[Haskell-cafe] Induction (help!)
daniel.is.fischer at web.de
Fri May 9 08:29:14 EDT 2008
Am Freitag, 9. Mai 2008 13:50 schrieb PR Stanley:
> 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.
Right. I only wanted to say that we might have chosen the wrong base case for
the proposition. If p(0) doesn't hold, then obviously [for all n. p(n)]
doesn't hold. But [for all n. p(n) implies p(n+1)] could still be true, and
in that case, if e.g. p(3) holds, then [for all n >= 3. p(n)] holds.
So if the base case fails, still a large portion of the proposition might be
saved, but if the induction step fails, that is not so.
> 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.
Sorry? The induction step consists of proving that
WHATEVER j is, IF p(j) holds, THEN p(j+1) holds, too.
If or when we have done that, it is tautological to say that we proved
(*) for all j. [p(j) implies p(j+1)].
In the notation used by Ryan Ingram yesterday, the induction step consists of
L, j :: Nat |- p(j) => p(j+1)
typically, that is done via proving
L, j :: Nat, p(j) |- p(j+1)
, i.e. proving p(j+1) under the assumption p(j), which by the logical rule of
deduction/implication introduction is equivalent to
L, j :: Nat |- p(j) => p(j+1).
> 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?
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