Construct a truth table for the given statement:

[tex]\[ \sim d \rightarrow \sim c \][/tex]

Fill in the truth table.
[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$d$ & $c$ & $\sim d$ & $\sim c$ & $\sim d \rightarrow \sim c$ \\
\hline
T & T & F & F & T \\
\hline
T & F & F & T & T \\
\hline
F & T & T & F & F \\
\hline
F & F & T & T & T \\
\hline
\end{tabular}
\][/tex]



Answer :

Sure, let's fill in the truth table step-by-step for the statement [tex]\(\sim d \rightarrow \sim c\)[/tex]:

1. List out all possible truth values for [tex]\(d\)[/tex] and [tex]\(c\)[/tex]:
- [tex]\(d = T\)[/tex], [tex]\(c = T\)[/tex]
- [tex]\(d = T\)[/tex], [tex]\(c = F\)[/tex]
- [tex]\(d = F\)[/tex], [tex]\(c = T\)[/tex]
- [tex]\(d = F\)[/tex], [tex]\(c = F\)[/tex]

2. Determine the values for [tex]\(\sim d\)[/tex] (not [tex]\(d\)[/tex]) and [tex]\(\sim c\)[/tex] (not [tex]\(c\)[/tex]) for each combination.

3. Evaluate the implication [tex]\(\sim d \rightarrow \sim c\)[/tex].

Here's the filled truth table:

[tex]\[ \renewcommand\arraystretch{1.5} \begin{array}{|c|c|c|c|c|} \hline d & c & \sim d & \sim c & \sim d \rightarrow \sim c \\ \hline T & T & F & F & T \\ \hline T & F & F & T & T \\ \hline F & T & T & F & F \\ \hline F & F & T & T & T \\ \hline \end{array} \][/tex]

Explanation for each row:

1. [tex]\(d = T, c = T\)[/tex]:
- [tex]\(\sim d = F\)[/tex]
- [tex]\(\sim c = F\)[/tex]
- [tex]\(\sim d \rightarrow \sim c = F \rightarrow F\)[/tex]: This is true because an implication is true if the antecedent ([tex]\(\sim d\)[/tex]) is false or both the antecedent and the consequent ([tex]\(\sim c\)[/tex]) are false.

2. [tex]\(d = T, c = F\)[/tex]:
- [tex]\(\sim d = F\)[/tex]
- [tex]\(\sim c = T\)[/tex]
- [tex]\(\sim d \rightarrow \sim c = F \rightarrow T\)[/tex]: This is true because an implication is true if the antecedent ([tex]\(\sim d\)[/tex]) is false.

3. [tex]\(d = F, c = T\)[/tex]:
- [tex]\(\sim d = T\)[/tex]
- [tex]\(\sim c = F\)[/tex]
- [tex]\(\sim d \rightarrow \sim c = T \rightarrow F\)[/tex]: This is false because an implication is false if the antecedent ([tex]\(\sim d\)[/tex]) is true and the consequent ([tex]\(\sim c\)[/tex]) is false.

4. [tex]\(d = F, c = F\)[/tex]:
- [tex]\(\sim d = T\)[/tex]
- [tex]\(\sim c = T\)[/tex]
- [tex]\(\sim d \rightarrow \sim c = T \rightarrow T\)[/tex]: This is true because an implication is true if both the antecedent ([tex]\(\sim d\)[/tex]) and the consequent ([tex]\(\sim c\)[/tex]) are true.

And thus, the final truth table is as shown.