The dual of statement p∨(q∧r)≡(p∨q)∧(p∨r) is
p∧(q∨r)≡(p∧q)∨(p∧r)
p∨(q∧r)≡(p∧q)∧r
p∧(q∧r)≡(p∧q)∧r
p∨(q∨r)=(p∧q)∧r
By definition of duality, replace ∧ by ∨ and ∨ by ∧ So, dual of p∧(q∨r)≡(p∧q)∨(p∧r)
Negation of the statement ∼p→(q∨r) is
If S(p,q,r)=∼p ∨∼(q ∨ r) is a compund statement, then S(~p,~q,~r) is