The tau level adds under the filtered tensor product

Write $t=\tau (C)$ and $u=\tau (D)$. For each degree $n$, let $B_n=\mathrm{im}({\mathrm{\partial }}_{n+1})$ and $Z_n=\ker ({\mathrm{\partial }}_n)$. Finite-dimensional linear algebra gives complements $H_n,L_n$ with $Z_n=B_n\oplus H_n$ and $C_n=Z_n\oplus L_n$; the differential maps $L_n$ isomorphically onto $B_{n-1}$. Thus each complex is the direct sum of homology representatives with contractible two-term complexes. If $h$ contracts one such summand $A$, then $h\otimes 1$ contracts $A\otimes E$: the two cross terms cancel by the graded tensor signs, while $(\mathrm{\partial }h+h\mathrm{\partial })\otimes 1=1$. Hence $H(C\otimes D)\cong H(C)\otimes H(D)\cong {\mathbb{F}}_2$. By [F1], choose cycles $x\in F_{t}C$ and $y\in F_{u}D$ representing the nonzero homology classes. Their tensor is a nonzero class in $H(C\otimes D)$ and lies in $F_{t+u}(C\otimes D)$, so $\tau (C\otimes D)\le t+u$.

Let $n$ be the homological degree of $[x]$. The image of $H(F_{t-1}C)$ in $H(C)$ is zero by minimality of $t$, so every cycle in $F_{t-1}{C}_n$ is a boundary in $C$. The cycle $x$ cannot lie in $B_n+F_{t-1}{C}_n$; otherwise $x=b+v$ would make $v=x-b$ a cycle in $F_{t-1}C$ representing the same nonzero class. Extend the linear functional that is zero on $B_n+F_{t-1}{C}_n$ and one on $x$ to a functional $\lambda :C_n\to {\mathbb{F}}_2$. It vanishes on boundaries, so is a cocycle. Construct $\mu$ on the homological degree of $y$ in $D$ in the same way. The tensor cocycle $\lambda \otimes \mu$ evaluates to one on $x\otimes y$ and vanishes on every $F_{t+u-1}(C\otimes D)$: each summand $F_{p}C\otimes F_{q}D$ with $p+q\le t+u-1$ has either $p<t$ or $q<u$. Therefore no class in that lower filtration maps to the nonzero total homology class, and $\tau (C\otimes D)\ge t+u$. Together with step 1.1 this proves equality.