Answer :
Let's complete the table step by step.
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \neg p & (\neg p \lor q) & (\neg p \lor q) \land p & [(\neg p \lor q) \land p] \to q \\ \hline \text{True} & \text{True} & \text{False} & \text{True} & \text{True} & \text{True} \\ \text{True} & \text{False} & \text{False} & \text{True} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{True} & \text{False} & \text{True} \\ \text{False} & \text{False} & \text{True} & \text{False} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
Here's the step-by-step breakdown of each column:
1. [tex]\( p \)[/tex] and [tex]\( q \)[/tex]: These are the given values for [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
[tex]\[ \begin{array}{c|c} p & q \\ \hline \text{True} & \text{True} \\ \text{True} & \text{False} \\ \text{False} & \text{True} \\ \text{False} & \text{False} \\ \end{array} \][/tex]
2. [tex]\( \neg p \)[/tex] (not [tex]\( p \)[/tex]):
- If [tex]\( p \)[/tex] is True, [tex]\( \neg p \)[/tex] is False.
- If [tex]\( p \)[/tex] is False, [tex]\( \neg p \)[/tex] is True.
[tex]\[ \begin{array}{c} \neg p \\ \hline \text{False} \\ \text{False} \\ \text{True} \\ \text{True} \\ \end{array} \][/tex]
3. [tex]\( (\neg p \lor q) \)[/tex] (not [tex]\( p \)[/tex] or [tex]\( q \)[/tex]):
- OR operation [tex]\(\lor\)[/tex] is True if at least one operand is True.
[tex]\[ \begin{array}{c} \neg p \lor q \\ \hline \text{True} \\ \text{True} \\ \text{True} \\ \text{False} \\ \end{array} \][/tex]
4. [tex]\( (\neg p \lor q) \land p \)[/tex] (not [tex]\( p \)[/tex] or [tex]\( q \)[/tex], and [tex]\( p \)[/tex]):
- AND operation [tex]\(\land\)[/tex] is True only if both operands are True.
[tex]\[ \begin{array}{c} (\neg p \lor q) \land p \\ \hline \text{True} \\ \text{True} \\ \text{False} \\ \text{False} \\ \end{array} \][/tex]
5. [tex]\([( \neg p \lor q) \land p] \to q\)[/tex] (if [tex]\(( \neg p \lor q) \land p\)[/tex], then [tex]\( q \)[/tex]):
- Implication [tex]\(\to\)[/tex] is False only if the antecedent is True and the consequent is False.
- In all other cases, the implication is True.
[tex]\[ \begin{array}{c} [(\neg p \lor q) \land p] \to q \\ \hline \text{True} \\ \text{False} \\ \text{True} \\ \text{True} \\ \end{array} \][/tex]
So the final completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \neg p & (\neg p \lor q) & (\neg p \lor q) \land p & [(\neg p \lor q) \land p] \to q \\ \hline \text{True} & \text{True} & \text{False} & \text{True} & \text{True} & \text{True} \\ \text{True} & \text{False} & \text{False} & \text{True} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{True} & \text{False} & \text{True} \\ \text{False} & \text{False} & \text{True} & \text{False} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \neg p & (\neg p \lor q) & (\neg p \lor q) \land p & [(\neg p \lor q) \land p] \to q \\ \hline \text{True} & \text{True} & \text{False} & \text{True} & \text{True} & \text{True} \\ \text{True} & \text{False} & \text{False} & \text{True} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{True} & \text{False} & \text{True} \\ \text{False} & \text{False} & \text{True} & \text{False} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
Here's the step-by-step breakdown of each column:
1. [tex]\( p \)[/tex] and [tex]\( q \)[/tex]: These are the given values for [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
[tex]\[ \begin{array}{c|c} p & q \\ \hline \text{True} & \text{True} \\ \text{True} & \text{False} \\ \text{False} & \text{True} \\ \text{False} & \text{False} \\ \end{array} \][/tex]
2. [tex]\( \neg p \)[/tex] (not [tex]\( p \)[/tex]):
- If [tex]\( p \)[/tex] is True, [tex]\( \neg p \)[/tex] is False.
- If [tex]\( p \)[/tex] is False, [tex]\( \neg p \)[/tex] is True.
[tex]\[ \begin{array}{c} \neg p \\ \hline \text{False} \\ \text{False} \\ \text{True} \\ \text{True} \\ \end{array} \][/tex]
3. [tex]\( (\neg p \lor q) \)[/tex] (not [tex]\( p \)[/tex] or [tex]\( q \)[/tex]):
- OR operation [tex]\(\lor\)[/tex] is True if at least one operand is True.
[tex]\[ \begin{array}{c} \neg p \lor q \\ \hline \text{True} \\ \text{True} \\ \text{True} \\ \text{False} \\ \end{array} \][/tex]
4. [tex]\( (\neg p \lor q) \land p \)[/tex] (not [tex]\( p \)[/tex] or [tex]\( q \)[/tex], and [tex]\( p \)[/tex]):
- AND operation [tex]\(\land\)[/tex] is True only if both operands are True.
[tex]\[ \begin{array}{c} (\neg p \lor q) \land p \\ \hline \text{True} \\ \text{True} \\ \text{False} \\ \text{False} \\ \end{array} \][/tex]
5. [tex]\([( \neg p \lor q) \land p] \to q\)[/tex] (if [tex]\(( \neg p \lor q) \land p\)[/tex], then [tex]\( q \)[/tex]):
- Implication [tex]\(\to\)[/tex] is False only if the antecedent is True and the consequent is False.
- In all other cases, the implication is True.
[tex]\[ \begin{array}{c} [(\neg p \lor q) \land p] \to q \\ \hline \text{True} \\ \text{False} \\ \text{True} \\ \text{True} \\ \end{array} \][/tex]
So the final completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \neg p & (\neg p \lor q) & (\neg p \lor q) \land p & [(\neg p \lor q) \land p] \to q \\ \hline \text{True} & \text{True} & \text{False} & \text{True} & \text{True} & \text{True} \\ \text{True} & \text{False} & \text{False} & \text{True} & \text{True} & \text{False} \\ \text{False} & \text{True} & \text{True} & \text{True} & \text{False} & \text{True} \\ \text{False} & \text{False} & \text{True} & \text{False} & \text{False} & \text{True} \\ \hline \end{array} \][/tex]
