Exercise 1.6 (5)
5. Show that each of the following statement pattern is a contingency.
i) (p ∧ ∼ q) → (∼ p ∧ ∼ q)
Ans:
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 |
Truth VALUES IN THE LAST Column are not identical. Hence, it is Contigency.
ii) (p → q) ↔ (∼ p ∨ q)
Ans:
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 value in the last column are T. Hence, it is tautology. Not Contigency.
iii) p ∧ [(p → ∼ q) → q]
Ans:
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 value in the last column are not identical. Hence, it is contingency.
iv) (p → q) ∧ (p → r)
Ans:
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 |
Truth value in the last column are not identical .Hence, it is contingency.