Answer :
To determine the correct truth table for the implication [tex]\( q \rightarrow p \)[/tex] (which reads "if [tex]\( q \)[/tex] then [tex]\( p \)[/tex]"), we need to understand how this logical connective behaves. The implication [tex]\( q \rightarrow p \)[/tex] is only false if [tex]\( q \)[/tex] is true and [tex]\( p \)[/tex] is false; in all other cases, it is true.
Let's go through each possible combination of truth values for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
1. When [tex]\( p \)[/tex] is true ([tex]\( T \)[/tex]) and [tex]\( q \)[/tex] is true ([tex]\( T \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because both [tex]\( q \)[/tex] and [tex]\( p \)[/tex] are true.
2. When [tex]\( p \)[/tex] is true ([tex]\( T \)[/tex]) and [tex]\( q \)[/tex] is false ([tex]\( F \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because if [tex]\( q \)[/tex] is false, the implication is always true regardless of [tex]\( p \)[/tex]'s value.
3. When [tex]\( p \)[/tex] is false ([tex]\( F \)[/tex]) and [tex]\( q \)[/tex] is true ([tex]\( T \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is false because [tex]\( q \)[/tex] being true should lead to [tex]\( p \)[/tex] being true, but [tex]\( p \)[/tex] is false.
4. When [tex]\( p \)[/tex] is false ([tex]\( F \)[/tex]) and [tex]\( q \)[/tex] is false ([tex]\( F \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because if [tex]\( q \)[/tex] is false, the implication is always true regardless of [tex]\( p \)[/tex]'s value.
So, the correct truth table for the statement [tex]\( q \rightarrow p \)[/tex] is:
[tex]\[ \begin{tabular}{|c|c|c|} \hline p & q & q \rightarrow p \\ \hline T & T & T \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{tabular} \][/tex]
Therefore, the correct truth table matches the second table provided:
[tex]\[ \begin{tabular}{|c|c|c|} \hline p & q & q \rightarrow p \\ \hline T & T & T \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{tabular} \][/tex]
Let's go through each possible combination of truth values for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
1. When [tex]\( p \)[/tex] is true ([tex]\( T \)[/tex]) and [tex]\( q \)[/tex] is true ([tex]\( T \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because both [tex]\( q \)[/tex] and [tex]\( p \)[/tex] are true.
2. When [tex]\( p \)[/tex] is true ([tex]\( T \)[/tex]) and [tex]\( q \)[/tex] is false ([tex]\( F \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because if [tex]\( q \)[/tex] is false, the implication is always true regardless of [tex]\( p \)[/tex]'s value.
3. When [tex]\( p \)[/tex] is false ([tex]\( F \)[/tex]) and [tex]\( q \)[/tex] is true ([tex]\( T \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is false because [tex]\( q \)[/tex] being true should lead to [tex]\( p \)[/tex] being true, but [tex]\( p \)[/tex] is false.
4. When [tex]\( p \)[/tex] is false ([tex]\( F \)[/tex]) and [tex]\( q \)[/tex] is false ([tex]\( F \)[/tex]):
- [tex]\( q \rightarrow p \)[/tex] is true because if [tex]\( q \)[/tex] is false, the implication is always true regardless of [tex]\( p \)[/tex]'s value.
So, the correct truth table for the statement [tex]\( q \rightarrow p \)[/tex] is:
[tex]\[ \begin{tabular}{|c|c|c|} \hline p & q & q \rightarrow p \\ \hline T & T & T \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{tabular} \][/tex]
Therefore, the correct truth table matches the second table provided:
[tex]\[ \begin{tabular}{|c|c|c|} \hline p & q & q \rightarrow p \\ \hline T & T & T \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{tabular} \][/tex]
