[FOM] FOM query

[Gregory Taylor]

Dear Professor Davis,

Please consider posting the following query under heading "Flawed inductive technique".  (I am a FOM subscriber.)  Thank you for your consideration.

****************BEGINNING OF QUERY
I wish to claim the following, omitting details.

CLAIM: where $\theta$ is a strong inaccessible and $\gamma < \theta$,
there exists a formula
and a sequence of domain elements
such that the expansion of said formula modulo saturation by said sequence is of rank no less than $\gamma$.  In other words, the ranks of formulae exceed any bound below $\theta$.

The simple proof proceeds by transfinite induction on $\gamma < \theta$.  The cases whereby $\gamma$ is $0$ or a successor are trivial.  So suppose that $\gamma$ is a limit ordinal.  Let $\delta$ be an arbitrary ordinal less than $\gamma$.  By induction hypothesis, there exists some formula
and some sequence such that
the expansion of said formula with respect to said sequence is of rank no less than $\delta$.
Moreover, without loss of generality, we may assume that, for $\delta < \gamma$, the assumed formulae
may be chosen so that
their expansions modulo saturation are pairwise distinct.
(For $\delta = 0$ choose some atomic formula, for $\delta = 1$ its negation, for $\delta = 2$ the negation of that negation, and so forth; for $\delta = \omega + 1$ choose the disjunction of all the foregoing, for $\delta = \omega + 2$ the negation of that disjunction, and so forth.)
In that case, the expansion of the infinite disjunction of all the assumed formulae
is of rank not less than $\gamma$ itself.

In connection with something like this proof, an anonymous referee of an earlier paper insisted, long ago, that this technique will not work beyond $\gamma =\omega^2$ but did not explain why. If that is correct, can anyone explain why this technique will not work?  Or was the referee wrong?

Gregory Taylor

****************************END OF QUERY?

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