I’ve blogged the 2-d version of the proof here: I think you’ll like it. I look forward to seeing your code!

Fixed!

Great catch regarding the Eckmann-Hilton bit. Encoding it into Haskell for bifunctors as we speak. =)

So, for those who are feeling confused: if * and + abide and have units 1 and 0, then 0=1, *=+, and * (and hence +) is commutative. The proof’s fun and reasonably straightforward, particularly if you use the 2-d notation. This is known as the Eckmann-Hilton argument, and it’s famous in category theory, having many beautiful consequences.

0 = 0|0 = 1|0 = 1|0 = 1 = 1
-|- — -
0|1 0|1 1

x|y = 1|y = 1|y = y = y|1 = y|1 = y|x = 1|x = 1|x = x
-|- — – — -|- -|- — -
x|1 x|1 x 1|x 1|x y|1 y|1 y

Let + and * be abiding binary operations, with identities 0 and * respectively. Then

0 = 0+0 = (0*1)+(1*0) = (0+1)*(1+0) = 1*1 = 1

And now:

x*y = (0+x)*(y+0) = (0*y)+(x*0) = (1*y)+(x*1) = y+x = (y*1)+(1*x) = (y+1)*(1+x) = (y+0)*(0+x) = y*x = (0+y)*(x+0) = (0*x)+(y*0) = (1*x)+(y*1) = x+y

This is the Eckmann-Hilton argument beloved of category theorists: among its implications are that a monoid object in the category of monoids is a commutative monoid, and that all higher homotopy groups are abelian. It becomes particularly elegant when you use the

x | y =(x*y)+(y*w)
z | w

notation – try it! The proof can be written in a circle, and it’s then called the Eckmann-Hilton clock.

The Catsters have done a video about the Eckmann-Hilton argument.