Δg(i) = g(0) if i = 0, else g(i) − g(i − 1)

Σg(i) = Σ{n: 0 → i} g(n)

Note that Δg(0) = g(0) by these definitions, and note that Σg(0) = g(0) as well. Call these the "zeroth element relations." (Also note that I am saying Σg(0) where perhaps [Σg](0) would be more clear.)

I will observe a very simple lemma which embeds Σg in a recursive form:

Σg(n + 1) = Σg(n) + g(n + 1) for any g, any n in N.

On to the proof. ΣΔF(0) = F(0) by composing those two zeroth element relations. Also, if ΣΔF(n) = F(n) for any n, then I claim that ΣΔF(n + 1) = F(n + 1). Proof:

ΣΔF(n + 1) = ΣΔF(n) + ΔF(n + 1), by the lemma.

ΣΔF(n + 1) = F(n) + ΔF(n + 1), because "if ΣΔF(n) = F(n)".

ΣΔF(n + 1) = F(n) + F(n + 1) − F(n), by definition of Δ.

ΣΔF(n + 1) = F(n + 1)

Since it holds for n=0, and whenever it holds for n, it holds for n + 1, by induction this holds for all n in N and therefore, by induction:

ΣΔF = F

The reverse argument is also pretty simple. It's also obviously true for n = 0 due to the zeroth element relations. For n > 0 we don't even need induction:

ΔΣF(n) = ΣF(n) − ΣF(n − 1), by definition of Δ

ΔΣF(n) = ΣF(n − 1) + F(n) − ΣF(n − 1), by lemma

ΔΣF(n) = F(n)

ΔΣF = F.

If you still don't believe me, here it all is in Python:

    >>> sigma = lambda seq: lambda i: sum(seq(u) for u in range(0, i + 1))
    >>> delta = lambda seq: lambda i: seq(0) if i == 0 else seq(i) - seq(i - 1)
    >>> expand = lambda seq: [seq(i) for i in range(0, 7)]
    >>> constant = lambda N: lambda i: N
    >>> expand(constant(3))
    [3, 3, 3, 3, 3, 3, 3]
    >>> expand(sigma(constant(3)))
    [3, 6, 9, 12, 15, 18, 21]
    >>> expand(delta(constant(3)))
    [3, 0, 0, 0, 0, 0, 0]
    >>> expand(sigma(delta(constant(3))))
    [3, 3, 3, 3, 3, 3, 3]
    >>> expand(delta(sigma(constant(3))))
    [3, 3, 3, 3, 3, 3, 3]