FOM: consistency equals existence

Robert Black Robert.Black at nottingham.ac.uk
Thu Feb 17 19:10:05 EST 2000

```Steve Simpson:

>There is an old dictum (Hilbert's?) that existence (of mathematical
>objects) is the same as consistency.
>
>[ Here is the argument.  Let S be a system of mathematical objects
>whose existence is in question.  Let T be a set of axioms describing
>all of the properties that S is to have.  Since mathematical objects
>are completely described and axiomatized by their properties, T ``is''
>S.  But according to the G"odel Completeness Theorem, T is consistent
>if and only if S ``exists'', in the sense that a model of T exists.
>So, existence (of S) equals consistency (of T). ]

I know two places where Hilbert explicitly says that for the mathematician
existence is consistency. One is in 'Ueber den Zahlbegriff' (first
published 1900 but delivered before that) and the clearer and more extended
is in the letter to Frege of 29 December 1899 (my translation):

'You write "From the truth of the axioms follows that they do not
contradict one another". It interested me greatly to read this sentence of
yours, because in fact for as long as I have been thinking, writing and
lecturing about such things, I have always said the very opposite: if
arbitrarily chosen axioms together with everything which follows from them
do not contradict one another, then they are true, and the things defined
through the axioms exist. For me that is the criterion of truth and
existence. The proposition 'every equation has a root' is true, or the
existence of roots is proved, as soon as the axiom 'every equation has a
root'  can be added to the other arithmetical axioms without it being
possible for a contradiction to arise by any deductions. This view is the
key not only for the understanding of my [Foundations of Geometry], but
also for example my recent [Ueber den Zahlbegriff], where I prove or at
least indicate that the system of all real numbers *exists*, while the
system of all Cantorean cardinalities or all Alefs - as Cantor himself
states in a similar way of thinking but in slightly different words - *does
not exist*.'

This is of course long before the Hilbert programme. It's at a time before
Hilbert himself got seriously interested in logic, a time when when nobody
had any idea of the significance of the distinction between first and
second order logic, a time when nobody had axiomatized set theory, let
alone axiomatized it in first-order logic, and a time when few people if
anyone had a clear enough idea of the distinction between semantic
entailment and formal derivability to even state the completeness theorem,
let alone prove it. So Hilbert's own reason for his claim can't have had
anything to do with the completeness theorem. Nor is it at all clear, once
we have made the various distinctions between first and second-order logic
and between entailment and derivability, what the claim amounts to. I would
be very interested to see a significantly later quote from Hilbert, one
which takes account of the distinction between derivability and entailment,
and still maintains the same doctrine in clarified form.

One way of taking the doctrine doesn't involve derivability at all. If we
think of mathematics as the study of all *possible* structures, then we can
say that in the mathematical universe posse is esse. So any system of
axioms which could be true is true. This is the credo of a kind of
mathematical platonist structuralism, and I suspect it may be roughly what
Hilbert meant in 1899, not having distinguished between the claim that a
system of axioms could be true and the claim that it doesn't lead to a
contradiction.  This way of thinking also provides a foundational role for
set theory as describing a structure, the universe of all sets, which is
designed (as far as possible) to be so rich that every possible structure
gets actualized in it (if it's *possible* for a system of axioms to have a
model, then it *actually* has a model in the universe of sets). In other
words, any system of axioms which *could* be true *is* true. Note also that
this doesn't depend on any distinction between first and second-order logic.

But suppose we think the doctrine does involve derivability, we stick to
first-order logic, and we want to give the completeness theorem a role.
It's easy to see how the set-theoretic realist does this. It's a theorem of
(first-order) set theory that every consistent first-order theory has a
model, but the axioms of first-order set theory are true (in the universe
of all sets), so every consistent first-order theory *does* have a model in
that universe.

What I can't see is how the completeness theorem plays a role if we're not
set-theoretic realists. Suppose we have a first-order theory which we know
or believe to be consistent, and we want to show that it has a model. Of
course it's a theorem of our set theory that any consistent theory has a
model. But we're not assuming our set theory to be true any more - only (at
most) consistent. But from the *consistency* of our set theory together
with the fact that it's provable in our set theory that every consistent
theory has a model, it doesn't follow even that our set theory has a model,
let alone that every consistent theory has a model. (e.g. perhaps there are
only finitely many objects, so that no theories satisfiable only in
infinite domains have models, so set theory will be a consistent theory
without a model.) If we help ourselves to Hilbert's doctrine, and argue
that since our set theory is consistent it must have a model, then we can
then conclude that every consistent first-order theory has a model. But as
a proof of Hilbert's doctrine that's flatly circular.

Robert Black
Dept of Philosophy
University of Nottingham
Nottingham NG7 2RD

tel. 0115-951 5845
home tel. 0115-947 5468
mobile 07974 675620

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