[proofplan] We prove by induction on $s = |U| = |V|$ that $\partial C_{UV}(\mathcal{A}) \subseteq C_{UV}(\partial \mathcal{A})$. The base case $s = 1$ is the [Simple Compression Reduces Shadow](/theorems/2592) theorem. For the inductive step, we take an arbitrary $B \in \partial C_{UV}(\mathcal{A})$ and find a parent $A' \in C_{UV}(\mathcal{A})$ with $B \subset A'$. The proof splits into two cases based on whether $B \cap (U \cup V) = V$ or not. When $B \cap (U \cup V) \neq V$, the compression fixes $B$, and we show $B \in \partial \mathcal{A}$ by constructing an explicit parent in $\mathcal{A}$, using an $(s-1)$-element sub-compression when the parent $A'$ is a compressed image rather than an original member of $\mathcal{A}$. When $B \cap (U \cup V) = V$, we show both $B$ and $C_{UV}(B)$ lie in $\partial \mathcal{A}$ by locating parents for each in $\mathcal{A}$, using the structure of $C_{UV}(\mathcal{A})$ and the sub-compression hypothesis. [/proofplan]

[step:Establish the base case $s = 1$ as the simple compression theorem]When $|U| = |V| = 1$, write $U = \{u\}$ and $V = \{v\}$. The compression $C_{UV}$ is the simple compression $C_{uv}$. The hypothesis requires that for every $u' \in U$, there exists $v' \in V$ such that $\mathcal{A}$ is $(U \setminus \{u'\}, V \setminus \{v'\})$-compressed. The only choice is $u' = u$, $v' = v$, giving $C_{\varnothing, \varnothing}$, which is the identity. Every family is $(\varnothing, \varnothing)$-compressed, so the hypothesis is vacuous. The conclusion $\partial C_{uv}(\mathcal{A}) \subseteq C_{uv}(\partial \mathcal{A})$ is the [Simple Compression Reduces Shadow](/theorems/2592) theorem.[/step]

[guided]When $|U| = |V| = 1$, write $U = \{u\}$ and $V = \{v\}$. The compression $C_{UV}$ is the simple compression $C_{uv}$, which acts on a set $A$ by replacing $v$ with $u$ when $v \in A$ and $u \notin A$, and fixing $A$ otherwise. The inductive hypothesis requires that for every $u' \in U$, there exists $v' \in V$ such that $\mathcal{A}$ is $(U \setminus \{u'\}, V \setminus \{v'\})$-compressed. Since $U$ and $V$ are singletons, the only possibility is $u' = u$ and $v' = v$, yielding the sub-compression $C_{\varnothing, \varnothing}$. This is the identity map on all families: the condition $A \cap (\varnothing \cup \varnothing) = \varnothing$ is vacuously satisfied, and $(A \setminus \varnothing) \cup \varnothing = A$, so $C_{\varnothing,\varnothing}(A) = A$ for every set $A$. Every family is therefore $(\varnothing, \varnothing)$-compressed, so the hypothesis places no constraint on $\mathcal{A}$. The conclusion $\partial C_{uv}(\mathcal{A}) \subseteq C_{uv}(\partial \mathcal{A})$ is precisely the statement of the [Simple Compression Reduces Shadow](/theorems/2592) theorem, which we take as established.[/guided]

[step:Take $B \in \partial C_{UV}(\mathcal{A})$ and set up the case analysis]Suppose $s \geq 2$ and the result holds for all compressions of size less than $s$. Let $B \in \partial C_{UV}(\mathcal{A})$. By definition of the lower shadow, there exists $A' \in C_{UV}(\mathcal{A})$ with $B \subset A'$ and $|A' \setminus B| = 1$. Write $A' = B \cup \{x\}$. We must show $B \in C_{UV}(\partial \mathcal{A})$. Membership $B \in C_{UV}(\partial \mathcal{A})$ means either (i) $B = C_{UV}(D)$ for some $D \in \partial \mathcal{A}$, or (ii) $B \in \partial \mathcal{A}$ and $C_{UV}(B) \in \partial \mathcal{A}$. We split into cases: either $B \cap (U \cup V) = V$ (so $C_{UV}(B) = (B \setminus V) \cup U \neq B$) or $B \cap (U \cup V) \neq V$ (so $C_{UV}(B) = B$).[/step]

[guided]Suppose $s \geq 2$ and the result holds for all compressions $C_{U', V'}$ with $|U'| = |V'| < s$. Let $B \in \partial C_{UV}(\mathcal{A})$. By definition of the lower shadow, there exists $A' \in C_{UV}(\mathcal{A})$ such that $B \subset A'$ and $|A' \setminus B| = 1$, so $A' = B \cup \{x\}$ for a unique element $x \notin B$. Our goal is to show $B \in C_{UV}(\partial \mathcal{A})$. Unpacking the definition of $C_{UV}$ applied to the family $\partial \mathcal{A}$, membership $B \in C_{UV}(\partial \mathcal{A})$ means one of: - There exists $D \in \partial \mathcal{A}$ such that $C_{UV}(D) = B$ (where possibly $D = B$ when $C_{UV}$ fixes $B$, or $D \neq B$ when $B$ is the compressed image of $D$). - $B \in \partial \mathcal{A}$ and $C_{UV}(B) \in \partial \mathcal{A}$ (the set $B$ is compressible, and both $B$ and its image belong to the shadow). The compression $C_{UV}$ changes a set $S$ exactly when $S \cap (U \cup V) = V$, i.e., $S$ contains all of $V$ and none of $U$. Otherwise $C_{UV}(S) = S$. This dichotomy on $B$ drives the two main cases: if $B \cap (U \cup V) \neq V$, then $C_{UV}(B) = B$ and we aim for option (ii) with both copies equal to $B$; if $B \cap (U \cup V) = V$, then $C_{UV}(B) \neq B$ and we aim for option (ii) by finding parents in $\mathcal{A}$ for both $B$ and $C_{UV}(B)$.[/guided]

[step:Handle $B \cap (U \cup V) \neq V$: the compression fixes $B$, so show $B \in \partial \mathcal{A}$]Suppose $B \cap (U \cup V) \neq V$, so $C_{UV}(B) = B$. It suffices to show $B \in \partial \mathcal{A}$. Since $A' \in C_{UV}(\mathcal{A})$, there exists $A \in \mathcal{A}$ with $A' \in \{A, C_{UV}(A)\}$. **Sub-case (a): $A' \in \mathcal{A}$.** Then $B \subset A' \in \mathcal{A}$ with $|A' \setminus B| = 1$, so $B \in \partial \mathcal{A}$. **Sub-case (b): $A' \notin \mathcal{A}$ and $A' = C_{UV}(A) = (A \setminus V) \cup U$ for some $A \in \mathcal{A}$ with $A \cap (U \cup V) = V$.** Here $U \subseteq A'$, $A' \cap V = \varnothing$, and $B \cap V = \varnothing$ (since $B \subset A'$). Since $U \subseteq A' = B \cup \{x\}$, either $U \subseteq B$ or $x \in U$. *If $x \notin U$:* Then $U \subseteq B$ and $x \in A' \setminus U = A \setminus V$, so $x \in A$ and $x \notin U \cup V$. Set $D = A \setminus \{x\}$. Then $D \in \partial \mathcal{A}$ (since $D \subset A \in \mathcal{A}$), and $D \cap (U \cup V) = V$ (since $x \notin U \cup V$), so $C_{UV}(D) = (D \setminus V) \cup U = A' \setminus \{x\} = B$. Hence $B = C_{UV}(D)$ with $D \in \partial \mathcal{A}$, giving $B \in C_{UV}(\partial \mathcal{A})$. *If $x \in U$:* Set $u_0 = x$. By hypothesis, there exists $v_0 \in V$ such that $\mathcal{A}$ is $(U', V')$-compressed, where $U' = U \setminus \{u_0\}$ and $V' = V \setminus \{v_0\}$. Since $A \cap (U \cup V) = V$, we have $A \cap (U' \cup V') = V'$, so $C_{U'V'}(A) = (A \setminus V') \cup U'$. Because $\mathcal{A}$ is $(U', V')$-compressed, $C_{U'V'}(A) \in \mathcal{A}$. Now $B = A' \setminus \{u_0\} = (A \setminus V) \cup (U \setminus \{u_0\}) = (A \setminus V) \cup U'$. And $C_{U'V'}(A) = (A \setminus V') \cup U' = (A \setminus V) \cup \{v_0\} \cup U'$ (since $v_0 \in V \subseteq A$ and $v_0 \notin V'$). Therefore $B = C_{U'V'}(A) \setminus \{v_0\}$. Since $C_{U'V'}(A) \in \mathcal{A}$ and $v_0 \in C_{U'V'}(A)$, we conclude $B \in \partial \mathcal{A}$, giving $B \in C_{UV}(\partial \mathcal{A})$.[/step]

[guided]Suppose $B \cap (U \cup V) \neq V$. Then $C_{UV}(B) = B$. To place $B$ in $C_{UV}(\partial \mathcal{A})$, it suffices to show $B \in \partial \mathcal{A}$ (since then $B \in \partial \mathcal{A}$ and $C_{UV}(B) = B \in \partial \mathcal{A}$, and the definition of $C_{UV}(\partial \mathcal{A})$ retains sets $D \in \partial \mathcal{A}$ with $C_{UV}(D) \in \partial \mathcal{A}$). Since $A' \in C_{UV}(\mathcal{A})$, there exists $A \in \mathcal{A}$ such that $A'$ is either $A$ itself (when $C_{UV}(A) \in \mathcal{A}$, so both $A$ and $C_{UV}(A)$ are retained) or $C_{UV}(A)$ (when $A$ is replaced by its compressed image). Write $A' = B \cup \{x\}$. **Sub-case (a): $A' \in \mathcal{A}$.** This is immediate: $B \subset A' \in \mathcal{A}$ with $|A' \setminus B| = 1$ gives $B \in \partial \mathcal{A}$ by definition of the lower shadow. **Sub-case (b): $A' \notin \mathcal{A}$ and $A' = C_{UV}(A) = (A \setminus V) \cup U$ for some $A \in \mathcal{A}$ with $A \cap (U \cup V) = V$.** The set $A$ contains all of $V$ and none of $U$, while $A' = (A \setminus V) \cup U$ contains all of $U$ and none of $V$. Since $B \subset A'$ and $A' \cap V = \varnothing$, we have $B \cap V = \varnothing$. Since $U \subseteq A' = B \cup \{x\}$, we split on whether $x \in U$. **If $x \notin U$:** Then $U \subseteq B$. The element $x$ satisfies $x \in A' \setminus U$. Since $A' = (A \setminus V) \cup U$, we have $A' \setminus U = A \setminus V$, so $x \in A \setminus V$ and in particular $x \in A$ and $x \notin U \cup V$. Define $D = A \setminus \{x\}$. Then $D \subset A \in \mathcal{A}$ with $|A \setminus D| = 1$, giving $D \in \partial \mathcal{A}$. Since $x \notin U \cup V$, removing $x$ does not affect the intersection with $U \cup V$: $D \cap (U \cup V) = A \cap (U \cup V) = V$. So $C_{UV}(D) = (D \setminus V) \cup U$. We compute: \begin{align*} C_{UV}(D) = (A \setminus \{x\} \setminus V) \cup U = ((A \setminus V) \setminus \{x\}) \cup U = ((A \setminus V) \cup U) \setminus \{x\} = A' \setminus \{x\} = B, \end{align*} where the penultimate equality uses $x \notin U$. Therefore $B = C_{UV}(D)$ with $D \in \partial \mathcal{A}$, so $B \in C_{UV}(\partial \mathcal{A})$. **If $x \in U$:** Set $u_0 = x \in U$, so $B = A' \setminus \{u_0\}$ and $U \setminus \{u_0\} \subseteq B$. By hypothesis, there exists $v_0 \in V$ such that $\mathcal{A}$ is $(U', V')$-compressed, where $U' = U \setminus \{u_0\}$ and $V' = V \setminus \{v_0\}$ with $|U'| = |V'| = s - 1$. The idea is to use the sub-compression to find a parent for $B$ in $\mathcal{A}$. Since $A \in \mathcal{A}$ and $A \cap (U \cup V) = V$, in particular $A \cap U = \varnothing$ (so $u_0 \notin A$) and $V \subseteq A$ (so $v_0 \in A$). We check whether $C_{U'V'}$ acts on $A$: $A \cap (U' \cup V') = A \cap V' = V'$ (using $A \cap U' = \varnothing$ and $V' \subset V \subseteq A$). So $C_{U'V'}(A) = (A \setminus V') \cup U'$. Since $\mathcal{A}$ is $(U', V')$-compressed, $C_{U'V'}(A) \in \mathcal{A}$. Now we express $B$ in terms of $C_{U'V'}(A)$: \begin{align*} B &= A' \setminus \{u_0\} = ((A \setminus V) \cup U) \setminus \{u_0\} = (A \setminus V) \cup U'. \end{align*} For $C_{U'V'}(A)$: since $A \setminus V' = A \setminus (V \setminus \{v_0\}) = (A \setminus V) \cup \{v_0\}$ (the element $v_0 \in V \subseteq A$ is kept when we remove only $V' = V \setminus \{v_0\}$), \begin{align*} C_{U'V'}(A) = (A \setminus V) \cup \{v_0\} \cup U'. \end{align*} Comparing: $B = (A \setminus V) \cup U' = C_{U'V'}(A) \setminus \{v_0\}$. Since $v_0 \in C_{U'V'}(A)$ and $C_{U'V'}(A) \in \mathcal{A}$, we get $B \in \partial \mathcal{A}$. Why does this work? The sub-compression hypothesis $\mathcal{A}$ is $(U', V')$-compressed guarantees $C_{U'V'}(A) \in \mathcal{A}$, which provides the parent set we need. Without this, we would have no way to find a member of $\mathcal{A}$ containing $B$, because $A'$ itself is not in $\mathcal{A}$ in this sub-case.[/guided]

[step:Handle $B \cap (U \cup V) = V$: show both $B$ and $C_{UV}(B)$ belong to $\partial \mathcal{A}$]Suppose $B \cap (U \cup V) = V$, so $V \subseteq B$, $U \cap B = \varnothing$, and $C_{UV}(B) = (B \setminus V) \cup U$. We show $B \in \partial \mathcal{A}$ and $C_{UV}(B) \in \partial \mathcal{A}$. **Showing $B \in \partial \mathcal{A}$:** Since $V \subseteq B \subset A'$, we have $V \subseteq A'$. We claim $A' \in \mathcal{A}$. If $A' \notin \mathcal{A}$, then $A' = C_{UV}(\tilde{A}) = (\tilde{A} \setminus V) \cup U$ for some $\tilde{A} \in \mathcal{A}$ with $\tilde{A} \cap (U \cup V) = V$, giving $A' \cap V = \varnothing$. But $V \subseteq A'$ forces $V = \varnothing$, contradicting $s \geq 2$. So $A' \in \mathcal{A}$, and $B \in \partial \mathcal{A}$. **Showing $C_{UV}(B) \in \partial \mathcal{A}$:** Since $V \subseteq B$ and $x \in A' \setminus B$, we have $x \notin V$. We split on $x \in U$ vs $x \notin U$. *Case $x \notin U$:* Then $A' \cap (U \cup V) = V$ (since $B \cap U = \varnothing$ and $x \notin U$). We verify $C_{UV}(A') \in \mathcal{A}$. Since $A' \in \mathcal{A}$ and $A' \in C_{UV}(\mathcal{A})$, and $A'$ cannot appear as $C_{UV}(\tilde{A})$ for any $\tilde{A}$ (that would give $A' \cap V = \varnothing$, contradicting $V \subseteq A'$), the set $A'$ enters $C_{UV}(\mathcal{A})$ via the second component $\{A \in \mathcal{A} : C_{UV}(A) \in \mathcal{A}\}$, which requires $C_{UV}(A') \in \mathcal{A}$. Now $C_{UV}(B) = (B \setminus V) \cup U$ and $C_{UV}(A') = (A' \setminus V) \cup U$. Since $B = A' \setminus \{x\}$ and $x \notin V$, we have $C_{UV}(B) = C_{UV}(A') \setminus \{x\}$. Since $x \in A' \setminus V$ gives $x \in C_{UV}(A')$ (and $x \notin U$ ensures $x$ is not already supplied by the $\cup U$ part), $C_{UV}(B) \subset C_{UV}(A') \in \mathcal{A}$ with $|C_{UV}(A') \setminus C_{UV}(B)| = 1$, so $C_{UV}(B) \in \partial \mathcal{A}$. *Case $x \in U$:* Set $u_0 = x$. Then $A' \cap U = \{u_0\}$, so $A' \cap (U \cup V) = V \cup \{u_0\} \neq V$ and $C_{UV}(A') = A'$. By hypothesis, there exists $v_0 \in V$ with $\mathcal{A}$ being $(U', V')$-compressed for $U' = U \setminus \{u_0\}$, $V' = V \setminus \{v_0\}$. Since $A' \cap (U' \cup V') = V'$ (because $A' \cap U' = \varnothing$ from $B \cap U = \varnothing$ and $u_0 \notin U'$, and $A' \cap V' = V'$ from $V \subseteq A'$), we get $C_{U'V'}(A') = (A' \setminus V') \cup U' \in \mathcal{A}$. We compute both sides. Since $V' = V \setminus \{v_0\}$ and $V \subseteq A'$, we have $A' \setminus V' = (A' \setminus V) \cup \{v_0\}$, giving $C_{U'V'}(A') = (A' \setminus V) \cup \{v_0\} \cup U'$. For $C_{UV}(B)$: since $B = A' \setminus \{u_0\}$ and $u_0 \in A' \setminus V$ (because $u_0 \in U$ and $U \cap V = \varnothing$), we get $C_{UV}(B) = ((A' \setminus V) \setminus \{u_0\}) \cup U = (A' \setminus V) \cup U'$ (since adding $U = U' \cup \{u_0\}$ restores $u_0$, while $u_0 \in A' \setminus V$ makes the removal redundant). Therefore $C_{UV}(B) = C_{U'V'}(A') \setminus \{v_0\}$, and since $v_0 \in C_{U'V'}(A')$ and $C_{U'V'}(A') \in \mathcal{A}$, we conclude $C_{UV}(B) \in \partial \mathcal{A}$. In both sub-cases, $B \in \partial \mathcal{A}$ and $C_{UV}(B) \in \partial \mathcal{A}$, so $B \in C_{UV}(\partial \mathcal{A})$.[/step]

[guided]Suppose $B \cap (U \cup V) = V$, so $B$ contains all of $V$ and none of $U$. The compression does not fix $B$: $C_{UV}(B) = (B \setminus V) \cup U \neq B$. To place $B$ in $C_{UV}(\partial \mathcal{A})$, we show both $B \in \partial \mathcal{A}$ and $C_{UV}(B) \in \partial \mathcal{A}$. (The definition of $C_{UV}(\partial \mathcal{A})$ retains any set $D \in \partial \mathcal{A}$ satisfying $C_{UV}(D) \in \partial \mathcal{A}$.) **Showing $B \in \partial \mathcal{A}$ by proving $A' \in \mathcal{A}$.** We have $V \subseteq B \subset A'$, so $V \subseteq A'$. We claim $A' \in \mathcal{A}$. Suppose for contradiction that $A' \notin \mathcal{A}$. Since $A' \in C_{UV}(\mathcal{A})$, it must appear as $C_{UV}(\tilde{A})$ for some $\tilde{A} \in \mathcal{A}$ with $\tilde{A} \cap (U \cup V) = V$. Then $A' = (\tilde{A} \setminus V) \cup U$, which satisfies $A' \cap V = \varnothing$ (the elements of $V$ were removed and the disjoint set $U$ was added). But $V \subseteq A'$ and $A' \cap V = \varnothing$ force $V = \varnothing$, contradicting $s = |V| \geq 2$. Therefore $A' \in \mathcal{A}$. Since $B \subset A' \in \mathcal{A}$ with $|A' \setminus B| = 1$, we get $B \in \partial \mathcal{A}$. This handles half of what we need. **Showing $C_{UV}(B) \in \partial \mathcal{A}$: finding a parent.** Write $A' = B \cup \{x\}$. Since $V \subseteq B$ and $x \in A' \setminus B$, we have $x \notin V$. We split on whether $x \in U$ or $x \notin U$, as this determines whether $C_{UV}$ acts on $A'$. **If $x \notin U$:** Then $A' \cap U = \varnothing$ (since $B \cap U = \varnothing$ and $x \notin U$), so $A' \cap (U \cup V) = V$ and $C_{UV}(A') = (A' \setminus V) \cup U \neq A'$. We verify $C_{UV}(A') \in \mathcal{A}$. The set $A'$ belongs to $C_{UV}(\mathcal{A})$ by hypothesis. How does $A'$ enter $C_{UV}(\mathcal{A})$? Recall $C_{UV}(\mathcal{A}) = \{C_{UV}(A) : A \in \mathcal{A}\} \cup \{A \in \mathcal{A} : C_{UV}(A) \in \mathcal{A}\}$. Could $A'$ appear in the first component, i.e., $A' = C_{UV}(\tilde{A})$ for some $\tilde{A} \in \mathcal{A}$? That would require $A' = (\tilde{A} \setminus V) \cup U$, giving $A' \cap V = \varnothing$. But $V \subseteq A'$, so $V = \varnothing$, contradicting $s \geq 2$. Therefore $A'$ enters $C_{UV}(\mathcal{A})$ only via the second component: $A' \in \mathcal{A}$ and $C_{UV}(A') \in \mathcal{A}$. Now we relate $C_{UV}(B)$ to $C_{UV}(A')$: \begin{align*} C_{UV}(B) &= (B \setminus V) \cup U, \\ C_{UV}(A') &= (A' \setminus V) \cup U = ((B \cup \{x\}) \setminus V) \cup U = (B \setminus V) \cup \{x\} \cup U, \end{align*} where the last equality uses $x \notin V$. Since $x \notin U$ either, the element $x$ belongs to $C_{UV}(A')$ but not to $U$, so $x \in C_{UV}(A') \setminus C_{UV}(B)$. Therefore $C_{UV}(B) = C_{UV}(A') \setminus \{x\}$, giving $C_{UV}(B) \subset C_{UV}(A') \in \mathcal{A}$ with $|C_{UV}(A') \setminus C_{UV}(B)| = 1$. Hence $C_{UV}(B) \in \partial \mathcal{A}$. **If $x \in U$:** Set $u_0 = x$. Then $A' \cap U = \{u_0\}$ (since $B \cap U = \varnothing$ and $u_0 \in A' \setminus B$), so $A' \cap (U \cup V) = V \cup \{u_0\} \neq V$ and $C_{UV}(A') = A' \in \mathcal{A}$. The compression $C_{UV}$ fixes $A'$, so $C_{UV}(A')$ does not give us a new set to use as a parent for $C_{UV}(B)$. Instead, we use the sub-compression hypothesis. By hypothesis, there exists $v_0 \in V$ such that $\mathcal{A}$ is $(U', V')$-compressed, where $U' = U \setminus \{u_0\}$ and $V' = V \setminus \{v_0\}$. We check that $C_{U'V'}$ acts on $A'$: $A' \cap U' = \varnothing$ (since $B \cap U = \varnothing$ and $u_0 \notin U'$) and $A' \cap V' = V'$ (since $V \subseteq B \subset A'$ and $V' \subset V$). So $A' \cap (U' \cup V') = V'$, and $C_{U'V'}(A') = (A' \setminus V') \cup U'$. Since $\mathcal{A}$ is $(U', V')$-compressed and $A' \in \mathcal{A}$, we get $C_{U'V'}(A') \in \mathcal{A}$. Now we compute $C_{U'V'}(A')$. Since $V' = V \setminus \{v_0\}$ and $V \subseteq A'$: \begin{align*} A' \setminus V' = (A' \setminus V) \cup \{v_0\}. \end{align*} (We remove $V \setminus \{v_0\}$ from $A'$, keeping $v_0$.) So: \begin{align*} C_{U'V'}(A') = (A' \setminus V) \cup \{v_0\} \cup U'. \end{align*} Next we compute $C_{UV}(B)$. We have $B = A' \setminus \{u_0\}$, so: \begin{align*} C_{UV}(B) = (B \setminus V) \cup U = ((A' \setminus \{u_0\}) \setminus V) \cup U. \end{align*} Since $u_0 \in U$ and $U \cap V = \varnothing$, we have $u_0 \notin V$. Also $u_0 \in A' \setminus V$ (since $u_0 \in A'$ and $u_0 \notin V$). So $(A' \setminus \{u_0\}) \setminus V = (A' \setminus V) \setminus \{u_0\}$, and: \begin{align*} C_{UV}(B) = ((A' \setminus V) \setminus \{u_0\}) \cup U = (A' \setminus V) \cup U, \end{align*} where the last equality holds because $u_0 \in U$ (removing $u_0$ from $A' \setminus V$ then taking the union with $U$ restores it). Comparing with $C_{U'V'}(A') = (A' \setminus V) \cup \{v_0\} \cup U'$: since $U = U' \cup \{u_0\}$ and $u_0 \in A' \setminus V$ (because $u_0 = x \in A'$ and $u_0 \notin V$), adding $\{u_0\}$ to $(A' \setminus V) \cup U'$ contributes nothing new. So $(A' \setminus V) \cup U = (A' \setminus V) \cup U'$, and therefore: \begin{align*} C_{UV}(B) = (A' \setminus V) \cup U' = C_{U'V'}(A') \setminus \{v_0\}. \end{align*} Since $v_0 \in C_{U'V'}(A')$ and $C_{U'V'}(A') \in \mathcal{A}$, we conclude $C_{UV}(B) \in \partial \mathcal{A}$. **Conclusion for this case:** We have shown $B \in \partial \mathcal{A}$ (from the fact that $A' \in \mathcal{A}$) and $C_{UV}(B) \in \partial \mathcal{A}$ (by finding a parent $C_{UV}(A')$ or $C_{U'V'}(A')$ in $\mathcal{A}$ depending on whether $x \in U$). So $B$ is retained in $C_{UV}(\partial \mathcal{A})$. Why is this case harder than the previous one? The difficulty is that showing $B \in \partial \mathcal{A}$ alone is not enough: since $C_{UV}(B) \neq B$, the definition of $C_{UV}(\partial \mathcal{A})$ retains $B$ only if *both* $B$ and $C_{UV}(B)$ belong to $\partial \mathcal{A}$. Finding a parent for $C_{UV}(B)$ in $\mathcal{A}$ requires either the structural fact that $C_{UV}(A') \in \mathcal{A}$ (which follows from $A' \in C_{UV}(\mathcal{A})$ when $x \notin U$) or the sub-compression hypothesis (when $x \in U$ and $C_{UV}$ fixes $A'$). This is where the inductive hypothesis is consumed.[/guided]

[step:Conclude $\partial C_{UV}(\mathcal{A}) \subseteq C_{UV}(\partial \mathcal{A})$] In every case, an arbitrary $B \in \partial C_{UV}(\mathcal{A})$ has been shown to belong to $C_{UV}(\partial \mathcal{A})$. This establishes \begin{align*} \partial C_{UV}(\mathcal{A}) \subseteq C_{UV}(\partial \mathcal{A}). \end{align*} Since $C_{UV}$ is an injection on $X^{(r-1)}$ (if $C_{UV}(D_1) = C_{UV}(D_2)$, then either both are fixed or both satisfy $D_i \cap (U \cup V) = V$, giving $(D_1 \setminus V) \cup U = (D_2 \setminus V) \cup U$ and hence $D_1 = D_2$), we have $|C_{UV}(\partial \mathcal{A})| = |\partial \mathcal{A}|$, and the inclusion gives \begin{align*} |\partial C_{UV}(\mathcal{A})| \leq |C_{UV}(\partial \mathcal{A})| = |\partial \mathcal{A}|. \end{align*} This completes the induction on $s$. [/step]