**Proof.**
Assume $R \to S$ and $S \to S'$ are syntomic. This implies that $R \to S'$ is flat by Lemma 10.38.4. It also implies that $R \to S'$ is of finite presentation by Lemma 10.6.2. Thus it suffices to show that the fibres of $R \to S'$ are local complete intersections. Choose a prime $\mathfrak p \subset R$. We have a factorization

\[ \kappa (\mathfrak p) \to S \otimes _ R \kappa (\mathfrak p) \to S' \otimes _ R \kappa (\mathfrak p). \]

By assumption $S \otimes _ R \kappa (\mathfrak p)$ is a local complete intersection, and by Lemma 10.134.3 we see that $S \otimes _ R \kappa (\mathfrak p)$ is syntomic over $S \otimes _ R \kappa (\mathfrak p)$. After replacing $S$ by $S \otimes _ R \kappa (\mathfrak p)$ and $S'$ by $S' \otimes _ R \kappa (\mathfrak p)$ we may assume that $R$ is a field. Say $R = k$.

Choose a prime $\mathfrak q' \subset S'$ lying over the prime $\mathfrak q$ of $S$. Our goal is to find a $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S'_{g'}$ is a global complete intersection over $k$. Choose a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a global complete intersection over $k$. Since $S_ g \to S'_ g$ is still syntomic also, and $g \not\in \mathfrak q'$ we may replace $S$ by $S_ g$ and $S'$ by $S'_ g$ and assume that $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a global complete intersection over $k$. Next we choose a $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S' = S[y_1, \ldots , y_ m]/(h_1, \ldots , h_ d)$ is a relative global complete intersection over $S$. Hence we have reduced to part (2) of the lemma.

Suppose that $R \to S$ and $S \to S'$ are relative global complete intersections. Say $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(h_1, \ldots , h_ d)$. Then

\[ S' \cong R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/(f_1, \ldots , f_ c, h'_1, \ldots , h'_ d) \]

for some lifts $h_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ of the $h_ j$. Hence it suffices to bound the dimensions of the fibres. Thus we may yet again assume $R = k$ is a field. In this case we see that we have a ring, namely $S$, which is of finite type over $k$ and equidimensional of dimension $n - c$, and a finite type ring map $S \to S'$ all of whose nonempty fibre rings are equidimensional of dimension $m - d$. Then, by Lemma 10.111.6 for example applied to localizations at maximal ideals of $S'$, we see that $\dim (S') \leq n - c + m - d$ as desired.
$\square$

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