Finite-diagonal cohomological double-complex spectral sequences

Replace $v$ on column $a$ by $(-1{)}^{a}v$. The mixed composites sum to $(-1{)}^{a}hv+(-1{)}^{a+1}vh=0$, and the new vertical arrow still squares to zero. Then regard the resulting cochain arrows as arrows in the opposite category. Thus the arrow from $C^{a-1,b}$ to $C^{a,b}$ becomes a homological arrow from bidegree $(a,b)$ to $(a-1,b)$ in the opposite category, with analogous vertical arrows. The square-zero and anticommuting identities are unchanged on reversing composition. The total complex there is the opposite of $T$. The increasing column cutoff through $p$ is the quotient $T\mathrm{/}{F}^{p+1}T$ viewed as a subobject in the opposite category; the same holds for the row cutoff.

Apply both homological theorems in that category. Opposite-category homology is original-category cohomology, since kernel and cokernel exchange. Reversing a page arrow of bidegree $(-r,r-1)$ gives bidegree $(r,1-r)$. For the first page, the within-column differential is $(-1{)}^{p}v$. This constant sign does not change its kernel, image, or their canonical quotient, so vertical cohomology has its canonical quotient identification and the $d_1$ on that quotient is induced by $h$. For the row filtration the within-row differential is $h$ and the next differential on horizontal degree $q$ is $(-1{)}^{q}v$.

Translate the abutment precisely. An image subobject of $H_n(T^{\mathrm{o}\mathrm{p}})$ in the opposite category corresponds to the quotient of $H^n(T)$ by $\ker (H^n(T)\to H^n(T\mathrm{/}{F}^{p+1}T))$. The long exact sequence for $0\to F^{p+1}T\to T\to T\mathrm{/}{F}^{p+1}T\to 0$ identifies this kernel with $F^{p+1}{H}^n(T)$. Thus the opposite of the successive quotient between cutoffs $p-1$ and $p$ is exactly $F^p{H}^n\mathrm{/}{F}^{p+1}{H}^n$, proving the asserted abutment rather than an unrelated filtration.

In total degree $n$ all summands have indices between zero and $n$, giving the stated endpoints; negative total degrees vanish. Finite filtrations are exhaustive and separated, and their quotient towers are eventually constant with value $H^n(T)$, so the completion map is an isomorphism. The first-quadrant bounds make both incident differentials eventually zero at each bidegree. These prove strong convergence, including zero and one-summand diagonals. If $a\ge c$, set $a^{\mathrm{\prime }}=a-c$ and replace $v$ by $(-1{)}^{c}v$; then $h+(-1{)}^{a^{\mathrm{\prime }}}((-1{)}^{c}v)=h+(-1{)}^{a}v$. The normalized total degree is $n-c$, and translation back preserves the original target degree. All constructions use finite biproducts and prescribed signs; no choice is used.