is solvable by AC, then it is solvable by LAAC.

By Theorem \[thm:ac-solves\], there is a homomorphism $f: {\wp}({\mathbf{B}}) \rightarrow {\mathbf{B}}$. Since the structure ${\mathbf{B}}$ has all constants, for each $b \in B$ there is a relation symbol $R_b$ with $R_b^{{\mathbf{B}}} = \{ (b) \}$. Since $(\{ b \}) \in R_b^{{\wp}({\mathbf{B}})}$, it must hold that $f( \{ b \} ) \in R_b^{{\mathbf{B}}}$, from which it follows that $f( \{ b \} ) = b$. The mapping $l$ defined by $l(S, b) = f(S)$ is then a homomorphism of the type described in Theorem \[thm:laac-solves\].

There exists a structure ${\mathbf{B}}$ having all constants such that ${{\mathsf{CSP}}}({\mathbf{B}})$ is solvable by LAAC but not by AC.

Take ${\mathbf{B}}$ to be the relational structure with universe $\{ 0, 1 \}$ over signature $\{ U_0, U_1, R_{(0, 0)}, R_{(1, 1)} \}$ where $$U_0^{\mathbf{B}}= \{ 0 \}$$ $$U_1^{\mathbf{B}}= \{ 1 \}$$ $$R_{(0, 0)}^{\mathbf{B}}= \{ 0, 1 \}^2 \setminus \{ (0, 0) \}$$ $$R_{(1, 1)}^{\mathbf{B}}= \{ 0, 1 \}^2 \setminus \{ (1, 1) \}.$$ It is straightforward to verify that the mapping $l$ defined by $l(\{ 0, 1 \}, b') = b'$, $l(\{ 0 \}, b') = 0$, and $l(\{ 1 \}, b') = 1$ for all 