Questions: Complete the row of each truth table. Use T for true and F for false. p q r r ∧ p q → (r ∧ p) --------------- T T F a b c ~b c → ~b b → a (c → ~b) ∧ (b → a) --------------------- T F T

Transcript text: Complete the row of each truth table. Use $T$ for true and $F$ for false. \begin{tabular}{|c|c|c|c|c|} \hline$p$ & $q$ & $r$ & $r \wedge p$ & $q \rightarrow(r \wedge p)$ \\ \hline T & T & $\square$ & F & $\square$ \\ \hline \end{tabular} \begin{tabular}{|c|c|c|c|c|c|c|} \hline$a$ & $b$ & $c$ & $\sim b$ & $c \rightarrow \sim b$ & $b \rightarrow a$ & $(c \rightarrow \sim b) \wedge(b \rightarrow a)$ \\ \hline T & F & T & $\square$ & $\square$ & $\square$ & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

To complete the truth tables, we need to evaluate the logical expressions for each row based on the given values of the variables. For the first table, determine the value of $r$ such that $r \wedge p = F$ , and then evaluate $q \rightarrow (r \wedge p)$ . For the second table, evaluate each logical expression step by step using the given values of $a$ , $b$ , and $c$ .

Step 1: Evaluate the First Truth Table

For the first truth table, we have:

$p = T$
$q = T$
To satisfy $r \wedge p = F$ , we must have $r = F$ .
Therefore, $r \wedge p = F$ results in $F$ .
Next, we evaluate $q \rightarrow (r \wedge p)$ : $q \rightarrow (r \wedge p) = T \rightarrow F = F$

Step 2: Evaluate the Second Truth Table

For the second truth table, we have:

$a = T$
$b = F$
$c = T$
We calculate $\sim b$ : $\sim b = T$
Next, we evaluate $c \rightarrow \sim b$ : $c \rightarrow \sim b = T \rightarrow T = T$
Then, we evaluate $b \rightarrow a$ : $b \rightarrow a = F \rightarrow T = T$
Finally, we compute $(c \rightarrow \sim b) \wedge (b \rightarrow a)$ : $(c \rightarrow \sim b) \wedge (b \rightarrow a) = T \wedge T = T$

Final Answer

The completed truth tables yield the following results:

For the first table: $r = F$ and $q \rightarrow (r \wedge p) = F$ .
For the second table: $\sim b = T$ , $c \rightarrow \sim b = T$ , $b \rightarrow a = T$ , and $(c \rightarrow \sim b) \wedge (b \rightarrow a) = T$ .

Thus, the final answers are:

First table: $r = F$ , $q \rightarrow (r \wedge p) = F$
Second table: $\sim b = T$ , $c \rightarrow \sim b = T$ , $b \rightarrow a = T$ , $(c \rightarrow \sim b) \wedge (b \rightarrow a) = T$

$\boxed{(r = F, q \rightarrow (r \wedge p) = F, \sim b = T, c \rightarrow \sim b = T, b \rightarrow a = T, (c \rightarrow \sim b) \wedge (b \rightarrow a) = T)}$

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