Simple BIBD are defined as those designs in which incindence relation is "is element". So effectively blocks are subsets of points. Equivalently there should be no "repeating blocks" ie. blocks that aren't uniquely determined by its 'elements' (here 'elements' means in sense of incidence relation).

For fixed block $B_0$ (Block-)residual designs (of symmetric $(v,k,\lambda)$-design $(\mathcal{P},\mathcal{B},\mathcal{I})$=(points,blocks,relation)) is defined as incidence structure with points $\mathcal{P}\setminus B_0$ and blocks $ \{ B \setminus B_0 : B\in \mathcal{B}, B\neq B_0 \} $ and incidence inherited from initial design.

This residual design is allowed to have repeted blocks (ie. not to be simple). In that case parameters of derived design are $(v-k,k-\lambda,\lambda)$.

But from a few simple examples it seems that if there are any repeated blocks, than there is fixed $f$ such that every block is repeated exactly $f$ times. So we can simply exclude those repated blocks and get SIMPLE $(v-k,k-\lambda,\frac{\lambda}{f})$ design.

Is this true? If not what is counter-example? If it is in fact true, how to prove it?

Additionaly we suppose that $k-\lambda\geq 2$.

EDIT (clarification): Maybe I wasn't clear enough. Bottom line is that if you start with SIMPLE SYMMETRIC BIBD $(\mathcal{P},\mathcal{B},\in)$ with $k-\lambda \geq 2$ is it true that "residual design" $(\mathcal{P}\setminus B_0,\{ B \setminus B_0 : B\in \mathcal{B}, B\neq B_0 \},\in)$ is DESIGN AT ALL?! (Note that here same blocks merge into one block since relation is '$\in$'). It will be if it is true that there is fixed $f$ such that every block is repeated exactly $f$ times if we interpret $\{ B \setminus B_0 : B\in \mathcal{B}, B\neq B_0 \}$ as multiset.

It isn't hard to prove that if $\lambda=1$ (since we have $k−\lambda\geq 2$) this is true and there aren't repeated blocks ie. $f=1$.

But what about $\lambda >1$?