If I define summation recursively, how do I prove formally that both of the following definitions are equivalent? First definition:
$\sum_{x=a}^{b} f(x)=f(b)+\sum_{x=a}^{b-1} f(x)$ if $a\leq b$, and equals $0$ otherwise
Second definition:
$\sum_{x=a}^{b} f(x)=f(a)+\sum_{x=a+1}^{b} f(x)$ if $a\leq b$, and equals $0$ otherwise
Intuitively, I have no doubt that both statements are always equal, but I am struggling to come up with a formal proof for it. I tried induction over $b$, but only got so far on working with the second definition:
$\sum_{x=a}^{b+1} f(x)=f(b+1)+\sum_{x=a}^{b}$

We do induction by the number $a-b+1$ of summands.
We start with the definition of the sum:
$$\sum_{x=a}^{b} f(x):=\sum_{x=a}^{b-1} f(x)+f(b)$$
and we want to prove
$$\sum_{x=a}^{b} f(x)=f(a)+\sum_{x=a+1}^{b} f(x)$$
For $b-a+1=1$ and $b-a+1=2$ the statement can be easily verified, so let us assume $b-a+1\gt3$. Then we have:
$$\begin{align}
\sum_{x=a}^{b} f(x)  
\\ \text{by definition of the sum we have} &
\\=f(b)+\sum_{x=a}^{b-1} f(x)  \tag 1
\\\text{so by the induction hypothesis we have}  
\\=f(b)+\left(f(a)+\sum_{x=a+1}^{b-1} f(x)\right) \tag 2
\\=f(a)+\left(f(b)+\sum_{x=a+1}^{b-1} f(x)\right) \tag 3
\\\text{and again by definition of the sum}
\\=f(a)+\sum_{x=a+1}^{b} f(x) \tag 4
\end{align}$$
We have chosen $b-a+1\ge3$ in the proof to guarantee that $\sum_{x=a+1}^{b-1} f(x)$ has at least one summand.
