# [FOM] 800: Beyond Perfectly Natural/6

Harvey Friedman hmflogic at gmail.com
Tue Apr 3 20:37:01 EDT 2018

```I have since placed the ms.

103. R<[A] = A’ – April 2, 2018, 5 pages. Improves over 102, but 102
is worth keeping.

where I rework 102 allowing me to put the independent statements in a
rather concise form. I will discuss first the infinite forms, as the
finite forms are obtained trivially by essentially merely replacing N
with [n] = {0,...,n}.

I now want to take up some TEMPLATING projects.

I have a history of this. When I push and push for a really simple
statement of a certain kind, and run into a brick wall after a while,
I then choose two strategies.

1. Come up with a very natural sufficient condition on the ad hoc
parts which make the statement true. THEN this results in a single
statement of the desired perfect naturalness. I succeeded with this in
Order Emulation Theory with its implicitly Pi01 sentences. See,
COMBINATORIAL MAXIMAL EMULATION/LOWER RELATIONS and COMBINATORIAL
MAXIMAL FUNCTIONS/LOWER FUNCTIONS page 8
#100

Of course, for many readers, this general statement in this case
(lower parameterizations) was not really needed, as

p < 1 implies S(p,1,...,k-1) iff S(p,2,...,k)

was actually perfectly natural. For others, it adds a lot. For yet
others, both are important. For me, having both the specific above and
the lower parameterization results are both very important, resonating
with different mathematical thinkers.

2. When I don't have 1, or where 1 is kind of problematic, I go for
TEMPLATING. This is the situation with the earlier developments in BRT
= Boolean Relation Theory, and templating with 6561 cases there (see
my website). In this strategy, I construct a family of statements
covering the independent statements nicely, where the family is itself
natural, and generally finite (finitely many statements), and claim or
hope to determine which statements are true, and necessary use large
cardinals. I.e., that every statement in the template is provable or
refutable using large cardinals, and large cardinals are needed to do
this.

HERE I am talking only about explicitly Pi01 in the integers, and in
this situation, we can seriously benefit from TEMPLATING. The
situation with 1 is more or less a bit problematic, and so I am
following the strategy in 2 above here.

Recall that I first introduce COMPLEMENTATION, an RCA_0 theorem that
is very basic, and then graduate up to the independent statement.

*TEMPLATING COMPLEMENTATION*
no expected challenge of ZFC

In , I used

COMPLEMENTATION. For all R containedin N^k x N^k there exists A
containedin N^k such that R<[A] = A'.

COMPLEMENTATION TEMPLATE. For all R containedin N^k x N^k there exists
A containedin N^k such that ALPHA = BETA.

Here ALPHA, BETA are supposed to be simple expressions in R and A,
drawn from a natural SPACE of expressions.

Note that there are a couple of other such equations that look a lot
like R<[A] = A'.

R<[A} = A'
R<[A]' = A
R<[A'] = A

The last two can be viewed as asserting fixed points. They say that A
is a fixed point of an operator. Obviously all three are equivalent in
that the first two are trivially equivalent, and the third comes from
the first by switching A and A'. So Complementation looked at in this
way is a consequence of the CONTRACTION MAPPING THEOREM, using the
usual metric on Cantor space of subsets of N^k.

A, R[A], R<[A], R[A'], R<[A'], N^k
and their complements

We can also look for inclusions instead of equations. Analyzing the
inclusion statements will give an analysis of the equation statements.

And while we are making life more difficult, why not allow systems of
inclusions.

After all that, we can instead ask for all BOOLEAN EQUATIONS in
A,R[A],R<[A], R[A'], R<[A'] with universal set N^k.

Thus we have a hierarchy of 5 ever more ambitious TEMPLATES to analyze.

CONJECTURE. These are completely analyzable - even the Boolean one -
and the analysis can be carried out in ACA_0, and probably, in RCA_0.

*DISCUSSION OF THE COMPLEX SITUATION*
ZFC known to be insufficient

In , we presented

2) For all order invariant R containedin N^k x N^k there exists A
containedin N^k without 2^(8k)!! - 1 such that ALPHA and BETA are
order equivalent.

EXOTIC COMPLEMENTATION. For order invariant R containedin N^k x N^k,
there exists A containedin N^k without 2^(8k)!! - 1 such that R<[A] x
A^2 x {2^(N)} and A' x A^2 x {2^(N)} are order equivalent.

So we have set

ALPHA = R<[A] x A^2 x {2^(N)}
BETA = A' x A^2 x {2^(N)}

We need to look at these as four fold Cartesian products:

ALPHA = R<[A] x A x A x {2^(N)}
BETA = A' x A x A x {2^(N)}

We now have a good challenging reasonably natural finite space of
expressions for ALPHA and BETA for 2) above. Namely,

ALPHA, BETA are finite Cartesian products from the set of expressions

A, R[A], R<[A], R[A'], R<[A'], {2^(N)}, N^k
and their complements

Obviously EXOTIC COMPLEMENTATION is a special case, and we know that
it is provably equivalent to Con(SMAH) over ACA'.

CONJECTURE. These are completely analyzable, and the analysis can be
carried out in SRP+. Every instance is provable in SMAH or refutable
in RCA_0.

Let's take another look at

2) For all order invariant R containedin N^k x N^k there exists A
containedin N^k without 2^(8k)!! - 1 such that ALPHA and BETA are
order equivalent.

We obviously want more and more. That number (in k) is just meant to
be a convenient very very large power of 2. So we can in some sense
make this prettier, but basically either adding a new variable n, or
complicating the statement in some other way. At this point, this does
not seem to be overall worth it for simplifying the associated
TEMPLATE.

HOWEVER, if I can find a neat way to genuinely simplify the template to this:

3) For all order invariant R containedin N^k x N^k there exists A
containedin N^k such that ALPHA and BETA are order equivalent.

while still having ALPHA and BETA drawn from s nice finite space, then
that would be a serious advance.

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My website is at https://u.osu.edu/friedman.8/ and my youtube site is at