Proof.
Fix $n$. Using the long exact cohomology sequence associated to the short exact sequence of complexes
\[ 0 \to F^{p + 1}K^\bullet \to F^ pK^\bullet \to \text{gr}^ pK^\bullet \to 0 \]
we find that $E_1^{p, n - p} = 0$ for $p \geq \max (p_0(n), p_0(n + 1))$ and $p < \min (p_1(n), p_1(n + 1))$. Hence the spectral sequence is bounded (Definition 12.24.7). This proves (1).
It is clear from the assumptions and Definition 12.24.5 that the filtration on $H^ n(K^\bullet )$ is finite. This proves (2).
Next we prove that the spectral sequence weakly converges to $H^*(K^\bullet )$ using Lemma 12.24.10. Let us show that we have equality in (12.24.6.1). Namely, for $p + r > p_0(n + 1)$ the map
\[ d : F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) \to F^{p + r}K^{n + 1} \]
ends up in the image of $d : F^{p + r}K^ n \to F^{p + r}K^{n + 1}$ because the complex $F^{p + r}K^\bullet $ is exact in degree $n + 1$. We conclude that $F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) = d(F^{p + r}K^ n) + \mathop{\mathrm{Ker}}(d) \cap F^ pK^ n$. Hence for such $r$ we have
\[ \mathop{\mathrm{Ker}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n} = F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n} \]
which proves the desired equality. To show that we have equality in (12.24.6.2) we use that for $p - r + 1 < p_1(n - 1)$ we have
\[ d(F^{p - r + 1}K^{n - 1}) = \mathop{\mathrm{Im}}(d) \cap F^{p - r + 1}K^ n \]
because the map $F^{p - r + 1}K^\bullet \to K^\bullet $ induces an isomorphism on cohomology in degree $n - 1$. This shows that we have
\[ F^ pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} = \mathop{\mathrm{Im}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n} \]
for such $r$ which proves the desired equality.
To see that the spectral sequence abuts to $H^*(K^\bullet )$ using Lemma 12.24.10 we have to show that $\bigcap _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Im}}(d) \cap K^ n$ and $\bigcup _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Ker}}(d) \cap K^ n$. For $p \geq p_0(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Im}}(d) \cap K^ n$ and for $p \leq p_1(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Ker}}(d) \cap K^ n$. Combining weak convergence, abutment, and boundedness we see that (2) and (3) are true.
$\square$
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