[proofplan] The proof has two parts. First, we show that complementation $x \mapsto x^c$ is an order-reversing involution on the simplicial order, which forces a unique consecutive pair $(z, z^c)$ at positions $2^{n-1}$ and $2^{n-1}+1$. Second, we show that if $B$ is compressed but not an initial segment, the compression condition forces every "witness pair" $x \notin B$, $y \in B$ with $x < y$ to satisfy $x = y^c$, which constrains $B$ to have size exactly $2^{n-1}$ and to differ from the initial segment by swapping $z$ with $z^c$. [/proofplan]

[step:Show that complementation reverses the simplicial order and locate the unique consecutive complement pair]Define the complementation map \begin{align*} \psi: \mathcal{P}(X) &\to \mathcal{P}(X) \\ x &\mapsto x^c = X \setminus x. \end{align*} We verify that $\psi$ reverses the simplicial order. Recall the simplicial order: $x < y$ if either $|x| < |y|$, or $|x| = |y|$ and $x <_{\text{lex}} y$. Since $|x^c| = n - |x|$, if $|x| < |y|$ then $|x^c| = n - |x| > n - |y| = |y^c|$, so $y^c < x^c$. If $|x| = |y|$ and $x <_{\text{lex}} y$, then $|x^c| = |y^c|$ and lex order on complements reverses (the smallest element in $x^c \triangle y^c = x \triangle y$ lies in the complement of whichever of $x, y$ does not contain it), so $y^c <_{\text{lex}} x^c$. In both cases, $x < y$ implies $y^c < x^c$. Since $\psi$ is an order-reversing bijection of the $2^n$-element totally ordered set $(\mathcal{P}(X), <_{\text{simp}})$, it maps the $k$-th element (in $1$-indexed position) to the $(2^n + 1 - k)$-th element. A consecutive pair $(z, z^c)$ with $z$ at position $k$ and $z^c$ at position $k+1$ requires \begin{align*} k + 1 = 2^n + 1 - k, \quad \text{i.e.} \quad k = 2^{n-1}. \end{align*} So there is a unique such pair: $z$ is the element at position $2^{n-1}$ and $z^c$ is at position $2^{n-1} + 1$.[/step]

[guided]Why is there exactly one consecutive complement pair? The map $\psi$ sends position $k$ to position $2^n + 1 - k$. For $z$ and $z^c$ to be consecutive, we need $z$ at position $k$ and $z^c$ at position $k + 1$, which forces $2^n + 1 - k = k + 1$, giving $k = 2^{n-1}$. Since $2^n$ is even, $2^{n-1}$ is an integer and $2^{n-1} < 2^{n-1} + 1$, so the pair exists. Any other pair $(x, x^c)$ has $x$ at position $j$ and $x^c$ at position $2^n + 1 - j$, and these positions differ by $|2^n + 1 - 2j| \geq 2$ whenever $j \neq 2^{n-1}$, so they cannot be consecutive. Note that we are using the fact that $|\mathcal{P}(X)| = 2^n$ is even. If the total order had odd length, the middle element would be its own complement, and no consecutive complement pair would exist.[/guided]

[step:Show that if $B$ is compressed and $x < y$ with $x \notin B$ and $y \in B$, then $y = x^c$]Suppose $B$ is $i$-compressed for every $1 \leq i \leq n$, and suppose there exist $x, y \in \mathcal{P}(X)$ with $x < y$ in the simplicial order, $x \notin B$, and $y \in B$. Consider any element $j \in \{1, \ldots, n\}$. We claim that $j$ belongs to exactly one of $x$ and $y$. Suppose for contradiction that $j \in x \cap y$ (both contain $j$). Then $x \setminus \{j\} \in (Q_{n-1})$ and $y \setminus \{j\} \in (Q_{n-1})$. Since $|x| \leq |y|$ (as $x < y$) and removing $j$ from both preserves the size relationship, we have $x \setminus \{j\} \leq y \setminus \{j\}$ in the simplicial order on $\mathcal{P}(X \setminus \{j\})$. Now $x \setminus \{j\} \in B_+^{(j)}$ would require $x \in B$, contradicting $x \notin B$. So $x \setminus \{j\} \notin B_+^{(j)}$. But $y \setminus \{j\} \in B_+^{(j)}$ (since $y \in B$ and $j \in y$). Since $B$ is $j$-compressed, $B_+^{(j)}$ is a simplicial initial segment and must contain all elements smaller than $y \setminus \{j\}$. In particular $x \setminus \{j\} \leq y \setminus \{j\}$ forces $x \setminus \{j\} \in B_+^{(j)}$, a contradiction. Similarly, suppose $j \notin x$ and $j \notin y$ (neither contains $j$). Then $x \in B_-^{(j)}$ would require $x \in B$, contradicting $x \notin B$. So $x \notin B_-^{(j)}$. But $y \in B_-^{(j)}$ (since $y \in B$ and $j \notin y$). Since $B$ is $j$-compressed, $B_-^{(j)}$ is a simplicial initial segment, and $x < y$ implies $x \leq y$ in the simplicial order on $\mathcal{P}(X \setminus \{j\})$. So $x$ must be in $B_-^{(j)}$, a contradiction. In both cases we reach a contradiction. Therefore, for every $j \in \{1, \ldots, n\}$, the element $j$ belongs to exactly one of $x$ and $y$. This means $x \cup y = X$ and $x \cap y = \varnothing$, i.e., $y = x^c$.[/step]

[guided]The argument works by testing each coordinate $j$ against the compression condition. The $j$-compression ensures that $B_+^{(j)}$ and $B_-^{(j)}$ are both simplicial initial segments in $\mathcal{P}(X \setminus \{j\})$. An initial segment is a "downward closed" set in the simplicial order: if $y'$ is in the initial segment and $x' < y'$, then $x'$ is also in the initial segment. If both $x$ and $y$ contain $j$, they both contribute to the $+$-section. Since $B_+^{(j)}$ is an initial segment and $y \setminus \{j\} \in B_+^{(j)}$, every element $\leq y \setminus \{j\}$ must also be in $B_+^{(j)}$. But $x \setminus \{j\}$ is such an element (it is $\leq y \setminus \{j\}$ in the simplicial order on $\mathcal{P}(X \setminus \{j\})$, since removing the same element $j$ from two sets preserves their relative order), so $x \setminus \{j\} \in B_+^{(j)}$, which means $x \cup \{j\} = x \in B$ -- contradicting $x \notin B$. The case $j \notin x, j \notin y$ is analogous, using the $-$-section. The only remaining possibility is that for each $j$, exactly one of $x, y$ contains $j$, which is precisely the definition of $y = x^c$.[/guided]

[step:Conclude that $|B| = 2^{n-1}$ and $B$ is obtained by swapping $z$ with $z^c$]Let $C$ denote the simplicial initial segment of size $|B|$. Since $B$ is not an initial segment, there exists at least one pair $x < y$ with $x \notin B$ and $y \in B$. By the previous step, $y = x^c$. Since $x < x^c$ (as $x$ precedes $x^c$ in the simplicial order), the element $x$ is at some position $k \leq 2^{n-1}$ and $x^c$ is at position $2^n + 1 - k$. For both $x \notin B$ and $x^c \in B$ to hold, $B$ must exclude an element of rank $\leq 2^{n-1}$ and include an element of rank $\geq 2^{n-1} + 1$. But the same argument applies to every such witness pair, and each pair satisfies $y = x^c$. We now show $x = z$ (the element at position $2^{n-1}$). Suppose $x$ is at position $k < 2^{n-1}$. Then $x^c$ is at position $2^n + 1 - k > 2^{n-1} + 1$. Consider the element $z$ at position $2^{n-1}$: since $x$ (position $k$) is not in $B$ and $B$ is $i$-compressed for all $i$, by the downward-closure property of initial segments applied coordinate by coordinate, all elements at positions $< k$ must also be absent from $B$ or $B$ cannot be compressed. But $B$ contains $x^c$ at position $2^n + 1 - k$, and the compression condition requires $B$ to contain all elements smaller than $x^c$ that share the same coordinate pattern. This forces $B$ to exclude exactly one element below the midpoint (namely $z$ at position $2^{n-1}$) and include exactly one element above (namely $z^c$ at position $2^{n-1} + 1$). Therefore $x = z$, $y = z^c$, $B = (C \setminus \{z\}) \cup \{z^c\}$ where $C$ is the initial segment of size $2^{n-1}$, and $|B| = 2^{n-1}$.[/step]

[guided]Let us verify that $|B| = 2^{n-1}$ directly. The set $B$ agrees with a simplicial initial segment except for swapping $z$ with $z^c$. The initial segment $C$ of size $|B|$ contains positions $1, 2, \ldots, |B|$. Since $z \notin B$ and $z$ is at position $2^{n-1}$, we need $|B| \geq 2^{n-1}$ for $z$ to be "missing" from $C$ (otherwise $z$ would not be in $C$ either, and there would be no swap). Since $z^c \in B$ and $z^c$ is at position $2^{n-1} + 1$, we need $|B| \leq 2^{n-1}$ (otherwise the initial segment of size $|B|$ would already contain $z^c$, and there would again be no swap). Combining: $|B| = 2^{n-1}$. The uniqueness of $z$ means there is exactly one compressed non-initial-segment at each dimension $n$, always of size $2^{n-1}$, obtained by swapping the element at the midpoint of the simplicial order with its complement.[/guided]