Concept:
P ⇒ Q means P̅ ∨ Q
Explanation:
P ∧ (P ⇒ Q) = P ∧ (P̅ ∨ Q) = P ∧ Q
Let us consider P.Q as A.
We have to find which of the given expression is implied by A, that is, (A => )
Construct the truth table for each option separately
Let A ≡ P ∧ (P ⇒ Q)
P
|
Q
|
P ⇒ Q
|
A
|
True
|
True
|
True
|
True
|
True
|
False
|
False
|
False
|
False
|
True
|
True
|
False
|
False
|
False
|
True
|
False
|
Consider all options one by one:
(i) False
P
|
Q
|
A
|
A ⇒ False
|
True
|
True
|
True
|
False
|
True
|
False
|
False
|
True
|
False
|
True
|
False
|
True
|
False
|
False
|
False
|
True
|
Here, all are not true. So, this option is incorrect.
(ii) Q
P
|
Q
|
A
|
A ⇒ Q
|
True
|
True
|
True
|
True
|
True
|
False
|
False
|
True
|
False
|
True
|
False
|
True
|
False
|
False
|
False
|
True
|
This option is implied by P ∧ (P ⇒ Q).
iii) True
P
|
Q
|
A
|
A ⇒ True
|
True
|
True
|
True
|
True
|
True
|
False
|
False
|
True
|
False
|
True
|
False
|
True
|
False
|
False
|
False
|
True
|
iv) P ∨ Q
P
|
Q
|
A
|
A ⇒ P ∨ Q
|
True
|
True
|
True
|
True
|
True
|
False
|
False
|
True
|
False
|
True
|
False
|
True
|
False
|
False
|
False
|
True
|
This is a tautology. So, option is correct.
v) –Q ∨ P
P
|
Q
|
A ≡ P ∧ (P ⇒ Q)
|
A ⇒ –Q ∨ P
|
True
|
True
|
True
|
True
|
True
|
False
|
False
|
True
|
False
|
True
|
False
|
True
|
False
|
False
|
False
|
True
|
This is also a tautology.
So, total 4 are implied by the given expression.