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 $k-\lambda\geq 2$.




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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 $\lambda >1$?



Residual design (BIBD) with repeated blocks
A necessary condition for residual designs not being simple is $k\le 2\lambda$. If $k>2\lambda$, repeated blocks in the residual design would correspond to blocks of the symmetric design intersecting in more than $\lambda$ points, which is not possible.


For $k\le 2\lambda$ repeated blocks are indeed possible, and the residual design need not be the multiple of a simple design. The smallest example I could come up with is a (15,8,4) design with the following incidence matrix:


111100011110000  

111100000001111  

110011011001100  

110011000110011  

101010111000011  

101010100111100  

100101110101010  

100101101010101  

011001110100101  

011001101011010  

010110110011001  

010110101100110  

001111010010110  

001111001101001  

000000011111111  



The residual with respect to the block corresponding to the first column of the matrix is a non-simple (7,4,4) design. Three of its blocks have multiplicity 2, and the remaining blocks are not repeated.


