Questions: Complete the truth table for the following compound statement. [ ~(p -> q) ∨ p] ∨ ~ q p q (p -> q) ~(p -> q) ~(p -> q) ∨ p ~(p -> q) ∨ p] ∨ ~ q T T ? ∨ ? ∨ ? ∨ ? ∨ v T F ? ∨ ? ∨ ? ∨ ? ∨ F T ? ∨ ? ∨ ? ∨ ? ∨ F F ? ∨ ? ∨ ? ∨ ? ∨ Is the compound statement a tautology? Yes ∨ Complete the truth table for the following compound statement. [ ~(p -> q) ∨ q] ∨ ~ p p q (p -> q) ~(p -> q) ~(p -> q) ∨ q [~(p -> q) ∨ q] ∨ ~ p T T ?v ?v ? v ?v T F ?v v ?v ?v ? v F T ?v ? ✔ ? ✔ ? F F ? v ? v ?v ? ∨ Is the compound statement a tautology? □

Solution Steps

To solve the problem of completing the truth tables for the given compound statements, we need to evaluate each component of the compound statement for all possible truth values of the variables \( p \) and \( q \). We will then determine if the entire compound statement is a tautology by checking if it is true for all possible truth value combinations.

1. For each row in the truth table, calculate the truth value of \( p \rightarrow q \).
2. Calculate the negation \(\sim(p \rightarrow q)\).
3. Evaluate the sub-expressions \(\sim(p \rightarrow q) \vee p\) and \(\sim(p \rightarrow q) \vee q\).
4. Finally, evaluate the entire compound statement \([\sim(p \rightarrow q) \vee p] \vee \sim q\) and \([\sim(p \rightarrow q) \vee q] \vee \sim p\).
5. Check if the compound statement is a tautology by verifying if the final column contains only true values.

We evaluate the compound statement \( [\sim(p \rightarrow q) \vee p] \vee \sim q \) for all combinations of truth values for \( p \) and \( q \):

• For \( (p, q) = (T, T) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee p = T \)
  • \( \sim q = F \)
  • Final result: \( T \)

• For \( (p, q) = (T, F) \):

  • \( p \rightarrow q = F \)
  • \( \sim(p \rightarrow q) = T \)
  • \( \sim(p \rightarrow q) \vee p = T \)
  • \( \sim q = T \)
  • Final result: \( T \)

• For \( (p, q) = (F, T) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee p = F \)
  • \( \sim q = F \)
  • Final result: \( F \)

• For \( (p, q) = (F, F) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee p = F \)
  • \( \sim q = T \)
  • Final result: \( T \)

The truth table for the first statement is: \[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & (p \rightarrow q) & \sim(p \rightarrow q) & \sim(p \rightarrow q) \vee p & [\sim(p \rightarrow q) \vee p] \vee \sim q \\ \hline T & T & T & F & T & T \\ T & F & F & T & T & T \\ F & T & T & F & F & F \\ F & F & T & F & F & T \\ \hline \end{array} \]

Since the final column contains \( F \) for one combination, the statement is not a tautology.

Next, we evaluate the compound statement \( [\sim(p \rightarrow q) \vee q] \vee \sim p \):

• For \( (p, q) = (T, T) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee q = T \)
  • \( \sim p = F \)
  • Final result: \( T \)

• For \( (p, q) = (T, F) \):

  • \( p \rightarrow q = F \)
  • \( \sim(p \rightarrow q) = T \)
  • \( \sim(p \rightarrow q) \vee q = T \)
  • \( \sim p = F \)
  • Final result: \( T \)

• For \( (p, q) = (F, T) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee q = T \)
  • \( \sim p = T \)
  • Final result: \( T \)

• For \( (p, q) = (F, F) \):

  • \( p \rightarrow q = T \)
  • \( \sim(p \rightarrow q) = F \)
  • \( \sim(p \rightarrow q) \vee q = F \)
  • \( \sim p = T \)
  • Final result: \( T \)

The truth table for the second statement is: \[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & (p \rightarrow q) & \sim(p \rightarrow q) & \sim(p \rightarrow q) \vee q & [\sim(p \rightarrow q) \vee q] \vee \sim p \\ \hline T & T & T & F & T & T \\ T & F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & F & F & T \\ \hline \end{array} \]

Since the final column contains only \( T \), the statement is a tautology.

Final Answer