# FOM: power shift axiom scheme

Harvey Friedman friedman at math.ohio-state.edu
Thu Apr 1 16:56:53 EST 1999

```This is a development that has just come out of my previous postings 4AM
4/1/99, 11:19AM 4/1/99, and 3:19PM 4/1/99.

NOTE: I forgot to say that I used S in these postings for the power set
operation.

The power shift axiom scheme is as follows. Let phi(x) be a formula in the
usual language of set theory with at most the free variable x.

(therexists x,y)(|x| = 2^|y| & phi(x) iff phi(y)).

BIG OPEN PROBLEM: Is the power shift axiom scheme consistent with Z or ZF?
If so, then NF is consistent. In fact, Z + the power shift axiom proves the
consistency of NF and more.

THEOREM 1. ZC refutes the power shift axiom scheme.

THEOREM 2. Z + "the power shift axiom scheme" proves the consistency of ZFC
+ sharps. Probably with fine structure arguments, this can be pushed up to
ZFC + "there exists Ramsey cardinals", and yet higher stuff nearing
measurable cardinals.

THEOREM 3. ZFC + "there exists a measurable cardinal" is translatable into
ZF + "the power shift axiom scheme."

Now consider this sequential shift axiom scheme: Let phi(x) be a formula of
set theory with only the free variable x.

\$) there exists sets x_1,x_2,... such that for all i, |x_i+1| >= 2^|x_i|
and phi(x_1,x_2,...) iff phi(x_2,x_3,...).

And

\$\$) there exists ordinals x_1 < x_2,... such that for all i,
phi(x_1,x_2,...) iff phi(x_2,x_3,...).

THEOREM 4. ZFC + \$\$), ZF + \$\$), ZFC + \$), ZF + \$), ZFC + "there exists a
measurable cardinal" are all intertranslatable. Also ZC + \$\$), Z + \$\$), ZC
+ \$), Z + \$), ZC + "there exists a measurable cardinal" are all
intertranslatable.

```