Theorem MED-2. Median law for Steiner quasigroups.

\{ \begin{matrix}x(yx) = y \\ (yz)z = y \end{matrix} \} =(rigidity)⇒ \{(xy)(zw) = (xz)(yw)\}.

Proof. Define the 5-ary composite operation f by

f(x, y, z, w, u) = ((xy)(zw))(u((xz)(yw))).

Now we have, by the law x(yx) = y, f(x, y, y, w, u) = u for all x. Thus by the rigidity property, the 5-ary expression f(x, y, z, w, u) does not depend upon x for all y, z, w, u. In particular, we have

f(x, y, z, w, u)  = f(x_{1}, y, z, w, u)  ∀x∀x_{1}
((xy)(zw))(u((xz)(yw)))= ((x_{1}y)(zw))(u((x_{1}z)(yw)))  ∀x∀x_{1}
= (((yz)y)(zw))(u(((yz)z)(yw)))  letting x_{1} = yz
= w(uw)  by the Steiner laws
= u
= ((xz)(yw))(u((xz)(yw))),

and hence one right-cancellation of the common term (u((xz)(yw))) immediately yields the desired median law (xy)(zw) = (xz)(yw). Q.E.D.