Question

The contrapositive of $p \to \left( {\neg q \to \neg r} \right)$
A.$\left( {\neg q \wedge r} \right) \to \neg p$
B.$\left( {q \wedge \neg r} \right) \to \neg p$
C.$p \to \left( {\neg r \vee q} \right)$
D.$p \wedge \left( {q \vee r} \right)$

Answer 1

Hint: If the conditional statement is $a \to b$, then the contrapositive statement is $\neg b \to \neg a$. Write the contrapositive statement for the given conditional statement. Then solve the bracket using the condition $a \to b \equiv \neg a \vee b$. To get the final answer, use to De Morgan’s law, $\neg \left( {p \wedge q} \right) = \neg p \vee \neg q$

Complete step-by-step answer:
We know that if the conditional statement is $a \to b$, then the contrapositive statement is $\neg b \to \neg a$
From the given conditional statement, $p \to \left( {\neg q \to \neg r} \right)$, we can write the contrapositive statement as,
$\neg \left( {\neg q \to \neg r} \right) \to \neg p$.
Now, we will solve the bracket.
As we know, $a \to b \equiv \neg a \vee b$
So, the contrapositive expression is equivalent to $\neg \left( {\neg q \to \neg r} \right) \to \neg p \equiv \neg \left( {q \vee \neg r} \right) \to \neg p$
Now, we will apply De Morgan’s law, which states that, $\neg \left( {a \wedge b} \right) = \neg a \vee \neg b$
Therefore, for $\neg \left( {q \vee \neg r} \right) \to \neg p$, we get,
$\neg \left( {q \vee \neg r} \right) \to \neg p \equiv \left( {\neg q \wedge r} \right) \to \neg p$
Hence, option A is correct.

Note: If the conditional statement is $p \to q$, then the converse is $q \to p$. If the conditional statement is $p \to q$, then the inverse is $\neg p \to \neg q$ and if the conditional statement is $p \to q$, then the contrapositive statement is $\neg q \to \neg p$. According to De Morgan’s law, $\neg \left( {p \wedge q} \right) = \neg p \vee \neg q$