(From MathOverflow). By Definition 2.24, one has the law

Specializing to \(y=x \diamond x\), we conclude

and hence by another application of Definition 2.24 we see that \(x \diamond x\) is idempotent:

Now, replacing \(x\) by \(x \diamond x\) in Equation 1 and then using Equation 2 we see that

so in particular \(x \diamond x\) commutes with \(y \diamond y\):

Also, from two applications of Equation 1 one has

Thus Equation 3 simplifies to \(x \diamond y = y \diamond x\), which is Definition 2.18.

By Lemma 4.5 it suffices to show that Definition 2.12 implies Definition 2.9. From Definition 2.12 one has

and also

giving \(x = y \diamond (x \diamond y)\), which is Definition 2.9.

By hypothesis, one has

for all \(x,y,z,w\). Various specializations of this give

5 gives Definition 2.31, while Equation 4, 5, 6 gives

which is Definition 2.23.

The implication of Definition 2.29 from Definition 2.2 is trivial. The converse is a surprisingly long chain of implications; see pages 326–327 of [ 3 ] . The initial law

is used to obtain, in turn,

Suppose that a magma \(G\) obeys Definition 2.27, thus

and

whence

which is Definition 2.30. This gives

while from Equation 7 one has

whence

This implies that \((x \diamond y) \diamond (x \diamond y)\) does not depend on \(x\), or on \(y\), hence is equal to some constant \(e\):

From Equation 7 the magma operation is surjective, hence

which gives Definition 2.15. Applying Equation 7 with \(x=y=z\) we conclude

while if we instead take \(y=z=e\) we have

hence

and then also

from which we readily conclude Definition 2.11, Definition 2.8; thus \(e\) is an identity element. From Equation 7 with \(z=e\) we now have

which is Definition 2.10. If instead we take \(y=e\) we have

which is Definition 2.9. So if we substitute \(z = x \diamond y\) and use Equation 9 we obtain

and hence

thanks to Equation 10. This gives Definition 2.18, thus \(G\) is now commutative. From Equation 7 once more one has

which one can simplify using commutativity and Equation 9 (or Equation 10) to eventually obtain

which is Definition 2.33. \(G\) is now commutative and associative, and every element is its own inverse and of exponent \(2\), hence is an abelian group thanks to Equation 8, so \(G\) is an abelian group of exponent \(2\) as claimed. The converse is easily verified.

It suffices to show that Definition 2.26 implies Definition 2.2. Pick an element \(0\) of \(G\) and define \(1 = 0 \diamond 0\) and \(2 = 1 \diamond 1\) (we do not require \(0,1,2\) to be distinct). From Definition 2.26 with \(x=z=0\) we have

If we then apply Definition 2.26 with \(z=1\) we conclude that

for all \(x,y\), from which one concludes \(x=x'\) for any \(x,x' \in G\), giving Definition 2.2.

See [ 7 ] . In fact this is the shortest law with this property.

A sketch of proof follows. One can easily verify that the Sheffer stroke operation obeys this law. Conversely, if this law holds, then automated theorem provers can show that the three Sheffer axioms

are satisfied. A classical result of Sheffer [ 9 ] then allows one to conclude.

See [ 5 , Theorem 5 ] . The proof is quite lengthy; a sketch is as follows. It is easy to see that natural central groupoids obey Equation 11. Conversely, if this law holds, then

so we have a central groupoid. Setting \(y = (t \diamond t) \diamond t\), \(z = t \diamond (t \diamond t)\), \(w = t \diamond t\) in Equation 11 we also obtain

Using the notation

we then have

A lengthy computer-assisted argument then gave the dual identity

Together, these give

Multiplying on the left by \(x = x^{(1)}\diamond x^{(2)}\), one can conclude that

One then has

and a similar argument gives

Since \((x \diamond x)^{(1)} = x^{(2)}\) and \((x \diamond x)^{(2)} = x^{(1)}\), we conclude that \(x^{(1)}\) and \(x^{(2)}\) are idempotent. Since \(x = x^{(1)} \diamond x^{(2)}\), we see that every \(x\) is the product of two idempotents. One can show that this representation is unique, and gives a canonical identification with a natural central groupoid.