(From MathOverflow). By E387, one has the law

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

and hence by another application of E387 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 E43.

By Lemma 3.5 it suffices to show that E29 implies E14. From E29 one has

and also

giving \(x = y \diamond (x \diamond y)\), which is E14.

By hypothesis, one has

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

5 gives E3722, while Equation 4, 5, 6 gives

which is E381.

The implication of E1689 from E2 is trivial. The converse is a surprisingly long chain of implications; see pages 326–327 of [ 5 ] . With some computer assistance, we found the following human-readable proof. We denote \(y^1=y\) and \(y^{n+1}=y^n\diamond y\) for \(n\geq 1\). We also introduce the notation

The initial equation states \(x = (y \diamond x) \diamond f(x,z)\). Our main step will be to prove that for all \(t\in M\) there exists \(w\in M\) such that \(f(t,w) = t\). The rest of the proof is then straightforward. Indeed, the initial equation gives \(t = (y \diamond t) \diamond f(t,w) = (y \diamond t) \diamond t = f(y,t)\) for any \(t,y\in M\). With such a simple expression of \(f\) the initial equation becomes \(x = (y \diamond x) \diamond z\), which easily implies the singleton law, for instance by writing \(x = ((y \diamond w) \diamond x) \diamond z = w \diamond z\) for any \(w,x,y,z \in M\).

There remains to prove \(f(t,w) = t\) for a well-chosen \(w \in M\), explicitly, \(w=g(t,t^5)=t\diamond t^7\). For any \(t,u,v \in M\), the combinations \(x = f(t,u)\) and \(y = v \diamond t\) satisfy \(y \diamond x = t\). Inserting these values into the initial equation yields the identity

Specialize to \(z=f(u,v)\) and note that \(f(t,u) \diamond z = (\cdots \diamond u) \diamond f(u,v) = u\) by the initial equation so that \(f(f(t,u),z) = (f(t,u) \diamond z) \diamond z = u \diamond z = g(u,v)\). Inserting this into 8 yields

On the one hand, 8 with \(z=u=t\) states that \(t^3 = t \diamond t^5\), so (using \(f(t^n,t)=t^{n+2}\))

and 9 with \((u,v)=(t^3,t)\) then implies \(f(t,t^3) = t \diamond g(t^3,t) = t \diamond f(t,t^5) = g(t,t^5)\). On the other hand, 9 with \((u,v)=(t,t^5)\) implies \(t^3 = t \diamond g(t,t^5)\). We deduce

Suppose that a magma \(G\) satisfies E1571, thus

and

whence

which is E2662. This gives

while from Equation 12 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 12 the magma operation is surjective, hence

which gives E40. Applying Equation 12 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 E23, E8; thus \(e\) is an identity element. From Equation 12 with \(z=e\) we now have

which is E16. If instead we take \(y=e\) we have

which is E14. So if we substitute \(z = x \diamond y\) and use Equation 14 we obtain

and hence

thanks to Equation 15. This gives E43, thus \(G\) is now commutative. From Equation 12 once more one has

which one can simplify using commutativity and Equation 14 (or Equation 15) to eventually obtain

which is E4512. \(G\) is now commutative and associative, and every element is its own inverse and of exponent \(2\), hence is an abelian group thanks to Equation 13, so \(G\) is an abelian group of exponent \(2\) as claimed. The converse is easily verified.

It suffices to show that E953 implies E2. 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 E953 with \(x=z=0\) we have

If we then apply E953 with \(z=1\) we conclude that

for all \(x,y\), from which one concludes \(x=x'\) for any \(x,x' \in G\), giving E2.

See [ 10 ] . In fact this is the shortest law with this property.

A sketch of proof follows. One can easily verify that the Sheffer stroke operation satisfies 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 [ 12 ] then allows one to conclude.

See [ 7 , Theorem 5 ] . The proof is quite lengthy; a sketch is as follows. It is easy to see that natural central groupoids satisfy E26302. 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 E26302 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.