[FOM] Large cardinals and finite combinatorics

Timothy Y. Chow tchow at alum.mit.edu
Mon Jun 18 14:25:05 EDT 2007

Joe Shipman wrote:
>For a given natural number n, there is a well-defined operation * on 
>the integers mod 2^n which can be calculated recursively using 2 rules:
>p*1 = (p+1) mod 2^n
>p*(q*r)=(p*q)*(p*r)  [self-distributive law]
>The 2^n x 2^n "multiplication table" is called the "nth Laver table".
>It can be shown with large cardinals (using an embedding from a rank 
>into a rank) that for any k there exists n such that, in the nth Laver 
>table, (1*1) does not equal (1*(2^k)).

Do you mean 1*(2^k+1) here?

This is interesting...I had not heard of it before.  Does anyone suspect 
that this is unprovable in ZFC?  Comparing it with Friedman's examples, I 
would be surprised if this required a rank-into-rank.

Here's a simple observation.  In the nth Laver table, relabel every 
integer i with the label "2^n - i."  Thus rows and columns are reversed 
and the entries are replaced with their distance from 2^n.  Then the top 
half of the nth Laver table consists of two copies of the (n-1)st 
relabeled table.  I don't know if this observation is worth anything---it 
might mess up all the nice algebraic properties---but I didn't see it 
explicitly mentioned when I briefly glanced at a couple of papers on the 


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