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Re: Your tetrit position paper



John, P1788,

Thank you.

Keep in mind the very important fact: by definition, a tetrit may never
assume the state (T,F) or (F,T) when an interval is empty. So for this
reason, I believe it is correct to deduce that tetrit (T,F) is "everywhere
true" and tetrit (F,T) is "everywhere false."

However, the paper could be updated to be more specific about this.

Nate


----- Original Message ----- From: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>
To: "P1788" <stds-1788@xxxxxxxxxxxxxxxxx>
Sent: Wednesday, May 19, 2010 5:34 AM
Subject: Your tetrit position paper


Nate, P1788

On 22 Apr 2010, at 17:49, Nate Hayes wrote:
Attached is a position paper I'd like to share with everyone for
discussion.

Thanks for writing this. I will shortly vote No to motion 15 and will say
that one thing needed to make me vote Yes is an extended Rationale listing
actual applications that show decoration properties in use, so as to
justify the definitions we make, etc. Your B&B example should be included
in full.

On p5, "such f(X') will be recursively bisected", I think f(X') should be
X'.

Your basic table in 2.2, and similarly in your emails about this:
  Rank    Tetrit        Description
  3         (T,F)        Everywhere true
  2         (T,T)        Somewhere true, somewhere false
  1         (F,T)        Everywhere false
  0         (F,F)        Nowhere true, nowhere false

This is not correct because "Everywhere true/false" doesn't exclude the
empty set. (Because (all x in Empty) P(x) is always true.) It needs to be
 Rank    Tetrit        Description
 3         (T,F)        Somewhere true, nowhere false
 2         (T,T)        Somewhere true, somewhere false
 1         (F,T)        Nowhere true, somewhere false
 0         (F,F)        Nowhere true, nowhere false

("nowhere Q" is the same as "everywhere !Q" so, arbitrarily, I've removed
"everywhere" in favour of "nowhere".)

You can't get away from each line containing TWO assertions.

John