To complete the truth table for the logical statement [tex]\(\sim q \leftrightarrow p\)[/tex], we need to determine the values for [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) and [tex]\(\sim q \leftrightarrow p\)[/tex] (not [tex]\(q\)[/tex] if and only if [tex]\(p\)[/tex]) for each combination of truth values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex].
Let's analyze each combination of [tex]\(p\)[/tex] and [tex]\(q\)[/tex] step-by-step:
### Step-by-Step Analysis:
1. When [tex]\(p = T\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is [tex]\(F\)[/tex] because [tex]\(q\)[/tex] is [tex]\(T\)[/tex].
- [tex]\(\sim q \leftrightarrow p\)[/tex] is [tex]\(F\)[/tex] because [tex]\(F\)[/tex] (not [tex]\(q\)[/tex]) is not equal to [tex]\(T\)[/tex] (p).
2. When [tex]\(p = T\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is [tex]\(T\)[/tex] because [tex]\(q\)[/tex] is [tex]\(F\)[/tex].
- [tex]\(\sim q \leftrightarrow p\)[/tex] is [tex]\(T\)[/tex] because [tex]\(T\)[/tex] (not [tex]\(q\)[/tex]) is equal to [tex]\(T\)[/tex] (p).
3. When [tex]\(p = F\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is [tex]\(F\)[/tex] because [tex]\(q\)[/tex] is [tex]\(T\)[/tex].
- [tex]\(\sim q \leftrightarrow p\)[/tex] is [tex]\(T\)[/tex] because [tex]\(F\)[/tex] (not [tex]\(q\)[/tex]) is equal to [tex]\(F\)[/tex] (p).
4. When [tex]\(p = F\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is [tex]\(T\)[/tex] because [tex]\(q\)[/tex] is [tex]\(F\)[/tex].
- [tex]\(\sim q \leftrightarrow p\)[/tex] is [tex]\(F\)[/tex] because [tex]\(T\)[/tex] (not [tex]\(q\)[/tex]) is not equal to [tex]\(F\)[/tex] (p).
Now, let us summarize our results in the truth table:
[tex]\[
\begin{array}{|c|c|c|c|}
\hline
p & q & \sim q & \sim q \leftrightarrow p \\
\hline
T & T & F & F \\
\hline
T & F & T & T \\
\hline
F & T & F & T \\
\hline
F & F & T & F \\
\hline
\end{array}
\][/tex]
This table corresponds to the correct answer.
