Decidability of "Is this regular expression prefix-free?"

Say that string $x$ is a prefix of a string $y$ if there exists a string $z$ such that $xz = y$, and say that $x$ is a
proper prefix of $y$ if in addition $x \not= y$. A language is prefix-free if it doesn’t contain a proper
prefix of any of its members.
$$
\text{Prefix-FreeREX} = \{(R) \mid R \text{ is a regular expression and $L(R)$ is prefix-free}\}
$$
I was wondering about how to prove that Prefix-FreeREX is decidable. Also why does a similar approach fail to show that Prefix-FreeCFG is decidable?

Let $L$ be a language on the alphabet $A$. Then $L$ is prefix-free if $L \cap LA^+$ is empty. If $L$ is regular and given by a regular expression, then $L \cap LA^+$ is also regular (and one can effectively compute a regular expression for it). Now testing whether a given regular language is empty is decidable.

For context-free languages, the situation is different. Above proof does not work because the context-free languages are not closed against intersection. In fact, there is no algorithm to decide whether a given context-free grammar generates a prefix-free language.

Here is a proof of this claim [Luc Boasson, personal communication].

Let $(u_1, v_1), \dotsm, (u_n,v_n)$ be two finite lists of words of $A^$ defining an instance of the Post Correspondence Problem (PCP) and let $X = \{x_1, \ldots, x_n\}$ be an alphabet of $n$ letters. Assume that $A$ and $X$ are disjoint and consider the homomorphims $u, v: X^ \to A^*$ defined by $u(x_i) = u_i$ and $v(x_i) = v_i$. Consider the languages
\begin{align}
U &= \{x^Ru(x) \mid x \in X^+ \} \\
V &= \{x^Rv(x) \mid x \in X^+ \}
\end{align}
Let $\sharp$ be a new letter. Then $U \cup V\sharp$ is a context-free language. It is prefix-free if and only if $U$ and $V$ are disjoint if and only if the Post Correspondence Problem has no solution. Thus deciding whether a given context-free language is prefix-free is undecidable.

StackExchange Computer Science Q#41184, answer score: 9