On Wed, May 7, 2008 at 8:01 PM, PR Stanley
So, when you apply the function to the first element in the set - e.g. Zero or Nil in the case of lists - you're actually testing to see the function works. Then in the inductive step you base everything on the assumption that p holds for some n and of course if that's true then p must hold for Succ n but you have to prove this by taking Succ from n and thus going bakc to its predecessor which is also the hypothesis p(n). So, to reiterate assumption: if hypothesis then conclusion if p(n) then p(Succ n) proof of assumption if p(Succ n) = Succ(p(n)) then we've won. If pn+1) = p(n) + p(1) then we have liftoff! I'm not going to go any further in case I'm once again on the wrong track. Cheers Paul
You've got the right idea. I should point out that it doesn't make sense to say p(Succ n) = Succ(p(n)), p(x) represents some statement that is either true or false, so it doesn't make sense to say Succ(p(n)). But I think you are on the right track. -Brent