Prove that if an algebra A has a discriminator term then every subdirectly irreducible algeba in V(A) has the same discriminator term.
Does this follow directly from the fact that every variety is generated by its SI algebras? Since if we suppose the SI algebras do not have this discriminator term, then A could not be in the variety generated by them.
First stage: observe that the statement that ``$d(x,y,z)$ is a discriminator term'' is expressible by a universal sentence. Thus, the members of $V({\mathbf A})$ for which some specific $d$ is a discriminator term is a universal subclass of $V({\mathbf A})$.
Second stage: observe that an algebra with a discriminator term is simple. Thus, the universal subclass of $V({\mathbf A})$ for which $d$ is discriminator consists of simple algebras and contains ${\mathbf A}$. In particular, the universal class generated by ${\mathbf A}$, $SP_U({\mathbf A})$, consists of simple algebras having $d$ as a discriminator.
Third stage: argue that a discriminator term is a 2/3-minority term for ${\mathbf A}$, hence for $V({\mathbf A})$. This implies that $V({\mathbf A})$ is congruence distributive.
Fourth stage: use Jonsson's Lemma to conclude that every SI in $V({\mathbf A})$ belongs to $HSP_U({\mathbf A})$. But since $SP_U({\mathbf A})$ consists of simple algebras, the $H$ operator has no effect. Thus all SI's in $V({\mathbf A})$ all lie in the universal class $SP_U({\mathbf A})$ in which $d$ is a discriminator.
