P ⇔ (Q ∨ ¬ Q) "P should be true because RHS will be TRUE always "
Q ⇔ R "when Q is true R is true" and "when Q is false R is false"
$(P ∧ Q) ⇒ ((P ∧ R) ∨ S)$
there can be only 2 cases (value of S doesn't matter)
1) P = True, Q = True and R = True
$(T ∧ T) ⇒ ((T ∧ T) ∨ S)$
so this case is True
2) P =True, Q = R = False
$(T ∧ F) ⇒ ((P ∧ R) ∨ S)$
In case implication if the premises is false then whole statement is true
this case is also true
The given expression is True in both the cases.
Answer is 1) True