Question

construct a truth table for the given compound statement.
$p \vee (\sim q \wedge r)$
complete the truth table.

p	q	r	$\sim q$	$\sim q \wedge r$	$p \vee (\sim q \wedge r)$
t	t	t	$\blacktriangledown$	$\blacktriangledown$	$\blacktriangledown$

Explanation:

Step1: Find $\sim q$

Given $q = T$, the negation of $T$ is $F$. So $\sim q = F$.

Step2: Find $\sim q \land r$

We know $\sim q = F$ and $r = T$. The conjunction ($\land$) of $F$ and $T$ is $F$ (since $F \land T = F$). So $\sim q \land r = F$.

Step3: Find $p \lor (\sim q \land r)$

Given $p = T$ and $\sim q \land r = F$. The disjunction ($\lor$) of $T$ and $F$ is $T$ (since $T \lor F = T$). So $p \lor (\sim q \land r) = T$.

Answer:

For the row where $p = T$, $q = T$, $r = T$:
$\sim q$: $F$
$\sim q \land r$: $F$
$p \lor (\sim q \land r)$: $T$

(If filling the table cells: the first dropdown (for $\sim q$) is $F$, the second (for $\sim q \land r$) is $F$, the third (for $p \lor (\sim q \land r)$) is $T$)