We have assumed that

In other words, \(a \overset {x^n}{\equiv } b\) (by the definition of the relation \(\overset {x^n}{\equiv }\)). Similarly, \(c \overset {x^n}{\equiv } d\). Hence, by the compatibility of \(\overset {x^n}{\equiv }\) with division (Theorem 1.328 (e)), we obtain \(\frac{a}{c} \overset {x^n}{\equiv } \frac{b}{d}\). In other words,

(by the definition of the relation \(\overset {x^n}{\equiv }\)). This proves Lemma A.27.

This can be proved similarly to Proposition 1.366, with the obvious changes (replacing multiplication by division, and requiring each \(\mathbf{b}_i\) to be invertible). Of course, instead of Lemma 1.346, we need to use Lemma A.27.

The set \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in I}\). In other words, \(U\) is a finite subset of \(I\) that determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in I}\) (by the definition of an \(x^n\)-approximator). Hence, in particular, \(U\) is finite. Moreover, \(U \subseteq I\) (since \(U\) is a subset of \(I\)).

Let \(M = U \cap J\). Thus, \(M = U \cap J \subseteq U\), so that the set \(M\) is finite (since \(U\) is finite). Moreover, \(M\) is a subset of \(J\) (since \(M = U \cap J \subseteq J\)).

Now, let \(N\) be a finite subset of \(J\) satisfying \(M \subseteq N \subseteq J\). We shall show that

Indeed, the set \(N \cup U\) is finite (since \(N\) and \(U\) are finite) and is a subset of \(I\) (since \(N \subseteq J \subseteq I\) and \(U \subseteq I\)); it also satisfies \(U \subseteq N \cup U \subseteq I\). Now, we recall that \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in I}\). Hence, Proposition 1.364 (a) (applied to \(U\) and \(N \cup U\) instead of \(M\) and \(J\)) yields

(since \(N \cup U\) is a finite subset of \(I\) satisfying \(U \subseteq N \cup U \subseteq I\)).

On the other hand, we have assumed that \(\left(\mathbf{a}_{i}\right)_{i \in I}\) is a family of invertible FPSs. Thus, for each \(i \in I\), the FPS \(\mathbf{a}_{i}\) is invertible, so that its inverse \(\mathbf{a}_{i}^{-1}\) is well-defined. We obviously have

(since the relation \(\overset {x^n}{\equiv }\) is reflexive).

Hence, by the compatibility of \(\overset {x^n}{\equiv }\) with multiplication (Theorem 1.328 (b)), applied to \(a = \prod _{i \in N \cup U} \mathbf{a}_{i}\) and \(b = \prod _{i \in U} \mathbf{a}_{i}\) and \(c = \prod _{i \in U \setminus J} \mathbf{a}_{i}^{-1}\) and \(d = \prod _{i \in U \setminus J} \mathbf{a}_{i}^{-1}\), we obtain

On the other hand, the sets \(N\) and \(U \setminus J\) are disjoint (since \(N \subseteq J\) and \(U \setminus J \subseteq J^c\)), and their union is \(N \cup (U \setminus J) = N \cup U\) (since \(U = (U \cap J) \cup (U \setminus J)\) and \(U \cap J = M \subseteq N\)). Hence, the set \(N \cup U\) is the union of its two disjoint subsets \(N\) and \(U \setminus J\). Thus, we can split the product \(\prod _{i \in N \cup U} \mathbf{a}_{i}\) as follows:

Multiplying both sides of this equality by \(\prod _{i \in U \setminus J} \mathbf{a}_{i}^{-1}\), we obtain

However, the set \(U\) is the union of its two disjoint subsets \(U \cap J\) and \(U \setminus J\). Thus, we can split the product \(\prod _{i \in U} \mathbf{a}_{i}\) as follows:

Multiplying both sides by \(\prod _{i \in U \setminus J} \mathbf{a}_{i}^{-1}\), we obtain

Substituting these two simplifications into the \(x^n\)-equivalence above, we obtain

In other words, each \(m \in \left\{ 0, 1, \ldots , n\right\} \) satisfies

(by the definition of \(\overset {x^n}{\equiv }\)).

Forget that we fixed \(N\). We have thus shown that every finite subset \(N\) of \(J\) satisfying \(M \subseteq N \subseteq J\) satisfies the above equality for each \(m \in \left\{ 0, 1, \ldots , n\right\} \).

Now, let \(m \in \left\{ 0, 1, \ldots , n\right\} \). Then, every finite subset \(N\) of \(J\) satisfying \(M \subseteq N \subseteq J\) satisfies

In other words, the set \(M\) determines the \(x^m\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) (by the definition of “determining the \(x^m\)-coefficient in a product”).

Forget that we fixed \(m\). We thus have shown that \(M\) determines the \(x^m\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) for each \(m \in \left\{ 0, 1, \ldots , n\right\} \). In other words, \(M\) determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\). In other words, \(M\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in J}\) (by the definition of an “\(x^n\)-approximator”, since \(M\) is a finite subset of \(J\)). In other words, \(U \cap J\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in J}\) (since \(M = U \cap J\)). This proves Lemma A.28.

Let \(J\) be a subset of \(I\). We shall show that the family \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is multipliable.

Fix \(n \in \mathbb {N}\). We know that the family \(\left(\mathbf{a}_{i}\right)_{i \in I}\) is multipliable. Hence, there exists an \(x^n\)-approximator \(U\) for \(\left(\mathbf{a}_{i}\right)_{i \in I}\) (by Lemma 1.363). Consider this \(U\).

Set \(M = U \cap J\). Lemma A.28 yields that \(U \cap J\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in J}\). In other words, \(M\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in J}\) (since \(M = U \cap J\)). In other words, \(M\) is a finite subset of \(J\) that determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) (by the definition of an \(x^n\)-approximator). Thus, the set \(M\) determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\). Hence, in particular, this set \(M\) determines the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\). Therefore, the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is finitely determined (by the definition of “finitely determined”, since \(M\) is a finite subset of \(J\)).

Forget that we fixed \(n\). We thus have proved that for each \(n \in \mathbb {N}\), the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is finitely determined. In other words, each coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is finitely determined. In other words, the family \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is multipliable (by the definition of “multipliable”).

Forget that we fixed \(J\). We thus have shown that the family \(\left(\mathbf{a}_{i}\right)_{i \in J}\) is multipliable whenever \(J\) is a subset of \(I\). In other words, any subfamily of \(\left(\mathbf{a}_{i}\right)_{i \in I}\) is multipliable. This proves Proposition 1.369.

This is just Theorem 1.328 (f) (the part about products), with the letters \(S\), \(s\), \(a_s\), and \(b_s\) renamed as \(V\), \(w\), \(c_w\), and \(d_w\).

We shall subdivide our proof into several claims.

Claim 1: Let \(w \in W\). Then, the family \(\left(\mathbf{a}_{s}\right)_{s \in S,\; f(s) = w}\) is multipliable.

[Proof of Claim 1: This was an assumption of Proposition 1.372.]

Let us set

This is well-defined, because for each \(w \in W\), the product \(\prod _{\substack {s \in S;\\ f(s) = w}} \mathbf{a}_{s}\) is well-defined (since Claim 1 shows that the family \(\left(\mathbf{a}_{s}\right)_{s \in S,\; f(s) = w}\) is multipliable).

Now, let \(n \in \mathbb {N}\). By Lemma 1.363 (applied to \(S\) and \(\left(\mathbf{a}_{s}\right)_{s \in S}\) instead of \(I\) and \(\left(\mathbf{a}_{i}\right)_{i \in I}\)), there exists an \(x^n\)-approximator for \(\left(\mathbf{a}_{s}\right)_{s \in S}\). Pick such an \(x^n\)-approximator, and call it \(U\). Then, \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{s}\right)_{s \in S}\); in other words, \(U\) is a finite subset of \(S\) that determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{s}\right)_{s \in S}\) (by the definition of an \(x^n\)-approximator).

The set \(U\) is finite. Thus, its image \(f(U) = \left\{ f(u) \mid u \in U\right\} \) is finite as well. Now, we claim the following:

Claim 2: For each \(w \in W\), we have

[Proof of Claim 2: Let \(w \in W\). Let \(J\) be the subset \(\left\{ s \in S \mid f(s) = w\right\} \) of \(S\). Then, the family \(\left(\mathbf{a}_{s}\right)_{s \in J}\) is just the family \(\left(\mathbf{a}_{s}\right)_{s \in S,\; f(s) = w}\), and thus is multipliable (by Claim 1). Furthermore, Lemma A.28 (applied to \(S\) and \(\left(\mathbf{a}_{s}\right)_{s \in S}\) instead of \(I\) and \(\left(\mathbf{a}_{i}\right)_{i \in I}\)) yields that \(U \cap J\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{s}\right)_{s \in J}\) (since \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{s}\right)_{s \in S}\)). Hence, Proposition 1.364 (b) (applied to \(J\), \(\left(\mathbf{a}_{s}\right)_{s \in J}\) and \(U \cap J\) instead of \(I\), \(\left(\mathbf{a}_{i}\right)_{i \in I}\) and \(M\)) yields

However, we have \(J = \left\{ s \in S \mid f(s) = w\right\} \). Thus, \(\prod _{s \in J} \mathbf{a}_{s} = \prod _{\substack {s \in S;\\ f(s) = w}} \mathbf{a}_{s} = \mathbf{b}_w\) and \(U \cap J = \left\{ s \in U \mid f(s) = w\right\} \). So the product sign “\(\prod _{s \in U \cap J}\)” is equivalent to “\(\prod _{\substack {s \in U;\\ f(s) = w}}\)”. Thus,

This proves Claim 2.]

Claim 3: The set \(f(U)\) determines the \(x^n\)-coefficient in the product of \(\left(\mathbf{b}_{w}\right)_{w \in W}\).

[Proof of Claim 3: Let \(T\) be a finite subset of \(W\) satisfying \(f(U) \subseteq T\). We must show that \(\left[x^n\right] \left(\prod _{w \in T} \mathbf{b}_w\right) = \left[x^n\right] \left(\prod _{w \in f(U)} \mathbf{b}_w\right)\).

By Claim 2, for each \(w \in W\) we have \(\mathbf{b}_{w} \overset {x^n}{\equiv } \prod _{\substack {s \in U;\\ f(s) = w}} \mathbf{a}_{s}\). Hence, by Lemma A.29, we get

and similarly

Now, for \(w \notin f(U)\), there are no \(s \in U\) with \(f(s) = w\), so the inner product is \(1\) (the empty product). Hence, for any finite set \(V \supseteq f(U)\),

(by the standard fiberwise reindexing of finite products). Applying this to both \(V = T\) and \(V = f(U)\), we conclude that both products equal \(\prod _{s \in U} \mathbf{a}_{s}\). Therefore,

This proves Claim 3.]

From Claim 3, the \(x^n\)-coefficient in the product of \(\left(\mathbf{b}_w\right)_{w \in W}\) is finitely determined (since \(f(U)\) is a finite subset of \(W\)). Since this holds for each \(n \in \mathbb {N}\), the family \(\left(\mathbf{b}_w\right)_{w \in W}\) is multipliable. This proves part (a) of Proposition 1.372.

For part (b), we observe that by Proposition 1.364 (b), the infinite product \(\prod _{s \in S} \mathbf{a}_s\) is \(x^n\)-equivalent to \(\prod _{s \in U} \mathbf{a}_s\). By the computation above, the infinite product \(\prod _{w \in W} \mathbf{b}_w\) is also \(x^n\)-equivalent to \(\prod _{s \in U} \mathbf{a}_s\). Since this holds for all \(n\), the two infinite products are equal.

(a) Fix \(n \in \mathbb {N}\). We know that the family \(\left(\mathbf{a}_{i}\right)_{i \in I}\) is multipliable. Hence, there exists an \(x^n\)-approximator \(U\) for \(\left(\mathbf{a}_{i}\right)_{i \in I}\) (by Lemma 1.363). Consider this \(U\).

We also know that the family \(\left(\mathbf{b}_{i}\right)_{i \in I}\) is multipliable. Hence, there exists an \(x^n\)-approximator \(V\) for \(\left(\mathbf{b}_{i}\right)_{i \in I}\) (by Lemma 1.363, applied to \(\mathbf{b}_{i}\) instead of \(\mathbf{a}_{i}\)). Consider this \(V\).

We know that \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in I}\). In other words, \(U\) is a finite subset of \(I\) that determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in I}\) (by the definition of an \(x^n\)-approximator). Hence, in particular, \(U\) is finite. Similarly, \(V\) is finite. Moreover, \(U \subseteq I\); similarly, \(V \subseteq I\).

Let \(M = U \cup V\). This set \(M\) is finite (since \(U\) and \(V\) are finite). Moreover, \(M = U \cup V \subseteq I \cup I = I\). Hence, \(M\) is a finite subset of \(I\).

Now, let \(N\) be a finite subset of \(I\) satisfying \(M \subseteq N \subseteq I\). We shall show that

Indeed, let \(m \in \left\{ 0, 1, \ldots , n\right\} \). We have \(U \subseteq U \cup V = M \subseteq N\). The set \(N\) is a finite subset of \(I\) and satisfies \(U \subseteq N\). Now, recall that \(U\) determines the first \(n+1\) coefficients in the product of \(\left(\mathbf{a}_{i}\right)_{i \in I}\). Hence, \(U\) determines the \(x^m\)-coefficient in the product of \(\left(\mathbf{a}_{i}\right)_{i \in I}\) (since \(m \in \left\{ 0, 1, \ldots , n\right\} \)). In other words, every finite subset \(T\) of \(I\) satisfying \(U \subseteq T \subseteq I\) satisfies

We can apply this to \(T = N\) (since \(N\) is a finite subset of \(I\) satisfying \(U \subseteq N \subseteq I\)), and thus obtain

However, we can also apply it to \(T = M\) (since \(M\) is a finite subset of \(I\) satisfying \(U \subseteq M \subseteq I\)), and thus obtain

Comparing these two equalities, we obtain

The same argument (applied to \(\mathbf{b}_{i}\) and \(V\) instead of \(\mathbf{a}_{i}\) and \(U\)) yields

Forget that we fixed \(m\). We thus have proved these equalities for each \(m \in \left\{ 0, 1, \ldots , n\right\} \). Hence, Lemma 1.346 (applied to \(a = \prod _{i \in N} \mathbf{a}_{i}\) and \(b = \prod _{i \in M} \mathbf{a}_{i}\) and \(c = \prod _{i \in N} \mathbf{b}_{i}\) and \(d = \prod _{i \in M} \mathbf{b}_{i}\)) yields that

for each \(m \in \left\{ 0, 1, \ldots , n\right\} \). Applying this to \(m = n\), we obtain

In view of

this rewrites as

Forget that we fixed \(N\). We thus have shown that every finite subset \(N\) of \(I\) satisfying \(M \subseteq N \subseteq I\) satisfies the above equality. In other words, \(M\) determines the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\) (by the definition of “determining the \(x^n\)-coefficient”). Hence, the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\) is finitely determined (since \(M\) is a finite subset of \(I\)).

Forget that we fixed \(n\). We thus have proved that for each \(n \in \mathbb {N}\), the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\) is finitely determined. In other words, the family \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\) is multipliable. This proves Proposition 1.366 (a).

(b) Proposition 1.366 (a) shows that the family \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\) is multipliable.

Let \(n \in \mathbb {N}\). In our above proof of Proposition 1.366 (a), we have seen the following:

• There exists an \(x^n\)-approximator \(U\) for \(\left(\mathbf{a}_{i}\right)_{i \in I}\).

• There exists an \(x^n\)-approximator \(V\) for \(\left(\mathbf{b}_{i}\right)_{i \in I}\).

Consider these \(U\) and \(V\). Let \(M = U \cup V\). Thus, \(M = U \cup V \supseteq U\) and \(M = U \cup V \supseteq V\). In our above proof, we have seen that \(M\) is a finite subset of \(I\) and \(M\) determines the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\).

We recall that the relation \(\overset {x^n}{\equiv }\) is an equivalence relation (by Theorem 1.328 (a)).

Now, the definition of the infinite product \(\prod _{i \in I} \left(\mathbf{a}_{i} \mathbf{b}_{i}\right)\) (namely, Definition 1.343 (b)) yields that

(since \(M\) is a finite subset of \(I\) that determines the \(x^n\)-coefficient in the product of \(\left(\mathbf{a}_{i} \mathbf{b}_{i}\right)_{i \in I}\)). On the other hand, the set \(U\) is an \(x^n\)-approximator for \(\left(\mathbf{a}_{i}\right)_{i \in I}\). Thus, Proposition 1.364 (b) (applied to \(U\) instead of \(M\)) yields

Since the relation \(\overset {x^n}{\equiv }\) is symmetric, we thus obtain \(\prod _{i \in U} \mathbf{a}_{i} \overset {x^n}{\equiv } \prod _{i \in I} \mathbf{a}_{i}\). Moreover, \(M\) is a finite subset of \(I\) satisfying \(U \subseteq M\); hence, Proposition 1.364 (a) yields

Hence, \(\prod _{i \in M} \mathbf{a}_{i} \overset {x^n}{\equiv } \prod _{i \in I} \mathbf{a}_{i}\) (since the relation \(\overset {x^n}{\equiv }\) is transitive). The same argument (applied to \(\left(\mathbf{b}_{i}\right)_{i \in I}\) and \(V\) instead of \(\left(\mathbf{a}_{i}\right)_{i \in I}\) and \(U\)) yields \(\prod _{i \in M} \mathbf{b}_{i} \overset {x^n}{\equiv } \prod _{i \in I} \mathbf{b}_{i}\). From these two \(x^n\)-equivalences, we obtain

(by Theorem 1.328 (b)). In view of

this rewrites as

In other words,

Applying this to \(m = n\), we obtain

In view of \(\left[x^{n}\right] \left(\prod _{i \in I} \left(\mathbf{a}_{i} \mathbf{b}_{i}\right)\right) = \left[x^{n}\right] \left(\prod _{i \in M} \left(\mathbf{a}_{i} \mathbf{b}_{i}\right)\right)\), this rewrites as

Now, forget that we fixed \(n\). We thus have proved that each \(n \in \mathbb {N}\) satisfies the above equality. In other words, each coefficient of the FPS \(\prod _{i \in I} \left(\mathbf{a}_{i} \mathbf{b}_{i}\right)\) equals the corresponding coefficient of \(\left(\prod _{i \in I} \mathbf{a}_{i}\right) \left(\prod _{i \in I} \mathbf{b}_{i}\right)\). Hence, we have

This proves Proposition 1.366 (b).

If \([x^m] p_{i,k} \neq 0\) for some \(m \leq n\), then \((i,k) \in T_m \subseteq T'_n\), contradicting the hypothesis.

All but finitely many \(i \in I\) satisfy \(k_i = 0\) (since \((k_i)_{i \in I}\) is essentially finite) and thus \(p_{i,k_i} = p_{i,0} = 1\). Hence, all but finitely many entries of the family \((p_{i,k_i})_{i \in I}\) equal \(1\). Thus, this family is multipliable (by Proposition 1.351).

The product \(\prod _{i \in s} p_{i,k_i}\) is a multiple of \(p_{j,k_j}\) (since \(p_{j,k_j}\) is one of the factors of this product). Moreover, by Claim 3, we have \([x^m] p_{j,k_j} = 0\) for each \(m \in \{ 0,1,\ldots ,n\} \). Hence, Lemma A.42 (applied to \(u = p_{j,k_j}\) and \(v = \prod _{i \in s} p_{i,k_i}\)) shows that \([x^m](\prod _{i \in s} p_{i,k_i}) = 0\) for each \(m \in \{ 0,1,\ldots ,n\} \). Applying this to \(m = n\) yields the claim.

We have \(k_j \in S_j\) and \(k_j \neq 0\), so \((j, k_j) \in \overline{S}\). Moreover, \((j, k_j) \notin T'_n\) (because if we had \((j, k_j) \in T'_n\), then we would have \(j \in I_n\) by the definition of \(I_n\); but this would contradict \(j \notin I_n\)). Hence, \((j, k_j) \in \overline{S} \setminus T'_n\). Therefore, Claim 5 yields \([x^n](\prod _{i \in s} p_{i,k_i}) = 0\).

Let \((k_i)_{i \in I} \in S^I_{\mathrm{fin}}\) satisfy \([x^n](\prod _{i \in I} p_{i,k_i}) \neq 0\). Then it must satisfy two properties: (1) all \(i \in I \setminus I_n\) satisfy \(k_i = 0\) (otherwise Claim 6 would give a contradiction); (2) all \(i \in I_n\) satisfy \(k_i \in K_n \cup \{ 0\} \) (otherwise \((j, k_j) \in \overline{S} \setminus T'_n\) for some \(j \in I_n\), and Claim 5 would give a contradiction). These two properties together restrict \((k_i)\) to finitely many possibilities (since the sets \(I_n\) and \(K_n \cup \{ 0\} \) are finite). Hence, all but finitely many families yield \([x^n](\prod _{i \in I} p_{i,k_i}) = 0\), proving summability.

Each term \(p_{i,k}\) with \(k \neq 0\) has \([x^m] p_{i,k} = 0\) by Claim 3 (since \((i,k) \notin T'_n\) because \(i \notin I_n\)).

For \(i \in J \setminus I_n\), we have \(\sum _{k \in S_i} p_{i,k} = 1 + \sum _{k \in S_i \setminus \{ 0\} } p_{i,k}\). By Claim 9, the remainder has zero coefficients up to degree \(n\), so multiplying by it does not change the coefficient at degree \(n\).

This is the standard finite distributive law (product of sums equals sum of products over all choice functions). See Proposition 1.383.

The proof establishes a bijection between essentially finite families supported on \(I_n\) and functions \(I_n \to S\), via restriction and extension by \(0\).

The coefficient \([x^k](f \circ g)\) involves a sum over \(d\) of \([x^d]f \cdot [x^k](g^d)\). For \(d {\gt} k\), \([x^k](g^d) = 0\) since \(\operatorname {ord}(g^d) \geq d {\gt} k\). For \(d \leq k \leq n\), \([x^d]f_1 = [x^d]f_2\) by \(x^n\)-equivalence.

By Lemma A.43, \(\prod _{i \in J} (f_i \circ g) = (\prod _{i \in J} f_i) \circ g\). The approximator condition gives \(\prod _{i \in J} f_i \overset {x^n}{\equiv } \prod _{i \in M} f_i\). By Lemma A.39, composing with \(g\) preserves the equivalence.

For each \(n\), combine the approximators for degrees \(0, 1, \ldots , n\) into a single approximator \(M\) (their union), then apply Lemma A.40.

Write \(v = u \cdot w\). Then \([x^m] v = \sum _{(i,j) \in \mathrm{antidiag}(m)} [x^i] u \cdot [x^j] w\). Since \(i \leq m \leq n\), we have \([x^i] u = 0\), so each term vanishes.