Sure, I can guide you step-by-step to construct the truth table for the compound statement \((q \wedge \sim p) \vee \sim q\).
Here is the step-by-step construction of the truth table:
1. Identify the possible values for \(p\) and \(q\).
2. Compute \(\sim p\): Negation of \(p\).
3. Compute \(q \wedge \sim p\): Logical AND between \(q\) and \(\sim p\).
4. Compute \(\sim q\): Negation of \(q\).
5. Compute \((q \wedge \sim p) \vee \sim q\): Logical OR between \(q \wedge \sim p\) and \(\sim q\).
Let's fill this in step-by-step:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
p & q & \sim p & q \wedge \sim p & \sim q & (q \wedge \sim p) \vee \sim q \\
\hline
T & T & F & F & F & F \\
T & F & F & F & T & T \\
F & T & T & T & F & T \\
F & F & T & F & T & T \\
\hline
\end{array}
\][/tex]
Here's a breakdown of the row-wise calculations:
- First Row \((p = T, q = T)\):
- \(\sim p\): \(F\)
- \(q \wedge \sim p\): \(T \wedge F = F\)
- \(\sim q\): \(F\)
- \((q \wedge \sim p) \vee \sim q\): \(F \vee F = F\)
- Second Row \((p = T, q = F)\):
- \(\sim p\): \(F\)
- \(q \wedge \sim p\): \(F \wedge F = F\)
- \(\sim q\): \(T\)
- \((q \wedge \sim p) \vee \sim q\): \(F \vee T = T\)
- Third Row \((p = F, q = T)\):
- \(\sim p\): \(T\)
- \(q \wedge \sim p\): \(T \wedge T = T\)
- \(\sim q\): \(F\)
- \((q \wedge \sim p) \vee \sim q\): \(T \vee F = T\)
- Fourth Row \((p = F, q = F)\):
- \(\sim p\): \(T\)
- \(q \wedge \sim p\): \(F \wedge T = F\)
- \(\sim q\): \(T\)
- \((q \wedge \sim p) \vee \sim q\): \(F \vee T = T\)
Thus, the complete truth table for the given compound statement \((q \wedge \sim p) \vee \sim q\) is:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
p & q & \sim p & q \wedge \sim p & \sim q & (q \wedge \sim p) \vee \sim q \\
\hline
T & T & F & F & F & F \\
T & F & F & F & T & T \\
F & T & T & T & F & T \\
F & F & T & F & T & T \\
\hline
\end{array}
\][/tex]
