IDNLearn.com is your go-to resource for finding expert answers and community support. Discover in-depth and trustworthy answers from our extensive network of knowledgeable professionals.
Sagot :
Sure, let's prove the logical equivalence [tex]\(( p \wedge q ) = \sim ( p \rightarrow \sim q )\)[/tex] using a truth table.
To do this, we will construct a truth table that includes columns for [tex]\(p\)[/tex], [tex]\(q\)[/tex], [tex]\(p \wedge q\)[/tex], [tex]\(\sim q\)[/tex], [tex]\(p \rightarrow \sim q\)[/tex], and [tex]\(\sim ( p \rightarrow \sim q )\)[/tex].
Here's the detailed step-by-step solution:
1. List the possible truth values for [tex]\(p\)[/tex] and [tex]\(q\)[/tex]:
There are four combinations of truth values for [tex]\(p\)[/tex] and [tex]\(q\)[/tex]:
- [tex]\(p = \text{False}, q = \text{False}\)[/tex]
- [tex]\(p = \text{False}, q = \text{True}\)[/tex]
- [tex]\(p = \text{True}, q = \text{False}\)[/tex]
- [tex]\(p = \text{True}, q = \text{True}\)[/tex]
2. Calculate [tex]\(p \wedge q\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & p \wedge q \\ \hline \text{False} & \text{False} & \text{False} \\ \text{False} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{False} \\ \text{True} & \text{True} & \text{True} \\ \hline \end{array} \][/tex]
3. Calculate [tex]\(\sim q\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & \sim q \\ \hline \text{False} & \text{False} & \text{True} \\ \text{False} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{True} \\ \text{True} & \text{True} & \text{False} \\ \hline \end{array} \][/tex]
4. Calculate [tex]\(p \rightarrow \sim q\)[/tex]:
Recall that [tex]\(p \rightarrow \sim q\)[/tex] is equivalent to [tex]\(\neg p \lor \sim q\)[/tex].
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & \sim q & p \rightarrow \sim q \\ \hline \text{False} & \text{False} & \text{True} & \text{True} \\ \text{False} & \text{True} & \text{False} & \text{True} \\ \text{True} & \text{False} & \text{True} & \text{True} \\ \text{True} & \text{True} & \text{False} & \text{False} \\ \hline \end{array} \][/tex]
5. Calculate [tex]\(\sim ( p \rightarrow \sim q )\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & p \rightarrow \sim q & \sim ( p \rightarrow \sim q ) \\ \hline \text{False} & \text{False} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{True} & \text{False} \\ \text{True} & \text{True} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
6. Compare [tex]\(p \wedge q\)[/tex] with [tex]\(\sim ( p \rightarrow \sim q )\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & p \wedge q & \sim ( p \rightarrow \sim q ) \\ \hline \text{False} & \text{False} & \text{False} & \text{False} \\ \text{False} & \text{True} & \text{False} & \text{False} \\ \text{True} & \text{False} & \text{False} & \text{False} \\ \text{True} & \text{True} & \text{True} & \text{True} \\ \hline \end{array} \][/tex]
From the table, we can see that the truth values for [tex]\(p \wedge q\)[/tex] and [tex]\(\sim ( p \rightarrow \sim q )\)[/tex] are always identical for all possible combinations of [tex]\(p\)[/tex] and [tex]\(q\)[/tex]. Hence, it is proven through the truth table that:
[tex]\[ ( p \wedge q ) = \sim ( p \rightarrow \sim q ) \][/tex]
This completes the proof of the logical equivalence.
To do this, we will construct a truth table that includes columns for [tex]\(p\)[/tex], [tex]\(q\)[/tex], [tex]\(p \wedge q\)[/tex], [tex]\(\sim q\)[/tex], [tex]\(p \rightarrow \sim q\)[/tex], and [tex]\(\sim ( p \rightarrow \sim q )\)[/tex].
Here's the detailed step-by-step solution:
1. List the possible truth values for [tex]\(p\)[/tex] and [tex]\(q\)[/tex]:
There are four combinations of truth values for [tex]\(p\)[/tex] and [tex]\(q\)[/tex]:
- [tex]\(p = \text{False}, q = \text{False}\)[/tex]
- [tex]\(p = \text{False}, q = \text{True}\)[/tex]
- [tex]\(p = \text{True}, q = \text{False}\)[/tex]
- [tex]\(p = \text{True}, q = \text{True}\)[/tex]
2. Calculate [tex]\(p \wedge q\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & p \wedge q \\ \hline \text{False} & \text{False} & \text{False} \\ \text{False} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{False} \\ \text{True} & \text{True} & \text{True} \\ \hline \end{array} \][/tex]
3. Calculate [tex]\(\sim q\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & \sim q \\ \hline \text{False} & \text{False} & \text{True} \\ \text{False} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{True} \\ \text{True} & \text{True} & \text{False} \\ \hline \end{array} \][/tex]
4. Calculate [tex]\(p \rightarrow \sim q\)[/tex]:
Recall that [tex]\(p \rightarrow \sim q\)[/tex] is equivalent to [tex]\(\neg p \lor \sim q\)[/tex].
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & \sim q & p \rightarrow \sim q \\ \hline \text{False} & \text{False} & \text{True} & \text{True} \\ \text{False} & \text{True} & \text{False} & \text{True} \\ \text{True} & \text{False} & \text{True} & \text{True} \\ \text{True} & \text{True} & \text{False} & \text{False} \\ \hline \end{array} \][/tex]
5. Calculate [tex]\(\sim ( p \rightarrow \sim q )\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & p \rightarrow \sim q & \sim ( p \rightarrow \sim q ) \\ \hline \text{False} & \text{False} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{False} \\ \text{True} & \text{False} & \text{True} & \text{False} \\ \text{True} & \text{True} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
6. Compare [tex]\(p \wedge q\)[/tex] with [tex]\(\sim ( p \rightarrow \sim q )\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|} \hline p & q & p \wedge q & \sim ( p \rightarrow \sim q ) \\ \hline \text{False} & \text{False} & \text{False} & \text{False} \\ \text{False} & \text{True} & \text{False} & \text{False} \\ \text{True} & \text{False} & \text{False} & \text{False} \\ \text{True} & \text{True} & \text{True} & \text{True} \\ \hline \end{array} \][/tex]
From the table, we can see that the truth values for [tex]\(p \wedge q\)[/tex] and [tex]\(\sim ( p \rightarrow \sim q )\)[/tex] are always identical for all possible combinations of [tex]\(p\)[/tex] and [tex]\(q\)[/tex]. Hence, it is proven through the truth table that:
[tex]\[ ( p \wedge q ) = \sim ( p \rightarrow \sim q ) \][/tex]
This completes the proof of the logical equivalence.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.
