Re: Petersen's magical argument


LauLuna wrote:

Call 'D' the following definition:

'the natural number which this definition defines (or 0 if it fails to
define) plus 1'

Now it seems evident that D defines some natural, for if it didn't,
then according to D itself, it would define 1 (0+1).

Call 'n' the number defined by D. According to D, n equals the number
defined by D plus 1. So n = n+1.

How and why should this argument be rejected? I'm interested in your
way to put it.

D doesn't define a number because it does not obey correct rules of
definition. One cannot then conclude that it defines 0, because it
would do so only if it followed the correct rules of definition.

Take another example. Suppose I define "plok" to mean "not plok". Then
this definition hasn't followed the correct rules of definition.
Therefore "plok" does not apply of anything, i.e. I am not plok. One
cannot conclude that I am therefore plok, because this inference only
applies when the correct rules of definition have been followed.

Put it formally as follows: suppose a definition is such that D = the
x (phi(x)). Then, if D is a correct definition, then phi(D). But if D
is not a correct definition, then not phi(D).

In this particular case, D = the x(if correct definition then x = D+1
and if not correct definition then x = 0). So phi(x) is "if correct
definition then x = D + 1 and if not correct definition then x = 0".
Now D is not a correct definition. So the attempt to conclude that
phi(D) is invalid, so one cannot conclude that D = 0. In fact, not
phi(D).

.

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