To show that the compound statements [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] are equivalent, we will construct and analyze the corresponding truth tables. This process involves evaluating the truth values for each compound statement under all possible truth values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex].
Let's start with the truth table for the first compound statement [tex]\(p \vee (q \wedge \sim p)\)[/tex]:
[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
p & q & \sim p & q \wedge \sim p & p \vee (q \wedge \sim p) \\
\hline
T & T & F & F & T \\
T & F & F & F & T \\
F & T & T & T & T \\
F & F & T & F & F \\
\hline
\end{array}
\][/tex]
Explanation:
- [tex]\(p\)[/tex] and [tex]\(q\)[/tex] represent the possible truth values.
- [tex]\(\sim p\)[/tex] is the negation of [tex]\(p\)[/tex].
- [tex]\(q \wedge \sim p\)[/tex] is the conjunction of [tex]\(q\)[/tex] and [tex]\(\sim p\)[/tex].
- [tex]\(p \vee (q \wedge \sim p)\)[/tex] is the disjunction of [tex]\(p\)[/tex] and [tex]\(q \wedge \sim p\)[/tex].
Next, let's build the truth table for the second compound statement [tex]\(p \vee q\)[/tex]:
[tex]\[
\begin{array}{|c|c|c|}
\hline
p & q & p \vee q \\
\hline
T & T & T \\
T & F & T \\
F & T & T \\
F & F & F \\
\hline
\end{array}
\][/tex]
Now, we compare the results of both truth tables:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
p & q & p \vee (q \wedge \sim p) & p \vee q \\
\hline
T & T & T & T \\
T & F & T & T \\
F & T & T & T \\
F & F & F & F \\
\hline
\end{array}
\][/tex]
As we can see:
- When [tex]\(p\)[/tex] is True and [tex]\(q\)[/tex] is True, both [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] evaluate to True.
- When [tex]\(p\)[/tex] is True and [tex]\(q\)[/tex] is False, both [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] evaluate to True.
- When [tex]\(p\)[/tex] is False and [tex]\(q\)[/tex] is True, both [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] evaluate to True.
- When [tex]\(p\)[/tex] is False and [tex]\(q\)[/tex] is False, both [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] evaluate to False.
Since the corresponding truth values of [tex]\(p \vee (q \wedge \sim p)\)[/tex] and [tex]\(p \vee q\)[/tex] match for all possible truth values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex], the compound statements are equivalent.
