Answer :
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.
