# [FOM] 546: New Pi01/solving CH

Harvey Friedman hmflogic at gmail.com
Fri Sep 26 12:05:23 EDT 2014

```*This research was partially supported by the John Templeton
Foundation grant ID #36297. The opinions expressed here are those of
the author and do not necessarily reflect the views of the John
Templeton Foundation.

There has been a major upgrade in the perfectly mathematically natural
Pi01 incompleteness project in almost every direction. See

Universal Properties and Incompleteness, September 26, 2014, 23 pages.
#85

I now want to take up a rather amusing approach to solving all
interesting set theoretic problems.

CONJECTURE. Something interesting will come out of suspending belief
and taking this temporarily seriously.

We first need to fix a linear ordering on sentences in the language of
ZFC. A < B means that A is simpler than B.

We only consider four categories of sentences. Let's call this

1. Sentences with one or more quantifiers, all of which range over
V(omega). These are essentially the arithmetic sentences.
2. Sentences with one or more quantifiers, all of which range over
V(omega + 1) . There are essentially the sentences of second order
arithmetic.
3. Sentences with one or more quantifiers, all of which range over
V(omega + 2). These are essentially the sentences of third order
arithmetic.
4. Sentences with one or more quantifiers, all of which range over V.

All of 1 are lower than all of 2 are lower than all of 3 are lower
than all of 4.

Now how do we compare within each category? In the standard way, we
assume prenex normal form, with the inside having only quantifiers
that are either bounded (by epsilon) are of lower "type". Then we
count the number of quantifier alterations of highest "type", as
usual, noting whether we have Sigma or Pi. If the number of
alternations is the same then the SIgma side is considered lower than
the Pi side.

To break ties, we stay within the common prefix class, and then
compare the total number of quantifiers, including those of smaller
type and bounded quantifiers. Further ties are not going to arise in
what I am proposing - and if they do, we will cross that bridge when
we come to it.

Now let T be an extension of ZFC and A be a sentence of ZFC.

We say that A is preferable to not(A) over T if and only if there is a
theorem of T + A, unprovable in T, in admissible form, which is lower
than any theorem of T + not(A), unprovable in T, in admissible form.

SOLVING EVERYTHING

1. Start by listing your favorite set theoretic statements
A1,A2,...,A8 in the order you would like them solved. We assume that
none are decided in ZFC.

2. Choose between A1 and not(A1) according to whether A1 is preferable
to  not(A1) over ZFC or not(A1) is preferable to A1 over ZFC. Add the
preferable one to ZFC to form T1. Mark any of A2,...,A8 that are
decided by T1, and put them away.

3. Choose between A2 and not(A2) according to whether A2 is preferable
to not(A2) over T1 or not(A2) is preferable to A2 over T1. Add the
preferable one to T1 to form T2. Mark any of A3,...,A8 that are
decided by T2, and put them away.

Continue this process all the way through the A's.

ROBUSTNESS?

Of course, maybe we get a different or even very different outcome
general, we probably want to front load stronger ones so that we have
to work out the smallest number of comparisons. I.e., if we front load
A1 appropriately, we could perhaps get a lot of the A2,...,A8 decided
by ZFC + A1 (or by ZFC + not(A1)).

STRATEGIC?

So suppose we fix on A1,...,A8, or even say A1,...,A16. We can ask
this question: find a permutation of A1,...,A16 such that the process
just described results in the smallest number of comparisons to be
made. I.e., that we have probably front loaded optimally, so that we
strategically carve out "powerful set theoretic space" early on.

CHOICE OF A's?

We could make a master list of published set theoretic questions, and
then put some subset of them according to the number of times they
occur in a literature search. We can of course just use these numbers
of hits not only for the selection of the A's, but their order. So it
is clear that CH = continuum hypothesis will be first.Thus we start of
with preferring not(CH) as is well known.

So I have thus solved CH. CH is false (smile).

If "there exists a measurable cardinal" is next, then obviously we are
starting with

not(CH) + "there exists a measurable cardinal".

Most set theorists obviously approve of this start.

Have fun.

****************************************
My website is at https://u.osu.edu/friedman.8/ and my youtube site is at
This is the 546th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-527 can be found at the FOM posting
http://www.cs.nyu.edu/pipermail/fom/2014-August/018092.html

528: More Perfect Pi01  8/16/14  5:19AM
529: Yet more Perfect Pi01 8/18/14  5:50AM
530: Friendlier Perfect Pi01
531: General Theory/Perfect Pi01  8/22/14  5:16PM
532: More General Theory/Perfect Pi01  8/23/14  7:32AM
533: Progress - General Theory/Perfect Pi01 8/25/14  1:17AM
534: Perfect Explicitly Pi01  8/27/14  10:40AM
535: Updated Perfect Explicitly Pi01  8/30/14  2:39PM
536: Pi01 Progress  9/1/14 11:31AM
537: Pi01/Flat Pics/Testing  9/6/14  12:49AM
538: Progress Pi01 9/6/14  11:31PM
539: Absolute Perfect Naturalness 9/7/14  9:00PM
540: SRM/Comparability  9/8/14  12:03AM
541: Master Templates  9/9/14  12:41AM