[proofplan] We regard the developed blocks as indexed by translation-and-base-block choices, so repeated subsets of $G$ are counted with multiplicity. Each translated block has size $k$ because every group translation is a bijection of $G$. To count the number of developed blocks through two distinct points, we compare such block indices with ordered internal differences in the original base blocks. The difference family hypothesis supplies exactly $\lambda$ occurrences of each nonzero group difference, giving the required pair incidence number. [/proofplan] [step:Count developed blocks with their translation indices] Define the index set \begin{align*} I := G \times \{1,\dots,t\}. \end{align*} For each index $(g,i) \in I$, define the block \begin{align*} D_{g,i} := g + B_i = \{g+x : x \in B_i\} \subset G. \end{align*} The collection $\mathcal{D}$ is the indexed family $(D_{g,i})_{(g,i)\in I}$, so if $D_{g,i}=D_{h,j}$ as subsets of $G$, the two occurrences are still counted separately unless $(g,i)=(h,j)$. [/step] [step:Verify that every developed block has size $k$] Fix $(g,i) \in I$. Define the translation map $\tau_g: G \to G$ by \begin{align*} \tau_g(x) := g+x \quad \text{for every } x \in G. \end{align*} The map $\tau_g$ is a bijection, with inverse $\tau_{-g}$. Therefore its restriction $\tau_g|_{B_i}: B_i \to D_{g,i}$ is also a bijection. Since $|B_i|=k$, it follows that \begin{align*} |D_{g,i}| = |B_i| = k. \end{align*} Thus every developed block has block size $k$. [/step] [step:Relate blocks through two fixed points to base-block differences] Fix distinct points $a,b \in G$. Since $a \neq b$, the element \begin{align*} d := b-a \end{align*} is nonzero in $G$. For each $i \in \{1,\dots,t\}$, define \begin{align*} P_i(d) := \{(x,y) \in B_i \times B_i : x \neq y,\ y-x=d\}. \end{align*} We claim that the set of indices of developed blocks containing both $a$ and $b$ is in bijection with the disjoint union \begin{align*} P(d) := \biguplus_{i=1}^{t} P_i(d). \end{align*} Define \begin{align*} \Phi: \{(g,i) \in I : a \in D_{g,i},\ b \in D_{g,i}\} &\to P(d) \end{align*} as follows. Given $(g,i)$ with $a,b \in D_{g,i}$, there exist unique elements $x,y \in B_i$ such that \begin{align*} a = g+x, \qquad b = g+y. \end{align*} The uniqueness follows because translation by $g$ is injective. Since $a \neq b$, we have $x \neq y$. Subtracting the two displayed identities gives \begin{align*} y-x = b-a = d. \end{align*} Set $\Phi(g,i) := (i,x,y)$, viewing $P(d)$ as the disjoint union indexed by $i$. Conversely, define \begin{align*} \Psi: P(d) &\to \{(g,i) \in I : a \in D_{g,i},\ b \in D_{g,i}\} \end{align*} by \begin{align*} \Psi(i,x,y) := (a-x,i). \end{align*} If $(i,x,y) \in P(d)$, then $x,y \in B_i$ and $y-x=d=b-a$. With $g:=a-x$, we have \begin{align*} a = g+x, \qquad g+y = a-x+y = a+(y-x)=a+d=b. \end{align*} Thus $a,b \in D_{g,i}$, so $\Psi$ is well-defined. The maps $\Phi$ and $\Psi$ are inverse to each other. Starting from $(g,i)$, the unique element $x \in B_i$ satisfying $a=g+x$ gives $g=a-x$, so $\Psi(\Phi(g,i))=(g,i)$. Starting from $(i,x,y) \in P(d)$, the block index $\Psi(i,x,y)$ is $(a-x,i)$, and the unique elements representing $a$ and $b$ in $(a-x)+B_i$ are precisely $x$ and $y$, so $\Phi(\Psi(i,x,y))=(i,x,y)$. [guided] We now fix the two points whose incidence number must be counted. Let $a,b \in G$ with $a \neq b$, and define \begin{align*} d := b-a. \end{align*} Because $a \neq b$, the group element $d$ is nonzero, so the difference family hypothesis applies to $d$. The central question is: when does a translated block $g+B_i$ contain both $a$ and $b$? This happens exactly when there are elements $x,y \in B_i$ such that \begin{align*} a = g+x, \qquad b = g+y. \end{align*} For a fixed $g$, these elements are unique, because translation by $g$ is injective. Subtracting the two equations eliminates the translation parameter: \begin{align*} b-a = (g+y)-(g+x)=y-x. \end{align*} Thus every developed block containing $a$ and $b$ determines an ordered pair $(x,y)$ in one of the base blocks with difference $y-x=d$. We make this correspondence precise. For each $i \in \{1,\dots,t\}$, define \begin{align*} P_i(d) := \{(x,y) \in B_i \times B_i : x \neq y,\ y-x=d\}, \end{align*} and define the disjoint union \begin{align*} P(d) := \biguplus_{i=1}^{t} P_i(d). \end{align*} The disjoint union records not only the ordered pair $(x,y)$ but also the base block $B_i$ in which it occurs. Define a map from developed blocks through $a$ and $b$ to these differences by sending an index $(g,i)$ to the unique triple $(i,x,y)$ satisfying \begin{align*} a = g+x, \qquad b = g+y. \end{align*} The condition $x \neq y$ follows from $a \neq b$, and the equation $y-x=d$ follows by subtraction. Hence this map lands in $P(d)$. Conversely, suppose $(i,x,y) \in P(d)$. To recover the translation, the equation $a=g+x$ forces \begin{align*} g = a-x. \end{align*} For this value of $g$, the point $a$ lies in $(a-x)+B_i$ because $x \in B_i$. Also, \begin{align*} (a-x)+y = a+(y-x)=a+d=b, \end{align*} so $b$ also lies in the same translated block. Therefore every ordered base-block difference $y-x=d$ gives exactly one indexed developed block containing $a$ and $b$. These two constructions undo each other: a block index $(g,i)$ recovers $x=a-g$ and $y=b-g$, while a triple $(i,x,y)$ recovers $g=a-x$. Hence the number of indexed developed blocks containing $a$ and $b$ is exactly the number of occurrences of $d=b-a$ in the difference multiset of the base family. [/guided] [/step] [step:Apply the difference family condition to obtain the pair incidence number] By the bijection constructed above, \begin{align*} |\{(g,i) \in I : a \in D_{g,i},\ b \in D_{g,i}\}| = |P(d)|. \end{align*} By definition of $P(d)$, \begin{align*} |P(d)| = \sum_{i=1}^{t} |\{(x,y) \in B_i \times B_i : x \neq y,\ y-x=d\}|. \end{align*} This is exactly the multiplicity of the nonzero element $d$ in the difference multiset $\Delta\mathcal{B}$. Since $\mathcal{B}$ is a $(v,k,\lambda)$ difference family, that multiplicity is $\lambda$. Therefore \begin{align*} |\{(g,i) \in I : a \in D_{g,i},\ b \in D_{g,i}\}| = \lambda. \end{align*} Since the distinct points $a,b \in G$ were arbitrary, every pair of distinct points of $G$ is contained in exactly $\lambda$ indexed developed blocks. Together with $|G|=v$ and the block-size verification above, this proves that $\mathcal{D}$ is a $2$-$(v,k,\lambda)$ design on point set $G$, with repeated blocks allowed. [/step]