LemmaStatement: AI-generatedProof: AI-generated

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Signed mixed binomial sums cancel at order one

Statement

For the real arrays $α, β$ of the signed spin-four isometry, define $w_{j, a, b} = (\binom{6}{j}) / \sqrt{(\binom{8}{j + a}) (\binom{8}{j + b})}$ for integers $j, a, b$ with $0 \leq j \leq 6$ and $0 \leq a, b \leq 2$ . Then $M_{a b} = \sum_{j = 0}^{6} w_{j, a, b} α_{j + a} β_{j + b} = 0$ for every $a, b$ . Equivalently one may sum to $8$ , declaring the terms with $j > 6$ zero before taking any quotient.

Facts & Assumptions

[F1]

The coefficient arrays, supports and negative amplitude are those of the signed spin-four isometry. Real coefficient arrays for the signed spin-four isometry

[F2]

For natural numbers $n, k$ , the binomial coefficient $(\binom{n}{k})$ counts the $k$ -element subsets of an $n$ -element set; in particular $(\binom{n}{0}) = (\binom{n}{n}) = 1$ and $(\binom{n}{k}) = 0$ for $k > n$ . The set [A]ᵏ of k-element subsets and the binomial coefficient C(n, k) := | [n]ᵏ|

Proof

Given: The real coefficient arrays, integer index ranges and zero-term convention in the Statement.

Goal: Establish the stated identities and implications.

1.1F1F2givenalgebra

For $0 \leq k \leq 8$ , an eight-element set has a $k$ -element subset, so $(\binom{8}{k}) > 0$ . The stated integer ranges give $0 \leq j + a, j + b \leq 8$ , hence all weights in the sum to $6$ are defined. A nonzero summand requires $j + a \in {0, 6}$ and $j + b \in {2, 8}$ . Their difference $b - a$ must belong to ${2, 8, - 4}$ and also to $[- 2, 2]$ . Thus $b - a = 2$ , forcing $(a, b) = (0, 2)$ . There are then exactly two possible indices, $j = 0$ and $j = 6$ , both inside the summation range. All other eight sums are termwise zero.

2.1F1F2step 1.1algebra∎

Empty and full subsets give $(\binom{6}{0}) = (\binom{6}{6}) = 1$ and $(\binom{8}{0}) = (\binom{8}{8}) = 1$ . In an eight-element set, each two-element subset is counted twice among the $8 \cdot 7$ ordered pairs of distinct elements, so $(\binom{8}{2}) = 8 \cdot 7 / 2 = 28$ . Complementation is a bijection between its six-element and two-element subsets, so $(\binom{8}{6}) = 28$ as well. Therefore $M_{02} = α_{0} β_{2} / \sqrt{28} + α_{6} β_{8} / \sqrt{28} = \sqrt{2} / (3 \sqrt{28}) - \sqrt{2} / (3 \sqrt{28}) = 0$ . This uses the sign before adding contributions. The omitted $j = 7, 8$ terms have zero numerator weight and contribute zero under the stated convention.

Depends on

Used by · 1

Finite signed-code KL product test

Dependency tree · 2 levels

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