Key Concepts and Formulas

• Implication: $p \to q \equiv \neg p \vee q$
• Tautology: A statement that is always true, regardless of the truth values of its components.
• Truth Tables: A method for determining the truth value of a compound statement based on the truth values of its components.
• De Morgan's Laws:
  • $\neg (p \wedge q) \equiv \neg p \vee \neg q$
  • $\neg (p \vee q) \equiv \neg p \wedge \neg q$

Step-by-Step Solution

Step 1: Express the implication using disjunction. We are given the expression $(p \to q) \Delta (p \nabla q)$. Using the implication equivalence, we can rewrite $p \to q$ as $\neg p \vee q$. Therefore, the expression becomes $(\neg p \vee q) \Delta (p \nabla q)$.

Step 2: Consider option (A): $\Delta = \vee, \nabla = \vee$. Substitute $\Delta = \vee$ and $\nabla = \vee$ into the expression. We get $(\neg p \vee q) \vee (p \vee q)$.

Step 3: Simplify the expression from Step 2. Using the associative property of disjunction, we have $(\neg p \vee q) \vee (p \vee q) \equiv \neg p \vee q \vee p \vee q$. Since $p \vee \neg p$ is always true (tautology), the expression becomes $(\neg p \vee p) \vee (q \vee q) \equiv T \vee q \equiv T$. Thus, when $\Delta = \vee$ and $\nabla = \vee$, the entire expression is a tautology.

Step 4: Consider option (B): $\Delta = \vee, \nabla = \wedge$. Substitute $\Delta = \vee$ and $\nabla = \wedge$ into the expression. We get $(\neg p \vee q) \vee (p \wedge q)$.

Step 5: Analyze the expression from Step 4 using cases. If $p$ is true and $q$ is false, then $(\neg p \vee q) \vee (p \wedge q) \equiv (F \vee F) \vee (T \wedge F) \equiv F \vee F \equiv F$. Since the expression is false for some truth values of $p$ and $q$, it's not a tautology.

Step 6: Consider option (C): $\Delta = \wedge, \nabla = \wedge$. Substitute $\Delta = \wedge$ and $\nabla = \wedge$ into the expression. We get $(\neg p \vee q) \wedge (p \wedge q)$.

Step 7: Analyze the expression from Step 6 using cases. If $p$ is false and $q$ is false, then $(\neg p \vee q) \wedge (p \wedge q) \equiv (T \vee F) \wedge (F \wedge F) \equiv T \wedge F \equiv F$. Since the expression is false for some truth values of $p$ and $q$, it's not a tautology.

Step 8: Consider option (D): $\Delta = \wedge, \nabla = \vee$. Substitute $\Delta = \wedge$ and $\nabla = \vee$ into the expression. We get $(\neg p \vee q) \wedge (p \vee q)$.

Step 9: Analyze the expression from Step 8 using cases. If $p$ is false and $q$ is false, then $(\neg p \vee q) \wedge (p \vee q) \equiv (T \vee F) \wedge (F \vee F) \equiv T \wedge F \equiv F$. Since the expression is false for some truth values of $p$ and $q$, it's not a tautology.

Step 10: Conclusion Only option (A) results in a tautology.

Common Mistakes & Tips

• Remember the correct equivalences for implication ($p \to q \equiv \neg p \vee q$).
• When checking if an expression is a tautology, it's often easier to find a case where it's false to disprove it.
• Using truth tables can be helpful, but simplification using logical equivalences is often faster.

Summary

We are given the expression $(p \to q) \Delta (p \nabla q)$ and asked to find the logical connectives $\Delta$ and $\nabla$ that make the expression a tautology. We first rewrite the implication as a disjunction: $(\neg p \vee q) \Delta (p \nabla q)$. We then test each of the given options. Option (A), where $\Delta = \vee$ and $\nabla = \vee$, leads to the expression $(\neg p \vee q) \vee (p \vee q)$, which simplifies to $T$, a tautology. The other options do not result in tautologies. Therefore, the correct answer is $\Delta = \vee$ and $\nabla = \vee$.

Final Answer

The final answer is \boxed{\Delta = \vee ,\nabla = \vee }, which corresponds to option (A).