I'm struggling with the following problem regarding designs:
Let D be a $(v,k,\lambda)$ symmetric design with point set $\Omega$ and set of block $\mathcal{B}$. That is $|\mathcal{B}|=v$.
The dual of D is $D^*$, the design with point set $\mathcal{B}$ and set of blocks $\Omega$ where a “point” B of $D^*$ is in the “block” x of $D^*$ if and only if the point x of D lies in the block B of D. Prove that $D^*$ is a $(v,k,\lambda)$-design.
What I thought to do was the following:
For $B\in\mathcal{B}$ and $x\in\Omega$ we have $B\in x \iff x\in B$ where LHS is in $D^*$ and RHS is in D.
Let every $x\in\Omega$ be in r blocks in $\mathcal{B}$. Since $v=b$, where b is $|\mathcal{B}|$ we have $r=k$. Now exactly k blocks will go to x when forming $D^*$ and therefore $|x|=k$ $\forall{x\in\Omega}$. So then $D^*$ is a $(v,k,\mu)$-design.
Let's assume that $\mu\neq\lambda$. Since $D^*$ is a design, we have $\mu(v-1)=r(k-1)=k^2-k$. But since D is a design we have $\lambda(v-1)=k^2-k$ and so it must be that $\mu = \lambda$. So each pair of "points" in $D^*$ appears in exactly $\lambda$ "blocks".
Hence $D^*$ is a $(v,k,\lambda)$-design.
Is this a sufficient proof or should I try another approach?