$ \newcommand{\lang}{L} \newcommand{\Nset}{\mathbb N} \newcommand{\Lset}{\mathcal L} \newcommand{\Rec}{\mathcal R} \newcommand{\RecEnum}{\Rec_\Nset} \newcommand{\accept}{\mathbf{a}} \newcommand{\reject}{\mathbf{r}} \newcommand{\halt}{\mathbf{h}} $ Let $\Rec$ be the set of recursive languages and $\RecEnum$ the set of recursively enumerable languages (hence the $\Nset$ in the subindex). If $\lang\in\RecEnum$ and $\lang^c \in \RecEnum$, where $\lang^c$ is the complement of $\lang$, is $L \in \Rec$?
Some thoughts
Ok, so the only tools I have available to me at the moment are Turing-machines $T$ and the fact that $\Rec\subset\RecEnum$. I'm aware that a language $L \in \RecEnum$, if some Turing machine accepts it, as in ends up in the special accept-state $\accept$ when reading input $l\in\lang$.
On the other hand, a language $L$ is simply recursive, if it is solved by a Turing-machine and at the same time the Turing-machine $T$ halts, given any other input. Solving a language means that a Turing-machine accepts any input $l \in L$ and rejects it, if $l \notin L$. Halting means a Turing-machine ends up in one of its special states $\accept$, $\reject$ or $\halt$, accept, reject and halt respectively, during its execution.
In other words, if $\lang \in \RecEnum$ and $\lang^c \in \RecEnum$, they should both be acceptable via some Turing machines $T$ and $T^c$, as in $\lang = \lang(T)$ and $\lang^c = \lang(T^c)$. However, this does not mean that the machines reject or halt given the respective input $l^c \notin \lang$ and $l\notin\lang^c$.
I would then be inclined to say that the claim is false, simply because to me it seems like having $\lang^c\in\RecEnum$ as well says nothing about the recursive nature of $\lang$. But are there any concrete counterexamples?