**Proof.**
Although the proof we give of this lemma is straightforward, it is a bit messy. To clarify the idea we explain what happens for low $n$ before giving the proof in general. For example, if $n = 0$, then (3) means that $P_1 = P_0 \times _ B P_0$. Since the ring map $P_0 \to B$ is surjective, this is of finite type over $A$ by More on Algebra, Lemma 15.5.1.

If $n = 1$, then (3) means that

\[ P_2 = \{ (f_0, f_1, f_2) \in P_1^3 \mid d_0f_0 = d_0f_1,\ d_1f_0 = d_0f_2,\ d_1f_1 = d_1f_2 \} \]

where the equalities take place in $P_0$. Observe that the triple

\[ (d_0f_0, d_1f_0, d_1f_1) = (d_0f_1, d_0f_2, d_1f_2) \]

is an element of the fibre product $P_0 \times _ B P_0 \times _ B P_0$ over $B$ because the maps $d_ i : P_1 \to P_0$ are morphisms over $B$. Thus we get a map

\[ \psi : P_2 \longrightarrow P_0 \times _ B P_0 \times _ B P_0 \]

The fibre of $\psi $ over an element $(g_0, g_1, g_2) \in P_0 \times _ B P_0 \times _ B P_0$ is the set of triples $(f_0, f_1, f_2)$ of $1$-simplices with $(d_0, d_1)(f_0) = (g_0, g_1)$, $(d_0, d_1)(f_1) = (g_0, g_2)$, and $(d_0, d_1)(f_2) = (g_1, g_2)$. As $P_\bullet \to B$ is a trivial Kan fibration the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus we see that $P_2$ fits into the cartesian diagram

\[ \xymatrix{ P_2 \ar[d] \ar[r] & P_1^3 \ar[d] \\ P_0 \times _ B P_0 \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^3 } \]

By More on Algebra, Lemma 15.5.2 we conclude. The general case is similar, but requires a bit more notation.

The case $n > 1$. By Simplicial, Lemma 14.19.14 the condition $P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet $ implies the same thing is true in the category of simplicial $A$-algebras and hence in the category of sets (as the forgetful functor from $A$-algebras to sets commutes with limits). Thus

\[ P_{n + 1} = \mathop{\mathrm{Mor}}\nolimits (\Delta [n + 1], P_\bullet ) = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ n \Delta [n + 1], \text{sk}_ n P_\bullet ) \]

by Simplicial, Lemma 14.11.3 and Equation (14.19.0.1). We will prove by induction on $1 \leq k < m \leq n + 1$ that the ring

\[ Q_{k, m} = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ k \Delta [m], \text{sk}_ k P_\bullet ) \]

is of finite type over $A$. The case $k = 1$, $1 < m \leq n + 1$ is entirely similar to the discussion above in the case $n = 1$. Namely, there is a cartesian diagram

\[ \xymatrix{ Q_{1, m} \ar[d] \ar[r] & P_1^ N \ar[d] \\ P_0 \times _ B \ldots \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^ N } \]

where $N = {m + 1 \choose 2}$. We conclude as before.

Let $1 \leq k_0 \leq n$ and assume $Q_{k, m}$ is of finite type over $A$ for all $1 \leq k \leq k_0$ and $k < m \leq n + 1$. For $k_0 + 1 < m \leq n + 1$ we claim there is a cartesian square

\[ \xymatrix{ Q_{k_0 + 1, m} \ar[d] \ar[r] & P_{k_0 + 1}^ N \ar[d] \\ Q_{k_0, m} \ar[r] & Q_{k_0, k_0 + 1}^ N } \]

where $N$ is the number of nondegenerate $(k_0 + 1)$-simplices of $\Delta [m]$. Namely, to see this is true, think of an element of $Q_{k_0 + 1, m}$ as a function $f$ from the $(k_0 + 1)$-skeleton of $\Delta [m]$ to $P_\bullet $. We can restrict $f$ to the $k_0$-skeleton which gives the left vertical map of the diagram. We can also restrict to each nondegenerate $(k_0 + 1)$-simplex which gives the top horizontal arrow. Moreover, to give such an $f$ is the same thing as giving its restriction to $k_0$-skeleton and to each nondegenerate $(k_0 + 1)$-face, provided these agree on the overlap, and this is exactly the content of the diagram. Moreover, the fact that $P_\bullet \to B$ is a trivial Kan fibration implies that the map

\[ P_{k_0} \to Q_{k_0, k_0 + 1} = \mathop{\mathrm{Mor}}\nolimits (\partial \Delta [k_0 + 1], P_\bullet ) \]

is surjective as every map $\partial \Delta [k_0 + 1] \to B$ can be extended to $\Delta [k_0 + 1] \to B$ for $k_0 \geq 1$ (small argument about constant simplicial sets omitted). Since by induction hypothesis the rings $Q_{k_0, m}$, $Q_{k_0, k_0 + 1}$ are finite type $A$-algebras, so is $Q_{k_0 + 1, m}$ by More on Algebra, Lemma 15.5.2 once more.
$\square$

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