IDNLearn.com: Your one-stop destination for reliable answers to diverse questions. Our Q&A platform is designed to provide quick and accurate answers to any questions you may have.
Sagot :
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]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
