Proof.
The statement that $[\mathcal{F}]_ r$ and $[\mathcal{G}]_ s$ intersect properly is immediate. Since we are proving an equality of cycles we may work locally on $X$. (Observe that the formation of the intersection product of cycles, the formation of $\text{Tor}$-sheaves, and forming the cycle associated to a coherent sheaf, each commute with restriction to open subschemes.) Thus we may and do assume that $X$ is affine.
Denote
\[ RHS(\mathcal{F}, \mathcal{G}) = [\mathcal{F}]_ r \cdot [\mathcal{G}]_ s \quad \text{and}\quad LHS(\mathcal{F}, \mathcal{G}) = \sum (-1)^ p [\text{Tor}_ p^{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})]_{r + s - \dim (X)} \]
Consider a short exact sequence
\[ 0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0 \]
of coherent sheaves on $X$ with $\text{Supp}(\mathcal{F}_ i) \subset \text{Supp}(\mathcal{F})$, then both $LHS(\mathcal{F}_ i, \mathcal{G})$ and $RHS(\mathcal{F}_ i, \mathcal{G})$ are defined for $i = 1, 2, 3$ and we have
\[ RHS(\mathcal{F}_2, \mathcal{G}) = RHS(\mathcal{F}_1, \mathcal{G}) + RHS(\mathcal{F}_3, \mathcal{G}) \]
and similarly for LHS. Namely, the support condition guarantees that everything is defined, the short exact sequence and additivity of lengths gives
\[ [\mathcal{F}_2]_ r = [\mathcal{F}_1]_ r + [\mathcal{F}_3]_ r \]
(Chow Homology, Lemma 42.10.4) which implies additivity for RHS. The long exact sequence of $\text{Tor}$s
\[ \ldots \to \text{Tor}_1(\mathcal{F}_3, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_1, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_2, \mathcal{G}) \to \text{Tor}_0(\mathcal{F}_3, \mathcal{G}) \to 0 \]
and additivity of lengths as before implies additivity for LHS.
By Algebra, Lemma 10.62.1 and the fact that $X$ is affine, we can find a filtration of $\mathcal{F}$ whose graded pieces are structure sheaves of closed subvarieties of $\text{Supp}(\mathcal{F})$. The additivity shown in the previous paragraph, implies that it suffices to prove $LHS = RHS$ with $\mathcal{F}$ replaced by $\mathcal{O}_ V$ where $V \subset \text{Supp}(\mathcal{F})$. By symmetry we can do the same for $\mathcal{G}$. This reduces us to proving that
\[ LHS(\mathcal{O}_ V, \mathcal{O}_ W) = RHS(\mathcal{O}_ V, \mathcal{O}_ W) \]
where $W \subset \text{Supp}(\mathcal{G})$ is a closed subvariety. If $\dim (V) = r$ and $\dim (W) = s$, then this equality is the definition of $V \cdot W$. On the other hand, if $\dim (V) < r$ or $\dim (W) < s$, i.e., $[V]_ r = 0$ or $[W]_ s = 0$, then we have to prove that $RHS(\mathcal{O}_ V, \mathcal{O}_ W) = 0$ 1.
Let $Z \subset V \cap W$ be an irreducible component of dimension $r + s - \dim (X)$. This is the maximal dimension of a component and it suffices to show that the coefficient of $Z$ in $RHS$ is zero. Let $\xi \in Z$ be the generic point. Write $A = \mathcal{O}_{X, \xi }$, $B = \mathcal{O}_{X \times X, \Delta (\xi )}$, and $C = \mathcal{O}_{V \times W, \Delta (\xi )}$. By Lemma 43.19.1 we have
\[ \text{coeff of }Z\text{ in } RHS(\mathcal{O}_ V, \mathcal{O}_ W) = \sum (-1)^ i \text{length}_ B \text{Tor}_ i^ B(A, C) \]
Since $\dim (V) < r$ or $\dim (W) < s$ we have $\dim (V \times W) < r + s$ which implies $\dim (C) < \dim (X)$ (small detail omitted). Moreover, the kernel $I$ of $B \to A$ is generated by a regular sequence of length $\dim (X)$ (Lemma 43.13.3). Hence vanishing by Lemma 43.16.2 because the Hilbert function of $C$ with respect to $I$ has degree $\dim (C) < n$ by Algebra, Proposition 10.60.9.
$\square$
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