$$
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$$
Discrete Faà di Bruno duality
The iterated forward difference \(\Delta\) E0004 of a composite
\(f \circ g\) expands as a sum over coverings \(\Cov(k)\) of the
direction set. Its Möbius dual is the Taylor composition, summed
over \(\KP_+^2(k)\).
(1) Theorem. Let \(X, Y, Z\) be abelian groups, \(g: X \to Y\) and
\(f: Y \to Z\) arbitrary maps, \(x \in X\),
\(u_1, \dots, u_k \in X\). Write \(y = g(x)\) and \(z = f(y)\).
-
Boolean Faà di Bruno (\(k \geq 1\)):
\[
\Delta(f \circ g; x; u_\bullet)
= \sum_{H \in \Cov(k)}
\Delta\bigl(f;\, y;\, (\Delta(g; x; u_T))_{T \in H}\bigr).
\]
-
Boolean Taylor composition:
\[
(f \circ g)(x + {\textstyle\sum_{i=1}^k} u_i)
= z + \sum_{H \in \KP_+^2(k)}
\Delta\bigl(f;\, y;\, (\Delta(g; x; u_T))_{T \in H}\bigr).
\]
All identities are exact with integer coefficients; no
regularity is assumed on \(f\) or \(g\).
Proof.
Write \(g_T := \Delta(g; x; u_T)\).
Ad 2) By Taylor duality E0005,
\(T(g; x; u_S) = y + \sum_{\emptyset \neq T \subseteq S} g_T\).
Applying Taylor duality to \(f\) at \(y\) in the directions
\((g_T)_{T \in \KP_+(S)}\) gives
\[
T(f \circ g; x; u_S)
= z + \sum_{H \in \KP_+^2(S)} \Delta(f; y; (g_T)_{T \in H}).
\]
Ad 1) Define
\(\varphi(S) := \sum_{H \in \Cov(S)} \Delta(f; y; (g_T)_{T \in H})\).
By Möbius inversion E0002 it suffices to show
\(\zeta(\varphi; S) = T(f \circ g; x; u_S) - z\): the
\(\mu\)-transform of the right side is \(\Delta(f \circ g; x; u_S)\)
for \(S \neq \emptyset\), the constant \(z\) cancelling by
\((1-1)^{|S|} = 0\). Indeed,
\[
\zeta(\varphi; S)
= \sum_{R \subseteq S} \sum_{H \in \Cov(R)}
\Delta(f; y; (g_T)_{T \in H})
= \sum_{H \in \KP_+^2(S)} \Delta(f; y; (g_T)_{T \in H}),
\]
since each \(H \in \KP_+^2(S)\) appears in exactly one summand,
namely the one for \(R = \bigcup H\). Now apply Ad 2).
(2) Validation (AI review, 2026-07-19, claude-fable-5, pass).
Proof logic, edge cases (\(k = 1\), \(k = 2\),
\(g = \mathrm{id}\), additive \(f\)), combinatorial bijection
\(\KP_+^2(S) = \bigsqcup_R \Cov(R)\), necessity of \(k \geq 1\).
Made \(z\)-cancellation in the Möbius step explicit in the proof.
No issues found.
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