Let $L$ be some language, and consider the operator $$ \min(L) := \{ u \in L \mid \mbox{no proper prefix of $u$ is in $L$} \} $$ where a word $u$ is called a prefix of $w$ if it is an initial segment of $w$, i.e. $w = ux$ for some $x \in X^{\ast}$, and it is proper if $u \ne w$. Now the prefix relation is a partial order, and we have that $\min(L)$ are precisely the minimal element in the partially ordered set $L$. For $u \in X^{\ast}$ denote by $A(u)$ the set of prefixes of $u$.
Now I am interested how $\min$ behaves with respect to some language operations, first it is easy to see that $\min(L^{\ast}) = \{\varepsilon\}$, as in general $\min(L) = \{\varepsilon\}$ if $\varepsilon \in L$. Now if we set for $U,V \subseteq X^{\ast}$ $$ c(U,V) := \{ u \in U \mid A(u) \setminus \{u\} \cap V \ne \emptyset \} $$ i.e. those $U$ which have no proper prefix in $V$, we have I guess $$ \min(U \cup V) = (\min(U) \cup \min(V)) \setminus ( c(U,V) \cup c(V,U) ) $$ Surely as if $u \in \min(U\cup V)$ and $u \in U$, then as it is minimal for all element from $U\cup V$, it is also minimal for the elements from just $U$, hence $u \in \min(U)$, and similar for $u \in V$, so $\min(U\cup V)\subseteq \min(U)\cup \min(V)$, and also if $u \in U$ it has no prefix in $V$ by definition of $\min(U\cup V)$ and similar if $v \in V$ it has no prefix in $U$, hence taken together we have $\min(U\cup V) \subseteq (\min(U) \cup \min(V)) \setminus ( c(U,V) \cup c(V,U) )$. Conversely suppose $u \in (\min(U) \cup \min(V)) \setminus ( c(U,V) \cup c(V,U) )$ and $u \in U$, then as it has no proper prefix in $U$ as $\min(U)$ and it has no proper prefix in $V$ as $u \notin c(U,v)$ we have $u \in \min(U\cup V)$. $\square$
But for concatenation I am stuck. Is there any expression for $$ \min(U\cdot V) = \quad ? $$ I see that we must have $UV \cap \min(U) \subseteq \min(UV)$, for if $x \in \min(U)$ and $uv \le x$ for $u \in U, v \in V$, we must have $u \le x$, hence $u = x$. Further as this implies $v = \varepsilon$, if $\varepsilon \notin V$ we could not have $uv \le x$, so it is also minimal in $UV$.
But this is all I see, the other inclusion does not hold, for example consider $U = \{a,ab\}, V = \{b\}$, then $\min(UV) = \{ab\}, \min U = \{a\}$ and $UV \cap \min U = \emptyset$.
So does anyone knows how the $\min$-operator behaves with respect to concatenation?
If $X$ is the alphabet, then $\min L = L-LX^+$. Thus $$ (1) \quad \min (UV) = UV - UVX^+. $$ If $U = \min U$, that is, if $U$ is a prefix code, then $\min (UV) = U\min V$, but I am afraid there is no interesting relation apart from (1) in the general case.