In [this][1] paper of **Asmuth–Bloom threshold SSS**, the algorithm is as follows:

**Shares Distribution**

To distribute n shares of a secret $K$ among the set of participants
$P = \{ p_i : 1 ≤ i ≤ n\}$, the dealer D does the following:
 
1) A set of integers $\{ p, m_1 < m_2 < · · · < m_n \}$, where $0 ≤ K < p$, is chosen subject to the following conditions:

 $\gcd(m_i , m_j)=1$ where for $i\neq j$
    
 $\gcd(p , m_i)=1$ ,for all $i$,
    
 $\prod \limits_{i=1}^{t}m_i > p \prod \limits_{i=1}^{t-1}m_{n-i+1}$
     
2) Let $M =\prod \limits_{i=1}^{t}m_i$. The dealer computes $$y = K+ap,$$
 where a is a positive integer generated randomly subject to the condition that $0 ≤ y < M$
	
3) The share of the $i^{th}$ participant,$1 ≤ i ≤ n$, is $$y_i = y~ mod ~m_i$$

**Secret  Construction**

Assume $C$ is a coalition of $t$ participants to construct the secret. 
Let $M_C =\prod \limits_{i=1}^{C}m_i$ 

1) Given the system $$y \equiv y_ i \mod m_ i $$  for $i ∈ C$, solve y in $GF(M_C  )$ uniquely using the CRT.

2) Compute the secret as
     $$K = y \mod p$$

According to the CRT, y can be determined uniquely in $GF(M_ C)$ . Since $y < M ≤ M_C$ , the solution is also unique in $GF(M)$.



In the [this] [2] original paper the author given to  recover $K$ , it clearly suffices to find y. If $y_{1},y_{2}.......y_{t}$ are known, then by the Chinese remainder theorem modulo $M_C =\prod \limits_{i=1}^{C}m_i$ .
 As  $M ≤ M_C$ this uniquely determines $y$ and thus $K$ . On the other 
hand, if   only $t-1$ shares were known, essentially no information about  the secret can be recovered

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*How to prove formally that the number of participants **lesser than threshold number $t$** cannot get secret?*


[2]:https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1056651
[1]: http://pdfs.semanticscholar.org/440d/fa8ab99301d4ea3a8b19c0748575915aef15.pdf