# 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,,,,  Reply With Quote

2. you have to upload here.  Reply With Quote

3. Aoa  Reply With Quote

4. you posted it in wrong forum  Reply With Quote

5. Where is the Solution ... tonight is the list night???  Reply With Quote

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

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.  Reply With Quote

There are currently 1 users browsing this thread. (0 members and 1 guests)

assignment 