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Sagot :
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.
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.
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