Re: Peano's Axioms are Inconsistent

On Aug 14, 4:47 am, Ben Bacarisse <ben.use...@xxxxxxxxx> wrote:
RussellE <reaste...@xxxxxxxxx> writes:
On Aug 13, 4:24 pm, Ben Bacarisse <ben.use...@xxxxxxxxx> wrote:
RussellE <reaste...@xxxxxxxxx> writes:

Do you mean that you agree that AxEy T(y,x) = '1' is what you mean by
"BTNM writes to every position in X"?  I ask because "yes" would have
done.  I can't help but thing you are hedging here.

I am saying induction proves every position in X
is written. Induction doesn't say anything about
the order of the quantifiers.

It took me a while to be sure the induction
proof doesn't require the assumption PA
is consistent. Now, I am pretty sure it doesn't.

This is bad news for you. It means I can
show PA proves its own consistency.
Using Godel, this proves PA is inconsistent.

You replied to a detail and not the part that addresses the heart of
your argument.

From this reply it sounds as if you now have a new argument that does
not rely on the leap that lets you deduce En Bn = {}.  I want to be
clear.  It that argument now finished?

There exists a step when the set of blank positions is non-empty.
I can prove the set of blank positions is empty using induction.
Therefore, there exists a step the set of blank positions is empty.

I have shown why there is a step when the set of
blank positions is empty many times. You can
deny my proof if you want.

You want to say the set of blank positions is never
empty at any step, but the set is empty in the "limit".

There are no limits in PA and I can prove the
set of blank positions is empty without assuming
there is a "limit".

You want to assume there is a limit.
You want to assume there is a step after "never".

If we are going to nitpick, your proof that the
set of blank positions is never empty at any
step is flawed because you use induction.

You claim if the set of blank positions (the set)
is non-empty at step n implies the set is non-empty
at step n+1. Why is this implied?

Consider a tape with two positions.
On step one, the set is non-empty.
On step two, the set is empty.
The only way you can "prove" the set
is non-empty on step n+1 is by
assuming this is true.

PA proves its own consistency.
PA proves BNTM writes to all x in X.


Russell
- Integers are an illusion.
.

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