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

EXERCISE 1.6 [PAGE 16]

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]

EXERCISE 1.6Q 1.1   PAGE 16
Exercise 1.6 | Q 1.1 | Page 16
Prepare truth tables for the following statement pattern.  p → (~ p ∨ q)  p → (~ p ∨ q)

Prepare truth tables for the following statement pattern.

p → (~ p ∨ q)

SOLUTION

p → (~ p ∨ q)

pq~p~ p ∨ qp → (~ p ∨ q)
TTFTT
TFFFF
FTTTT
FFTTT
EXERCISE 1.6Q 1.2   PAGE 16
Exercise 1.6 | Q 1.2 | Page 16

Prepare truth tables for the following statement pattern.

(~ p ∨ q) ∧ (~ p ∨ ~ q)

SOLUTION

(~ p ∨ q) ∧ (~ p ∨ ~ q)

pq~p~q~p∨q~p∨~q(~p∨q)∧(~p∨~q)
TTFFTFF
TFFTFTF
FTTFTTT
FFTTTTT
EXERCISE 1.6Q 1.3   PAGE 16
Exercise 1.6 | Q 1.3 | Page 16

Prepare truth tables for the following statement pattern.

(p ∧ r) → (p ∨ ~ q)

SOLUTION

(p ∧ r) → (p ∨ ~ q)

pqr~qp ∧ rp∨~q(p ∧ r) → (p ∨ ~ q)
TTTFTTT
TTFFFTT
TFTTTTT
TFFTFTT
FTTFFFT
FTFFFFT
FFTTFTT
FFFTFTT
EXERCISE 1.6Q 1.4   PAGE 16
Exercise 1.6 | Q 1.4 | Page 16

Prepare truth tables for the following statement pattern.

(p ∧ q) ∨ ~ r

SOLUTION

(p ∧ q) ∨ ~ r

pqr~rp ∧ q(p ∧ q) ∨ ~ r
TTTFTT
TTFTTT
TFTFFF
TFFTFT
FTTFFF
FTFTFT
FFTFFF
FFFTFT
EXERCISE 1.6Q 2.1   PAGE 16
Exercise 1.6 | Q 2.1 | Page 16

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

q ∨ [~ (p ∧ q)]

SOLUTION

pqp ∧ q~ (p ∧ q)q ∨ [~ (p ∧ q)]
TTTFT
TFFTT
FTFTT
FFFTT

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

EXERCISE 1.6Q 2.2   PAGE 16
Exercise 1.6 | Q 2.2 | Page 16
Examine whether the following statement pattern is a tautology, a contradiction or a contingency.  (~ q ∧ p) ∧ (p ∧ ~ p)
Examine whether the following statement pattern is a tautology, a contradiction or a contingency.  (~ q ∧ p) ∧ (p ∧ ~ p)

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

(~ q ∧ p) ∧ (p ∧ ~ p)

SOLUTION

pq~p~q(~q∧p)(p∧~p)(~q∧p)∧(p∧~p)
TTFFFFF
TFFTTFF
FTTFFFF
FFTTFFF

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

EXERCISE 1.6Q 2.3   PAGE 16
Exercise 1.6 | Q 2.3 | Page 16

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

(p ∧ ~ q) → (~ p ∧ ~ q)

SOLUTION

pq~p~qp∧~q~p∧~q(p∧~q)→(~p∧~q)
TTFFFFT
TFFTTFF
FTTFFFT
FFTTFTT

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

EXERCISE 1.6Q 2.4   PAGE 16
Exercise 1.6 | Q 2.4 | Page 16

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

~ p → (p → ~ q)

SOLUTION

pq~p~qp→~q~p→(p→~q)
TTFFFT
TFFTTT
FTTFTT
FFTTTT

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

EXERCISE 1.6Q 3.1   PAGE 16
Exercise 1.6 | Q 3.1 | Page 16
Prove that the following statement pattern is a tautology.  (p ∧ q) → q

Prove that the following statement pattern is a tautology.

(p ∧ q) → q

SOLUTION

pqp ∧ q(p∧q)→q
TTTT
TFFT
FTFT
FFFT

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

EXERCISE 1.6Q 3.2   PAGE 16
Exercise 1.6 | Q 3.2 | Page 16

Prove that the following statement pattern is a tautology.

(p → q) ↔ (~ q → ~ p)

SOLUTION

pq~p~qp→q~q→~p(p→q)↔(~q→~p)
TTFFTTT
TFFTFFT
FTTFTTT
FFTTTTT

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

EXERCISE 1.6Q 3.3   PAGE 16
Exercise 1.6 | Q 3.3 | Page 16

Prove that the following statement pattern is a tautology.

(~p ∧ ~q ) → (p → q)

SOLUTION

pq~p~q~p∧~qp→q(~p∧~q)→(p→q)
TTFFFTT
TFFTFFT
FTTFFTT
FFTTTTT

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

EXERCISE 1.6Q 3.4   PAGE 16
Exercise 1.6 | Q 3.4 | Page 16

Prove that the following statement pattern is a tautology.

(~ p ∨ ~ q) ↔ ~ (p ∧ q)

SOLUTION

pq~p~q~p∨~qp∧q~p∨~q(~p∨~q↔~(p ∧ q)
TTFFFTFT
TFFTTFTT
FTTFTFTT
FFTTTFTT

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

EXERCISE 1.6Q 4.1   PAGE 16
Exercise 1.6 | Q 4.1 | Page 16
Prove that the following statement pattern is a contradiction.  (p ∨ q) ∧ (~p ∧ ~q)

Prove that the following statement pattern is a contradiction.

(p ∨ q) ∧ (~p ∧ ~q)

SOLUTION

pq~p~qp∨q~p∧~q(p∨q)∧(~p∧~q)
TTFFTFF
TFFTTFF
FTTFTFF
FFTTFTF

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

EXERCISE 1.6Q 4.2   PAGE 16
Exercise 1.6 | Q 4.2 | Page 16

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ ~p

SOLUTION

pq~pp∧q(p∧q)∧~p
TTFTF
TFFFF
FTTFF
FFTFF

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

EXERCISE 1.6Q 4.3   PAGE 16
Exercise 1.6 | Q 4.3 | Page 16

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ (~p ∨ ~q)

SOLUTION

pq~p~qp∧q~p∨~q(p∧q)∧(~p∨~q)
TTFFTFF
TFFTFTF
FTTFFTF
FFTTFTF

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

EXERCISE 1.6Q 4.4   PAGE 16
Exercise 1.6 | Q 4.4 | Page 16

Prove that the following statement pattern is a contradiction.

(p → q) ∧ (p ∧ ~ q)

SOLUTION

pq~qp→qp∧~q(p→q)∧(p∧~q)
TTFTFF
TFTFTF
FTFTFF
FFTTFF

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

EXERCISE 1.6Q 5.1   PAGE 16
Exercise 1.6 | Q 5.1 | Page 16
Show that the following statement pattern is contingency.  (p∧~q) → (~p∧~q)

Show that the following statement pattern is contingency.

(p∧~q) → (~p∧~q)

SOLUTION

pq~p~qp∧~q~p∧~q(p∧~q)→(~p∧~q)
TTFFFFT
TFFTTFF
FTTFFFT
FFTTFTT

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

EXERCISE 1.6Q 5.2   PAGE 16
Exercise 1.6 | Q 5.2 | Page 16

Show that the following statement pattern is contingency.

(p → q) ↔ (~ p ∨ q)

SOLUTION

pq~pp→q~p∨q(p→q)↔(~p∨q)
TTFTTT
TFFFFT
FTTTTT
FFTTTT

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

EXERCISE 1.6Q 5.3   PAGE 16
Exercise 1.6 | Q 5.3 | Page 16

Show that the following statement pattern is contingency.

p ∧ [(p → ~ q) → q]

SOLUTION

pq~qp→~q(p→~q)→qp∧[(p→~q)→q]
TTFFTT
TFTTFF
FTFTTF
FFTTFF

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

EXERCISE 1.6Q 5.4   PAGE 16
Exercise 1.6 | Q 5.4 | Page 16

Show that the following statement pattern is contingency.

(p → q) ∧ (p → r)

SOLUTION

pqrp→qp→r(p→q)∧(p→r)
TTTTTT
TTFTFF
TFTFTF
TFFFFF
FTTTTT
FTFTTT
FFTTTT
FFFTTT

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

EXERCISE 1.6Q 6.1   PAGE 16
Using the truth table, verify  p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Exercise 1.6 | Q 6.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

SOLUTION

12345678
pqrq∧rp∨(q∧r)p∨qp∨r(p∨q)∧(p∨r)
TTTTTTTT
TTFFTTTT
TFTFTTTT
TFFFTTTT
FTTTTTTT
FTFFFTFF
FFTFFFTF
FFFFFFFF

The entries in columns 5 and 8 are identical.

∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

EXERCISE 1.6Q 6.2   PAGE 16
Exercise 1.6 | Q 6.2 | Page 16

Using the truth table, verify

p → (p → q) ≡ ~ q → (p → q)

SOLUTION

123456
pq~qp→qp→(p→q)~q→(p→q)
TTFTTT
TFTFFF
FTFTTT
FFTTTT

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

∴ p → (p → q) ≡ ~ q → (p → q)

EXERCISE 1.6Q 6.3   PAGE 16
Exercise 1.6 | Q 6.3 | Page 16

Using the truth table, verify

~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

SOLUTION

12345678
pq~qp→~q

~(p→~q)

~(~q)p∧~(~q)p∧q
TTFFTTTT
TFTTFFFF
FTFTFTFF
FFTTFFFF

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

∴ ~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

EXERCISE 1.6Q 6.4   PAGE 16
Exercise 1.6 | Q 6.4 | Page 16

Using the truth table, verify

~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

SOLUTION

1234567
pq~p(p∨q)~(p∨q)~p∧q~(p∨q)∨(~p∧q)
TTFTFFF
TFFTFFF
FTTTFTT
FFTFTFT

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

∴ ~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

EXERCISE 1.6Q 7.1   PAGE 16
Using the truth table, verify  p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Exercise 1.6 | Q 7.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

SOLUTION

12345678
pqrq∧rp∨(q∧r)p∨qp∨r(p∨q)∧(p∨r)
TTTTTTTT
TTFFTTTT
TFTFTTTT
TFFFTTTT
FTTTTTTT
FTFFFTFF
FFTFFFTF
FFFFFFFF

The entries in columns 5 and 8 are identical.

 ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

EXERCISE 1.6Q 7.2   PAGE 16
Exercise 1.6 | Q 7.2 | Page 16

Prove that the following pair of statement pattern is equivalent.

p ↔ q and (p → q) ∧ (q → p)

SOLUTION

123456
pqp↔qp→qq→p(p→q)∧(q→p)
TTTTTT
TFFFTF
FTFTFF
FFTTTT

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

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

EXERCISE 1.6Q 7.3   PAGE 16
Exercise 1.6 | Q 7.3 | Page 16

Prove that the following pair of statement pattern is equivalent.

p → q and ~ q → ~ p and ~ p ∨ q

SOLUTION

1234567
pq~p~qp→q~q→~p~p∨q
TTFFTTT
TFFTFFF
FTTFTTT
FFTTTTT

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

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

EXERCISE 1.6Q 7.4   PAGE 16
Exercise 1.6 | Q 7.4 | Page 16

Prove that the following pair of statement pattern is equivalent.

~(p ∧ q) and ~p ∨ ~q

SOLUTION

1234567
pq~p~qp∧q~(p∧q)~p∨~q
TTFFTFF
TFFTFTT
FTTFFTT
FFTTFTT

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

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


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