Sure, let's consider the given conditions. We have [tex]\( p = \text{True} \)[/tex] (T) and [tex]\( q = \text{False} \)[/tex] (F).
The expressions to evaluate are:
1. [tex]\( p \rightarrow q \)[/tex]
2. [tex]\( q \rightarrow p \)[/tex]
3. [tex]\( \sim p \rightarrow \sim q \)[/tex]
4. [tex]\( \sim q \rightarrow \sim p \)[/tex]
We apply the definitions of each logical implication:
- The implication [tex]\( p \rightarrow q \)[/tex] is equivalent to [tex]\( \sim p \lor q \)[/tex].
- The implication [tex]\( q \rightarrow p \)[/tex] is equivalent to [tex]\( \sim q \lor p \)[/tex].
- The implication [tex]\( \sim p \rightarrow \sim q \)[/tex] is equivalent to [tex]\( p \lor \sim q \)[/tex].
- The implication [tex]\( \sim q \rightarrow \sim p \)[/tex] is equivalent to [tex]\( q \lor \sim p \)[/tex].
Given that [tex]\( p = \text{True} \)[/tex] and [tex]\( q = \text{False} \)[/tex], we substitute these values into the above expressions:
1. [tex]\( p \rightarrow q \)[/tex]: This is [tex]\( \sim p \lor q \)[/tex]. Since [tex]\( p \)[/tex] is True, [tex]\( \sim p \)[/tex] is False. Thus, [tex]\( \sim p \lor q \)[/tex] becomes False [tex]\(\lor\)[/tex] False, which is False. So, [tex]\( p \rightarrow q \)[/tex] is False.
2. [tex]\( q \rightarrow p \)[/tex]: This is [tex]\( \sim q \lor p \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True. Thus, [tex]\( \sim q \lor p \)[/tex] becomes True [tex]\(\lor\)[/tex] True, which is True. So, [tex]\( q \rightarrow p \)[/tex] is True.
3. [tex]\( \sim p \rightarrow \sim q \)[/tex]: This is [tex]\( p \lor \sim q \)[/tex]. Since [tex]\( q \)[/tex] is False, [tex]\( \sim q \)[/tex] is True. Thus, [tex]\( p \lor \sim q \)[/tex] becomes True [tex]\(\lor\)[/tex] True, which is True. So, [tex]\( \sim p \rightarrow \sim q \)[/tex] is True.
4. [tex]\( \sim q \rightarrow \sim p \)[/tex]: This is [tex]\( q \lor \sim p \)[/tex]. Since [tex]\( p \)[/tex] is True, [tex]\( \sim p \)[/tex] is False. Thus, [tex]\( q \lor \sim p \)[/tex] becomes False [tex]\(\lor\)[/tex] False, which is False. So, [tex]\( \sim q \rightarrow \sim p \)[/tex] is False.
Therefore, we can complete the table as follows:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$p$[/tex] & [tex]$q$[/tex] & [tex]$p \rightarrow q$[/tex] & [tex]$q \rightarrow p$[/tex] & [tex]$\sim p \rightarrow \sim q$[/tex] & [tex]$\sim q \rightarrow \sim p$[/tex] \\
\hline
T & F & F & T & T & F \\
\hline
\end{tabular}
