Question
Question: Identify the false statement (a) \( \sim \left[ {p \vee \left( { \sim q} \right)} \right] \equi...
Identify the false statement
(a) ∼[p∨(∼q)]≡(∼p)∨q
(b) [p∨q]∨(∼p)is a tautology
(c) [p∧q]∧(∼p)is a contradiction
(d) ∼[p∨q]≡(∼p)∨(∼q)
Solution
To identify the false statement from the given four options, we will proceed by checking the options one by one. To check if the statements are true, one can also make truth tables of the statements.
Complete step-by-step answer:
(a) ∼[p∨(∼q)]
Since by De Morgan’s Law, ∼(p∨q)≡∼p∧∼q, we get
≡(∼p)∧∼(∼q)
≡(∼p)∧q
Therefore ∼[p∨(∼q)]≡(∼p)∨q
Hence, (a) is a false statement.
(b) Next, we are to check if [p∨q]∨(∼p)is a tautology.
The truth table for [p∨q]∨(∼p)is given by
p | q | [p∨q]∨(∼p) |
---|---|---|
F | F | T |
F | T | T |
T | F | T |
T | T | T |
Therefore [p∨q]∨(∼p)is a Tautology.
Hence, (b) is true.
(c) Similarly, the truth table for [p∧q]∧(∼p) is given by
p | q | [p∧q]∧(∼p) |
---|---|---|
F | F | F |
F | T | F |
T | F | F |
T | T | F |
Therefore, [p∧q]∧(∼p) is a contradiction.
Hence, (c) is true.
(d) ∼[p∨q]
≡(∼p)∧(∼q) (by De Morgan’s Law)
Therefore ∼[p∨q]≡(∼p)∨(∼q)
Hence, (d) is a false statement.
Therefore, the false statements are ∼[p∨(∼q)]≡(∼p)∨qand ∼[p∨q]≡(∼p)∨(∼q).
Hence, the correct options are (a) and (d).
Note: Remember De Morgan’s Law:
∼(p∨q)≡∼p∧∼q
∼(p∧q)≡∼p∨∼q
Try to make a truth table for easy calculation.