Part (1) is essentially true by definition. Namely, if $y \in Z$, then we can take $k = 0$ and $V_0 = W_0 = \overline{\{y\}}$. \medskip\noindent Proof of (2). Let $y \leadsto y'$ be a nontrivial specialization and let $V_0 \subset W_0 \supset V_1 \subset W_1 \supset \ldots \subset W_k$ is a system for $y$. Here there are two cases. Case I: $V_k = W_k$, i.e., $c_k = 0$. In this case we can set $V'_k = W'_k = \overline{\{y'\}}$. An easy computation shows that $\delta(V_0, W_0, \ldots, V'_k, W'_k) \leq \delta(V_0, W_0, \ldots, V_k, W_k)$ because only $b_{k - 1}$ is changed into a bigger integer. Case II: $V_k \not = W_k$, i.e., $c_k > 0$. Observe that in this case $\max_{i = 0, 1, \ldots, k} (c_i + c_{i + 1} + \ldots + c_k - b_i - b_{i + 1} - \ldots - b_{k - 1}) > 0$. Hence if we set $V'_{k + 1} = W_{k + 1} = \overline{\{y'\}}$, then although $k$ is replaced by $k + 1$, the maximum now looks like $$ \max_{i = 0, 1, \ldots, k + 1} (c_i + c_{i + 1} + \ldots + c_k + c_{k + 1} - b_i - b_{i + 1} - \ldots - b_{k - 1} - b_k) $$ with $c_{k + 1} = 0$ and $b_k = \text{codim}(V_{k + 1}, W_k) > 0$. This is strictly smaller than $\max_{i = 0, 1, \ldots, k} (c_i + c_{i + 1} + \ldots + c_k - b_i - b_{i + 1} - \ldots - b_{k - 1})$ and hence $\delta(V_0, W_0, \ldots, V'_{k + 1}, W'_{k + 1}) \leq \delta(V_0, W_0, \ldots, V_k, W_k)$ as desired. \medskip\noindent Proof of (3). Given $y \in Y$ and $z \in \overline{\{y\}} \cap Z$ we get the system $$ V_0 = \overline{\{z\}} \subset W_0 = \overline{\{y\}} $$ and $c_0 = \text{codim}(V_0, W_0) = \dim(\mathcal{O}_{\overline{\{y\}}, z})$ by Properties, Lemma \ref{properties-lemma-codimension-local-ring}. Thus we see that $\delta(V_0, W_0) = 0 + c_0 = c_0$ which proves what we want. \medskip\noindent Proof of (4). Let $\delta$ be a dimension function on $Y$. Let $V_0 \subset W_0 \supset V_1 \subset W_1 \supset \ldots \subset W_k$ be a system for $y$. Let $y'_i \in W_i$ and $y_i \in V_i$ be the generic points, so $y_0 \in Z$ and $y_k = y$. Then we see that $$ \delta(y_i) - \delta(y_{i - 1}) = \delta(y'_{i - 1}) - \delta(y_{i - 1}) - \delta(y'_{i - 1}) + \delta(y_i) = c_{i - 1} - b_{i - 1} $$ Finally, we have $\delta(y'_k) - \delta(y_{k - 1}) = c_k$. Thus we see that $$ \delta(y) - \delta(y_0) = c_0 + \ldots + c_k - b_0 - \ldots - b_{k - 1} $$ We conclude $\delta(V_0, W_0, \ldots, W_k) \geq k + \delta(y) - \delta(y_0)$ which proves what we want. \medskip\noindent Proof of (5). The function $\delta(y) = \dim(\overline{\{y\}})$ is a dimension function. Hence $\delta(y) \leq \delta_Z(y)$ by part (4). By part (3) we have $\delta_Z(y) \leq \delta(y)$ and we are done. \medskip\noindent Proof of (6). Given such a sequence of points, we may assume all the specializations $y'_i \leadsto y_{i + 1}$ are nontrivial (otherwise we can shorten the chain of specializations). Then we set $V_i = \overline{\{y_i\}}$ and $W_i = \overline{\{y'_i\}}$ and we compute $\delta(V_0, W_1, V_1, \ldots, W_{k - 1}) = k$ because all the codimensions $c_i$ of $V_i \subset W_i$ are $1$ and all $b_i > 0$. This implies $\delta_Z(y'_{k - 1}) \leq k$ as $y'_{k - 1}$ is the generic point of $W_k$. Then $\delta_Z(y) \leq k$ by part (2) as $y$ is a specialization of $y_{k - 1}$. \medskip\noindent Proof of (7). This is clear as their are fewer systems to consider in the computation of $\delta^{Y'}_{Y' \cap Z}$.