LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Off-shell free Dirac Ward telescoping and tree currents

Statement

Use the $(- + + +)$ coordinate convention, $h = - η = diag (1, - 1, - 1, - 1)$ , and the Clifford matrices of Dirac Clifford matrices and energy projections; the shell parameter $μ > 0$ has inverse-length units as in Scalar future mass shell, invariant measure and test restriction. For $p = (p^{0}, p) \in R^{4}$ , write $Γ (p) : = γ^{0} p^{0} - \sum_{j = 1}^{3} γ^{j} p_{j}, K (p) : = Γ (p) - μ I_{4} .$ When $h (p, p) \neq μ^{2}$ , define the finite rational inverse $S_{0} (p) : = \frac{Γ (p) + μ I_{4}}{h (p, p) - μ^{2}} .$ For $p, p + k$ with both denominators nonzero, $S_{0} (p + k) Γ (k) S_{0} (p) = S_{0} (p) - S_{0} (p + k) .$ For any finite chain of off-shell momenta, the corresponding insertion differences telescope with both endpoint terms retained. If $u (p), u (p^{'})$ satisfy $(Γ (p) - μ) u (p) = 0$ , $(Γ (p^{'}) - μ) u (p^{'}) = 0$ and $p, p^{'}$ lie on the future shell, then $\overset{ˉ}{u} (p^{'}) Γ (p - p^{'}) u (p) = 0$ , where $\overset{ˉ}{u} = u^{†} β$ . If $v (q)$ satisfies $(Γ (q) + μ) v (q) = 0$ on the future shell, then $\overset{ˉ}{v} (q) Γ (p + q) u (p) = 0$ . These are finite rational/tree identities; no Feynman boundary value, singular distribution product, loop theorem, or all-diagram gauge-independence claim is included.

Facts & Assumptions

Given: $μ > 0$ , the real $(- + + +)$ momentum convention, and the finite matrix and spinor data in the Statement.

[A1]

Countable Choice is included because Dirac Clifford matrices and energy projections states its mathematical convention under that hypothesis; no additional choice is used in the finite matrix calculations. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[A2]

The external spinors in the current clauses satisfy exactly the displayed positive- and negative-energy equations; these are hypotheses, not claims about a constructed scattering state.

[F1]

Dirac Clifford matrices and energy projections defines $β, γ^{0}, γ^{j}$ , $\overset{ˉ}{z} = z^{†} β$ , and $Γ (p) = γ^{0} p^{0} - \sum_{j} γ^{j} p_{j}$ with the stated $(- + + +)$ convention.

[F2]

Scalar future mass shell, invariant measure and test restriction fixes $η = diag (- 1, 1, 1, 1)$ , $h = - η$ , $μ$ and momentum in inverse-length units, and the future-shell condition $p^{0} = \sqrt{∣ p ∣^{2} + μ^{2}}$ .

[F3]

Finite/formal gauge covariance and matrix Ward identity: finite invertible matrices obey the resolvent identity $B^{- 1} (B - A) A^{- 1} = A^{- 1} - B^{- 1}$ , and the formal inverse variation is coefficientwise.

Proof

1.1A1F1F2algebra

From the displayed Pauli matrices in [F1], direct multiplication gives $σ_{j}^{†} = σ_{j}$ , $σ_{j}^{2} = I_{2}$ , and $σ_{i} σ_{j} + σ_{j} σ_{i} = 0$ for $i \neq j$ . The block formulas then give $β^{2} = I_{4}$ , $β^{†} = β$ , $γ^{0 †} = γ^{0}$ , $γ^{j †} = - γ^{j}$ , and ${γ^{a}, γ^{b}} = 2 h^{a b} I_{4}$ . Consequently $Γ (p)^{2} = h (p, p) I_{4}$ and $β Γ (p)^{†} β = Γ (p)$ for every real $p$ .

2.1F1F2step 1.1algebra

Multiplying $Γ (p) - μ I_{4}$ and $Γ (p) + μ I_{4}$ in either order gives $(h (p, p) - μ^{2}) I_{4}$ by [step 1.1]. Thus $K (p)$ is invertible off shell and its inverse is exactly $S_{0} (p)$ .

2.2A2F1step 1.1algebra

If $K (p) u (p) = 0$ , then taking the adjoint and using $β K (p)^{†} β = K (p)$ from [step 1.1] gives $\overset{ˉ}{u} (p) K (p) = 0$ . The same calculation applied to $(Γ (q) + μ) v (q) = 0$ gives $\overset{ˉ}{v} (q) (Γ (q) + μ) = 0$ .

3.1F3step 2.1algebra

Since $Γ (p + k) - Γ (p) = Γ (k)$ , the resolvent identity [F3] with $A = K (p)$ , $B = K (p + k)$ gives $S_{0} (p + k) Γ (k) S_{0} (p) = S_{0} (p) - S_{0} (p + k)$ .

4.1step 3.1algebra

For off-shell $p_{0}, \dots, p_{N}$ , put $k_{j} = p_{j + 1} - p_{j}$ . Applying [step 3.1] at each adjacent pair yields $S_{0} (p_{j + 1}) Γ (k_{j}) S_{0} (p_{j}) = S_{0} (p_{j}) - S_{0} (p_{j + 1})$ ; summing $j = 0, \dots, N - 1$ cancels each interior term and leaves $S_{0} (p_{0}) - S_{0} (p_{N})$ .

5.1A2F1step 2.2algebra∎

For $k = p - p^{'}$ , $Γ (k) = Γ (p) - Γ (p^{'})$ ; the right equation for $u (p)$ and left equation for $\overset{ˉ}{u} (p^{'})$ from [step 2.2] therefore give $\overset{ˉ}{u} (p^{'}) Γ (k) u (p) = μ \overset{ˉ}{u} (p^{'}) u (p) - μ \overset{ˉ}{u} (p^{'}) u (p) = 0$ . For $k = p + q$ , $Γ (k) = Γ (p) + Γ (q)$ ; the right $u$ equation and left $v$ equation give $\overset{ˉ}{v} (q) Γ (k) u (p) = μ \overset{ˉ}{v} (q) u (p) - μ \overset{ˉ}{v} (q) u (p) = 0$ .

Off-shell free Dirac Ward telescoping and tree currents

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources