Here is the proof that ΣΔF = F. First we define our "sequences", which are just functions from N to R; N includes 0. Next we define operators, which I will say

Here is the proof that ΣΔF = F. First we define our "sequences", which are just functions from N to R; N includes 0. Next we define operators, which I will say always take precedence over function arguments. We define two operators in particular: 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] Enjoy. :-)