[proofplan] The statement fixes pairwise disjoint fibres $(F_x)_{x \in X}$ with $|F_x|=w(x)$ and $Y=\bigcup_{x\in X}F_x$, so the weights determine the sizes of the global groups and empty fibres correspond exactly to weight zero. For each master block $B \in \mathcal B$, choose one of the assumed ingredient [group divisible designs](/page/Group%20Divisible%20Design) on the nonempty fibres above $B$. The global block family is obtained by taking all ingredient blocks together. The only point to check is the pair condition: a pair from two different global groups lies over a unique pair $\{x,y\} \subset X$, and the master [pairwise balanced design](/page/Pairwise%20Balanced%20Design) property gives a unique master block containing $\{x,y\}$. The ingredient design on that master block then covers the lifted pair exactly once, while all other ingredients cannot contain it. [/proofplan] [step:Choose the ingredient designs and assemble the global block family] For each $x \in X$, the theorem statement gives a fibre $F_x$ with $|F_x|=w(x)$, and these fibres are pairwise disjoint with union $Y$. For each $B \in \mathcal B$, define \begin{align*} Y_B := \bigcup_{x \in B} F_x. \end{align*} By the hypothesis on the block $B$, choose an ingredient [group divisible design](/page/Group%20Divisible%20Design) \begin{align*} (Y_B,\mathcal G_B,\mathcal C_B), \qquad \mathcal G_B := \{F_x : x \in B,\ F_x \neq \varnothing\}, \end{align*} whose block family $\mathcal C_B$ has all block sizes in $K'$. The exclusion of empty fibres is exactly the convention required when $w(x)=0$, since then $|F_x|=0$ and $F_x=\varnothing$ cannot be a nonempty group. Define the global group family and global block family by \begin{align*} \mathcal G := \{F_x : x \in X,\ F_x \neq \varnothing\}, \qquad \mathcal C := \bigcup_{B \in \mathcal B} \mathcal C_B. \end{align*} Since the sets $(F_x)_{x \in X}$ are pairwise disjoint and their union is $Y$, the family $\mathcal G$ is a partition of $Y$ after omitting empty fibres. Every block $C \in \mathcal C$ lies in some $\mathcal C_B$, so $|C| \in K'$. [/step] [step:Verify that no global block contains two points from one global group] Let $C \in \mathcal C$. Then $C \in \mathcal C_B$ for some $B \in \mathcal B$. Since $(Y_B,\mathcal G_B,\mathcal C_B)$ is a group divisible design, the block $C$ contains at most one point from each group in $\mathcal G_B$. Now let $x \in X$ with $F_x \neq \varnothing$. If $x \in B$, then $F_x \in \mathcal G_B$, so $|C \cap F_x| \le 1$. If $x \notin B$, then $C \subset Y_B$ and the fibres are pairwise disjoint, so $C \cap F_x = \varnothing$. Hence every block in $\mathcal C$ contains at most one point from each group in $\mathcal G$. [/step] [step:Cover every pair from two distinct global groups exactly once] Let $p,q \in Y$ lie in distinct groups of $\mathcal G$. Then there exist unique distinct points $x,y \in X$ such that \begin{align*} p \in F_x, \qquad q \in F_y. \end{align*} The uniqueness follows because the fibres $(F_z)_{z \in X}$ are pairwise disjoint. Since $(X,\mathcal B)$ is a [pairwise balanced design](/page/Pairwise%20Balanced%20Design) with parameters $\operatorname{PBD}(v,K,1)$, there is a unique block $B_{xy} \in \mathcal B$ such that \begin{align*} \{x,y\} \subset B_{xy}. \end{align*} Inside the ingredient design on $B_{xy}$, the fibres $F_x$ and $F_y$ are distinct groups. Therefore the pair $\{p,q\}$ is contained in exactly one block of $\mathcal C_{B_{xy}}$. It remains to show that no ingredient design over another master block also contains $\{p,q\}$. Suppose $B \in \mathcal B$ and some block $C \in \mathcal C_B$ contains both $p$ and $q$. Since $C \subset Y_B$, we must have $p,q \in Y_B$. Because $p \in F_x$ and $q \in F_y$, this implies $x,y \in B$. Thus $B$ contains the two-element subset $\{x,y\}$, so by uniqueness in the master $\operatorname{PBD}$, $B = B_{xy}$. Hence the unique block of $\mathcal C_{B_{xy}}$ containing $\{p,q\}$ is the unique global block of $\mathcal C$ containing $\{p,q\}$. [guided] We check the defining pair condition for the assembled design. Take two points $p,q \in Y$ that lie in different global groups. By definition of the global groups, there are unique $x,y \in X$ such that \begin{align*} p \in F_x, \qquad q \in F_y. \end{align*} The points $x$ and $y$ are distinct because $p$ and $q$ are assumed to lie in different global groups. The master design is a [pairwise balanced design](/page/Pairwise%20Balanced%20Design) with parameters $\operatorname{PBD}(v,K,1)$, so the two-point set $\{x,y\}$ is contained in exactly one master block. Denote this block by $B_{xy}$. The ingredient design placed on $B_{xy}$ is a [group divisible design](/page/Group%20Divisible%20Design) whose groups are the nonempty fibres above points of $B_{xy}$. Since $p \in F_x$ and $q \in F_y$, both $F_x$ and $F_y$ are nonempty; since $x \neq y$, they are distinct groups in that ingredient design. Therefore the group divisible design pair axiom inside the ingredient gives exactly one ingredient block $C_{pq} \in \mathcal C_{B_{xy}}$ such that \begin{align*} \{p,q\} \subset C_{pq}. \end{align*} Now we rule out any second occurrence coming from another master block. Suppose $B \in \mathcal B$ and $C \in \mathcal C_B$ also contains $p$ and $q$. Since every block in $\mathcal C_B$ is contained in \begin{align*} Y_B = \bigcup_{z \in B} F_z, \end{align*} the inclusion $p \in C \subset Y_B$ forces $x \in B$, because $p$ belongs only to the fibre $F_x$. Similarly, $q \in C \subset Y_B$ forces $y \in B$. Hence $\{x,y\} \subset B$. But the master [pairwise balanced design](/page/Pairwise%20Balanced%20Design) contains the pair $\{x,y\}$ in exactly one block, namely $B_{xy}$. Therefore $B = B_{xy}$. Thus the pair $\{p,q\}$ occurs once inside the ingredient over $B_{xy}$ and cannot occur in any ingredient over a different master block. This proves that every pair of points from distinct global groups is contained in exactly one global block. [/guided] [/step] [step:Conclude that the assembled incidence structure is a group divisible design] The family $\mathcal G$ partitions $Y$, every block in $\mathcal C$ has size in $K'$, no block in $\mathcal C$ contains two points from the same group in $\mathcal G$, and every pair of points from distinct groups in $\mathcal G$ is contained in exactly one block of $\mathcal C$. These are precisely the axioms for a [group divisible design](/page/Group%20Divisible%20Design) on $Y$ with group set $\mathcal G$ and block sizes in $K'$. Therefore the assembled design is the desired group divisible design. [/step]