Since $c_r(\mathcal{E})$ is a bivariant class (Lemma \ref{lemma-cap-cp-bivariant}) we may assume $X = X'$ and we have to show that $c_r(\mathcal{E}) \cap [X]_n = [Z(s)]_{n - r}$ in $\CH_{n - r}(X)$. We will prove the lemma by induction on $r \geq 0$. (The case $r = 0$ is trivial.) The case $r = 1$ is handled by Lemma \ref{lemma-geometric-cap}. Assume $r > 1$. \medskip\noindent Let $\pi : P \to X$ be the projective space bundle associated to $\mathcal{E}$ and consider the short exact sequence $$ 0 \to \mathcal{E}' \to \pi^*\mathcal{E} \to \mathcal{O}_P(1) \to 0 $$ By the projective space bundle formula (Lemma \ref{lemma-chow-ring-projective-bundle}) it suffices to prove the equality after pulling back by $\pi$. Observe that $\pi^{-1}Z(s) = Z(\pi^*s)$ has $\delta$-dimension $\leq n - 1$ and that the assumption on regular sequences at generic points of $\delta$-dimension $n - 1$ holds by flat pullback, see Algebra, Lemma \ref{algebra-lemma-flat-increases-depth}. Let $t \in \Gamma(P, \mathcal{O}_P(1))$ be the image of $\pi^*s$. We claim $$ [Z(t)]_{n + r - 2} = c_1(\mathcal{O}_P(1)) \cap [P]_{n + r - 1} $$ Assuming the claim we finish the proof as follows. The restriction $\pi^*s|_{Z(t)}$ maps to zero in $\mathcal{O}_P(1)|_{Z(t)}$ hence comes from a unique element $s' \in \Gamma(Z(t), \mathcal{E}'|_{Z(t)})$. Note that $Z(s') = Z(\pi^*s)$ as closed subschemes of $P$. If $\xi \in Z(s')$ is a generic point with $\delta(\xi) = n - 1$, then the ideal of $Z(s')$ in $\mathcal{O}_{Z(t), \xi}$ can be generated by a regular sequence of length $r - 1$: it is generated by $r - 1$ elements which are the images of $r - 1$ elements in $\mathcal{O}_{P, \xi}$ which together with a generator of the ideal of $Z(t)$ in $\mathcal{O}_{P, \xi}$ form a regular sequence of length $r$ in $\mathcal{O}_{P, \xi}$. Hence we can apply the induction hypothesis to $s'$ on $Z(t)$ to get $c_{r - 1}(\mathcal{E}') \cap [Z(t)]_{n + r - 2} = [Z(s')]_{n - 1}$. Combining all of the above we obtain \begin{align*} c_r(\pi^*\mathcal{E}) \cap [P]_{n + r - 1} & = c_{r - 1}(\mathcal{E}') \cap c_1(\mathcal{O}_P(1)) \cap [P]_{n + r - 1} \\ & = c_{r - 1}(\mathcal{E}') \cap [Z(t)]_{n + r - 2} \\ & = [Z(s')]_{n - 1} \\ & = [Z(\pi^*s)]_{n - 1} \end{align*} which is what we had to show. \medskip\noindent Proof of the claim. This will follow from an application of the already used Lemma \ref{lemma-geometric-cap}. We have $\pi^{-1}(Z(s)) = Z(\pi^*s) \subset Z(t)$. On the other hand, for $x \in X$ if $P_x \subset Z(t)$, then $t|_{P_x} = 0$ which implies that $s$ is zero in the fibre $\mathcal{E} \otimes \kappa(x)$, which implies $x \in Z(s)$. It follows that $\dim_\delta(Z(t)) \leq n + (r - 1) - 1$. Finally, let $\xi \in Z(t)$ be a generic point with $\delta(\xi) = n + r - 2$. If $\xi$ is not the generic point of the fibre of $P \to X$ it is immediate that a local equation of $Z(t)$ is a nonzerodivisor in $\mathcal{O}_{P, \xi}$ (because we can check this on the fibre by Algebra, Lemma \ref{algebra-lemma-grothendieck}). If $\xi$ is the generic point of a fibre, then $x = \pi(\xi) \in Z(s)$ and $\delta(x) = n + r - 2 - (r - 1) = n - 1$. This is a contradiction with $\dim_\delta(Z(s)) \leq n - r$ because $r > 1$ so this case doesn't happen.