On Wed, May 7, 2008 at 8:01 PM, PR Stanley <[EMAIL PROTECTED]> wrote:
> 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
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