Generalise the notation of some constraints I would like your help to write the following set of constraints using a notation that works for any generic $K\in \{3,4,5,6,...\}$.
Consider $K$ real numbers $\mu_1<\mu_2<...<\mu_K$. Let $\alpha_j\equiv \mu_{j+1}-\mu_j$ for $j=1,...,K-1$.
Let me report below the constraints that I want to impose. RHS stays for right-hand-side. LHS stays for left-hand-side.

For $K=3$: 
$$
\begin{cases}
\alpha_1\neq \alpha_2 & \text{(1): [first difference $\neq$ from last difference]}\\
\end{cases}
$$

For $K=4$:
$$
\begin{cases}
\alpha_1\neq \alpha_3 & \text{(1): [first difference $\neq$ from last difference]}\\
------\\
\alpha_1+\alpha_2\neq \alpha_3 & \text{(2): [add one adjacent difference to the LHS of (1)]}\\
\alpha_1\neq \alpha_2+ \alpha_3& \text{(3): [add one adjacent difference to the RHS of (1)]}\\
\end{cases}
$$

For $K=5$:
$$
{\small
\begin{cases}
\alpha_1\neq \alpha_4 & \text{(1): [first difference $\neq$ from last difference]}\\
------\\
\alpha_1+\alpha_2\neq \alpha_4 & \text{(2): [add one adjacent difference to the LHS of (1)]}\\
\alpha_1\neq \alpha_3+ \alpha_4& \text{(3): [add one adjacent difference to the RHS of (1)]}\\
------\\
\alpha_1+\alpha_2+\alpha_3\neq \alpha_4 & \text{(4): [add one adjacent difference to the LHS of (2)]}\\
\alpha_1+\alpha_2\neq \alpha_4+\alpha_3 & \text{(5): [add one adjacent difference to the RHS of (2)]}\\
------\\
\alpha_1+\alpha_2\neq \alpha_3+ \alpha_4& \text{ [add one adjacent difference to the RHS of (3)] [redundant]}\\
\alpha_1\neq \alpha_2+ \alpha_3+ \alpha_4& \text{(6): [add one adjacent difference to the RHS of (3)]}\\
\end{cases}}
$$
For $K=6$:
$$
\begin{cases}
\alpha_1\neq \alpha_5 & \text{(1): [first difference $\neq$ from last difference]}\\
-------\\
\alpha_1+\alpha_2\neq \alpha_5 &  \text{(2): [add one adjacent difference to the LHS of (1)]}\\
\alpha_1\neq \alpha_5+\alpha_4 & \text{(3): [add one adjacent difference to the RHS of (1)]}\\
-------\\
\alpha_1+\alpha_2+\alpha_3\neq \alpha_5 &  \text{(4): [add one adjacent difference to the LHS of (2)]}\\
\alpha_1+\alpha_2\neq \alpha_5+\alpha_4 &  \text{(5): [add one adjacent difference to the LHS of (2)]}\\
-------\\
\alpha_1+\alpha_2\neq \alpha_5+\alpha_4 & \text{ [add one adjacent difference to the RHS of (3)] [redundant]}\\
\alpha_1\neq \alpha_5+\alpha_4+\alpha_3 & \text{(6): [add one adjacent difference to the RHS of (3)]}\\
-------\\
\alpha_1+\alpha_2+\alpha_3+\alpha_4\neq \alpha_5 &  \text{(7): [add one adjacent difference to the LHS of (4)]}\\
\alpha_1+\alpha_2+\alpha_3\neq \alpha_5+\alpha_4 &  \text{(8): [add one adjacent difference to the LHS of (4)]}\\
-------\\
\alpha_1+\alpha_2+\alpha_3\neq \alpha_5+\alpha_4 &  \text{[add one adjacent difference to the LHS of (5)] [redundant]}\\
\alpha_1+\alpha_2\neq \alpha_5+\alpha_4+\alpha_3 &  \text{(9): [add one adjacent difference to the LHS of (5)]}\\
-------\\
\alpha_1+\alpha_2\neq \alpha_5+\alpha_4+\alpha_3& \text{[add one adjacent difference to the RHS of (6)] [redundant]}\\
\alpha_1\neq \alpha_5+\alpha_4+\alpha_3+\alpha_2 & \text{(10): [add one adjacent difference to the RHS of (6)]}\\
\end{cases}
$$

Question: There is a mechanic behind these constraints that is replicable for any $K$. I can see some kind of circular argument, but I find it hard to formalise. Could you suggest a notation to represent these constraints that works for any $K$?

Update: 
I found this way of writing the desired constraints: for $h=2,...,K$
$$
\begin{aligned}
&\alpha_1+\alpha_2+...+\alpha_{h-3}+\alpha_{h-2}+ \alpha_{h-1}  \neq  \alpha_{K-1}\\
&\alpha_1+\alpha_2+...+\alpha_{h-3}+\alpha_{h-2}\hspace{1.2cm}  \neq \alpha_{K-1}+\alpha_{K-2}\\
&\alpha_1+\alpha_2+...+\alpha_{h-3}\hspace{2.4cm}  \neq \alpha_{K-1}+\alpha_{K-2}+\alpha_{K-3}\\
&...\\
&\alpha_1 \hspace{5.2cm}\neq \alpha_{K-1}+\alpha_{K-2}+\alpha_{K-3}+...+\alpha_{K-h-1}\\
\end{aligned}
$$
Is it correct? Are there simpler ways?
 A: We consider   the case  $K=6$, rearrange  the constraints  conveniently and derive  a formula which describes all the constraints. From this setting we derive a formula for the general case easily.

Case: $K=6$
We have
\begin{align*}
\alpha_1&\ne\alpha_5&\\
&\ne  \alpha_4+\alpha_5&\\
&\ne \alpha_3+\alpha_4+\alpha_5&\\
&\ne  \alpha_2+\alpha_3+\alpha_4+\alpha_5&\color{blue}{\alpha_1\ne  \alpha_N+\cdots+\alpha_5}\\
&&\qquad\qquad  \color{blue}{   (2\leq  N\leq   5)}\\
\alpha_1+\alpha_2&\ne\alpha_5&\\
&\ne  \alpha_4+\alpha_5&\\
&\ne \alpha_3+\alpha_4+\alpha_5&\color{blue}{\alpha_1+\alpha_2\ne  \alpha_N+\cdots+\alpha_5}\\
&&\qquad\qquad \color{blue}{    (3\leq  N\leq   5)}\\
\alpha_1+\alpha_2+\alpha_3&\ne\alpha_5&\\
&\ne  \alpha_4+\alpha_5&\color{blue}{\alpha_1+\alpha_2+\alpha_3\ne  \alpha_N+\cdots+\alpha_5}\\
&&\qquad\qquad \color{blue}{    (4\leq  N\leq   5)}\\
\alpha_1+\alpha_2+\alpha_3+\alpha_4&\ne\alpha_5
&\color{blue}{\alpha_1+\alpha_2+\alpha_3+\alpha_4\ne  \alpha_N+\cdots+\alpha_5}\\
&&\qquad\qquad \color{blue}{    (5\leq  N\leq   5)}\\
\end{align*}

We     see  in  case  $K=6$   we  have  $4$    groups,  each  group has  constraints  with equal  terms   at     the left-hand   side. On   the right-hand side we have blue   marked  formulas, one for each group.
We can now derive  a formula which   describes all the constraints, namely
\begin{align*}
\alpha_1+\cdots+\alpha_M\ne  \alpha_N+\cdots+\alpha_5\qquad\qquad     1\leq M<N\leq   5\tag{1}
\end{align*}

We obtain from (1) a formula describing all constraints  for general  $K\geq 3$, namely
\begin{align*}
\color{blue}{\alpha_1+\cdots+\alpha_M\ne  \alpha_N+\cdots+\alpha_{K-1}\qquad\qquad     1\leq M<N\leq   K-1}
\end{align*}
or using sigma notation:
\begin{align*}
\color{blue}{\qquad\qquad\sum_{m=1}^M\alpha_m\ne\sum_{n=N}^{K-1}  \alpha_n\qquad\qquad\qquad\qquad     1\leq M<N\leq   K-1}
\end{align*}

