Final answer:
To prove the logical equivalence of the given pairs, we can apply the laws and theorems taught in class. The first pair p ↔ q can be simplified using the distributive law and the law of excluded middle. The second pair ¬p → (q → r) and q → (p ∨ r) can be transformed using the law of contrapositive and the law of implication. The third expression can be verified as a tautology by constructing a truth table.
Step-by-step explanation:
To prove the logical equivalence of the given pairs using the laws and theorems taught in class:
1. Starting with the pair p ↔ q and the expression (p ∧ q) ∨ (¬p ∧ ¬q), we can use the distributive law to rewrite the expression as (p ∨ ¬p) ∧ (p ∨ ¬q) ∧ (q ∨ ¬p) ∧ (q ∨ ¬q). Applying the law of excluded middle, we know that p ∨ ¬p and q ∨ ¬q are always true, giving us the simplified expression p ∧ ¬q ∨ ¬p ∧ q. Using the commutative law, we can reorder the terms to obtain ¬q ∧ p ∨ q ∧ ¬p, which is equivalent to p ↔ q.
2. For the pair ¬p → (q → r) and q → (p ∨ r), we can use the law of contrapositive to rewrite ¬p → (q → r) as p ∨ (q → r). Then, using the law of implication, we can transform the expression as p ∨ ¬q ∨ r. Similarly, for q → (p ∨ r), we can use the law of contrapositive to rewrite q → (p ∨ r) as ¬q ∨ (p ∨ r). By commutativity and association laws, we can rearrange the terms to obtain p ∨ ¬q ∨ r, which is equivalent to ¬p → (q → r).
3. To prove that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology, we can construct a truth table. Assuming p, q, and r can take either true or false values, we evaluate the given expression for all possible combinations of truth values. If the expression evaluates to true in all cases, then it is a tautology. By constructing the truth table, we can see that the expression evaluates to true for all cases, confirming that it is a tautology.