Le 14-juil.-06, à 18:52, Tom Caylor a écrit :

> Here is where I believe the crux is:  "..." means you can continue to
> add the "I" as many times as you want.  Actually, this is equivalent
> to: "..." means you can continue to add the "I" as many times as you
> want and you can.  It's just a little redundant to say it that way.

> Now A and B *know*, as well as anyone can even know, what finite means.

True, but unprovable. With comp you are betting here.

>  All they have to do is perform some experimentation to get the idea
> that, after a while of adding "I" they eventually get tired and/or
> loose interest, so they have to *stop*.

Yes but my friend B, which is an angel, a cousin of the analytical 
second order arithmetic with the omega rule. He is tired after counting 
up to number like |||...|||...|||...  ...||||||||.

> What's so difficult about
> understanding what stopping is?

I am not denying we have some intuition of that. Just pointing that 
mathematicians can show we cannot define what finite means through 
first order logic, and then second order logic builds on that 
intuition, so that really "finite" is not a notion we can define in any 
finite way. Nor can we define "NOT finite", that is what "infinite" 

> Even the word "finite" has "fin" in
> it, i.e. "end".  The notion is defined by invariance.

Relative one. You can imagine something stopping compare to something 
which does not stop.

> Something
> similar (invariant) is happening (adding "I" at one step is considered
> the same action as adding "I" in another step)

Actually adding | at the end of ||||..... giving |||....| is different 
from adding | at the end of |||.

> and then the invariance
> disappears, i.e. the adding of the "I" is no longer happening.

Yes but when? I know you and me know that. The point is that we cannot 
explain it without admitting at the start that we know that. "that" has 
the type of a quale.


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