The rank-one tilting quotient is the truncated fusion category

By [F1] every object decomposes into finitely many indecomposable $T(m)$. By [F2], all $T(m)$ with $m\ge \mathrm{\ell }-1$ have negligible identities and become zero. For retained $a$, $T(a)=\mathrm{\Delta }(a)=\mathrm{\nabla }(a)$ is simple; Hom orthogonality in [F1] gives $\mathrm{Hom}(T(a),T(b))={\delta }_{ab}\mathbb{C}$. Thus the subcategory of retained sums has rectangular matrix Hom spaces. Its trace on each endomorphism block is $[a+1]$ times the ordinary matrix trace, by [F2,F5]. These nonzero weights make the pairing with the reverse Hom space nondegenerate. Consequently a morphism is negligible exactly when its retained-to-retained block vanishes; all its other blocks factor through negligible summands and are negligible by [F3]. Hence the quotient is precisely the semisimple category of retained sums, with the asserted complete simple list.

The tensor product and composition descend by the tensor-ideal property [F3]. The unit $T(0)$ has scalar endomorphisms and dimension $1$, so is simple and survives. The dual of $T(m)$ is an indecomposable tilting of the same highest weight: its character is symmetric, duality preserves indecomposability, and [F1] classifies it as $T(m)$. Thus duality preserves both the retained and negligible parts of step 1.1; it descends, and the evaluation/coevaluation snakes remain identities in the quotient. The invertible braidings, pivotal maps and dual-compatible twist of [F4] descend by naturality and [F3]. The quotient is therefore rigid and ribbon. Left and right pivotal traces on retained simples are respectively the sums of their $K$ and $K^{-1}$ weight eigenvalues; symmetric strings make them equal to $[a+1]$. Additivity and the matrix-block description prove equality on all endomorphisms, so the quotient is spherical.

Put $k=\mathrm{\ell }-2$, $S_a=[T(a)]$ for $0\le a\le k$, and $x=S_1$. The tensor filtration [F1] for $V\otimes \mathrm{\Delta }(a)$ has factors $\mathrm{\Delta }(a+1),\mathrm{\Delta }(a-1)$, omitting the negative label at $a=0$. For $a\le k$ these factors are simple tiltings, including $\mathrm{\Delta }(\mathrm{\ell }-1)$ at the boundary, by [F2]. Their extensions split by [F1]'s Ext orthogonality, since each is both Weyl and dual Weyl. Hence in the quotient $x{S}_a=S_{a-1}+S_{a+1}$ with $S_{-1}=S_{k+1}=0$, $S_0=1$. For $k=1$ this says $x^2=1$. Define integer polynomials $P_0=1$, $P_1=x$, $P_{a+1}=x{P}_a-P_{a-1}$. Then $S_a=P_a(x)$ and $P_{k+1}(x)=0$.

Induction with this recurrence gives the polynomial identity $P_a{P}_b={\sum }_{c=\mathrm{\mid }a-b\mathrm{\mid },\mathrm{\mid }a-b\mathrm{\mid }+2,\dots ,a+b}{P}_c$: for $b=0$ it is immediate, and subtracting the preceding $b-1$ identity from $x$ times the $b$ identity cancels its two interior shifted sums and leaves the $b+1$ endpoints. Modulo $P_{k+1}=0$, the same recurrence gives $P_{k+1+j}=-P_{k+1-j}$. The terms above $k$ in the product sum therefore cancel the retained terms from $2k+2-a-b$ upwards, with matching parity; a possible term $P_{k+1}$ is zero. The remaining indices are precisely $\mathrm{\mid }a-b\mathrm{\mid }\le c\le \min (a+b,2k-a-b)$ of parity $a+b$. The simple classes are independent by step 1.1, so this class identity is the actual direct-sum multiplicity rule, not merely a character prediction. Since $2k=2\mathrm{\ell }-4$, it is the displayed formula. All arguments preserve every $\mathrm{\ell }\ge 3$ and complete the theorem.