Re:Syntax Independence?
Alessio Guglielmi <[email protected]> Mon, 8 Mar 2010 11:01:13 +0100
| Newsgroups | gmane.science.mathematics.frogs |
|---|---|
| Message-ID | <[email protected]> |
Hello,
At 14:14 +0000 3/3/10, Jon Awbrey wrote:
>Could you say something a little more exoteric about syntax independence,
>what it means in general terms?
More or less, a robustness theorem establishes the polynomial
equivalence of all proof systems belonging to a certain formalism,
independently of two syntactic factors:
1) the language of the formulae;
2) the axioms and inference rules adopted.
In this sense, we have syntax independence.
Of course, we require some conditions: concerning (1), we ask for the
chosen set of connectives to be adequate (i.e., every boolean
function is representable); concerning (2), we ask for the axioms and
rules to form an implicationally complete system, (i.e., if A1,...,An
|= B, then there is a proof of B from hypotheses A1, ..., An). Note
that these conditions are only based on semantics.
For example, let's take the formalism of Frege proof systems, i.e.,
the usual Hilbert proof systems restricted to propositional logic. By
definition, Frege systems are implicationally complete, and we want
to prove a robustness theorem. So, we want to prove that each Frege
system is able to simulate any other Frege system in polynomial time.
We address the language (1) and the axioms and rules (2) separately.
For the language part, we choose a canonical base of connectives, and
we show that any formula in any language can be reduced to a
canonical one in polynomial time. This requires a rather
simple-minded and straightforward analysis of the formula tree, and
is an argument that can be performed in every formalism, provided
that they use formulae. This argument is due to Reckhow.
Let us now suppose that we want to polynomially simulate a proof
system P by using another proof system Q.
For the axioms, we observe that every axiom scheme in P must be
provable in Q by some proof (because Q is complete). Each instance of
the axiom in P will then be simulated by an instance of the
corresponding proof in Q, and the size of this instance depends
linearly on the size of the formulae used to instantiate P and Q.
This is so, because the size of the proof in Q simulating the axiom
scheme in P is fixed.
For the rules, implicational completeness comes into play. Any rule scheme
A1 ... An
-------------
B
in P can be simulated by a proof scheme in Q of fixed size (this
exists because Q is implicationally complete). Analogously to the
case of axioms, the instances of these schemes only depend linearly
on the size of formulae used to instantiate.
From all this, the robustness theorem follows (Cook & Reckhow). Note
that the argument is straightforward and simple, thanks to the rather
strong requirement of implicational completeness.
Now, if we move to Gentzen formalisms or deep inference, we
immediately see that implicational completeness is given (or taken
away) by the presence (or absence) of the cut rule.
In fact, if A1,...,An |= B, completeness ensures that we can prove
(A1 ^ ... ^ An) -> B, but only a cut rule (or something morally
equivalent) can guarantee a proof of B from hypotheses A1, ..., An.
For example, in the case of one hypothesis, we can build
__
||
A A -> B
cut ---------- .
B
So, robustness can be immediately proved for Gentzen formalisms as it
has been done for Frege. If we extend Frege or deep inference with
substitution or Tseitin's extension, for example, again we can prove
robustness, because we don't touch implicational completeness.
However, what happens when we remove the cut, and so we lose
implicational completeness? We lose, at the same time, a way to
define the *formalism* of cut-free systems, and a way to prove
robustness for it. In other words, we lose syntax-independence. This
is the problem that I have with Lutz: he shot his foot without even
noticing it. Insensitive (we're friends, eh?).
I believe that, actually, syntax-independence, and so, robustness,
can be obtained even for cut-free systems, but this requires: (1) a
syntax-independent way to characterise cut-free systems, and (2) a
technique to prove their polynomial equivalence.
As I said, a starting point might be the atomic flows, but there are
other ways that come to mind. In any case, much has to be done before
we can claim to deal with robustness for cut-free systems in the same
sense as we do for Frege, and Gentzen and deep-inference systems with
cut. BTW, this is a nice research subject, I think.
All we can do for now, which is what Lutz really is doing, is
reasoning on single proof systems. This is, of course, interesting,
but not syntax-independent and not robust.
Ciao,
-Alessio