[Haskell-cafe] Induction (help!)

PR Stanley prstanley at ntlworld.com
Fri May 9 07:50:56 EDT 2008

	Paul: okay, da capo: We prove/test through case analysis
 > that the predicate p holds true for the first/starting case/element
 > in the sequence. When dealing with natural numbers this could be 0 or
 > 1. We try the formula with 0 and if it returns the desired result we
 > move onto the next stage. If the formula doesn't work with 0 and so
 > the predicate does not hold true for the base case then we've proved
 > that it's a nonstarter.

	Well, it might hold for all n >= 3. But you're right, if p doesn't hold for
the base case, then it doesn't hold for _all_ cases.
	Paul: I don't understand the point you're contending. We've chosen 0 
as our base case and if p(0) doesn't hold then nothing else will for 
our proof. Granted, you may want to start from 3 or 4 as your base 
case but we're not doing that here and for all we know forall n >= 3 
p(n) but this isn't relevant to our proof, surely.
	Paul: In the inductive step we'll make a couple of assumptions: we'll
 > imagine that p(j). We'll also assume that p holds true for the
 > successor of j - p(j+1).

	Daniel: No. In the induction step, we prove that
IF p(j) holds, THEN p(j+1) holds, too.
p(j) is the assumption, and we prove that *given that assumption*, p(j+1)
Then we have proved
(*) p(j) implies p(j+1), for all j.
	Paul: No, you haven't proved anything! I'm sorry but your assertion 
fails to make much sense.

	Daniel: If we already have established the base case, p(0), we have
p(0) and (p(0) implies p(1)) - the latter is a special case of (*) - from that
follows p(1).
Then we have
p(1) and (p(1) implies p(2), again a special case of (*), therefore p(2).
Now we have p(2) and (p(2) implies p(3)), hence p(3) and so on.

	Paul: Then with the help of rules and the protocol available to us we'll
 > try to establish whether the formula (f) gives us f(j) = f(j+1) - f(1)
 > So, we know that the predicate holds for 0 or at least one element.
 > By the way, could we have 3 or 4 or any element other than 0?

	Daniel: Sure, anything. Start with proving p(1073) and the induction 
proves p(n) for
all n >= 1073, it does not say anything about n <= 1072.

	Paul: p(0). Then we set out to find out if p holds for the successor of 0
 > followed by the successor of the successor of 0 and so forth.
 > However, rather than laboriously applying p to every natural number
 > we innstead try to find out if f(j+1) - f(1) will take us back to
 > fj). I think this was the bit I wasn't getting. The assumptions in
 > the inductive step and the algebraic procedures are not to prove the
 > formula or premise per se. That's sort of been done with the base
 > case. Rather, they help us to illustrate that f remains consistent
 > while allowing for any random element to be succeeded or broken down
 > a successive step at a time until we reach the base/starting element/value.
 > Okay so far?


More information about the Haskell-Cafe mailing list