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
At 22:43 07/05/2008, you wrote:
On Wed, May 7, 2008 at 9:27 PM, PR Stanley <[EMAIL PROTECTED]> wrote:
> Hi
> One of you chaps mentioned the Nat data type
>
> data Nat = Zero | Succ Nat
>
> Let's have
> add :: Nat -> Nat -> Nat
> add Zero n = n
> add (Succ m)n = Succ (add m n)
>
> Prove
> add m Zero = m
To prove this by induction on m, you would need to show:
1) add Zero Zero = Zero
2) If "add m Zero = m", then "add (Succ m) Zero = Succ m"
Number (1) is completely trivial, nothing more needs to be said. (2)
is easy, after expanding the definition.
Here the P I used was P(x) := add m Zero = m, the thing we were trying
to prove. (1) is a base case, P(Zero). (2) is the inductive step,
"If P(m) then P(Succ m)".
Hoping I don't sound patronizing: if you're still having trouble, then
I suspect you haven't heard what it means to prove an "if-then"
statement. Here's a silly example.
We want to prove: If y = 10, then y - 10 = 0.
First we *assume* the condition of the if. We can consider it true.
Assume y = 10.
Show y - 10 = 0.
Well, y = 10, so that's equivalent to 10 - 10 = 0, which is true.
Luke
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