[Haskell-cafe] Re: Clarification on proof section of HS: The Craft
japple at freeshell.org
Mon May 2 10:21:14 EDT 2005
Echo Nolan wrote:
> 1) Prove that the property holds for the null list
> 2) Prove that the property holds for (x:xs) under the assumption
> that the property holds for xs.
As someone else said, this works for finite lists.
> My question is this: what is the proof of (P() and ( P(xs) => P(x:xs))) => P(xs)?
Assume there is some finite list ys for which this statement is not
true. In other words, assume that
p() && (p(ys) => p(x:ys))
Then there is then a smallest list ys with such property.
ys /= , since P().
Consider tail ys. It is smaller than ys, so the implication holds for
it. However, that means that
p(tail ys) => p(head ys : tail ys)
or p(ys). That's a contradiction, so our assumption that there was some
finite list ys for which the implication ws false is false.
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