How do I show that [p ? (q ? r)] = [p ^(¬r) ? (¬q)]

Question

[p ⇒ (q ⇒ r)] = [p ^(¬r) ⇒ (¬q)]

Answer 1

There are two ways of answering this question. The most comprehensive would be to use a truth table (which I cannot do in this system). The other is to use the rules of inference to show how you can get from one form to the other, as follows:

 

(p -> (q -> r))

(p -> (~q v r)) Material Implication

(~p v (~q v r)) Material Implication

(~p v (r v ~q)) Commutativity

((~p v r) v ~q) Associativity

(~(~p v r) -> ~q) Material Implication

((p ^ ~r) -> ~q) DeMorgan's Rule

 

And there you have it. There may be a more direct way to get there, but that is what came to me this morning. :)