Re: Question about Quine's New Foundations

Aatu Koskensilta wrote:
lugita15@xxxxxxxxx wrote:
Aatu Koskensilta wrote:
lugita15@xxxxxxxxx wrote:
That is, can you add a typical ambiguity scheme to ZFC which states
that every wff phi is equivalent to the corresponding formula phi+ which
results from increasing the ranks of all the variables in phi by 1?
No. Setting aside the fact that it's not clear what is meant by the
"rank of a variable" in general,
It is very clear what the rank of a set is. This is defined by
transfinite recursion.

Certainly. What isn't clear is what the "rank" of a *variable* occurring
in a formula in the language of set theory is.

the formula "all sets of rank 0 are empty" is certainly not equivalent
to "all sets of rank 1 are empty".

This is not what is meant by a typical ambiguity scheme. We do not
replace all uses of the term "rank n" in a wff with "rank n+1."
Rather, we raise the ranks of all the set variables in a wff by 1.

How is the rank of a set variable in a formula in the language of set
theory determined?

That's the point. Instead of considering TST which is an obvious typed
theory, let us instead consider Zermelo set theory with the seperation
scheme restricted to formulas wih bounded quantification. The same
difficulty as in ZFC is how to assign types (ranks) to the variables.
That is the motivation for a typical ambiguity scheme; it merely relies
on the obvious intuition we have that as long as types (ranks) can be
assigned consistently for the variables in a formula, it does not
matter what particular type (rank) each variable in each formula is
assigned.
A correct example of typical ambiguity is that for each ordinal n, the
following is a theorem: there exists a set which contains all sets of
rank n.

Given an arbitrary ordinal alpha there is in general no sentence
expressing "there exists a set which contains all sets of rank alpha".

Yes there is. That set is called V_alpha.
In any case, the scheme introduced by Specker,

Phi <--> Phi+

says that every Phi is equivalent to Phi+, not only that if Phi is
provable so is Phi+ (and hence any formula obtained from Phi raising the
types by a fixed amount). Indeed, while Specker's scheme is not provable
in TST, the inference rule Phi |- Phi+ is conservative over TST.

I'm not precisely sure how to answer your objection concerning "all sets of
rank 0 are empty," but if your objections are correct then won't there be the
same objections to adding a typical ambiguity scheme to TST?

My objections do not apply to the simple theory of types in any obvious
sense. For any type n the formula "all sets of type n are empty" is false.

Then by similar reasoning, shouldn't "all sets of type 0 are empty"
also be false?
--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

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Relevant Pages

  • Re: Question about Quines New Foundations
    ... that every wff phi is equivalent to the corresponding formula phi+ which results from increasing the ranks of all the variables in phi by 1? ... It is very clear what the rank of a set is. ... rank 0 are empty," but if your objections are correct then won't there be the same objections to adding a typical ambiguity scheme to TST? ...
    (sci.logic)
  • Re: Question about Quines New Foundations
    ... that every wff phi is equivalent to the corresponding formula phi+ which ... It is very clear what the rank of a set is. ... This is not what is meant by a typical ambiguity scheme. ... Instead of considering TST which is an obvious typed ...
    (sci.logic)
  • Re: Question about Quines New Foundations
    ... that every wff phi is equivalent to the corresponding formula phi+ which results from ... It is very clear what the rank of a set is. ... This is not what is meant by a typical ambiguity scheme. ...
    (sci.logic)
  • Re: Question about Quines New Foundations
    ... that every wff phi is equivalent to the corresponding formula phi+ which results from ... Setting aside the fact that it's not clear what is meant by the "rank of a variable" in general, the formula "all sets of rank 0 are empty" is certainly not equivalent to "all sets of rank 1 are empty". ...
    (sci.logic)