Question

$(p \rightarrow q) \land \neg q ] \rightarrow p$ is Totology, is equivalent to

• A. $p \land \neg q$
• B. $q \vee p$
• C. $p \land q$
• D. $p \land q$

Answer 1

Answer: (B)

$q \vee p$

Answer 2

Solution:

1. Rewrite $p\to q$ as $\neg p \lor q$.
   Thus, the antecedent becomes:

   $$(\neg p \lor q) \land \neg q.$$

2. Apply distribution:

   $$(\neg p \lor q) \land \neg q = (\neg p \land \neg q) \lor (q \land \neg q).$$

   Since $q \land \neg q$ is always false, we have:

   $$(\neg p \lor q) \land \neg q = \neg p \land \neg q.$$

3. Now, the original expression is:

   $$(\neg p \land \neg q) \to p.$$

   An implication $A\to B$ is equivalent to $\neg A \lor B$, so:

   $$\neg(\neg p \land \neg q) \lor p.$$

4. Using De Morgan's law:

   $$\neg(\neg p \land \neg q) = p \lor q.$$

   Therefore, the expression simplifies to:

   $$(p \lor q) \lor p = p\lor q.$$

Thus, the given formula is equivalent to $p \lor q$, which is the same as $q \vee p$.