and hence AC solves ${{\mathsf{CSP}}}({\mathbf{B}})$ by Theorem \[thm:ac-solves\].

To show that the problem ${{\mathsf{CSP}}}({\mathbf{B}})$ is not solvable by LAAC, let $f$ be an arbitrary mapping from $({\wp}(B) \setminus \{ \emptyset \}) \times B$ to $B$ that satisfies $f(\{ b \}, b') = b$ for all $b, b' \in B$. We show that $f$ cannot be a homomorphism from ${\wp}({\mathbf{B}}) \times {\mathbf{B}}$ to ${\mathbf{B}}$, which suffices by Theorem \[thm:laac-solves\]. We consider two cases depending on the value of $f( \{ 0, 1 \}, 0)$.

-   If $f( \{ 0, 1 \}, 0) = 1$, then we use the facts that $(\{ 0 \}, \{ 0, 1 \}, \{ 0, 1 \}) \in R^{{\wp}({\mathbf{B}})}$ and that $(0, 0, 0) \in R^{{\mathbf{B}}}$; we have that $(f( \{0 \}, 0), f( \{ 0, 1 \}, 0), f( \{ 0, 1 \}, 0)) = (0, 1, 1)$, which is not contained in $R^{{\mathbf{B}}}$, implying that $f$ is not a homomorphism of the desired type.

-   If $f( \{ 0, 1 \}, 0) = 0$, then we use the facts that $(\{ 1 \}, \{ 0, 1 \}, \{ 0, 1 \}) \in S^{{\wp}({\mathbf{B}})}$ and that $(0, 0, 0) \in S^{{\mathbf{B}}}$; we have that $(f( \{ 