The paper’s closing Remark (6-term/3-step operator) FAILS for the two \(q\)-Stirling first-kind conventions (Milestone 13, cas/check_remark_convention.py). Under the classical signed \(\mathrm{S_1}\) reading (\(\mathrm{S_1}(n,k)=\mathrm{S_1}(n{-}1,k{-}1) -(n{-}1)\, \mathrm{S_1}(n{-}1,k)\)), however, the Remark is correct exactly as printed: the printed 6-term recurrence holds on \(F\), the printed certificate (inner factor \(q^m\)) satisfies the pointwise identity with \(g(m,n,m)=0\) and the edge-shift cancellation \(g(m,n,n{+}1)=-E(m,n)\) at the upper boundary (cas/check_remark_classical.py, Milestone 14). The external gemini-review draft’s inner factor \(q^{m+k}\) is a mis-transcription. The classical reading is formalized in Lean (Milestone 14 + follow-up): the pointwise certificate identity and the telescoping step are proved, and the rescaled (\(q^n\)-scaled, denominator-free) 6-term series form vanishes on the entire \((q,m,n)\) grid with no hypotheses (remarkOpSeriesR_eq_zero_full): the empty rows \(n{\lt}m\) and the diagonal \(n=m\) directly (the diagonal closes through the \(q\)-independent singleton values \(-(m+2)+2(m+1)-m=0\)), and the main rows \(m{\lt}n\) for every rational \(q\) by a polynomial lift from \(0{\lt}q{\lt}1\). Consequently the printed recurrence requires only the hypothesis \(q\neq 0\) (remark_recurrence) — no row condition \(m+1\le n\) and no certificate hypotheses \(q^{m+1},q^{m+2}\neq 1\) (the CAS audit covers the same degenerate rows in its part 1b: diagonal, empty, and \(q\in \{ 0,1\} \)). The printed-object transport layer is also complete: the printed certificate \(g\) (remarkGcert, over the denominator \(-q^{m+1}-q^{m+2}+q^{2m+3}+1=(1-q^{m+1})(1-q^{m+2})\)), the printed six-term operator remarkOpPrinted and edge-shift remarkPrintedEdge are defined, and the bridge remarkGp_eq connects them to the proved rescaled layer on the certificate domain (\(q\neq 0\), \(q^{m+1},q^{m+2}\neq 1\), \(k\le n{+}1\)): \(G_p=q^{n+2m}\cdot \mathrm{ds}\cdot g\) through the sentinel \(k=n{+}1\) (weight identity cert_weight_id). From the bridge and the transports remarkOpR_eq/remarkEdgeR_eq follow the printed pointwise identity remark_pointwise (\(\mathtt{op}(k)=g(k{+}1)-g(k)\), with the certificate-domain small case remark_pointwise_small_printed at \(k=1\)), both boundaries remark_certificate_lower (\(g(m)=0\), all \(q\)) and remark_certificate_upper (\(g(n{+}1)=-E(m,n)\)), and the telescoping theorem remark_telescoping: \(\sum _{k=m+1}^{n}\mathtt{op}(k)=g(n{+}1)-g(m{+}1)\). The printed \(k=m\) endpoint is also covered: the hypothesis-free rescaled column identity remark_pointwise_at_m (\(q^{2m}\cdot \mathrm{ds}\cdot \mathtt{opR}(m)=G_p(m{+}1)-G_p(m)\), CAS part 4c) transports through the bridge to remark_pointwise_at_m_printed (\(\mathtt{op}(m)=g(m{+}1)-g(m)\) for \(m\le n\); the row condition is needed because the bridge only covers \(k\le n{+}1\), which fails exactly on the empty rows \(m{\gt}n\)), the full printed pointwise identity holds on the whole range remark_pointwise_full (\(m\le k\le n\)), and the full telescoping theorem remark_telescoping_full sums over \(\sum _{k=m}^{n}\mathtt{op}(k)=g(n{+}1)-g(m)\) — including the lower endpoint contribution (nonzero in general: \(\mathtt{op}(1)=-\tfrac 12\) at \((q,m,n)=(2,1,2)\)) and the diagonal rows \(m=n\) — closing the printed certificate proof of the Remark exactly as in the CAS audit (check_remark_classical.py).