Answer :
Sure! Let's start by filling in the truth table step-by-step.
### Step 1: Fill in the Negation Columns
1. For [tex]\( P = T \)[/tex] (True) and [tex]\( Q = T \)[/tex] (True):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( F \)[/tex] (False)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( F \)[/tex] (False)
2. For [tex]\( P = T \)[/tex] (True) and [tex]\( Q = F \)[/tex] (False):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( F \)[/tex] (False)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( T \)[/tex] (True)
3. For [tex]\( P = F \)[/tex] (False) and [tex]\( Q = T \)[/tex] (True):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( T \)[/tex] (True)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( F \)[/tex] (False)
4. For [tex]\( P = F \)[/tex] (False) and [tex]\( Q = F \)[/tex] (False):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( T \)[/tex] (True)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( T \)[/tex] (True)
Let's update the truth table with these values:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & \\ \hline $T$ & $F$ & $F$ & $T$ & \\ \hline $F$ & $T$ & $T$ & $F$ & \\ \hline $F$ & $F$ & $T$ & $T$ & \\ \hline \end{tabular} \][/tex]
### Step 2: Compute the Implication ([tex]\( \sim P \rightarrow \sim Q \)[/tex])
An implication [tex]\( p \rightarrow q \)[/tex] is False only if [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is False; otherwise, it is True.
1. For [tex]\( \sim P = F \)[/tex] and [tex]\( \sim Q = F \)[/tex]:
- The implication [tex]\( F \rightarrow F \)[/tex] is [tex]\( T \)[/tex] (True).
2. For [tex]\( \sim P = F \)[/tex] and [tex]\( \sim Q = T \)[/tex]:
- The implication [tex]\( F \rightarrow T \)[/tex] is [tex]\( T \)[/tex] (True).
3. For [tex]\( \sim P = T \)[/tex] and [tex]\( \sim Q = F \)[/tex]:
- The implication [tex]\( T \rightarrow F \)[/tex] is [tex]\( F \)[/tex] (False).
4. For [tex]\( \sim P = T \)[/tex] and [tex]\( \sim Q = T \)[/tex]:
- The implication [tex]\( T \rightarrow T \)[/tex] is [tex]\( T \)[/tex] (True).
Let's update the truth table with these values:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & $T$ \\ \hline $T$ & $F$ & $F$ & $T$ & $T$ \\ \hline $F$ & $T$ & $T$ & $F$ & $F$ \\ \hline $F$ & $F$ & $T$ & $T$ & $T$ \\ \hline \end{tabular} \][/tex]
So, the completed truth table is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & $T$ \\ \hline $T$ & $F$ & $F$ & $T$ & $T$ \\ \hline $F$ & $T$ & $T$ & $F$ & $F$ \\ \hline $F$ & $F$ & $T$ & $T$ & $T$ \\ \hline \end{tabular} \][/tex]
This completes the truth table for the given logical expression [tex]\( \sim P \rightarrow \sim Q \)[/tex].
### Step 1: Fill in the Negation Columns
1. For [tex]\( P = T \)[/tex] (True) and [tex]\( Q = T \)[/tex] (True):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( F \)[/tex] (False)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( F \)[/tex] (False)
2. For [tex]\( P = T \)[/tex] (True) and [tex]\( Q = F \)[/tex] (False):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( F \)[/tex] (False)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( T \)[/tex] (True)
3. For [tex]\( P = F \)[/tex] (False) and [tex]\( Q = T \)[/tex] (True):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( T \)[/tex] (True)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( F \)[/tex] (False)
4. For [tex]\( P = F \)[/tex] (False) and [tex]\( Q = F \)[/tex] (False):
- [tex]\( \sim P \)[/tex] (not [tex]\( P \)[/tex]) is [tex]\( T \)[/tex] (True)
- [tex]\( \sim Q \)[/tex] (not [tex]\( Q \)[/tex]) is [tex]\( T \)[/tex] (True)
Let's update the truth table with these values:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & \\ \hline $T$ & $F$ & $F$ & $T$ & \\ \hline $F$ & $T$ & $T$ & $F$ & \\ \hline $F$ & $F$ & $T$ & $T$ & \\ \hline \end{tabular} \][/tex]
### Step 2: Compute the Implication ([tex]\( \sim P \rightarrow \sim Q \)[/tex])
An implication [tex]\( p \rightarrow q \)[/tex] is False only if [tex]\( p \)[/tex] is True and [tex]\( q \)[/tex] is False; otherwise, it is True.
1. For [tex]\( \sim P = F \)[/tex] and [tex]\( \sim Q = F \)[/tex]:
- The implication [tex]\( F \rightarrow F \)[/tex] is [tex]\( T \)[/tex] (True).
2. For [tex]\( \sim P = F \)[/tex] and [tex]\( \sim Q = T \)[/tex]:
- The implication [tex]\( F \rightarrow T \)[/tex] is [tex]\( T \)[/tex] (True).
3. For [tex]\( \sim P = T \)[/tex] and [tex]\( \sim Q = F \)[/tex]:
- The implication [tex]\( T \rightarrow F \)[/tex] is [tex]\( F \)[/tex] (False).
4. For [tex]\( \sim P = T \)[/tex] and [tex]\( \sim Q = T \)[/tex]:
- The implication [tex]\( T \rightarrow T \)[/tex] is [tex]\( T \)[/tex] (True).
Let's update the truth table with these values:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & $T$ \\ \hline $T$ & $F$ & $F$ & $T$ & $T$ \\ \hline $F$ & $T$ & $T$ & $F$ & $F$ \\ \hline $F$ & $F$ & $T$ & $T$ & $T$ \\ \hline \end{tabular} \][/tex]
So, the completed truth table is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline $P$ & $Q$ & $\sim P$ & $\sim Q$ & $\sim P \rightarrow \sim Q$ \\ \hline $T$ & $T$ & $F$ & $F$ & $T$ \\ \hline $T$ & $F$ & $F$ & $T$ & $T$ \\ \hline $F$ & $T$ & $T$ & $F$ & $F$ \\ \hline $F$ & $F$ & $T$ & $T$ & $T$ \\ \hline \end{tabular} \][/tex]
This completes the truth table for the given logical expression [tex]\( \sim P \rightarrow \sim Q \)[/tex].
