Final answer:
To show that (p → (q → r)) ∧ (∼ r) is equivalent to (p → ∼ q) ∧ (∼ r), we can use logical equivalences and the equivalence of → to a disjunction to simplify the equation step by step. We start by rewriting the equation using the equivalence of → to a disjunction, then apply DeMorgan's Law and distribute the terms. We continue simplifying by using the equivalence of → to a disjunction and applying DeMorgan's Law again. Finally, we apply the equivalence of ∧ to a conjunction and simplify the equation to show that both sides are equivalent.
Step-by-step explanation:
We can start by simplifying the left side of the equation:
(p → (q → r)) ∧ (∼ r)
We can use the equivalence of → to a disjunction to rewrite the first part as:
(∼ p ∨ (q → r)) ∧ (∼ r)
Now, we can apply DeMorgan's Law to the left side of the equation:
((∼ p ∨ q) ∧ (∼ p ∨ r)) ∧ (∼ r)
Next, we can distribute (∼ p ∨ q) ∧ (∼ p ∨ r):
((∼ p ∧ (∼ p ∨ r)) ∨ (q ∧ (∼ p ∨ r))) ∧ (∼ r)
Simplifying (∼ p ∧ (∼ p ∨ r)) ∨ (q ∧ (∼ p ∨ r)):
((∼ p) ∨ (q ∧ (∼ p ∨ r))) ∧ (∼ r)
Next, we can use the equivalence of → to a disjunction to rewrite (∼ p ∨ r) as (p → r):
((∼ p) ∨ (q ∧ (p → r))) ∧ (∼ r)
Now, we can apply DeMorgan's Law to (∼ p ∨ (q ∧ (p → r))):
((∼ (∼ p)) ∧ (∼ (q ∧ (p → r)))) ∧ (∼ r)
Next, we simplify (∼ (∼ p)) to p:
(p ∧ (∼ (q ∧ (p → r)))) ∧ (∼ r)
Now, we can simplify (∼ (q ∧ (p → r))) to (∼ q ∨ (∼ p ∨ r)):
(p ∧ (∼ q ∨ (∼ p ∨ r))) ∧ (∼ r)
Lastly, we can apply the equivalence of ∧ to a conjunction to rewrite (p ∧ (∼ q ∨ (∼ p ∨ r))) as (p ∧ (∼ p ∨ r) ∧ (∼ q ∨ (∼ p ∨ r))):
(p ∧ (∼ p ∨ r) ∧ (∼ q ∨ (∼ p ∨ r))) ∧ (∼ r)
Now, we have simplified the left side of the equation. We can see that it is equivalent to the right side, (p → ∼ q) ∧ (∼ r), through a series of logical equivalences and applications of the laws of logic.