\operatorname{Abs}(C)$. Then, $\ok{\tuple{A,f^\sharp}}$ is a sound abstract interpretation when $\ok{\alpha \circ f \sqsubseteq f^\sharp\circ \alpha}$. Moreover, the abstract function $\ok{f^A {\triangleq}\alpha \circ f \circ \gamma: A\rightarrow A}$ is called the *best correct approximation* (b.c.a.) of $f$ on $A$ because any abstract interpretation $\ok{\tuple{A,f^\sharp}}$ is sound iff $\ok{f^A \sqsubseteq f^\sharp}$. Hence, for any abstraction $A$, $\ok{f^A}$ plays the role of the best possible approximation of $f$ on $A$.

The Problem
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Given a semantic function $f:C\rightarrow C$ on some concrete domain $C$ and an abstraction $A \in \operatorname{Abs}(C)$, does there exist the *most abstract domain* that induces the same best correct approximation of $f$ as $A$ does?

Let us formalize the above question. Consider two abstractions $A,B\in \operatorname{Abs}(C)$. We say that $A$ and $B$ induce the same best correct approximation of $f$ when $f^A$ and $f^B$ are the same function up to isomorphic representations of abstract values. If $\mu_A$ and $\mu_B$ are the corresponding ucos then this boils down to: $$\mu_A {\circ}f {\circ}\mu_A = \mu_B {\circ}f {\circ}\mu_B .$$ In order to keep the notation easy, this is denoted simply by $\ok{f^A = f^B}$. Also, if $F\subseteq C{\rightarrow}C$ is a set of concrete functions then $\ok{F^A = F^B}$ means that for any 