alex @alexthecamel@schelling.pt

@rogers @johncarlosbaez
ok i've updated my code to just brute force check + and * are well defined, so there's no problems here
if i'm doing the free rig on n variables i'd have to be more subtle I assume.
https://github.com/agunning/freerig

@rogers @johncarlosbaez
I guess I've not accounted for relations of the form
x(y+z)^2=x(y+z)? I don't think I'm missing any which are nontrivial but that is a bit of an oversight

@rogers @johncarlosbaez
i think >2 term relations don't add any more constraints: eg
x+y+z=x+y+xy+yx+z=x+y+z+xy+yx+xz+zx=x+y+z+xy+yx+xz+zx+zy+yz=(x+y+z)^2

@highergeometer@mathstodon.xy@rogers@lipn.info @johncarlosbaez
it's a bit code-golfy I'm afraid
https://github.com/agunning/freerig

@rogers @johncarlosbaez
the a+b vs - and 1+a+b vs - entries should be 0, giving us a figure of 284, and i seem to have gotten some python code to spit out the same number, so I'll call that probably correct

@rogers @johncarlosbaez
of course 4^7 is a small enough number that you could just computationally build a graph with vertices connected with the x+y=x+xy+yx+y eq relation and find connected components

@rogers @johncarlosbaez
hopefully it is not too unclear what I mean here

@rogers @johncarlosbaez
will get a picture of the categories I have when my phone charges enough

@rogers
should be two terms of the form xab+yaba for x,y>0 (depending on parity of x+y)
adding multiples of 1 and a shouldn't change this

@rogers @johncarlosbaez
you do have ab+ba=(ab+ba)^2=ab+ba+bab+aba=(aba+bab)^2=aba+bab, right?

@rogers
(way I went about doing this generally is trying to find all equalities/equivalence classes from linear combinations of ab,ba,aba,bab, then tried to see how these reduced when i added 1,a,b)

@rogers oops thats a typo it should be (ab+bab)^2+2aba+ba

@rogers
also true that
ab+aba+ba=(ab+ba)^2+aba
=ab+2aba+bab+ba=(ab+bab)^2+aba+ba
=2ab+2aba+2bab+ba
=2ab+aba+bab+ba
=2ab+ba

@julianquast @johncarlosbaez
[oops: i have discovered a calculation error, my current figure is 305]

@julianquast @johncarlosbaez but this is essentially how I came up with these invariants

@julianquast @johncarlosbaez
unless I've done something wrong both lower and upper bound

probably the easiest way to prove these invariants is to construct homs to simpler rigs e.g
-to F_2 with f(a)=f(b)=f(1)=1
-setting ab=ba=0
-the rig {0,1,a,b,ab,ba} with a+b=ab+ba=1, a+a=a,b+b=b

@johncarlosbaez
but i think you only need to square binomial things (eg ab+aba=(ab+aba)^2=2ab+2aba) in order to generate the relations we're interested in.

@johncarlosbaez
there are only a few things that distinguish elements
-the number of 1s, as,bs (adjusting for 1+a=1+3a)
- the parity of the number of terms
- whether there is a term with an initial/final a/b
(and also any a/b at all)
only elements these don't distinguish is ab vs 3ab, aba vs 3aba and symmetrically

@johncarlosbaez
you end up getting weird things like

a+ab+b=(a+b)^2+ab=
a+2ab+ba+b=a+3ab+2ba+b=
a+4ab+3ab+b=a+2ab+3ab+b (as 4=2) = ... =a+ba+b

@johncarlosbaez
I believe the number is small enough that it's hand-countable (the figure I've gotten is 278 though it's bashy enough I've likely made a mistake)