Type the correct answer in each box. Use T for true and F for false.

Complete the truth table for the contrapositive of a conditional statement.

\begin{tabular}{|c|c|c|c|}
\hline
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$\sim q \rightarrow \sim p$[/tex] \\
\hline
T & T & T & [tex]$\square$[/tex] \\
\hline
T & F & F & [tex]$\square$[/tex] \\
\hline
F & T & T & [tex]$\square$[/tex] \\
\hline
F & F & T & [tex]$\square$[/tex] \\
\hline
\end{tabular}



Answer :

To fill in the truth table for the contrapositive of the conditional statement, we need to evaluate the logical expression for each scenario.

Here is the detailed step-by-step solution:

1. For [tex]\( p = T \)[/tex] and [tex]\( q = T \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] (which means [tex]\( p \)[/tex] implies [tex]\( q \)[/tex]) is True ([tex]\(T\)[/tex]) because if [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is True, the implication is True.
- The contrapositive of [tex]\( p \rightarrow q \)[/tex] is [tex]\( \sim q \rightarrow \sim p \)[/tex] (which means not [tex]\( q \)[/tex] implies not [tex]\( p \)[/tex]).
- [tex]\( \sim q \)[/tex] is False ([tex]\(F\)[/tex]), and [tex]\( \sim p \)[/tex] is False ([tex]\(F\)[/tex]).
- [tex]\( \sim q \rightarrow \sim p \)[/tex] is True ([tex]\(T\)[/tex]) because False implies False is True.

2. For [tex]\( p = T \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is False ([tex]\(F\)[/tex]) because if [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is False, the implication is False.
- The contrapositive [tex]\( \sim q \rightarrow \sim p \)[/tex]:
- [tex]\( \sim q \)[/tex] is True ([tex]\(T\)[/tex]), and [tex]\( \sim p \)[/tex] is False ([tex]\(F\)[/tex]).
- [tex]\( \sim q \rightarrow \sim p \)[/tex] is False ([tex]\(F\)[/tex]) because True implies False is False.

3. For [tex]\( p = F \)[/tex] and [tex]\( q = T \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True ([tex]\(T\)[/tex]) because if [tex]\( p \)[/tex] is False, the implication is True regardless of the value of [tex]\( q \)[/tex].
- The contrapositive [tex]\( \sim q \rightarrow \sim p \)[/tex]:
- [tex]\( \sim q \)[/tex] is False ([tex]\(F\)[/tex]), and [tex]\( \sim p \)[/tex] is True ([tex]\(T\)[/tex]).
- [tex]\( \sim q \rightarrow \sim p \)[/tex] is True ([tex]\(T\)[/tex]) because False implies True is True.

4. For [tex]\( p = F \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True ([tex]\(T\)[/tex]) because if [tex]\( p \)[/tex] is False, the implication is True regardless of the value of [tex]\( q \)[/tex].
- The contrapositive [tex]\( \sim q \rightarrow \sim p \)[/tex]:
- [tex]\( \sim q \)[/tex] is True ([tex]\(T\)[/tex]), and [tex]\( \sim p \)[/tex] is True ([tex]\(T\)[/tex]).
- [tex]\( \sim q \rightarrow \sim p \)[/tex] is True ([tex]\(T\)[/tex]) because True implies True is True.

So, the completed truth table is:

\begin{tabular}{|c|c|c|c|}
\hline [tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$\sim q \rightarrow \sim p$[/tex] \\
\hline T & T & T & T \\
\hline T & F & F & F \\
\hline F & T & T & T \\
\hline F & F & T & T \\
\hline
\end{tabular}