[proofplan]
The hypotheses of hypersimplicity already give that $A$ is computably enumerable and has infinite complement, so only immunity of the complement must be proved. Assume an infinite computably enumerable set $B$ lies inside $\mathbb{N} \setminus A$. From an enumeration of $B$, extract a computable one-to-one sequence $(b_n)_{n \in \mathbb{N}}$ of elements of $B$, and let $D_n=\{b_n\}$. These singleton sets form a pairwise disjoint strong array meeting $\mathbb{N} \setminus A$ at every index, contradicting hypersimplicity.
[/proofplan]
[step:Reduce simplicity to ruling out infinite computably enumerable subsets of the complement]
Since $A$ is hypersimple, by definition $A$ is computably enumerable and $\mathbb{N} \setminus A$ is infinite. Therefore, to prove that $A$ is simple, it remains only to prove that $\mathbb{N} \setminus A$ is immune, meaning that there is no infinite computably enumerable set $B \subset \mathbb{N} \setminus A$.
[/step]
[step:Assume an infinite computably enumerable subset of the complement exists]
Suppose, toward contradiction, that there exists an infinite computably enumerable set $B \subset \mathbb{N} \setminus A$. Since $B$ is computably enumerable, there is a total computable map
\begin{align*}
e: \mathbb{N} &\to \mathbb{N}
\end{align*}
whose range is $B$. Since $B$ is infinite, for each $n \in \mathbb{N}$ the enumeration $e(1),e(2),\dots$ contains at least $n$ distinct values.
[/step]
[step:Extract a computable injective sequence from the enumeration of $B$]
Define a map
\begin{align*}
b: \mathbb{N} &\to \mathbb{N}
\end{align*}
as follows. For each $n \in \mathbb{N}$, run the computable enumeration $e$ until $n$ distinct values have appeared, and let $b_n$ be the $n$th distinct value that appears. This procedure halts for every $n$ because $B$ is infinite, and it is computable because one only compares finitely many already observed natural numbers at each finite stage. Thus $(b_n)_{n \in \mathbb{N}}$ is a computable injective sequence with $b_n \in B$ for every $n \in \mathbb{N}$.
[guided]
We need to turn the assumed infinite computably enumerable subset $B$ into a strong array of finite sets. The cleanest way is to extract one new element of $B$ at each index. Let
\begin{align*}
e: \mathbb{N} &\to \mathbb{N}
\end{align*}
be a total computable enumeration with range $B$.
For each $n \in \mathbb{N}$, we compute $b_n$ by scanning the finite list
\begin{align*}
e(1), e(2), \dots, e(s)
\end{align*}
until this list contains $n$ distinct values. Since $B$ is infinite and the range of $e$ is exactly $B$, such a stage $s$ exists. The algorithm then outputs the $n$th distinct value discovered. This proves that the map
\begin{align*}
b: \mathbb{N} &\to \mathbb{N}
\end{align*}
is computable.
The sequence is injective because $b_n$ is defined to be a fresh value not previously counted among $b_1,\dots,b_{n-1}$. Also, every $b_n$ belongs to $B$, since each $b_n$ is some value of the enumeration $e$. Therefore $(b_n)_{n \in \mathbb{N}}$ is a computable one-to-one sequence of elements of $B$.
[/guided]
[/step]
[step:Build a strong array of singleton sets meeting the complement]
For each $n \in \mathbb{N}$, define the finite set
\begin{align*}
D_n := \{b_n\}.
\end{align*}
The sequence $(D_n)_{n \in \mathbb{N}}$ is uniformly computable: given $n,x \in \mathbb{N}$, compute $b_n$ and decide
\begin{align*}
x \in D_n \iff x=b_n.
\end{align*}
Each $D_n$ is finite, and the sets are pairwise disjoint because the sequence $(b_n)_{n \in \mathbb{N}}$ is injective. Hence $(D_n)_{n \in \mathbb{N}}$ is a strong array of pairwise disjoint finite sets.
Moreover, since $b_n \in B \subset \mathbb{N} \setminus A$, we have
\begin{align*}
D_n \cap (\mathbb{N} \setminus A) = \{b_n\} \neq \varnothing
\end{align*}
for every $n \in \mathbb{N}$.
[/step]
[step:Contradict hypersimplicity and conclude simplicity]
The strong array $(D_n)_{n \in \mathbb{N}}$ is pairwise disjoint and satisfies $D_n \cap (\mathbb{N} \setminus A) \neq \varnothing$ for every $n \in \mathbb{N}$. This contradicts hypersimplicity, which requires that every pairwise disjoint strong array have some member $D_n \subset A$.
Therefore no infinite computably enumerable set $B \subset \mathbb{N} \setminus A$ exists. Since hypersimplicity already gives that $A$ is computably enumerable and that $\mathbb{N} \setminus A$ is infinite, the set $A$ is simple.
[/step]
