Writing E2 implies H proves convergence

Put $C^0={\mathbb{F}}_2$, $C^n=0$ otherwise and $d=0$. Set $F^{p}C=C$ for every integer $p$. This decreasing filtration is exhaustive but not separated. All its adjacent quotients are zero, and all page cycle/boundary quotients in F2 are zero: the denominator already contains the entire preceding filtration level. Thus $E_2=E_{\mathrm{\infty }}=0$, while $H^0(C)={\mathbb{F}}_2$. The induced filtration on $H^0$ is constantly ${\mathbb{F}}_2$ and is not separated; its quotient completion is zero. A double arrow cannot make this strongly convergent to the nonzero target with a finite normalized filtration. Indeed zero graded pieces and finite zero/full endpoints would force that target to be zero.

There is a separate reconstruction omission even when strong convergence is proved. The finite filtration $0\subset 2\mathbb{Z}\mathrm{/}4\subset \mathbb{Z}\mathrm{/}4$ has two $\mathbb{Z}\mathrm{/}2$ quotients, as does $0\subset (\mathbb{Z}\mathrm{/}2)\oplus 0\subset (\mathbb{Z}\mathrm{/}2{)}^2$. Their targets differ because the first has an element of order four and the second has none. F3 realizes this phenomenon in a collapsed sequence. Accordingly the record in F1 requires actual stabilization and graded identifications, verified filtration conditions, and resolution or explicit retention of extensions. The first witness uses no choice; the two finite-group calculations do not use the optional AC splitting branch of F3.