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Sagot :
To solve the truth table values for the inverse of a conditional statement, we need to evaluate each case:
The inverse of [tex]\( p \rightarrow q \)[/tex] is [tex]\( \sim p \rightarrow \sim q \)[/tex].
Consider the given values:
- [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are the basic statements.
- [tex]\( p \rightarrow q \)[/tex] is the conditional statement "if [tex]\( p \)[/tex], then [tex]\( q \)[/tex]".
- [tex]\( \sim p \rightarrow \sim q \)[/tex] is the inverse "if not [tex]\( p \)[/tex], then not [tex]\( q \)[/tex]".
Let's fill in the missing values for each case:
1. For [tex]\( p = T \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is False because when [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] fails.
- [tex]\( \sim p \)[/tex] is False (negation of True), and [tex]\( \sim q \)[/tex] is True (negation of False).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to True because [tex]\( \sim p \)[/tex] False implies [tex]\( \sim q \)[/tex] True.
2. For [tex]\( p = F \)[/tex] and [tex]\( q = T \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True because when [tex]\( p \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] is always True.
- [tex]\( \sim p \)[/tex] is True (negation of False), and [tex]\( \sim q \)[/tex] is False (negation of True).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to False because [tex]\( \sim p \)[/tex] True does not imply [tex]\( \sim q \)[/tex] False.
3. For [tex]\( p = F \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True because when [tex]\( p \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] is always True.
- [tex]\( \sim p \)[/tex] is True (negation of False), and [tex]\( \sim q \)[/tex] is True (negation of False).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to True because [tex]\( \sim p \)[/tex] True implies [tex]\( \sim q \)[/tex] True.
Therefore, the completed truth table is as follows:
\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 \hline [tex]$T$[/tex] & [tex]$T$[/tex] & [tex]$T$[/tex] & F \\
\hline \hline [tex]$T$[/tex] & [tex]$F$[/tex] & [tex]$F$[/tex] & T \\
\hline \hline [tex]$F$[/tex] & [tex]$T$[/tex] & T & F \\
\hline \hline [tex]$F$[/tex] & [tex]$F$[/tex] & T & T \\
\hline
\end{tabular}
The inverse of [tex]\( p \rightarrow q \)[/tex] is [tex]\( \sim p \rightarrow \sim q \)[/tex].
Consider the given values:
- [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are the basic statements.
- [tex]\( p \rightarrow q \)[/tex] is the conditional statement "if [tex]\( p \)[/tex], then [tex]\( q \)[/tex]".
- [tex]\( \sim p \rightarrow \sim q \)[/tex] is the inverse "if not [tex]\( p \)[/tex], then not [tex]\( q \)[/tex]".
Let's fill in the missing values for each case:
1. For [tex]\( p = T \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is False because when [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] fails.
- [tex]\( \sim p \)[/tex] is False (negation of True), and [tex]\( \sim q \)[/tex] is True (negation of False).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to True because [tex]\( \sim p \)[/tex] False implies [tex]\( \sim q \)[/tex] True.
2. For [tex]\( p = F \)[/tex] and [tex]\( q = T \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True because when [tex]\( p \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] is always True.
- [tex]\( \sim p \)[/tex] is True (negation of False), and [tex]\( \sim q \)[/tex] is False (negation of True).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to False because [tex]\( \sim p \)[/tex] True does not imply [tex]\( \sim q \)[/tex] False.
3. For [tex]\( p = F \)[/tex] and [tex]\( q = F \)[/tex]:
- [tex]\( p \rightarrow q \)[/tex] is True because when [tex]\( p \)[/tex] is False, the implication [tex]\( p \rightarrow q \)[/tex] is always True.
- [tex]\( \sim p \)[/tex] is True (negation of False), and [tex]\( \sim q \)[/tex] is True (negation of False).
- [tex]\( \sim p \rightarrow \sim q \)[/tex] evaluates to True because [tex]\( \sim p \)[/tex] True implies [tex]\( \sim q \)[/tex] True.
Therefore, the completed truth table is as follows:
\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 \hline [tex]$T$[/tex] & [tex]$T$[/tex] & [tex]$T$[/tex] & F \\
\hline \hline [tex]$T$[/tex] & [tex]$F$[/tex] & [tex]$F$[/tex] & T \\
\hline \hline [tex]$F$[/tex] & [tex]$T$[/tex] & T & F \\
\hline \hline [tex]$F$[/tex] & [tex]$F$[/tex] & T & T \\
\hline
\end{tabular}
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