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Complete the truth table for the inverse of a conditional statement.

Type the correct answer in each box. Use T for true and F for false.

\begin{tabular}{|c|c|c|c|}
\hline
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$\sim p \rightarrow \sim q$[/tex] \\
\hline
T & T & T & \_\_\_ \\
\hline
T & F & F & \_\_\_ \\
\hline
F & T & T & \_\_\_ \\
\hline
F & F & T & \_\_\_ \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
Sure, let's complete the truth table step-by-step.

We need to fill in the column for the inverse of the conditional statement ([tex]\(\sim p \rightarrow \sim q\)[/tex]).

Here is the truth table format that we need to fill in:

\begin{tabular}{|c|c|c|c|}
\hline
p & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$\sim p \rightarrow \sim q$[/tex] \\
\hline
T & T & T & [ ] \\
\hline
T & F & F & [ ] \\
\hline
F & T & T & [ ] \\
\hline
F & F & T & [ ] \\
\hline
\end{tabular}

1. For the case when [tex]\(p\)[/tex] is True (T) and [tex]\(q\)[/tex] is True (T):
- [tex]\(\sim p\)[/tex] (not [tex]\(p\)[/tex]) is False (F) and [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is False (F)
- Thus, [tex]\(\sim p \rightarrow \sim q\)[/tex] is True (T)

2. For the case when [tex]\(p\)[/tex] is True (T) and [tex]\(q\)[/tex] is False (F):
- [tex]\(\sim p\)[/tex] (not [tex]\(p\)[/tex]) is False (F) and [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is True (T)
- Thus, [tex]\(\sim p \rightarrow \sim q\)[/tex] is True (T)

3. For the case when [tex]\(p\)[/tex] is False (F) and [tex]\(q\)[/tex] is True (T):
- [tex]\(\sim p\)[/tex] (not [tex]\(p\)[/tex]) is True (T) and [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is False (F)
- Thus, [tex]\(\sim p \rightarrow \sim q\)[/tex] is False (F)

4. For the case when [tex]\(p\)[/tex] is False (F) and [tex]\(q\)[/tex] is False (F):
- [tex]\(\sim p\)[/tex] (not [tex]\(p\)[/tex]) is True (T) and [tex]\(\sim q\)[/tex] (not [tex]\(q\)[/tex]) is True (T)
- Thus, [tex]\(\sim p \rightarrow \sim q\)[/tex] is True (T)

So, the completed truth table is:

\begin{tabular}{|c|c|c|c|}
\hline
p & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$\sim p \rightarrow \sim q$[/tex] \\
\hline
T & T & T & T \\
\hline
T & F & F & T \\
\hline
F & T & T & F \\
\hline
F & F & T & T \\
\hline
\end{tabular}

Therefore, filling in the blanks we get:

1. For [tex]$p = T$[/tex] and [tex]$q = T$[/tex], [tex]\(\sim p \rightarrow \sim q\)[/tex] = T
2. For [tex]$p = T$[/tex] and [tex]$q = F$[/tex], [tex]\(\sim p \rightarrow \sim q\)[/tex] = T
3. For [tex]$p = F$[/tex] and [tex]$q = T$[/tex], [tex]\(\sim p \rightarrow \sim q\)[/tex] = F
4. For [tex]$p = F$[/tex] and [tex]$q = F$[/tex], [tex]\(\sim p \rightarrow \sim q\)[/tex] = T
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