Thread: CS607 Assignment No 3 Solution

1. CS607 Assignment No 3 Solution

I have need CS607 Assignment No 3 Solution.please reply me urgent. 4 Jan is last date of assignment,,,,

2. you have to upload here.

3. Aoa

4. you posted it in wrong forum

5. Where is the Solution ... tonight is the list night???

6. no body haven't discuss solution with us. so how we can provide it.

7. 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.

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