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Sagot :
To determine the correct truth values for the expression [tex]\(\sim( p \rightarrow q ) \rightarrow( p \wedge \sim q)\)[/tex], we need to create a truth table and evaluate each combination of [tex]\(p\)[/tex] and [tex]\(q\)[/tex].
Let's examine each row step by step:
1. For [tex]\(p = T\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(T \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(T \wedge \sim T\)[/tex], which is [tex]\(T \wedge F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
2. For [tex]\(p = T\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(T \rightarrow F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim F\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(T \wedge \sim F\)[/tex], which is [tex]\(T \wedge T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(T \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
3. For [tex]\(p = F\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(F \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(F \wedge \sim T\)[/tex], which is [tex]\(F \wedge F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
4. For [tex]\(p = F\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(F \wedge \sim F\)[/tex], which is [tex]\(F \wedge T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
Given these evaluations, the correct statement for the truth table should be:
\begin{tabular}{ccc}
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$\sim( p \rightarrow q ) \rightarrow( p \wedge \sim q)$[/tex] \\
\hline
[tex]$T$[/tex] & [tex]$T$[/tex] & [tex]$F$[/tex] \\
[tex]$T$[/tex] & [tex]$F$[/tex] & [tex]$T$[/tex] \\
[tex]$F$[/tex] & [tex]$T$[/tex] & [tex]$F$[/tex] \\
[tex]$F$[/tex] & [tex]$F$[/tex] & [tex]$F$[/tex] \\
\end{tabular}
Thus, the answer is:
[tex]\[(T, T, F), (T, F, T), (F, T, F), (F, F, F)\][/tex]
Let's examine each row step by step:
1. For [tex]\(p = T\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(T \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(T \wedge \sim T\)[/tex], which is [tex]\(T \wedge F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
2. For [tex]\(p = T\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(T \rightarrow F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim F\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(T \wedge \sim F\)[/tex], which is [tex]\(T \wedge T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(T \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
3. For [tex]\(p = F\)[/tex] and [tex]\(q = T\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(F \rightarrow T\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(F \wedge \sim T\)[/tex], which is [tex]\(F \wedge F\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
4. For [tex]\(p = F\)[/tex] and [tex]\(q = F\)[/tex]:
- [tex]\(p \rightarrow q\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
- [tex]\(\sim (p \rightarrow q)\)[/tex] is [tex]\(\sim T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(p \wedge \sim q\)[/tex] is [tex]\(F \wedge \sim F\)[/tex], which is [tex]\(F \wedge T\)[/tex], which is [tex]\(F\)[/tex].
- [tex]\(\sim (p \rightarrow q) \rightarrow (p \wedge \sim q)\)[/tex] is [tex]\(F \rightarrow F\)[/tex], which is [tex]\(T\)[/tex].
Given these evaluations, the correct statement for the truth table should be:
\begin{tabular}{ccc}
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$\sim( p \rightarrow q ) \rightarrow( p \wedge \sim q)$[/tex] \\
\hline
[tex]$T$[/tex] & [tex]$T$[/tex] & [tex]$F$[/tex] \\
[tex]$T$[/tex] & [tex]$F$[/tex] & [tex]$T$[/tex] \\
[tex]$F$[/tex] & [tex]$T$[/tex] & [tex]$F$[/tex] \\
[tex]$F$[/tex] & [tex]$F$[/tex] & [tex]$F$[/tex] \\
\end{tabular}
Thus, the answer is:
[tex]\[(T, T, F), (T, F, T), (F, T, F), (F, F, F)\][/tex]
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