Let $(A,F)$ be an algebra, let $F^{\star}$ be the centralizer of $F$ and $F^{\star\star}:=(F^{\star})^{\star}$ the bi-centralizer. Let $[F]$ denote the clone of operations generated by $F$.
We say that:
1) $F$ acts bicentrally on $A$ when $[F]=F^{\star\star}$ ($[F]\subseteq F^{\star\star}$ is true in any case)
2) $F$ acts fully on $A$ if for each $\phi\in F^{\star\star}$ and each $c\in A^n$, where $n$ is the arity of $\phi$, there exists $\phi'\in[F]$ such that $c\phi=c\phi'$
3) $F$ acts densely on $A$ when for any finite set of $n$-tuples $c_1,\ldots ,c_r\in A^n$ there exists $\phi'\in [F]$ such that $c_i\phi=c_i\phi'$ for all $i$.
Clearly, bicentrally implies densely implies fully. I am trying to prove that bicentrally is equivalent to densely in the following particular case: When for all $n$, the $F^{\star}$-algebra $A^n$ is finitely generated. I found a proof of this on "Universal Algebra", P.M.Cohn, at page 129 but unfortunately is wrong. Can someone provide a correct proof?