= Solution
Write the negative filtration as $F_0R=R\supseteq F_{-1}R\supseteq F_{-2}R\supseteq\cdots$, with $F_iRF_jR\subseteq F_{i+j}R$. Completeness is used in the separated sense $R\cong\varprojlim R/F_{-n}R$; in particular the intersection of all filtration terms is zero.
For any <left ideal> $J$, its induced associated graded ideal is a <left ideal> of $\operatorname{gr}R$. Choose finite homogeneous generators and lift them to $j_1,\ldots,j_r\in J$, with orders $n_i\ge0$, meaning $j_i\in F_{-n_i}R\setminus F_{-n_i-1}R$. Let $j\in J$. At each order $s$, the leading symbol of the current residual is a sum of the generator symbols with homogeneous coefficients. Lift these coefficients to $c_{i,s}\in F_{-(s-n_i)}R$ (zero when $n_i>s$) and subtract $\sum_i c_{i,s}j_i$. The new residual lies in $F_{-(s+1)}R$.
Iterating expresses $j$ as partial sums of $\sum_i c_i j_i$, with residuals tending to zero. For fixed $i$ the successive coefficient increments have orders tending to infinity. Separated completeness therefore supplies $c_i=\sum_s c_{i,s}\in R$. Multiplication is continuous by the multiplicative filtration, so
$$
\boxed{j=\sum_{i=1}^r c_i j_i}.
$$
The sum on the right is finite, despite the convergent process used to obtain its coefficients. It belongs to the actual <left ideal> generated by the lifts; no assumption that $J$ is closed was made. Every <left ideal> is consequently finitely generated, and \b[$R$ is a <left Noetherian ring>]. This is the left-handed <complete negative filtered-graded transfer of Noetherianity>; the separated-completeness hypothesis is essential to the limiting argument.
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