ (~T ∧ F)]
≡ ~[(F ∨ F) ∧ (F ∧ F)]
≡ ~(F ∧ F)
≡ ~F
≡ T
Hence, the truth value of the given statement is true.

(iv) (p → q) ↔ ~(p ∨ q)
Solution:
(p → q) ↔ ~(p ∨ q) = (T → T) ↔ ~(T ∨ T)
≡ T ↔ ~ (T)
≡ T ↔ F
≡ F
Hence, the truth value of the given statement is false.

(v) [(p ∨ s) → r] ∨ [~(p → q) ∨ s]
Solution:
[(p ∨ s) → r] ∨ ~[~(p → q) ∨ s]
≡ [(T ∨ F) → F] ∨ ~[ ~(T → T) ∨ F]
≡ (T → F) ∨ ~(~T ∨ F)
≡ F ∨ ~ (F ∨ F)
≡ F ∨ ~F
≡ F ∨ T
≡ T
Hence, the truth value of the given statement is true.

(vi) ~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
Solution:
~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
≡ ~[T ∨ (F ∧ F)] ∧ ~[(F ∧ ~F) ∧ T]
≡ ~[T ∨ F] ∧ ~[(F ∧ T) ∧ T]
≡ ~T ∧ ~(F ∧ T)
≡ F ∧ ~F
≡ F ∧ T
≡ F
Hence, the truth value of the given statement is false.

Question 4.
Assuming that the following statements are true:
p : Sunday is a holiday.
q : Ram does not study on holiday.
Find the truth values of the following statements:
(i) Sunday is not holiday or Ram studies on holiday.
Solution:
The symbolic form of the statement is ~p ∨ ~q.

∴ the truth value of the given statement is F.

(ii) If Sunday is not a holiday, then Ram studies on holiday.
Solution:
The symbolic form of the given statement is ~p → ~q.

∴ the truth value of the given statement is T.

(iii) Sunday is a holiday and Ram studies on holiday.
Solution:
The symbolic form of the given statement is p ∧ q.

∴ the truth value of the given statement is F.

Question 5.
If p : He swims.
q : Water is warm.
Give the verbal statements for the following symbolic statements:
(i) p ↔ ~q
Solution:
p ↔ ~ q
He swims if and only if the water is not warm.

(ii) ~(p ∨ q)
Solution:
~(p ∨ q)
It is not true that he swims or water is warm.

(iii) q → p
Solution:
q → p
If water is warm, then he swims.

(iv) q ∧ ~p
Solution:
q ∧ ~p
The water is warm and he does not swim.