
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
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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, \[\neg \left( {p \wedge q} \right) = \neg p \vee \neg q\]
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, \[\neg \left( {p \wedge q} \right) = \neg p \vee \neg q\]
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