Chapter 1: Mathematical Logic

Exercise 1.1

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

#### EXERCISE 1.1Q 1 PAGE 2

Exercise 1.1 | Q 1 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## A triangle has ‘n’ sides

### SOLUTION

It is an open sentence. Hence, it is not a statement.

[Note: Answer given in the textbook is ‘it is a statement’. However, we found that ‘It is not a statement’.]

#### EXERCISE 1.1Q 2 PAGE 2

Exercise 1.1 | Q 2 | Page 2

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## The sum of interior angles of a triangle is 180°

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 3 PAGE 2

Exercise 1.1 | Q 3 | Page 2

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## You are amazing!

### SOLUTION

It is an exclamatory sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 4 PAGE 2

Exercise 1.1 | Q 4 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## Please grant me a loan.

### SOLUTION

It is a request. Hence, it is not a statement.

#### EXERCISE 1.1Q 5 PAGE 2

Exercise 1.1 | Q 5 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## √-4 is an irrational number.

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 6 PAGE 2

Exercise 1.1 | Q 6 | Page 2

## x2 − 6x + 8 = 0 implies x = −4 or x = −2.

### SOLUTION

It is a statement which is false. Hence, it’s truth value if F.

#### EXERCISE 1.1Q 7 PAGE 3

Exercise 1.1 | Q 7 | Page 3

## He is an actor.

### SOLUTION

It is an open sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 8 PAGE 3

Exercise 1.1 | Q 8 | Page 3

## Did you eat lunch yet?

### SOLUTION

It is an interrogative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 9 PAGE 3

Exercise 1.1 | Q 9 | Page 3

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## Have a cup of cappuccino.

### SOLUTION

It is an imperative sentence, hence it is not a statement.

#### EXERCISE 1.1Q 10 PAGE 3

Exercise 1.1 | Q 10 | Page 3

## (x + y)2 = x2 + 2xy + y2 for all x, y ∈ R.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 11 PAGE 3

Exercise 1.1 | Q 11 | Page 3

## Every real number is a complex number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 12 PAGE 3

Exercise 1.1 | Q 12 | Page 3

## 1 is a prime number.

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 13 PAGE 3

Exercise 1.1 | Q 13 | Page 3

## With the sunset the day ends.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 14 PAGE 3

Exercise 1.1 | Q 14 | Page 3

## 1 ! = 0

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

[Note: Answer in the textbook is incorrect]

#### EXERCISE 1.1Q 15 PAGE 3

Exercise 1.1 | Q 15 | Page 3

## 3 + 5 > 11

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 16 PAGE 3

Exercise 1.1 | Q 16 | Page 3

## The number π is an irrational number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

[Note: Answer in the textbook is incorrect]

#### EXERCISE 1.1Q 17 PAGE 3

Exercise 1.1 | Q 17 | Page 3

## x2 - y2 = (x + y)(x - y) for all x, y ∈ R.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 18 PAGE 3

Exercise 1.1 | Q 18 | Page 3

## The number 2 is the only even prime number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 19 PAGE 3

Exercise 1.1 | Q 19 | Page 3

## Two co-planar lines are either parallel or intersecting.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 20 PAGE 3

Exercise 1.1 | Q 20 | Page 3

## The number of arrangements of 7 girls in a row for a photograph is 7!.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 21 PAGE 3

Exercise 1.1 | Q 21 | Page 3

## Give me a compass box.

### SOLUTION

It is an imperative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 22 PAGE 3

Exercise 1.1 | Q 22 | Page 3

## Bring the motor car here.

### SOLUTION

It is an imperative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 23 PAGE 3

Exercise 1.1 | Q 23 | Page 3

## It may rain today.

### SOLUTION

It is an open sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 24 PAGE 3

Exercise 1.1 | Q 24 | Page 3

## If a + b < 7, where a ≥ 0 and b ≥ 0 then a < 7 and b < 7.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 25 PAGE 3

Exercise 1.1 | Q 25 | Page 3

## Can you speak in English?

### SOLUTION

### It is an interrogative sentence. Hence, it is not a statement.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.2 [Page 6]

#### EXERCISE 1.2Q 1.1 PAGE 6

Express the following statement in symbolic form.

e is a vowel or 2 + 3 = 5

#### SOLUTION

Let p : e is a vowel.

q : 2 + 3 = 5

The symbolic form is p ∨ q.

#### EXERCISE 1.2Q 1.2 PAGE 6

Express the following statement in symbolic form.

Mango is a fruit but potato is a vegetable.

#### SOLUTION

Let p : Mango is a fruit.

q : Potato is a vegetable.

The symbolic form is p Λ q.

#### EXERCISE 1.2Q 1.3 PAGE 6

Express the following statement in symbolic form.

Milk is white or grass is green.

#### SOLUTION

Let p : Milk is white.

q : Grass is green.

The symbolic form is p ∨ q.

#### EXERCISE 1.2Q 1.4 PAGE 6

Express the following statement in symbolic form.

I like playing but not singing.

#### SOLUTION

Let p : I like playing.

q : I do not like singing.

The symbolic form is p ∧ q.

#### EXERCISE 1.2Q 1.5 PAGE 6

Express the following statement in symbolic form.

Even though it is cloudy, it is still raining.

#### SOLUTION

Let p : It is cloudy.

q : It is still raining.

The symbolic form is p ∧ q.

#### EXERCISE 1.2Q 2.1 PAGE 6

Write the truth value of the following statement.

Earth is a planet and Moon is a star.

#### SOLUTION

Let p : Earth is a planet.

q : Moon is a star.

The truth values of p and q are T and F respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.2 PAGE 6

Write the truth value of the following statement.

16 is an even number and 8 is a perfect square.

#### SOLUTION

Let p : 16 is an even number.

q : 8 is a perfect square.

The truth values of p and q are T and F respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.3 PAGE 6

Write the truth value of the following statement.

A quadratic equation has two distinct roots or 6 has three prime factors.

#### SOLUTION

Let p : A quadratic equation has two distinct roots.

q : 6 has three prime factors.

The truth values of p and q are F and F respectively.

The given statement in symbolic form is p ∨ q.

∴ p ∨ q ≡ F ∨ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.4 PAGE 6

Write the truth value of the following statement.

The Himalayas are the highest mountains but they are part of India in the North East.

#### SOLUTION

Let p : Himalayas are the highest mountains.

q : Himalayas are the part of India in the north east.

The truth values of p and q are T and T respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.3 [Page 7]

#### EXERCISE 1.3Q 1.1 PAGE 7

Write the negation of the following statement.

All men are animals.

#### SOLUTION

Some men are not animals.

#### EXERCISE 1.3Q 1.2 PAGE 7

Write the negation of the following statement.

− 3 is a natural number.

#### SOLUTION

– 3 is not a natural number.

#### EXERCISE 1.3Q 1.3 PAGE 7

Write the negation of the following statement.

It is false that Nagpur is capital of Maharashtra

#### SOLUTION

Nagpur is capital of Maharashtra.

#### EXERCISE 1.3Q 1.4 PAGE 7

Write the negation of the following statement.

2 + 3 ≠ 5

#### SOLUTION

2 + 3 = 5

#### EXERCISE 1.3Q 2.1 PAGE 7

Write the truth value of the negation of the following statement.

## √5 is an irrational number.

#### SOLUTION

Truth value of the given statement is T.

∴ Truth value of its negation is F.

#### EXERCISE 1.3Q 2.2 PAGE 7

Write the truth value of the negation of the following statement.

London is in England.

#### SOLUTION

Truth value of the given statement is T.

∴ Truth value of its negation is F.

#### EXERCISE 1.3Q 2.3 PAGE 7

Write the truth value of the negation of the following statement.

For every x ∈ N, x + 3 < 8.

#### SOLUTION

Truth value of the given statement is F.

∴ Truth value of its negation is T.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.4 [Pages 10 - 11]

Write the following statement in symbolic form.

If triangle is equilateral then it is equiangular.

#### SOLUTION

Let p : Triangle is equilateral.

q : Triangle is equiangular.

The symbolic form is p → q.

Write the following statement in symbolic form.

It is not true that “i” is a real number.

#### SOLUTION

Let p : i is a real number.

The symbolic form is ~ p.

Write the following statement in symbolic form.

Even though it is not cloudy, it is still raining.

#### SOLUTION

Let p : It is cloudy.

q : It is raining.

The symbolic form is ~p ∧ q.

Write the following statement in symbolic form.

Milk is white if and only if the sky is not blue.

#### SOLUTION

Let p : Milk is white.

q : Sky is blue.

The symbolic form is p ↔ ~ q.

Write the following statement in symbolic form.

Stock prices are high if and only if stocks are rising.

#### SOLUTION

Let p : Stock prices are high.

q : Stock are rising

The symbolic form is p ↔ q.

Write the following statement in symbolic form.

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.

The symbolic form is p → q.

Find the truth value of the following statement.

It is not true that 3 − 7i is a real number.

#### SOLUTION

Let p : 3 – 7i is a real number.

The truth value of p is F.

The given statement in symbolic form is ~p.

∴ ~ p ≡ ~ F ≡ T

∴ Truth value of the given statement is T.

Find the truth value of the following statement.

If a joint venture is a temporary partnership, then discount on purchase is credited to the supplier.

#### SOLUTION

Let p : A joint venture is a temporary partnership. q : Discount on purchase is credited to the supplier.

The truth value of p and q are T and F respectively.

The given statement in symbolic form is p → q.

∴ p → q ≡ T → F ≡ F

∴ Truth value of the given statement is F.

Find the truth value of the following statement.

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.

The truth values of p and q are F and T respectively.

The given statement in symbolic form is p ↔ q.

∴ p ↔ q ≡ F ↔ T ≡ F

∴ Truth value of the given statement is F.

Find the truth value of the following statement.

Neither 27 is a prime number nor divisible by 4.

#### SOLUTION

Let p : 27 is a prime number.

q : 27 is divisible by 4.

The truth values of p and q are F and F respectively.

The given statement in symbolic form is ~ p ∧ ~ q.

∴ ~ p ∧ ~ q ≡ ~ F ∧ ~ F ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

Find the truth value of the following statement.

3 is a prime number and an odd number.

#### SOLUTION

Let p : 3 is a prime number.

q : 3 is an odd number.

The truth values of p and q are T and T respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

p ∧ (q ∧ r)

#### SOLUTION

p ∧ (q ∧ r) ≡ T ∧ (T ∧ F)

≡ T ∧ F

≡ F

Hence, truth value if F.

If p and q are true and r and s are false, find the truth value of the following compound statement.

(p → q) ∨ (r ∧ s)

#### SOLUTION

(p → q) ∨ (r ∧ s) ≡ (T → T) ∨ (F ∧ F)

≡ T ∨ F

≡ T

Hence, truth value if T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

~ [(~ p ∨ s) ∧ (~ q ∧ r)]

#### SOLUTION

~ [(~ p ∨ s) ∧ (~ q ∧ r)] ≡ ~[(~T ∨ F) ∧ (~T ∧ F)]

≡ ~[(F ∨ F) ∧ (F ∧ F)

≡ ~ (F ∧ F)

≡ ~ F

≡ T

Hence, truth value if T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

(p → q) ↔ ~(p ∨ q)

#### SOLUTION

(p → q) ↔ ~(p ∨ q) ≡ (T → T) ↔ (T ∨ T)

≡ T ↔ ~ T

≡ T ↔ F

≡ F

Hence, truth value if F.

If p and q are true and r and s are false, find the truth value of the following compound statement.

[(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, truth value is T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

~ [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, truth value is F.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

Sunday is not holiday or Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is ~ p ∨~q

∴ ~ p ∨~ q ≡ ~ T ∨ ~ T

≡ F ∨ F

≡ F

Hence, truth value is F.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

If Sunday is not holiday then Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is

~ p → ~ q

∴ ~ p → ~ q ≡ ~ T → ~ T

≡ F → F

≡ T

Hence, truth value is T.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

Sunday is a holiday and Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is p ∧ ~ q

∴ p ∧ ~ q ≡ T ∧ ~ T

≡ T ∧ F

≡ F

Hence, truth value is F.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

p ↔ ~ q

#### SOLUTION

He swims if and only if water is not warm.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

~ (p ∨ q)

#### SOLUTION

It is not true that he swims or water is warm.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

q → p

#### SOLUTION

If water is warm then he swims.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

q ∧ ~ p

#### SOLUTION

Water is warm and he does not swim.

### EXERCISE 1.5PAGE 12

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.5 [Page 12]

Use quantifiers to convert the following open sentences defined on N, into a true statement.

x^{2} + 3x - 10 = 0

#### SOLUTION

∃ x ∈ N, such that x^{2} + 3x – 10 = 0

It is true statement, since x = 2 ∈ N satisfies it.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

3x - 4 < 9

#### SOLUTION

∃ x ∈ N, such that 3x – 4 < 9

It is true statement, since

x = 2, 3, 4 ∈ N satisfies 3x - 4 < 9.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

n^{2} ≥ 1

#### SOLUTION

∀ n ∈ N, n^{2} ≥ 1

It is true statement, since all n ∈ N satisfy it.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

2n - 1 = 5

#### SOLUTION

∃ n ∈ N, such that 2n - 1 = 5

It is a true statement since all n = 3 ∈ N satisfy 2n - 1 = 5.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

y + 4 > 6

#### SOLUTION

∃ y ∈ N, such that y + 4 > 6

It is a true statement since y = 3, 4, ... ∈ N satisfy y + 4 > 6.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

3y - 2 ≤ 9

#### SOLUTION

∃ y ∈ N, such that 3y - 2 ≤ 9

It is a true statement since y = 1, 2, 3 ∈ N satisfy it.

If B = {2, 3, 5, 6, 7} determine the truth value of ∀ x ∈ B such that x is prime number.

#### SOLUTION

For x = 6, x is not a prime number.

∴ x = 6 does not satisfies the given statement.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∃ n ∈ B, such that n + 6 > 12.

#### SOLUTION

For n = 7, n + 6 = 7 + 6 = 13 > 12

∴ n = 7 satisfies the equation n + 6 > 12.

∴ The given statement is true.

∴ It’s truth value is T.

If B = {2, 3, 5, 6, 7} determine the truth value of

∃ n ∈ B, such that 2n + 2 < 4.

#### SOLUTION

There is no n in B which satisfies 2n + 2 < 4.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∀ y ∈ B, such that y^{2} is negative.

#### SOLUTION

There is no y in B which satisfies y^{2} < 0.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∀ y ∈ B, such that (y - 5) ∈ N

#### SOLUTION

For y = 2, y – 5 = 2 – 5 = –3 ∉ N.

∴ y = 2 does not satisfies the equation (y – 5) ∈ N.

∴ The given statement is false.

∴ It’s truth value is F.

### EXERCISE 1.6PAGE 16

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.6 [Page 16]

Prepare truth tables for the following statement pattern.

p → (~ p ∨ q)

#### SOLUTION

p → (~ p ∨ q)

p | q | ~p | ~ p ∨ q | p → (~ p ∨ q) |

T | T | F | T | T |

T | F | F | F | F |

F | T | T | T | T |

F | F | T | T | T |

Prepare truth tables for the following statement pattern.

(~ p ∨ q) ∧ (~ p ∨ ~ q)

#### SOLUTION

(~ p ∨ q) ∧ (~ p ∨ ~ q)

p | q | ~p | ~q | ~p∨q | ~p∨~q | (~p∨q)∧(~p∨~q) |

T | T | F | F | T | F | F |

T | F | F | T | F | T | F |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

Prepare truth tables for the following statement pattern.

(p ∧ r) → (p ∨ ~ q)

#### SOLUTION

(p ∧ r) → (p ∨ ~ q)

p | q | r | ~q | p ∧ r | p∨~q | (p ∧ r) → (p ∨ ~ q) |

T | T | T | F | T | T | T |

T | T | F | F | F | T | T |

T | F | T | T | T | T | T |

T | F | F | T | F | T | T |

F | T | T | F | F | F | T |

F | T | F | F | F | F | T |

F | F | T | T | F | T | T |

F | F | F | T | F | T | T |

Prepare truth tables for the following statement pattern.

(p ∧ q) ∨ ~ r

#### SOLUTION

(p ∧ q) ∨ ~ r

p | q | r | ~r | p ∧ q | (p ∧ q) ∨ ~ r |

T | T | T | F | T | T |

T | T | F | T | T | T |

T | F | T | F | F | F |

T | F | F | T | F | T |

F | T | T | F | F | F |

F | T | F | T | F | T |

F | F | T | F | F | F |

F | F | F | T | F | T |

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

q ∨ [~ (p ∧ q)]

#### SOLUTION

p | q | p ∧ q | ~ (p ∧ q) | q ∨ [~ (p ∧ q)] |

T | T | T | F | T |

T | F | F | T | T |

F | T | F | T | T |

F | F | F | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

(~ q ∧ p) ∧ (p ∧ ~ p)

#### SOLUTION

p | q | ~p | ~q | (~q∧p) | (p∧~p) | (~q∧p)∧(p∧~p) |

T | T | F | F | F | F | F |

T | F | F | T | T | F | F |

F | T | T | F | F | F | F |

F | F | T | T | F | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

(p ∧ ~ q) → (~ p ∧ ~ q)

#### SOLUTION

p | q | ~p | ~q | p∧~q | ~p∧~q | (p∧~q)→(~p∧~q) |

T | T | F | F | F | F | T |

T | F | F | T | T | F | F |

F | T | T | F | F | F | T |

F | F | T | T | F | T | T |

The truth values in the last column are not identical. Hence, it is contingency.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

~ p → (p → ~ q)

#### SOLUTION

p | q | ~p | ~q | p→~q | ~p→(p→~q) |

T | T | F | F | F | T |

T | F | F | T | T | T |

F | T | T | F | T | T |

F | F | T | T | T | T |

All the truth values in the last column are T. Hence, it is tautology.

Prove that the following statement pattern is a tautology.

(p ∧ q) → q

#### SOLUTION

p | q | p ∧ q | (p∧q)→q |

T | T | T | T |

T | F | F | T |

F | T | F | T |

F | F | F | T |

All the truth values in the last column are T. Hence, it is tautology.

Prove that the following statement pattern is a tautology.

(p → q) ↔ (~ q → ~ p)

#### SOLUTION

p | q | ~p | ~q | p→q | ~q→~p | (p→q)↔(~q→~p) |

T | T | F | F | T | T | T |

T | F | F | T | F | F | T |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a tautology.

(~p ∧ ~q ) → (p → q)

#### SOLUTION

p | q | ~p | ~q | ~p∧~q | p→q | (~p∧~q)→(p→q) |

T | T | F | F | F | T | T |

T | F | F | T | F | F | T |

F | T | T | F | F | T | T |

F | F | T | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a tautology.

(~ p ∨ ~ q) ↔ ~ (p ∧ q)

#### SOLUTION

p | q | ~p | ~q | ~p∨~q | p∧q | ~p∨~q | (~p∨~q↔~(p ∧ q) |

T | T | F | F | F | T | F | T |

T | F | F | T | T | F | T | T |

F | T | T | F | T | F | T | T |

F | F | T | T | T | F | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a contradiction.

(p ∨ q) ∧ (~p ∧ ~q)

#### SOLUTION

p | q | ~p | ~q | p∨q | ~p∧~q | (p∨q)∧(~p∧~q) |

T | T | F | F | T | F | F |

T | F | F | T | T | F | F |

F | T | T | F | T | F | F |

F | F | T | T | F | T | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ ~p

#### SOLUTION

p | q | ~p | p∧q | (p∧q)∧~p |

T | T | F | T | F |

T | F | F | F | F |

F | T | T | F | F |

F | F | T | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ (~p ∨ ~q)

#### SOLUTION

p | q | ~p | ~q | p∧q | ~p∨~q | (p∧q)∧(~p∨~q) |

T | T | F | F | T | F | F |

T | F | F | T | F | T | F |

F | T | T | F | F | T | F |

F | F | T | T | F | T | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p → q) ∧ (p ∧ ~ q)

#### SOLUTION

p | q | ~q | p→q | p∧~q | (p→q)∧(p∧~q) |

T | T | F | T | F | F |

T | F | T | F | T | F |

F | T | F | T | F | F |

F | F | T | T | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Show that the following statement pattern is contingency.

(p∧~q) → (~p∧~q)

#### SOLUTION

p | q | ~p | ~q | p∧~q | ~p∧~q | (p∧~q)→(~p∧~q) |

T | T | F | F | F | F | T |

T | F | F | T | T | F | F |

F | T | T | F | F | F | T |

F | F | T | T | F | T | T |

The truth values in the last column are not identical. Hence, it is contingency.

Show that the following statement pattern is contingency.

(p → q) ↔ (~ p ∨ q)

#### SOLUTION

p | q | ~p | p→q | ~p∨q | (p→q)↔(~p∨q) |

T | T | F | T | T | T |

T | F | F | F | F | T |

F | T | T | T | T | T |

F | F | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology. Not contingency.

Show that the following statement pattern is contingency.

p ∧ [(p → ~ q) → q]

#### SOLUTION

p | q | ~q | p→~q | (p→~q)→q | p∧[(p→~q)→q] |

T | T | F | F | T | T |

T | F | T | T | F | F |

F | T | F | T | T | F |

F | F | T | T | F | F |

Truth values in the last column are not identical. Hence, it is contingency.

Show that the following statement pattern is contingency.

(p → q) ∧ (p → r)

#### SOLUTION

p | q | r | p→q | p→r | (p→q)∧(p→r) |

T | T | T | T | T | T |

T | T | F | T | F | F |

T | F | T | F | T | F |

T | F | F | F | F | F |

F | T | T | T | T | T |

F | T | F | T | T | T |

F | F | T | T | T | T |

F | F | F | T | T | T |

The truth values in the last column are not identical. Hence, it is contingency.

Exercise 1.6 | Q 6.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | r | q∧r | p∨(q∧r) | p∨q | p∨r | (p∨q)∧(p∨r) |

T | T | T | T | T | T | T | T |

T | T | F | F | T | T | T | T |

T | F | T | F | T | T | T | T |

T | F | F | F | T | T | T | T |

F | T | T | T | T | T | T | T |

F | T | F | F | F | T | F | F |

F | F | T | F | F | F | T | F |

F | F | F | F | F | F | F | F |

The entries in columns 5 and 8 are identical.

∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Using the truth table, verify

p → (p → q) ≡ ~ q → (p → q)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 |

p | q | ~q | p→q | p→(p→q) | ~q→(p→q) |

T | T | F | T | T | T |

T | F | T | F | F | F |

F | T | F | T | T | T |

F | F | T | T | T | T |

In the above truth table, entries in columns 5 and 6 are identical.

∴ p → (p → q) ≡ ~ q → (p → q)

Using the truth table, verify

~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | ~q | p→~q | ~(p→~q) | ~(~q) | p∧~(~q) | p∧q |

T | T | F | F | T | T | T | T |

T | F | T | T | F | F | F | F |

F | T | F | T | F | T | F | F |

F | F | T | T | F | F | F | F |

In the above table, entries in columns 5, 7, and 8 are identical.

∴ ~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

Using the truth table, verify

~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

p | q | ~p | (p∨q) | ~(p∨q) | ~p∧q | ~(p∨q)∨(~p∧q) |

T | T | F | T | F | F | F |

T | F | F | T | F | F | F |

F | T | T | T | F | T | T |

F | F | T | F | T | F | T |

In the above truth table, the entries in columns 3 and 7 are identical.

∴ ~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

Exercise 1.6 | Q 7.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | r | q∧r | p∨(q∧r) | p∨q | p∨r | (p∨q)∧(p∨r) |

T | T | T | T | T | T | T | T |

T | T | F | F | T | T | T | T |

T | F | T | F | T | T | T | T |

T | F | F | F | T | T | T | T |

F | T | T | T | T | T | T | T |

F | T | F | F | F | T | F | F |

F | F | T | F | F | F | T | F |

F | F | F | F | F | F | F | F |

The entries in columns 5 and 8 are identical.

∴ ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Prove that the following pair of statement pattern is equivalent.

p ↔ q and (p → q) ∧ (q → p)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 |

p | q | p↔q | p→q | q→p | (p→q)∧(q→p) |

T | T | T | T | T | T |

T | F | F | F | T | F |

F | T | F | T | F | F |

F | F | T | T | T | T |

In the above table, entries in columns 3 and 6 are identical.

∴ Statement p ↔ q and (p → q) ∧ (q → p) are equivalent.

Prove that the following pair of statement pattern is equivalent.

p → q and ~ q → ~ p and ~ p ∨ q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

p | q | ~p | ~q | p→q | ~q→~p | ~p∨q |

T | T | F | F | T | T | T |

T | F | F | T | F | F | F |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

In the above table, entries in columns 5, 6 and 7 are identical.

∴ Statement p → q and ~q → ~p and ~p ∨ q are equivalent.

Prove that the following pair of statement pattern is equivalent.

~(p ∧ q) and ~p ∨ ~q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

p | q | ~p | ~q | p∧q | ~(p∧q) | ~p∨~q |

T | T | F | F | T | F | F |

T | F | F | T | F | T | T |

F | T | T | F | F | T | T |

F | F | T | T | F | T | T |

In the above table, entries in columns 6 and 7 are identical.

∴ Statement ~(p ∧ q) and ~p ∨ ~q are equivalent.

### EXERCISE 1.7PAGE 17

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.7 [Page 17]

Write the dual of the following:

(p ∨ q) ∨ r

#### SOLUTION

(p ∧ q) ∧ r

Write the dual of the following:

~(p ∨ q) ∧ [p ∨ ~ (q ∧ ~ r)]

#### SOLUTION

~(p ∧ q) ∨ [p ∧ ~ (q ∨ ~ r)]

Write the dual of the following:

p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r

#### SOLUTION

p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

Write the dual of the following:

~(p ∧ q) ≡ ~ p ∨ ~ q

#### SOLUTION

~(p ∨ q) ≡ ~ p ∧ ~ q

Write the dual statement of the following compound statement.

13 is a prime number and India is a democratic country.

#### SOLUTION

13 is a prime number or India is a democratic country.

Write the dual statement of the following compound statement.

Karina is very good or everybody likes her.

#### SOLUTION

Karina is very good and everybody likes her.

Write the dual statement of the following compound statement.

Radha and Sushmita cannot read Urdu.

#### SOLUTION

Radha or Sushmita cannot read Urdu.

Write the dual statement of the following compound statement.

A number is a real number and the square of the number is non-negative.

#### SOLUTION

A number is a real number or the square of the number is non-negative.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.8 [Page 21]

Write the negation of the following statement.

All the stars are shining if it is night.

#### SOLUTION

Let q : All stars are shining.

p : It is night.

The given statement in symbolic form is p → q. It’s negation is ~ (p → q) ≡ p ∧ ~ q

∴ The negation of a given statement is ‘It is night and some stars are not shining’.

Write the negation of the following statement.

∀ n ∈ N, n + 1 > 0

#### SOLUTION

∃ n ∈ N such that n + 1 ≤ 0.

Write the negation of the following statement.

∃ n ∈ N, (n^{2} + 2) is odd number.

#### SOLUTION

∀ n ∈ N, (n^{2} + 2) is not odd number.

Write the negation of the following statement.

Some continuous functions are differentiable.

#### SOLUTION

All continuous functions are not differentiable.

Using the rules of negation, write the negation of the following:

(p → r) ∧ q

#### SOLUTION

~ [(p → r) ∧ q] ≡ ~(p → r) ∨ ~q ....[Negation of conjunction]

≡ (p ∧ ~ r) ∨ ~q .....[Negation of implication]

Using the rules of negation, write the negation of the following:

~(p ∨ q) → r

#### SOLUTION

~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ ~r ....[Negation of implication]

≡ (~p ∧ ~q) ∧ ~r .....[Negation of disjunction]

Using the rules of negation, write the negation of the following:

(~p ∧ q) ∧ (~q ∨ ~r)

#### SOLUTION

~[(~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 on negation]

Write the converse, inverse, and contrapositive of the following statement.

If it snows, then they do not drive the car.

#### SOLUTION

Let p : It snows.

q : They do not drive the car.

∴ The given statement is p → q.

Its converse is q → p.

If they do not drive the car then it snows.

Its inverse is ~p → ~q.

If it does not snow then they drive the car.

Its contrapositive is ~q → ~p.

If they drive the car then it does not snow.

Write the converse, inverse, and contrapositive of the following statement.

If he studies, then he will go to college.

#### SOLUTION

Let p : He studies.

q : He will go to college.

∴ The given statement is p → q.

Its converse is q → p.

If he will go to college then he studies.

Its inverse is ~p → ~q.

If he does not study then he will not go to college.

Its contrapositive is ~q → ~p.

If he will not go to college then he does not study.

With proper justification, state the negation of the following.

(p → q) ∨ (p → r)

#### SOLUTION

~[(p → q) ∨ (p → r)]

≡ ~(p → q) ∧ ~(p → r) ...[Negation of disjunction]

≡ (p ∧ ~ q) ∧ (p ∧ ~r) ....[Negation of implication]

With proper justification, state the negation of the following.

(p ↔ q) ∨ (~q → ~r)

#### SOLUTION

~[(p ↔ q) ∨ (~q → ~r)]

≡ ~(p ↔ q) ∧ (~q → ~r) ....[Negation of disjunction]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ ~(~q → ~r) ....[Negation of double implication]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~ q ∧ ~(~r)] ....[Negation of implication]

≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~ q ∧ r) ....[Negation of negation]

With proper justification, state the negation of the following.

(p → q) ∧ r

#### SOLUTION

~[(p → q) ∧ r]

≡ ~ (p → q) ∨ ~ r ....[Negation of conjunction]

≡ (p ∧ ~q) ∨ ~ r ....[Negation of implication]

### EXERCISE 1.9PAGE 22

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.9 [Page 22]

Without using truth table, show that

p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)

#### SOLUTION

L.H.S.

≡ p ↔ q

≡ (p → q) ∧ (q → p)

≡ (~p ∨ q) ∧ (~q ∨ p)

≡ [~ p ∧ (~ q ∨ p)] ∨ [q ∧ (~ q ∨ p)] ....[Distributive law]

≡ [(~ p ∧ ~ q) ∨ (~ p ∧ p)] ∨ [(q ∧ ~ q) ∨ (q ∧ p)] .....[Distributive Law]

≡ [(~ p ∧ ~ q) ∨ F] ∨ [F ∨ (q ∧ p)] ....[Complement Law]

≡ (~ p ∧ ~ q) ∨ (q ∧ p) ....[Identity Law]

≡ (p ∧ q) ∨ (~ p ∧ ~ q) ....[Commutative Law]

≡ R.H.S.

Without using truth table, show that

p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p

#### SOLUTION

L.H.S.

≡ p ∧ [(~ p ∨ q) ∨ ~ q]

≡ p ∧ [(~ p ∨ (q ∨ ~ q)] .....[Associative law]

≡ p ∧ (~ p ∨ T) .....[Complement law]

≡ p ∧ T .....[Identity law]

≡ p .....[Identity law]

≡ R.H.S.

Without using truth table, show that

~ [(p ∧ q) → ~ q] ≡ p ∧ q

#### SOLUTION

L.H.S.

≡ ~ [(p ∧ q) → ~ q]

≡ (p ∧ q) ∧ ~ (~ q) ....[Negation of implication]

≡ (p ∧ q) ∧ q .....[Negation of a negation]

≡ p ∧ (q ∧ q) ....[Associative law]

≡ p ∧ q .....[Identity law]

≡ R.H.S.

Without using truth table, show that

~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p

#### SOLUTION

L.H.S.

≡ ~r → ~ (p ∧ q)

≡ ~(~ r) ∨ ~ (p ∧ q) ....[p → q ≡ ~ p ∨ q]

≡ r ∨ ~(p ∧ q) ....[Negation of negation]

≡ r ∨ (~p ∨ ~q) ....[De Morgan’s law]

≡ ~p ∨ (~q ∨ r) .....[Commutative and associative law]

≡ ~p ∨ (q → r) ....[p → q ≡ ~ p ∨ q]

≡ (q → r) ∨ ~p ......[Commutative law]

≡ ~[~ (q → r)] ∨ ~ p ......[Negation of negation]

≡ [~ (q → r)] → ~ p .....[p → q ≡ ~ p ∨ q]

= R.H.S.

Without using truth table, show that

(p ∨ q) → r ≡ (p → r) ∧ (q → r)

#### SOLUTION

L.H.S.

≡ (p ∨ q) → r

≡ ~ (p ∨ q) ∨ r ....[p → q → ~ p ∨ q]

≡ (~ p ∧ ~ q) ∨ r ....[De Morgan’s law]

≡ (~ p ∨ r) ∧ (~ q ∨ r) .....[Distributive law]

≡ (p → r) ∧ (q → r) .....[p → q → ~ p ∨ q]

= R.H.S.

Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p

#### SOLUTION

L.H.S.

= [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]

= R.H.S.

Using the algebra of statement, prove that

(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)

#### SOLUTION

L.H.S.

= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)

≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)] ....[Associative Law]

≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~ q ∧ ~ p)] ....[Commutative Law]

≡ (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]

= R.H.S.

Using the algebra of statement, prove that

#### SOLUTION

(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

L.H.S.

= (p ∨ q) ∧ (~ p ∨ ~ q)

≡ [(p ∨ q) ∧ ~ p] ∨ [(p ∨ q) ∧ ~ q] .....[Distributive law]

≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)] .....[Distributive law]

≡ [F ∨ (q ∧ ~p)] ∨ [(p ∧ ~ q) ∨ F] .....[Complement law]

≡ (q ∧ ~p) ∨ (p ∧ ~ q) .....[Identity law]

≡ (p ∧ ~ q) ∨ (~ p ∧ q) ....[Commutative law]

= R.H.S.

### EXERCISE 1.10PAGES 22 - 27

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.10 [Pages 22 - 27]

Represent the truth of the following statement by the Venn diagram.

If a quadrilateral is a rhombus, then it is a parallelogram.

#### SOLUTION

Let U : The set of all quadrilaterals.

P : The set of all parallelograms.

R : The set of all rhombuses.

The above Venn diagram represents truth of the given statement, R ⊂ P.

Draw a Venn diagram for the truth of the following statement.

Some share brokers are chartered accountants.

#### SOLUTION

Let U : The set of all human beings.

S : The set of all share brokers.

C : The set of all chartered accountants.

The above Venn diagram represents the truth of the given statement i.e., S ∩ C ≠ φ.

Represent the following statement by the Venn diagram.

If n is a prime number and n ≠ 2, then it is odd.

#### SOLUTION

Let, U : The set of all real numbers.

P : The set of all prime numbers n and n ≠ 2.

O : The set of all odd numbers.

The above Venn diagram represents the truth of the given statement i.e., P ⊂ O.

## MISCELLANEOUS EXERCISE 1 [PAGES 29 - 34]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Miscellaneous Exercise 1 [Pages 29 - 34]

Choose the correct alternative :

Which of the following is not a statement?

Smoking is injuries to health

2 + 2 = 4

2 is the only even prime number.

Come here

#### SOLUTION

Come here

#### QUESTION

Choose the correct alternative :

Which of the following is an open statement?

x is a natural number.

Give answer a glass of water.

WIsh you best of luck.

Good morning to all.

#### SOLUTION

#### QUESTION

Choose the correct alternative :

Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as.

commutative law

associative law

De-Morgan's law

distributive law

#### SOLUTION

#### QUESTION

Choose the correct alternative :

The false statement in the following is

p ∧ (∼ p) is contradiction

(p → q) ↔ (∼ q → ∼ p) is a contradiction.

~ (∼ p) ↔ p is a tautology

p ∨ (∼ p) ↔ p is a tautology

#### SOLUTION

#### QUESTION

Choose the correct alternative :

For the following three statements

p : 2 is an even number.

q : 2 is a prime number.

r : Sum of two prime numbers is always even.

Then, the symbolic statement (p ∧ q) → ∼ r means.

2 is an even and prime number and the sum of two prime numbers is always even.

2 is an even and prime number and the sum of two prime numbers is not always even.

If 2 is an even and prime number, then the sum of two prime numbers is not always even.

If 2 is an even and prime number, then the sum of two prime numbers is also even.

#### SOLUTION

#### QUESTION

Choose the correct alternative :

If p : He is intelligent.

q : He is strong

Then, symbolic form of statement “It is wrong that, he is intelligent or strong” is

∼p ∨ ∼ p

∼ (p ∧ q)

∼ (p ∨ q)

p ∨ ∼ q

#### SOLUTION

#### QUESTION

Choose the correct alternative :

The negation of the proposition “If 2 is prime, then 3 is odd”, is

If 2 is not prime, then 3 is not odd.

2 is prime and 3 is not odd.

2 is not prime and 3 is odd.

If 2 is not prime, then 3 is odd.

#### SOLUTION

#### QUESTION

Choose the correct alternative :

The statement (∼ p ∧ q) ∨∼ q is

p ∨ q

p ∧ q

∼ (p ∨ q)

∼ (p ∧ q)

#### SOLUTION

#### QUESTION

Choose the correct alternative :

Which of the following is always true?

(p → q) ≡ ∼ q → ∼ p

∼ (p ∨ q) ≡ ∼ p ∨ ∼ q

∼ (p → q) ≡ p ∧ ∼ q

∼ (p ∨ q) ≡ ∼ p ∧ ∼ q

#### SOLUTION

#### QUESTION

Choose the correct alternative :

∼ (p ∨ q) ∨ (∼ p ∧ q) is logically equivalent to

∼ p

p

q

∼ q

#### SOLUTION

#### QUESTION

Choose the correct alternative :

If p and q are two statements then (p → q) ↔ (∼ q → ∼ p) is

contradiction

tautology

Neither (i) not (ii)

None of the these

#### SOLUTION

#### QUESTION

Choose the correct alternative :

If p is the sentence ‘This statement is false’ then

truth value of p is T

truth value of p is F

p is both true and false

p is neither true nor false

#### SOLUTION

#### QUESTION

Choose the correct alternative :

Conditional p → q is equivalent to

p → ∼ q

∼ p ∨ q

∼ p → ∼ q

p ∨∼q

#### SOLUTION

#### QUESTION

Choose the correct alternative :

Negation of the statement “This is false or That is true” is

That is true or This is false

That is true and This is false

That is true and That is false

That is false and That is true

#### SOLUTION

#### QUESTION

Choose the correct alternative :

If p is any statement then (p ∨ ∼ p) is a

contingency

contradiction

tautology

None of them

#### SOLUTION

#### QUESTION

Fill in the blanks :

The statement q → p is called as the ––––––––– of the statement p → q.

#### SOLUTION

The statement q → p is called as the Converse of the statement p → q.

#### QUESTION

Fill in the blanks :

Conjunction of two statement p and q is symbolically written as –––––––––.

#### SOLUTION

#### Conjunction of two statement p and q is symbolically written as p ∧ q.

#### QUESTION

Fill in the blanks :

If p ∨ q is true then truth value of ∼ p ∨ ∼ q is –––––––––.

#### SOLUTION

If p ∨ q is true then truth value of ∼ p ∨ ∼ q is F.

#### QUESTION

Fill in the blanks :

Negation of “some men are animal” is –––––––––.

#### SOLUTION

Negation of “some men are animal” is No men are animals.

#### QUESTION

Fill in the blanks :

Truth value of if x = 2, then x2 = − 4 is –––––––––.

#### SOLUTION

Truth value of if x = 2, then x^{2} = − 4 is F.

#### QUESTION

Fill in the blanks :

Inverse of statement pattern p ↔ q is given by –––––––––.

#### SOLUTION

Inverse of statement pattern p ↔ q is given by ∼ p → ∼ q.

#### QUESTION

Fill in the blanks :

p ↔ q is false when p and q have ––––––––– truth values.

#### SOLUTION

p ↔ q is false when p and q have different truth values.

#### QUESTION

Fill in the blanks :

Let p : the problem is easy. r : It is not challenging then verbal form of ∼ p → r is –––––––––.

#### SOLUTION

Let p : the problem is easy. r : It is not challenging then verbal form of ∼ p → r is If the problem is not easy them it is not challenging.

#### QUESTION

Fill in the blanks :

Truth value of 2 + 3 = 5 if and only if − 3 > − 9 is –––––––––.

#### SOLUTION

Truth value of 2 + 3 = 5 if and only if − 3 > − 9 is T.

#### QUESTION

State whether the following statement is True or False :

Truth value of 2 + 3 < 6 is F.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

There are 24 months in year is a statement.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

p ∨ q has truth value F is both p and q has truth value F.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

The negation of 10 + 20 = 30 is, it is false that 10 + 20 ≠ 30.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

Dual of (p ∧ ∼ q) ∨ t is (p ∨ ∼ q) ∨ C.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

Dual of “John and Ayub went to the forest” is “John and Ayub went to the forest”.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

“His birthday is on 29th February” is not a statement.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

x^{2} = 25 is true statement.

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

Truth value of

True

False

#### SOLUTION

#### QUESTION

State whether the following statement is True or False :

p ∧ t = p.

True

False

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

Ice cream Sundaes are my favourite.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

x + 3 = 8 ; x is variable.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

Read a lot to improve your writing skill.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

z is a positive number.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

(a + b)^{2} = a^{2} + 2ab + b^{2} for all a, b ∈ R.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

(2 + 1)^{2} = 9.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

Why are you sad?

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

How beautiful the flower is!

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

The square of any odd number is even.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

All integers are natural numbers.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

If x is real number then x2 ≥ 0.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

Do not come inside the room.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Solve the following :

State which of the following sentences are statements in logic.

What a horrible sight it was!

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The square of every real number is positive.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

Every parallelogram is a rhombus.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

a^{2} − b^{2} = (a + b) (a − b) for all a, b ∈ R.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

Please carry out my instruction.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The Himalayas is the highest mountain range.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

(x − 2) (x − 3) = x^{2} − 5x + 6 for all x∈R.

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

What are the causes of rural unemployment?

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

0! = 1

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

The quadratic equation ax^{2} + bx + c = 0 (a ≠ 0) always has two real roots.

Is a statement

Is not a statement

#### SOLUTION

The quadratic equation ax^{2} + bx + c = 0 (a ≠ 0) always has two real roots is a statement.

Hence, its truth value is F.

#### QUESTION

Which of the following sentence is a statement? In case of a statement, write down the truth value.

What is happy ending?

Is a statement

Is not a statement

#### SOLUTION

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The Sun has set and Moon has risen.

#### SOLUTION

Let p : The sun has set.

q : The moon has risen

The symbolic form is p ∧ q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

Mona likes Mathematics and Physics.

#### SOLUTION

Let p : Mona likes Mathematics

q : Mona likes Physics

The symbolic form is p ∧ q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

3 is prime number if 3 is perfect square number.

#### SOLUTION

Let p : 3 is a prime number.

q : 3 is a perfect square number.

The symbolic form is p ↔ q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

Kavita is brilliant and brave.

#### SOLUTION

Let p : Kavita is brilliant.

q : Kavita is brave.

The symbolic form is p ∧ q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If Kiran drives the car, then Sameer will walk.

#### SOLUTION

Let p : Kiran drives the car.

q : Sameer will walk.

The symbolic form is p → q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The necessary condition for existence of a tangent to the curve of the function is continuity.

#### SOLUTION

The given statement can also be expressed as ‘If the function is continuous, then the tangent to the curve exists’.

Let p : The function is continuous

q : The tangent to the curve exists.

∴ p → q is the symbolic form of the given statement.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

To be brave is necessary and sufficient condition to climb the Mount Everest.

#### SOLUTION

Let p : To be brave

q : climb the Mount Everest

∴ p ↔ q is the symbolic form of the given statement.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

x^{3} + y^{3} = (x + y)^{3} if xy = 0.

#### SOLUTION

Let p : x^{3} + y^{3} = (x + y)^{3}

q : xy = 0

∴ p ↔ q is the symbolic form of the given statement.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

The drug is effective though it has side effects.

#### SOLUTION

The given statement can also be expressed as “The drug is effective and it has side effects”

Let p : The drug is effective.

q : It has side effects.

∴ p ∧ q is the symbolic form of the given statement.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If a real number is not rational, then it must be irrational.

#### SOLUTION

Let p : A real number is not rational.

q : A real number must be irrational.

The symbolic form is p → q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

It is not true that Ram is tall and handsome.

#### SOLUTION

Let p : Ram is tall.

q : Ram is handsome.

The symbolic form is ∼(p ∧ q).

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

Even though it is not cloudy, it is still raining.

#### SOLUTION

Let p : it is cloudy.

q : It is still raining.

The symbolic form is ~ p ∧ q.

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

It is not true that intelligent persons are neither polite nor helpful.

#### SOLUTION

Let p : Intelligent persons are neither polite nor helpful

The symbolic form is ∼ p.

Alternate method:

Let p : Intelligent persons are polite.

q : Intelligent persons are helpful.

The symbolic form is ~(~ p ∧ ~ q).

#### QUESTION

Assuming the first statement p and second as q. Write the following statement in symbolic form.

If the question paper is not easy then we shall not pass.

#### SOLUTION

Let p : The question paper is not easy.

q : We shall not pass.

The symbolic form is p → q.

#### QUESTION

If p : Proof is lengthy.

q : It is interesting.

Express the following statement in symbolic form.

Proof is lengthy and it is not interesting.

#### SOLUTION

p ∧ ∼ q

#### QUESTION

If p : Proof is lengthy.

q : It is interesting.

Express the following statement in symbolic form.

If proof is lengthy then it is interesting.

#### SOLUTION

p → q

#### QUESTION

If p : Proof is lengthy.

q : It is interesting.

Express the following statement in symbolic form.

It is not true that the proof is lengthy but it is interesting.

#### SOLUTION

∼(p ∧ q)

#### QUESTION

If p : Proof is lengthy.

q : It is interesting.

Express the following statement in symbolic form.

It is interesting iff the proof is lengthy.

#### SOLUTION

q ↔ p

#### QUESTION

Let p : Sachin wins the match.

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

(p ∧ q) ∨ r

#### SOLUTION

Sachin wins the match or he is the member of Rajya Sabha or Sachin is happy.

#### QUESTION

Let p : Sachin wins the match.

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

p → r

#### SOLUTION

If Sachin wins the match then he is happy.

#### QUESTION

Let p : Sachin wins the match.

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

∼ p ∨ q

#### SOLUTION

Sachin does not win the match or he is the member of Rajya Sabha.

#### QUESTION

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

p → (p ∧ r)

#### SOLUTION

If sachin wins the match, then he is the member of Rajyasabha or he is happy.

#### QUESTION

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

p → q

#### SOLUTION

If Sachin wins the match then he is a member of Rajyasabha.

#### QUESTION

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

(p ∧ q) ∧ ∼ r

#### SOLUTION

Sachin wins the match and he is the member of Rajyasabha but he is not happy.

#### QUESTION

q : Sachin is a member of Rajya Sabha.

r : Sachin is happy.

Write the verbal statement of the following.

∼ (p ∨ q) ∧ r

#### SOLUTION

It is false that Sachin wins the match or he is the member of Rajyasabha but he is happy.

#### QUESTION

Determine the truth value of the following statement.

4 + 5 = 7 or 9 − 2 = 5

#### SOLUTION

Let p : 4 + 5 = 7

q : 9 – 2 = 5

The truth values of p and q are F and F respectively. The given statement in symbolic form is p ∨ q.

∴ p ∨ q ≡ F ∨ F ≡ F

∴ Truth value of the given statement is F.

#### QUESTION

Determine the truth value of the following statement.

If 9 > 1 then x^{2} − 2x + 1 = 0 for x = 1

#### SOLUTION

Let p : 9 > 1

q : x^{2} – 2x + 1 = 0 for x = 1

The truth values of p and q are T and T respectively. The given statement in symbolic form is p → q.

∴ p → q ≡ T → T ≡ T

∴ Truth value of the given statement is T.

#### QUESTION

Determine the truth value of the following statement.

x + y = 0 is the equation of a straight line if and only if y^{2} = 4x is the equation of the parabola.

#### SOLUTION

Let p : x + y = 0 is the equation of a straight line.

q : y^{2} = 4x is the equation of the parabola.

The truth values of p and q are T and T respectively.

The given statement in symbolic form is p ↔ q.

∴ p ↔ q ≡ T ↔ T ≡ T

∴ Truth value of the given statement is T.

#### QUESTION

Determine the truth value of the following statement.

It is not true that 2 + 3 = 6 or 12 + 3 =5

#### SOLUTION

Let p : 2 + 3 = 6

q : 12 + 3 = 5

The truth values of p and q are F and F respectively.

The given statement in symbolic form is ~(p ∨ q).

∴ ~(p ∨ q) ≡ ~(F ∨ F) ≡ ~F ≡ T

∴ Truth value of the given statement is T.

#### QUESTION

Assuming the following statement.

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

Stock prices are not high or stocks are rising.

#### SOLUTION

Given that the truth values of both p and q are T.

The symbolic form of the given statement is ~ p ∨ q.

∴ ~ p ∨ q ≡ ~ T ∨ T ≡ F ∨ T

Hence, truth value is T.

#### QUESTION

Assuming the following statement.

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

Stock prices are high and stocks are rising if and only if stock prices are high.

#### SOLUTION

The symbolic form of the given statement is

(p ∧ q) ↔ p.

∴ (p ∧ q) ↔ p ≡ (T ∧ T) ↔ T

≡ T ↔ T

≡ T

Hence, truth value is T.

#### QUESTION

Assuming the following statement.

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

If stock prices are high then stocks are not rising.

#### SOLUTION

The Symbolic form of the given statement is p → ~ q.

∴ p → ~ q ≡ T → ~ T ≡ T → F ≡ F

Hence, truth value is F.

#### QUESTION

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

It is false that stocks are rising and stock prices are high.

#### SOLUTION

The symbolic form of the given statement is ~(q ∧ p).

∴ ~(q ∧ p) ≡ ~(T ∧ T) ≡ ~T ≡ F

Hence, truth value is F.

#### QUESTION

Assuming the following statement.

p : Stock prices are high.

q : Stocks are rising.

to be true, find the truth value of the following.

Stock prices are high or stocks are not rising iff stocks are rising.

#### SOLUTION

The symbolic form of the given statement is (p ∨ ~q) ↔ q.

∴ (p ∨ ~q) ↔ q ≡ (T ∨ ~T) ↔ T

≡ (T ∨ F) ↔ T

≡ T ↔ T

≡ T

Hence, truth value is T.

#### QUESTION

Rewrite the following statement without using conditional –

(Hint : p → q ≡ ∼ p ∨ q)

If price increases, then demand falls.

#### SOLUTION

Let p : Prince increases.

q : demand falls.

The given statement is p → q.

But p → q ≡ ~p ∨ q.

The given statement can be written as ‘Price does not increase or demand falls’.

#### QUESTION

Rewrite the following statement without using conditional –

(Hint : p → q ≡ ∼ p ∨ q)

If demand falls, then price does not increase.

#### SOLUTION

Let p : demand falls.

q : Price does not increase.

The given statement is p → q.

But p → q ≡ ~ p ∨ q.

∴ The given statement can be written as ‘Demand does not fall or price does not increase’.

#### QUESTION

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

(p ∧ q) → ∼ p.

#### SOLUTION

(p ∧ q) → ∼ p ≡ (T ∧ T) → ∼ T

≡ T → F

≡ F.

Hence, truth value is F.

#### QUESTION

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

p ↔ (q → ∼ p)

#### SOLUTION

p ↔ (q → ∼ p) ≡ T ↔ (T → ∼ T)

≡ T ↔ (T → F)

≡ T ↔ F

≡ F

Hence, truth value is F.

#### QUESTION

If p, q, r are statements with truth values T, T, F respectively determine the truth values of the following.

(p ∧ ∼ q) ∨ (∼ p ∧ q)

#### SOLUTION

(p ∧ ∼ q) ∨ (∼ p ∧ q) ≡ (T ∧ ∼ T) ∨ (∼ T ∧ T)

≡ (T ∧ F) ∨ (F ∧ T)

≡ F ∨ F

≡ F

Hence, truth value is F.

#### QUESTION

∼ (p ∧ q) → ∼ (q ∧ p)

#### SOLUTION

∼ (p ∧ q) → ∼ (q ∧ p) ≡ ∼ (T ∧ T) → ∼ (T ∧ T)

≡ ~ T → ~ T

≡ F → F

≡ T

Hence, truth value is T.

#### QUESTION

∼ [(p → q) ↔ (p ∧ ∼ q)]

#### SOLUTION

∼[(p → q) ↔ (p ∧ ∼q)] ≡ ∼ [(T → T) ↔ (T ∧ ∼ T)]

≡ ~[T ↔ (T ∧ F)]

≡ ~(T ↔ F)

≡ ~ F

≡ T

Hence, truth value is T.

#### QUESTION

Write the negation of the following.

If ∆ABC is not equilateral, then it is not equiangular.

#### SOLUTION

Let p : ∆ ABC is not equilateral.

q : ∆ ABC is not equiangular.

The given statement is p → q.

Its negation is ~(p → q) ≡ p ∧ ~ q

∴ The negation of given statement is '∆ ABC is not equilateral and it is equiangular'.

#### QUESTION

Write the negation of the following.

Ramesh is intelligent and he is hard working.

#### SOLUTION

Let p : Ramesh is intelligent.

q : Ramesh is hard working.

The given statement is p ∧ q.

Its negation is ~(p ∧ q) ≡ ~ p ∨ ~ q

∴ The negation of the given statement is ‘Ramesh is not intelligent or he is not hard-working.’

#### QUESTION

Write the negation of the following.

An angle is a right angle if and only if it is of measure 90°.

#### SOLUTION

Let p : An angle is a right angle.

q : An angle is of measure 90°.

The given statement is p ↔ q.

Its negation is ~(p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)

∴ The negation of the given statement is ‘An angle is a right angle and it is not of measure 90° or an angle is of measure 90° and it is not a right angle.’

#### QUESTION

Write the negation of the following.

Kanchanganga is in India and Everest is in Nepal.

#### SOLUTION

Let p : Kanchanganga is in India.

q : Everest is in Nepal.

The given statement is p ∧ q.

Its negation is ~(p ∧ q) ≡ ~ p ∨ ~ q.

The negation of a given statement is ‘Kanchanganga is not in India or Everest is not in Nepal’.

#### QUESTION

Write the negation of the following.

If x ∈ A ∩ B, then x ∈ A and x ∈ B.

#### SOLUTION

Let p : x ∈ A ∩ B

q : x ∈ A

r : x ∈ B

The given statement is p → (q ∧ r).

Its negation is ~[p → (q ∧ r)], and

~[p → (q ∧ r)] ≡ p ∧ ~ (q ∧ r) ≡ p ∧ ~ q ∨ ~ r

∴ The negation of given statement is x ∈ A ∩ B and x ∉ A or x ∉ B.

#### QUESTION

Construct the truth table for the following statement pattern.

(p ∧ ~q) ↔ (q → p)

#### SOLUTION

p | q | ~q | p∧~q | q→p | (p∧~q)↔(q→p) |

T | T | F | F | T | F |

T | F | T | T | T | T |

F | T | F | F | F | T |

F | F | T | F | T | F |

#### QUESTION

Construct the truth table for the following statement pattern.

(~p ∨ q) ∧ (~p ∧ ~q)

#### SOLUTION

p | q | ~p | ~q | ~p∨q | ~p∧~q | (~p∨q)∧(~p∧~q) |

T | T | F | F | T | F | F |

T | F | F | T | F | F | F |

F | T | T | F | T | F | F |

F | F |