Proposition: Reduction of the candidate identifications

Let $I\leftrightarrow (ab)$ be a bijection of index sets. If ${\mathsf{A}}^{\lambda \mu ,I}={\mathfrak{a}}^{\lambda \mu ,ab}$, $R={E_0}^{\mathrm{R}}$ as distributions on ${\mathbb{R}}^4$, and the two symbols ${\sigma }_{\epsilon }$ denote the same function, then $L^{\mu ,I}(s,a)={\Lambda }^{\mu ,ab}(s,a)$ for all $(s,a)$. None of the three equalities is supplied by the record. After the fifth pass each, and each further candidate, reduces to one sentence to be located:

1. $R={E_0}^{\mathrm{R}}$ holds if and only if ${E_0}^{\mathrm{R}}$ is a fundamental solution of ${{\partial }_t}^2-\Delta$ with support in a past-bounded half-space (Lemma lem:R(iii)–(iv)); the reported normalization $\square {E_0}^{\mathrm{R}}=\mathrm{id}$ supplies the first half, the support sentence is missing.

2. $\mathsf{A}=\mathfrak{a}$ holds if and only if XIV-H's $a_1$ in (eq:a1H) is XII-D's $a_1$ in (eq:a1); then $a_1=-\mathcal{P}$ (Lemma lem:fiber) and $X^{\mu }[h]$ in (eq:L) has the form of XII-D's ${X_1}^{\mu }$.

3. ${\sigma }_{\epsilon }={\sigma }_{\epsilon }$ requires XIV-H's definition of ${\sigma }_{\epsilon }(a)$ as a function on ${\mathbb{R}}^4$; XII-B displays the XII sheet's ramp ${\sigma }_{\epsilon }(t)=\chi (t/\epsilon )$ XIIB.

4. $f$: XIV-H's $f$ is a compactly supported test weight if and only if it is XII-D's $f\in {C^{\infty }}_c({\mathring{M}}_T)$ XIIDweight; XIV-H's definition is not reproduced.

5. ${\dot{a}}_2$, $\mathsf{D}$: XIV-H's ${\dot{a}}_2$ is (eq:a2dot) if and only if XIV-H's $a_2$ is XII-E's (eq:a2); the construction of ${\mathsf{D}}^{\alpha \mu ,I,J}$ from ${\dot{a}}_2$ is XIV-H's and not reproduced, so even then $\mathsf{D}\stackrel{?}{=}\mathfrak{d}$ is a second step.

6. ${E^{\mathrm{A}}}_m$ is the kernel of Lemma lem:advanced if and only if XIV declares it the advanced inverse of $\square +m^2$; no such declaration is in the record.

7. $T^{\mu \nu }(\varphi ,\psi )$ in (eq:hess) is $T^{\mathrm{pol}}$ or $T^{\mathrm{sym}}$ of (eq:Tpol); the choice fixes a factor two (Lemma lem:polar).

Proof. Substituting the three equalities into (eq:L) gives ${\sigma }_{\epsilon }(a){\mathfrak{a}}^{\lambda \mu ,ab}{\partial }_{a^{\lambda }}{E_0}^{\mathrm{R}}(s-a)$, which is (eq:XIIA) with $(x,z)=(s,a)$. Items (a)–(g) cite the lemmas that reduce each to a single sentence, and the manifest records that the sentence is absent. The reading rule of record forbids identification by notation.