Please type the answers. **Problem 5: Logical Evaluation Using Truth Tables**Evaluate the following logical formulas using truth tables to categorize each as a tautology, a contradiction, or neither:a. \( (M \land \neg N) \lor (\neg M \land N) \)b. \( (M \land \neg N) \lor (\neg M \land N) \)c. \( (\neg M \land \neg N) \lor (\neg M \lor N) \lor (M \land \neg N) \)

Consider the variables M and N. Since there are only 2 variables, the total number of combinations of truth values of these variables is 22=4. We know that if p is true its negation ¬p is false and if p is false, then ¬p is true. The conjunction p∧q is true only when both p and q are true. Otherwise, it is false. The disjunction p∨q is false only when both p and q are false. Otherwise, it is true. A statement is a tautology, if it is true for all combinations of the variables. A statement is a contradiction, if it is false for all combinations of the variables. Construct the following truth table. M N ¬M ¬N M∧¬N ¬M∧N ¬M∧¬N ¬M∨N M∧¬N∨¬M∧N M∧¬N∧¬M∧N ¬M∧¬N∨¬M∨N ¬M∧¬N∨¬M∨N∨M∧¬N T T F F F F F T F F T T T F F T T F F F T F F T F T T F F T F T T F T T F F T T F F T T F F T T (a) From the 9th column of the above truth table, we can see that M∧¬N∨¬M∧N takes truth value T for some combinations while it take truth value F for other combinations. Hence, M∧¬N∨¬M∧N is neither a tautology nor a contradiction. (b) The 10th column shows that M∧¬N∧¬M∧N always takes the truth value F. Hence, M∧¬N∧¬M∧N is a contradiction. (c) The 12th column shows that ¬M∧¬N∨¬M∨N∨M∧¬N always takes the truth value T. Hence, ¬M∧¬N∨¬M∨N∨M∧¬N is a tautology.