Key Concepts and Formulas

• Tautology: A compound statement that is always true, regardless of the truth values of its individual components.
• Logical Operators:
  • $\wedge$ (AND): $p \wedge q$ is true only if both $p$ and $q$ are true.
  • $\vee$ (OR): $p \vee q$ is true if either $p$ or $q$ or both are true.
  • $\sim$ (NOT): $\sim p$ is true if $p$ is false, and vice versa.

Step-by-Step Solution

Step 1: Understand the problem We are given that $*, \odot \in\{\wedge, \vee\}$, meaning that each of these operators can be either AND or OR. We are also given the expression $(p * q) \odot(p \odot \sim q)$ and told that it is a tautology. We need to find the ordered pair $(*, \odot)$ that makes this expression a tautology. We will test each of the given options.

Step 2: Test option (A): $(\vee, \wedge)$ If $* = \vee$ and $\odot = \wedge$, the expression becomes $(p \vee q) \wedge (p \wedge \sim q)$. Let's analyze this expression. $(p \vee q) \wedge (p \wedge \sim q) \equiv (p \wedge p) \wedge (p \wedge \sim q) \vee (q \wedge p) \wedge (q \wedge \sim q)$ $\equiv p \wedge \sim q \wedge p \vee q \wedge p \wedge (q \wedge \sim q)$ $\equiv (p \wedge \sim q) \vee (q \wedge p \wedge F)$ $\equiv (p \wedge \sim q) \vee F$ $\equiv p \wedge \sim q$ This expression is not a tautology because it is false when $p$ is false or $q$ is true. Therefore, option (A) is incorrect.

Step 3: Test option (B): $(\vee, \vee)$ If $* = \vee$ and $\odot = \vee$, the expression becomes $(p \vee q) \vee (p \vee \sim q)$. $(p \vee q) \vee (p \vee \sim q) \equiv p \vee q \vee p \vee \sim q \equiv p \vee p \vee q \vee \sim q \equiv p \vee (q \vee \sim q)$ Since $q \vee \sim q$ is always true (a tautology), we have: $p \vee T \equiv T$ Thus, the expression is a tautology. Therefore, option (B) is correct.

Step 4: Test option (C): $(\wedge, \wedge)$ If $* = \wedge$ and $\odot = \wedge$, the expression becomes $(p \wedge q) \wedge (p \wedge \sim q)$. $(p \wedge q) \wedge (p \wedge \sim q) \equiv p \wedge q \wedge p \wedge \sim q \equiv (p \wedge p) \wedge (q \wedge \sim q) \equiv p \wedge F \equiv F$ This expression is always false (a contradiction), so it is not a tautology. Therefore, option (C) is incorrect.

Step 5: Test option (D): $(\wedge, \vee)$ If $* = \wedge$ and $\odot = \vee$, the expression becomes $(p \wedge q) \vee (p \vee \sim q)$. $(p \wedge q) \vee (p \vee \sim q) \equiv (p \wedge q) \vee p \vee \sim q$ $\equiv ((p \wedge q) \vee p) \vee \sim q \equiv p \vee \sim q$ This expression is not a tautology because it is false when $p$ is false and $q$ is true. Therefore, option (D) is incorrect.

Common Mistakes & Tips

• Careless application of logical equivalences: Make sure to apply the correct logical equivalences when simplifying the expressions. For example, the distributive law can be helpful.
• Not checking all possibilities: Remember to test each of the given options systematically.
• Confusing tautology with contradiction: A tautology is always true, while a contradiction is always false.

Summary We are given that the expression $(p * q) \odot(p \odot \sim q)$ is a tautology, and we need to find the correct ordered pair $(*, \odot)$. By substituting each of the given options into the expression and simplifying using logical equivalences, we found that the expression is a tautology only when $* = \vee$ and $\odot = \vee$.

Final Answer The final answer is \boxed{(\vee, \vee)}, which corresponds to option (B).