Discriminator variety

Remarks

Example: the variety of Boolean algebras is generated by the two-element Boolean algebra $\mathbf{2}$. On $\mathbf{2}$, the term $s(x, y, z) = (x \wedge (x \oplus y)) \vee (z \wedge \neg(x \oplus y))$, with $x \oplus y = (x \wedge \neg y) \vee (\neg x \wedge y)$, is the discriminator. Check: if $x = y$ then $x \oplus y = 0$ and $s = z$; if $x \neq y$ then $x \oplus y = 1$ and $s = x$. Non-example: the variety of lattices is not a discriminator variety. A discriminator term $s$ satisfies $s(x, y, y) \approx x$ and $s(x, x, y) \approx y$ on $\mathcal{K}$, hence on $V(\mathcal{K})$. So a discriminator variety is congruence-permutable, while the three-element chain is a lattice with non-permuting congruences. The identities $s(x, y, y) \approx x \approx s(x, y, x)$ and $s(x, x, y) \approx y$ show that $s$ is a Pixley term, so every discriminator variety is an arithmetical variety. The term $s$ need not act as the discriminator on every member of $V(\mathcal{K})$; for Boolean algebras it does so only on $\mathbf{2}$ and the trivial algebra.