Verification : Prove every variety is generated by the class of subdirectly irreducible algebras in the class.

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Prove every variety is generated by the class of subdirectly irreducible algebras in the class.

Let $\mathcal{V(K)}$ be a variety for the class $\mathcal{K}$ of similar type algebras and let $\mathcal{K'}$ be the class made of all the subdirectly irreducible algebras in $\mathcal{K}$. Now, every algebra $A$ is isomorphic to a subdirect product of subdirectly irreducible algebras. AS $\mathcal{V(K)}$ is a variety we get $\forall A \in \mathcal{V(K)}$, $A $ is isomorphic to a subdirect product of subdirectly irreducible algebras $B_{i \in I} \in \mathcal{V(K)}$. Hence by taking all possible subdirect products of members of $\mathcal{K'}$. We are able to generate isomorphic copies of each $A \in \mathcal{V(K)}$. So we get $\mathcal{V(K)} \le \mathcal{V(K')}$ so we need to make sure we do not generate any more than that. However, we do not as $\mathcal{K'} \subseteq \mathcal{K}$ thus $\mathcal{V(K')} \subseteq \mathcal{V(K)}$. Hence, $\mathcal{V(K')} = \mathcal{V(K)}$.

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Your approach to the problem is a little convoluted. I assume that you are trying to prove that

every variety $\mathcal{W}$ is generated by the subdirectly irreducible algebras in $\mathcal{W}$,

but at a first glance you seem to try proving the (false) statement every variety generated by a class $\mathcal{K}$ is generated by the subdirectly irreducible algebras in the class. To see this is false, take the two-element lattice $\boldsymbol{2}$ and some non-distributive lattice $\mathbf{L}$ and $\mathcal{K}:=\{\boldsymbol{2},\mathbf{L}\times\mathbf{L}\}$ as your class.

Hint to prove the result: if $\mathbf{A}$ is the subdirect product of $\mathbf{A}_i$, then $\mathbf{A}_i\in H(\{\mathbf{A}\})$ for all $i$.

Hover for the rest of the proof.

Call $\mathcal{W}_{SI}$ the class of subdirectly irreducibles of $\mathcal{W}$. Let $\mathbf{A}\in\mathcal{W}$. Then since $\mathcal{W}$ is a variety, it is closed under taking homomorphic images. Hence every subdirect factor of $\mathbf{A}$ is a member of $\mathcal{W}_{SI}$. Hence every $\mathbf{A}\in\mathcal{W}$ is a subalgebra of a product of elements of $\mathcal{W}_{SI}$, and a fortiori belongs to the variety generated by it.