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.
