QUESTION IMAGE
Question
construct a truth table for the given statement.
(¬q ∧ p) ∨ (q ∧ p)
complete the truth table.
| p | q | ¬q | ¬q ∧ p | q ∧ p | (¬q ∧ p) ∨ (q ∧ p) |
| t | t | f | f | t | t |
| t | f | t | t | f | t |
| f | t | ▼ | ▼ | ▼ | ▼ |
Step1: Analyze $\sim q$ when $p = F, q = T$
The negation of $q$ (denoted as $\sim q$) is the opposite truth value of $q$. Since $q = T$, then $\sim q = F$.
Step2: Analyze $\sim q \wedge p$ when $p = F, q = T$
The conjunction ($\wedge$) is true only when both components are true. Here, $\sim q = F$ and $p = F$, so $\sim q \wedge p = F \wedge F = F$.
Step3: Analyze $q \wedge p$ when $p = F, q = T$
The conjunction of $q$ and $p$: $q = T$ and $p = F$, so $q \wedge p = T \wedge F = F$.
Step4: Analyze $(\sim q \wedge p) \vee (q \wedge p)$ when $p = F, q = T$
The disjunction ($\vee$) is true if at least one component is true. Here, $\sim q \wedge p = F$ and $q \wedge p = F$, so $(\sim q \wedge p) \vee (q \wedge p) = F \vee F = F$.
Now let's fill the table row by row for the case $p = F, q = T$:
- $\sim q$: $F$ (from Step1)
- $\sim q \wedge p$: $F$ (from Step2)
- $q \wedge p$: $F$ (from Step3)
- $(\sim q \wedge p) \vee (q \wedge p)$: $F$ (from Step4)
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For the row where $p = F$ and $q = T$:
- $\sim q$: $\boldsymbol{F}$
- $\sim q \wedge p$: $\boldsymbol{F}$
- $q \wedge p$: $\boldsymbol{F}$
- $(\sim q \wedge p) \vee (q \wedge p)$: $\boldsymbol{F}$
The completed row (third row) is:
| $p$ | $q$ | $\sim q$ | $\sim q \wedge p$ | $q \wedge p$ | $(\sim q \wedge p) \vee (q \wedge p)$ |
|---|