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Let's solve the logical expression [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex] using the given truth values for [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & \square \\ \hline T & F & \square \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
For each combination of [tex]\( p \)[/tex] and [tex]\( q \)[/tex], we evaluate the expression step by step.
1. [tex]\( p = T \)[/tex], [tex]\( q = T \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is True, [tex]\( \sim q \)[/tex] is False.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is True and [tex]\( \sim q \)[/tex] is False, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is True and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is True and [tex]\( r \)[/tex] is False, the result is False.
- Therefore, the result is False.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & \square \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
2. [tex]\( p = T \)[/tex], [tex]\( q = F \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is True and [tex]\( \sim q \)[/tex] is True, so [tex]\( p \wedge \sim q \)[/tex] is True.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is False and [tex]\( p \wedge \sim q \)[/tex] is True. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is False, the result is always True regardless of the value of [tex]\( r \)[/tex].
- Therefore, the result is True.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
3. [tex]\( p = F \)[/tex], [tex]\( q = T \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is True, [tex]\( \sim q \)[/tex] is False.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is False and [tex]\( \sim q \)[/tex] is False, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is True and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is True and [tex]\( r \)[/tex] is False, the result is False.
- Therefore, the result is False.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
4. [tex]\( p = F \)[/tex], [tex]\( q = F \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is False and [tex]\( \sim q \)[/tex] is True, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is False and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is False, the result is always True regardless of the value of [tex]\( r \)[/tex].
- Therefore, the result is True.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{array} \][/tex]
So, the complete truth table for [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex] is:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{array} \][/tex]
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & \square \\ \hline T & F & \square \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
For each combination of [tex]\( p \)[/tex] and [tex]\( q \)[/tex], we evaluate the expression step by step.
1. [tex]\( p = T \)[/tex], [tex]\( q = T \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is True, [tex]\( \sim q \)[/tex] is False.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is True and [tex]\( \sim q \)[/tex] is False, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is True and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is True and [tex]\( r \)[/tex] is False, the result is False.
- Therefore, the result is False.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & \square \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
2. [tex]\( p = T \)[/tex], [tex]\( q = F \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is True and [tex]\( \sim q \)[/tex] is True, so [tex]\( p \wedge \sim q \)[/tex] is True.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is False and [tex]\( p \wedge \sim q \)[/tex] is True. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is False, the result is always True regardless of the value of [tex]\( r \)[/tex].
- Therefore, the result is True.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & \square \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
3. [tex]\( p = F \)[/tex], [tex]\( q = T \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is True, [tex]\( \sim q \)[/tex] is False.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is False and [tex]\( \sim q \)[/tex] is False, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is True and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is True and [tex]\( r \)[/tex] is False, the result is False.
- Therefore, the result is False.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & \square \\ \hline \end{array} \][/tex]
4. [tex]\( p = F \)[/tex], [tex]\( q = F \)[/tex]:
- First, we find [tex]\( \sim q \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True.
- Next, evaluate [tex]\( p \wedge \sim q \)[/tex]. [tex]\( p \)[/tex] is False and [tex]\( \sim q \)[/tex] is True, so [tex]\( p \wedge \sim q \)[/tex] is False.
- Now, [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex]. Here, [tex]\( q \)[/tex] is False and [tex]\( p \wedge \sim q \)[/tex] is False. In an implication [tex]\( q \rightarrow r \)[/tex], if [tex]\( q \)[/tex] is False, the result is always True regardless of the value of [tex]\( r \)[/tex].
- Therefore, the result is True.
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{array} \][/tex]
So, the complete truth table for [tex]\( q \rightarrow (p \wedge \sim q) \)[/tex] is:
[tex]\[ \begin{array}{|c|c|c|} \hline p & q & q \rightarrow (p \wedge \sim q) \\ \hline T & T & F \\ \hline T & F & T \\ \hline F & T & F \\ \hline F & F & T \\ \hline \end{array} \][/tex]
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