Re: Can you find anything wrong with this solution to the Halting Problem?

From: Alex Hunsley (lard_at_tardis.ed.ac.molar.uk)
Date: 07/13/04 

Date: Tue, 13 Jul 2004 16:08:03 +0100

Peter Olcott wrote:
> Only direct refutation or confirmation of this message will
> be replied to, anything else will be considered off-topic and
> ignored.

Oh boo. Call mine a refutation.
However, it's a refutation of what you understand proof by contradiction to
be.. Hans has it right: you've missed the meaning.
When using proof by contradiction, any one contradiction throws your assumption
or theory out of the window, no matter how many 'supporting' cases you can find.
Here's an easy peasy example:

theorem:
  *every* real number, x, has a negative counterpart, -x, which is not
  equal to x.

counter-example:
   for x = 0, we have - x = -0 = 0, which is equal to x. theorem disproved.

Now, it doesn't matter how much you wail on about how the theorem holds for 3,
- 11.7, 9999.123, or any other number: my counter example has nailed the
theorem as bogus. End of story.

And remember, the halting problem does not say "it is impossible to decide
haltability for any machines at all", it says "it is impossible to decide the
haltability for *all* possible machines" - therefore, it is impossible to make
a T.M. that takes another T.M. as input and decides on its haltability.

alex

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