CS607 Assignment No 3 Solution

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Idea Solution :

(v) (p → q)∧¬(p∨(¬q∧r)); (¬p∨q)∧(¬p∧¬(¬q∧r)); (¬p∨q)∧(¬p∧(q∨¬r));
(¬p ∨ q) ∧ ((¬p ∧ q) ∨ (¬p ∧ ¬r)); ((¬p ∨ q) ∧ (¬p ∧ q)) ∨ ((¬p ∨ q) ∧ (¬p ∧ ¬r));
((¬p ∧ (¬p ∧ q)) ∨ (q ∧ (¬p ∧ q))) ∨ ((¬p ∧ (¬p ∧ ¬r)) ∨ (q ∧ (¬p ∧ ¬r)));
(¬p∧q)∨(¬p∧q)∨(¬p∧¬r)∨(q ∧¬p∧¬r); (¬p∧q)∨(¬p∧¬r)∨(q ∧¬p∧¬r).
So a disjunctive normal form is (¬p ∧ q) ∨ (¬p ∧ ¬r) ∨ (q ∧ ¬p ∧ ¬r).
We’ll directly find a conjunctive normal form from the original term:
(p → q) ∧ ¬(p ∨ (¬q ∧ r)); (¬p ∨ q) ∧ ¬p ∧ ¬(¬q ∧ r); (¬p ∨ q) ∧ ¬p ∧ (q ∨ ¬r),
which is already a conjunctive normal form.