Easily verfied, hard to show: $\big(K-|V-a|\big)^+$ is equal to linear combination of $\big(V-c_i\big)^+$ for $c_i \in \mathbb{R}$.

Question

Let $K>0$, $a \in \mathbb{R}$, and $V$ be a random variable.

I want to prove the following identity $$\big(K-|V-a|\big)^+ = \sum_{i=1}^n a_i\big(V-c_i\big)^+,$$ for some $n \in \mathbb{N}, a_i\in\mathbb{Z}, c_i \in \mathbb{R}$, where $\big(\cdot\big)^+ := \max\{\cdot, 0\}$.

I know that the solution is $$\big(K-|V-a|\big)^+ = \big(V-(a+K)\big)^+ + \big(V-(a-K)\big)^+ - 2\big(V-a\big)^+$$

It is easily verified by looking at the following cases separately:

1. $V < a-K$,
2. $a-K \le V < a$,
3. $a \le V \le a+K$,
4. $a+K < V$.

One way to come up with the bounds is

$$ \begin{align} & (K - |V - a|)^+ \\ =& \big( K- (a-V)^+-(V-a)^+\big)^+\\ =& \big( (\mathbb{1}_{\{V \le a\}}+\mathbb{1}_{\{a<V\}})K- \mathbb{1}_{\{V \le a\}}[a-V]-\mathbb{1}_{\{a<V\}}[V-a]\big)^+\\ =& \big(\mathbb{1}_{\{V \le a\}}[ V-(a-K)]+\mathbb{1}_{\{a<V\}}[(a+K)-V]\big)^+\\ =& \big( (\mathbb{1}_{\{V < a-K\}}+\mathbb{1}_{\{ a-K\le V \le a\}})[ V-(a-K)] \\&+ (\mathbb{1}_{\{a<V\le a+K\}}+\mathbb{1}_{\{a+K<V\}})[(a+K)-V]\big)^+\\ =& \big( \mathbb{1}_{\{V < a-K\}}[ V-(a-K)]\big)^+ + \big(\mathbb{1}_{\{ a-K\le V \le a\}}[ V-(a-K)]\big)^+ \\&+ \big(\mathbb{1}_{\{a<V\le a+K\}}[(a+K)-V]\big)^+ +\big(\mathbb{1}_{\{a+K<V\}})[(a+K)-V]\big)^+\\ =& \mathbb{1}_{\{ a-K\le V \le a\}}[ V-(a-K)] + \mathbb{1}_{\{a<V\le a+K\}}[(a+K)-V] \end{align} $$

How is one supposed to come up with the righthand side of the solution constructively without knowing the exact form?

Thanks for reading.