FOM: chess challenge to experts

Kanovei kanovei at
Sun Nov 12 07:36:42 EST 2000

> Date: Fri, 10 Nov 2000 18:43:39 -0500
> From: Harvey Friedman <friedman at>
> Reply to Kanovei 6:42PM 11/10/00:
> >I began by the claim that (ontologically) there is NO
> >mathematical statement true but not provable.
> Are you claiming that "every true mathematical statement is provable"?  

No, I like my version, it is not a formal statement, hence, 
not-E has not necessarily the same meaning as A-not. 
In particular your version would assume that I am obliged 
to demonstrate WHY it is so, why in my version it is a duty 
of an opponent to give a counterexample.

> you are making this claim, then you should clarify what provability means
> here. Provable by what means?

By mathematical means. Nowadays it is to be understood as 
in ZFC. This reservation I do not understand (unless you hint 
on new axioms to be added to ZFC). Anyway, let it be: by 
methods accepted as mathematical means by experts. Today this 
is ZFC tomorrow maybe something else (I doubt).

> If you are instead claiming that
> "every mathematical statement proved to be true is provable"
No, just as above. Every math. statmnt which is (ontologically) 
true is (mathematically) provable. Any counterexample ? 

CHESS CHALLENGE to experts. 

By chess laws a chess game cannot be longer than some number 
of moves which can be explicitly estimated. 
It is, therefore, a PA theorem that 
*either* W has a winning strategy *or* B has a non-losing strategy. 

Is this really true in the physical word ? 

I will not challenge this in matters of "tonns of GB of memory" 
that either strategy may need to be formulated. 
However, it is certain that possible strategies will definitely 
need astronomical memory to count, to store intermediate data, 
etc. Is there a counting method which, absolutely independently of 
the size of counting material, is stable against quantum-mechanical 
effects that yield mistakes even with a perfecr hardware 
(whatever hardware may be in this case) ? 

If chess is too complicated, the following is another version. 
Let k(1)=1 and k(n+1)=10^{k(n)}. 
Is there a counting algorithm which allows to check whether 
a+b=b+a holds for all numbers a,b < k(1000), with stability 
against QM spoiling guaranteed ?


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