Re: Yet another contribution to the P-NP question

I apologize for the delay in writing this. I was in the process of
typing my reply when it occurred to me that I had not fully appreciated
what deciding a language using this scheme meant.

With the assertion** that any recursion that decides that a
computation satisfies some property

>
What does that mean?  So far we all we know is that a recursion (let's
call one R to save overusing the letter M) defines, for any string s, a
sequence R^i(s).  What does it mean to say that R decides anything?
Explain what you mean with a simple trivial case.  Don't jump right into
explaining what "decides that a computation satisfies some property"
means.
>
My guess: if a "recursion" is indeed the map from one TM configuration
to the next, then R decides membership of a set S like this:
>
  s in S iff exists(i) such that R^i(s) = ...q_accept...
  s not in S iff exists(i) such that R^i(s) = ...q_reject...
>
Is my guess right?  If not, you need to say a lot more.
>

Indeed R does not decide anything, you would still have to use a machine
D to observe R and decide S. In the paper, note that we don't work with
P and NP at the start. Rather, we work with Pr.
If I may re-render your guess,
   s in S iff S.(R(s)) satisfies P1 = {q_accept}
   s not in S iff S.(R(s)) satisfies P2 = {q_reject}
But what we have instead is if S is in Pr, then
   f(t) in S iff S.(R(t)) satisfies P1 after it becomes stationary
   f(t) not in S iff S.(R(t)) and satisfies P2 after it becomes
stationary
where f is a computable surjection, the runtime of R is polynomially
bounded, and P1 and P2 are generalizations of {q_accept} and {q_reject}.
To decide that s is in S in the usual sense, we'd need a machine D to
enumerate inputs to R, observe what R satisfies at stationarity, compute
f, and then check for equality between the result of f and the input s.
This is more inefficient, but we have defined this so that we know that
R (and D) will always halt, and the relevant polynomial bound, anyway,
is on the runtime of R.

Yes, this might be a fatal flaw, but I am not yet sure there is even an
argument to be fatally flawed because I don't yet know what your central
term, "decides that a computation satisfies some property means".  This
would seem, at the very least, to require that both "recursions" /and/
properties be encoded.  How do you plan to do that?
>

Now, here is why we no longer need to make that assertion.
Considering L_P again, remember that we've defined things so that
s is not in L_P <=> not A1 and not A2 and not A3 ...
where A1, A2, A3 ... are statements that are a priori undecided (and we
have a different set of such statements for every s).
If L_P was in Pr, then we could construct a D as above observing L_P and
deciding the left hand side.
But we may also construct E to decide the right hand side using D. We
have to decide A1, A2 and A3 "independently" because any outcome of
those statements is logically consistent with the inputs to E (i.e., say
there are a 100 of such statements and any 99 have been decided, it
would still mean that there was 1 that could be either true or false,
which means it's undecided).
Because the only other operations in E and D are string enumeration and
computing f, all of these takes place in the recursion that D observes,
and we set things up so that the number of operations to be performed
escapes any polynomial bound.

And that's really the core of the idea. The definitions serve to handle
the technicalities that arise in the details, and to make sure that
existing concepts can be subsumed into the new scheme we have. If this
path doesn't sound promising, then it is unlikely that I have a proof
after all.

>
It might have a fatal "the only way to do X is Y" assumption in it.

Of course, there might still be unjustified implicit assumptions I'm
making, but I feel it reads better now (and at least section 16 seems
clearer to me). I am also more confident about it, but again, I submit
everything to your judgement.

>
But I have another question.  Why all the new terms?  Unless I am way
off-base with what I think you mean, all this can be written using the
accepted and familiar terminology of Turing machines.  If you simply
don't know enough about how to talk about TMs, I would be happy to help
you re-write what you are saying in more conventional language (time
permitting).
>

Thank you very much. I would be immensely grateful for any assistance
offered to me throughout this process. If there's a clearer and more
natural way to say the same things I'm saying, then I all for re-writing
the document.
Also, I have a new version of the document up where I insert a paragraph
between sections 15 and 16 (numbered 15.1) to explain what deciding a
language means in this context, and I have completely rewritten section
16 by building out what I sketched above in this reply.
I wouldn't call it a different argument, it's just that I now have a
reason for why the assertion is true (or at least, I think I do).
Thank you again for reading thus far.