$$
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Shift identities
The forward difference \(\Delta\) E0004 expands into subset sums
when the basepoint, or a direction slot, carries a sum of
vectors.
(1) Lemma. For \(f: X \to Y\), \(x \in X\), \(u_1, \dots, u_k \in X\),
and \(w_1, \dots, w_m \in X\):
-
Basepoint shift.
\[
\Delta(f; x + {\textstyle\sum_j} w_j; u_1, \dots, u_k)
= \sum_{E \subseteq [m]}
\Delta(f; x; (w_j)_{j \in E}, u_1, \dots, u_k).
\]
Each shift \(w_j\) is independently "in" or "out" as an
additional direction; \(E = \emptyset\) recovers the
unshifted term.
-
Direction shift. For \(m \geq 1\),
\[
\Delta(f; x; {\textstyle\sum_j} w_j, u_2, \dots, u_k)
= \sum_{\emptyset \neq E \subseteq [m]}
\Delta(f; x; (w_j)_{j \in E}, u_2, \dots, u_k).
\]
A slot carrying a sum expands over the nonempty subsets
of its summands; by symmetry of \(\Delta\) E0004 the same
holds in any slot. The simplest instance is
\(\Delta(f; x; u + w) = \Delta(f; x; u) + \Delta(f; x; w)
+ \Delta(f; x; u, w)\).
Proof.
Ad basepoint shift) For \(m = 1\): by definition,
\(\Delta(f; x; w, u_1, \dots, u_k)\) is the alternating sum
over \(S \subseteq \set{0} \cup [k]\), which splits as the
terms with \(0 \in S\) (giving \(\Delta(f; x + w; u_\bullet)\))
minus those with \(0 \notin S\) (giving
\(\Delta(f; x; u_\bullet)\)). Rearranging:
\(\Delta(f; x + w; u_\bullet) = \Delta(f; x; u_\bullet)
+ \Delta(f; x; w, u_\bullet)\). For general \(m\): induct on
\(m\), shifting by one \(w_j\) at a time. Each step doubles the
number of terms (each existing \(E\) spawns \(E\) and
\(E \cup \set{j}\)), giving the sum over all \(E \subseteq [m]\).
Ad direction shift) Let \(G(\cdot) := \Delta(f; \cdot;
u_2, \dots, u_k)\), so that \(\Delta(f; x; v, u_2, \dots, u_k)
= G(x + v) - G(x)\) for any \(v\). Then
\[
\Delta(f; x; {\textstyle\sum_j} w_j, u_2, \dots, u_k)
= G(x + {\textstyle\sum_j} w_j) - G(x)
= \sum_{E \subseteq [m]} \Delta(G; x; (w_j)_{j \in E})
- G(x),
\]
by the basepoint shift applied to \(G\). The \(E = \emptyset\)
term is the \(0\)-fold difference \(G(x)\) and cancels; each
remaining term is
\(\Delta(G; x; (w_j)_{j \in E})
= \Delta(f; x; (w_j)_{j \in E}, u_2, \dots, u_k)\):
subsets of the disjoint union \(E \sqcup \set{2, \dots, k}\)
factor as pairs, so both sides unfold to the same alternating
sum.
(2) Validation (AI review, 2026-07-19, claude-fable-5, pass).
Verified both identities against the defining sum in E0004 (signs
in the m = 1 split, induction step, slot-1 co-recursion); probed
m = 0, k = 0, and the m ≥ 1 hypothesis (E = ∅ term cancellation);
numeric check with \(f(x) = x^3\), m = 2, k = 2. Added a one-line
justification for the composition step
\(\Delta(G; x; w_E) = \Delta(f; x; w_E, u_2, \dots, u_k)\).
Used by: