Given a basis element $[w]\times\bs\beta$ with a composition component $[w]$ and an $\mathbb R\la\mathbb B\ra$-component $\bs\beta$,
where $w$ is a composition of $n\in\mathbb N$ and $\bs\beta$ has length $2^n$, we call any odd-degree
monomial of signature expansions of the form $[\bs a_1]\,[\bs a_2]\,\cdots\,[\bs a_{2m+1}]$ that produces
$[w]\times\bs\beta$ as one of the terms in its expansion, a monomial generator or just generator of $[w]\times\bs\beta$.

At this stage and in this step, it is useful to give a brief coarse overview of our overall strategy,
which we implement in detail in the subsequent steps below. We show in this step how, for any given composition component,
we can identify, for the associated basis elements, specific collections of generators. We call the sets of
basis elements and corresponding collections of generators “coefficient blocks” or simply “blocks”.
We show that such blocks are necessarily square.
To start, consider any one-part composition $w$ of $n$, so the two corresponding basis elements
are $[\bs0 n\bs0]\rightsquigarrow[n]\times(1,0)$ and $[\bs0 n\bs0^\dag]\rightsquigarrow[n]\times(0,1)$.
The basis element $[n]\times(1,0)$ is generated by signature expansion $[\bs n]$, while
$[n]\times(0,1)$ is not. On the other hand, both basis elements are generated by $[\bs0]\,[\bs{n-2}]\,[\bs0]$.
This exhausts all the possible odd-degree monomials in $\pi_n$ that could generate
$[n]\times(1,0)$ and  $[n]\times(0,1)$. Thus for a one-part composition, the possible odd-degree generators
have the form,

\begin{equation*}
\end{equation*}

where $[\star]$ represents the appropriate generic signature expansion, i.e. in the first instance it is $[\bs n]$
and in the second instance, i.e. for $[\bs0]\,[\star]\,([\bs0])$, the middle $[\star]$ factor is $[\bs{n-2}]$.
Note, we allow $[\star]=[\bs0]$.
The only other possibilities are $[\star]\,[\bs0]\,[\bs0]$ and $[\bs0]\,[\bs0]\,[\star]$.
However for $n\geqslant3$, we can rule these two possibilities out as $[\bs0]\,[\bs0]=2\cdot\{\bs01\bs0\}$ and any subsequent Pöppe product
of this term with $[\bs0]$ would generate a basis element $[w]\times\bs\beta$ where the composition $w$ has two parts.
Note of course, the triple Pöppe product $[\bs a]\,[\bs b]\,[\bs c]$ is naturally associative.

Now consider any two part composition $w=a_1a_2$ of $n$. We observe that basis elements with such a two-part
composition component can in principle be generated by $[\star]$ and $[\bs0]\,[\star]\,[\bs0]$, which we have already come across just above.
However such basis elements can also be generated by any of the following four generators of the form,

\begin{equation*}
\end{equation*}

We observe that each possible generator above contains only two `$[\star]$' factors, consistent with the
two-part composition component of the basis elements we are aiming to generate. Further note that
we can also see that the four generators above can be constructed from the previous two generators
$[\star]$ and $[\bs0]\,[\star]\,([\bs0])$ corresponding to one-part compositions, by applying
one of the three actions $[\star]\,[\star]\,(\cdot)$, $[\star]\,[\bs0]\,(\cdot)$ or $[\bs0]\,[\star]\,(\cdot)$ to them.
For example, the first generator above is constucted by applying the action $[\star]\,[\star]\,(\cdot)$ to $[\star]=[\bs0]$,
where we must set the argument $[\star]=[\bs0]$ to preserve the two-part composition component of the basis element
we wish to generate. The next two generators above are constructed by applying the actions
$[\star]\,[\bs0]\,(\cdot)$ or $[\bs0]\,[\star]\,(\cdot)$ to $[\star]$. Now consider the final quintic generator above.
Applying the action $[\star]\,[\star]\,(\cdot)$ to $[\bs0]\,[\star]\,([\bs0])$ would produce a generator with too many `$[\star]$'
factors, while in principle, either of the actions $[\star]\,[\bs0]\,(\cdot)$ or $[\bs0]\,[\star]\,(\cdot)$ could be
applied to $[\bs0]\,[\star]\,([\bs0])$. However the action of $[\star]\,[\bs0]\,(\cdot)$ on $[\bs0]\,[\star]\,(\cdot)$
is nilpotent. We demonstrate this below in Lemma <ref>. Hence the action
$[\star]\,[\bs0]\,(\cdot)$ rigorously applied to $[\bs0]\,[\star]\,([\bs0])$ produces zero. Thus the
only viable action is $[\bs0]\,[\star]\,(\cdot)$ on $[\bs0]\,[\star]\,([\bs0])$ producing the quintic generator shown.
From another perspective, for the quintic generator, in the case $n\geqslant5$, any other quintic arrangment with
two $[\star]$- and three $[\bs0]$-factors, would necessitate a consecutive pair `$[\bs0]\,[\bs0]$' that would result
in generating a basis element whose composition component has more than two parts.

In the case of basis elements with a three-part composition component $w=a_1a_2a_3$,
the possible generators are, in principle, any of the generators we have already seen, as well as,
the generators of the form,

\begin{align*}
\end{align*}

We remark that each possible generator above contains only three `$[\star]$' factors.
We see that the first two generators are constructed by applying the action $[\star]\,[\star]\,(\cdot)$
to the generators for basis elements with one-part composition components.
The next set of generators are constructed by applying the action $[\star]\,[\bs0]\,(\cdot)$ to
the generators for basis elements with two-part composition components, taking into account
the nilpotent action of $[\star]\,[\bs0]\,(\cdot)$ on $[\bs0]\,[\star]\,(\cdot)$.
This accounts for the next two generators. Then the final four generators are constructed
by applying the action $[\bs0]\,[\star]\,(\cdot)$ to all four of the generators
for basis elements with two-part composition components.

For the case of basis elements with a four-part composition component $w=a_1a_2a_3a_4$,
the possible generators are, in principle, besides any of the generators we have already seen,
generators of the form,

\begin{align*}
\end{align*}

Again, each possible generator above contains only four `$[\star]$' factors.
They are constructed by applying the action
$[\star]\,[\star]\,(\cdot)$ to the generators for basis elements with two-part composition components,
applying the action $[\star]\,[\bs0]\,(\cdot)$ to
the generators for basis elements with three-part composition components, taking into account
the nilpotent action of $[\star]\,[\bs0]\,(\cdot)$ on $[\bs0]\,[\star]\,(\cdot)$, and
then also applying the action $[\bs0]\,[\star]\,(\cdot)$ to all of the generators
for basis elements with three-part composition components.

We have seen that for basis elements with a composition component with $k=1,2,3$ or $4$ parts,
the number of generators that might produce such a basis element is $2^{k}$.
We have not shown that corresponding to a given basis element, the generators
constructed in the manner indicated are unique at leading order. We demonstrate
this below in Steps $7$ and $8$. Assuming this is the case for the moment,
we have the following.

For a given basis element with composition component $w$, the number of
monomial generators that can generate that basis element at leading order is $2^{|w|}$.

As observed, the result is true for $|w|=1,2,3,4$. Assume the result is
true for $|w|=1,2,\ldots,k$ for some $k\in\mathbb N$.
The set of generators for basis elements with composition components of $k+1$ parts
are constructed by: (i) Applying the action  $[\star]\,[\star]\,(\cdot)$
to the generators for basis elements with $(k-1)$-part composition components of which there are $2^{k-1}$ by assumption;
(ii) Applying the action $[\star]\,[\bs0]\,(\cdot)$ to the generators for basis elements with $k$-part composition components, taking into account
the nilpotent action of $[\star]\,[\bs0]\,(\cdot)$ on $[\bs0]\,[\star]\,(\cdot)$. Since there are $2^k$ generators corresponding to any
basis element with a composition of $k$-parts, and half of these start with the factor `$\,[\bs0]\,\,[\star]$', there are $2^{k-1}$
generators constructed in this way; and then finally (iii) Applying the action $[\bs0]\,[\star]\,(\cdot)$ to all of the generators
for basis elements with $k$-part composition components, of which there are $2^k$. Adding these three contributions up, $2^{k-1}+2^{k-1}+2^k=2^{k+1}$,
and the result follows by induction.

We can also view this last result from another perspective as follows.
For each $k$-part composition, when $k$ is odd, the set of new generators are characterised
as follows. First, we include the degree $k$ monomial $[\star]\,[\star]\,\cdots\,[\star]$,
of which there is only one choice. We also include the degree $k+2$ monomials which
contain two non-adjacent `$\,[\bs0]$' factors; there are $k+1$ choose $2$ possible monomials of this form.
Then we can also include degree $k+4$ monomials which contain four non-adjacent `$\,[\bs0]$' factors;
there are $k+1$ choose $4$ possible monomials of this form. And so forth until we reach the single
degree $2k+1$ monomial of the form $[\bs0]\,[\star]\,[\bs0]\,[\star]\,[\bs0]\,\cdots\,[\star]\,[\bs0]$.
Here we have implicitly used that the number of ways to place $r$ objects in non-adjacent slots whose
total number is $m$, is given by $m-r+1$ choose $r$. In the examples just presented, we
considered the number of possible ways of placing $2\ell$ factors of the form `$\,[\bs0]$' in
a monomial of degree $k+2\ell$, for $\ell=0,1,\ldots,(k+1)/2$. Hence the total number of
monomials of each degree outlined being $k+1$ choose $2\ell$. Thus with $k$ odd, the total number of
such new odd-degree monomials is given by the sum over $\ell=0,1,\ldots,(k+1)/2$ of $k+1$ choose $2\ell$,
i.e. the sum on the left shown in Lemma <ref>.
Suppose now $k$ is even. The lowest degree monomials that might generate the corresponding basis element
are those of degree $k+1$ with a single factor `$\,[\bs0]$'. There are $k+1$ such monomials.
We can also include degree $k+3$ monomials with three non-adjacent factors `$\,[\bs0]$'; there
are $k+1$ choose $3$ such possible monomials, and so forth. In the final highest degree monomial,
of degree $2k+1$ has the single form $[\bs0]\,[\star]\,[\bs0]\,[\star]\,[\bs0]\,\cdots\,[\star]\,[\bs0]$.
Thus with $k$ even, the total number of such new odd-degree monomials is given by the sum
over $\ell=0,1,\ldots,k/2$ of $k+1$ choose $2\ell+1$, i.e. the sum on the right shown in Lemma <ref>.
In consequence we have the following important result.

The aforementioned sums, in the respective $k$ is odd and then even cases, are equal to $2^k$.
In other words, respectively, when $k$ is odd and then even, we have,

\begin{equation*}
\sum_{\ell=0}^{(k+1)/2}\begin{pmatrix} k+1\\ 2\ell\end{pmatrix}=2^k\qquad\text{and}\qquad
\sum_{\ell=0}^{k/2}\begin{pmatrix} k+1\\ 2\ell+1\end{pmatrix}=2^k.
\end{equation*}

Suppose $k$ is odd. Then by direct computation, we observe,

\begin{align*}
\sum_{\ell=0}^{(k+1)/2}\!\frac{(k+1)!}{(k+1-2\ell)!(2\ell)!}
=&\;1+\begin{pmatrix} k\\ 1 \end{pmatrix}+\begin{pmatrix} k\\ 2\end{pmatrix}+\begin{pmatrix} k\\ 3\end{pmatrix}+\cdots+\begin{pmatrix} k\\ k-1\end{pmatrix}+1,
\end{align*}

where we matched up respective pairs from the sums and then used that $2^k=(1+1)^k$. This gives the first result.
When $k$ is even, we again use that $2^k=(1+1)^k$, and observe,

\begin{equation*}
2^k=\begin{pmatrix} k\\ 0 \end{pmatrix}+\begin{pmatrix} k\\ 1\end{pmatrix}+\begin{pmatrix} k\\ 2\end{pmatrix}+\cdots
+\begin{pmatrix} k\\ k\end{pmatrix}=\sum_{\ell=0}^{k/2}\frac{(k+1)!}{(k-2\ell)!(2\ell+1)!},
\end{equation*}

where we paired up successive terms and parameterised the pairs by $\ell=0,1,\ldots,k/2$.
This gives the second result.

The crucial observation from the result of Lemmas <ref> and <ref> is the following.

For a given composition component $w$ of a block set of basis elements $[w]\times\bs\beta$ parameterised
by the tuples $\bs\beta\in\mathbb R\la\mathbb B\ra$,
the number of new monomial generators equals the dimension of the tuple block, i.e. $2^{|w|}$.

One of our main concerns now is to show that the resulting square block of signature coefficients has full rank.
The next three steps address this issue, making the analysis of this section more precise.

Step 4: The three standard triple actions.
We have seen that the triple product action in Corollary <ref>, in the full form given therein,
as well as in the special forms $[\bs a]\,[\bs0]\,\bigl(\cdot\bigr)$ and $[\bs0]\,[\bs b]\,\bigl(\cdot\bigr)$,
are used to construct the generators corresponding to a given basis element. We call these three actions the
standard triple actions.

We call the actions $[\bs a]\,[\bs b]\,\bigl(\cdot\bigr)$, $[\bs a]\,[\bs0]\,\bigl(\cdot\bigr)$
and $[\bs0]\,[\bs b]\,\bigl(\cdot\bigr)$ the three standard triple actions.

The result of the action $[\bs a]\,[\bs b]\,\bigl(\cdot\bigr)$ is given in Corollary <ref>.
As we use them frequently hereafter, we record the result of the standard actions
$[\bs a]\,[\bs0]\,(\cdot)$ and $[\bs0]\,[\bs b]\,(\cdot)$ in the following Corollary.
They are just special cases which we call the special actions.

The two special actions $[\bs a]\,[\bs0]\,\bigl(\cdot\bigr)$ and $[\bs0]\,[\bs b]\,\bigl(\cdot\bigr)$
are given at leading order by,

\begin{align*}
[\bs a]\,[\bs0]\,\bigl([cw]\times\bs\beta\bigr)
=&\;\chi(a\ot 0\ot cw)\!\cdot\![(a+1)(c+1)w]\times\bigl((1,-1)\lhd\bigl(\bs\beta+(-1)^{|w|}\bs\beta^\dag\bigr)\bigr)+\cdots,\\
[\bs0]\,[\bs b]\,\bigl([cw]\times\bs\beta\bigr)
=&\;\chi(0\ot b\ot cw)\!\cdot\!\Bigl([(b+2)cw]\times\bigl((2,0)\lhd\bs\beta+(-1)^{|w|}(0,2)\lhd\bs\beta^\dag\bigr)\\
\end{align*}

By direct computation using the Pöppe product rules in Lemma <ref>, we observe that
$[\bs a]\,[\bs0]\,\bigl([cw\times\bs\varphi]\bigr)$ equals,

\begin{align*}
\chi(a\ot 0\ot cw)\cdot\bigl(&\;[\bs0(a+1)\bs0(c+1)(w\times\hat{\bs\varphi})]+[\bs0(a+1)\bs0(c+1)(w\times\hat{\bs\varphi})^\dag]\\
\end{align*}

at leading order, giving the first result.
Then, by direct computation for the other case, we observe that $[\bs0]\,[\bs b]\,\bigl([cw\times\bs\varphi]\bigr)$ equals,

\begin{align*}
\chi(0\ot b\ot cw)\cdot\bigl(&\;2\cdot[\bs0(b+2)\bs0c)(w\times\hat{\bs\varphi})]+2\cdot[\bs0(b+2)\bs0^\dag c)(w\times\hat{\bs\varphi})^\dag]\\
\end{align*}

at leading order, giving the second result.

Comparing the results of Corollary <ref> with Corollary <ref> we emphasise
two observations, that there is: (i) A natural contraction of the action forms due to the `$\,[\bs0]$' factors in the action;
(ii) An apparent change of sign in the second term on the right in the first example. We can view both cases as
the consequence of substituting $[\nu\times\bs0]$ for $[\bs b]$ in the first case and then $[\nu\times\bs0]$ for $[\bs a]$
in the second case. The sign change, perhaps more easily observed from the corresponding result in Lemma <ref>,
is a consquence of the fact that to make the appropriate substitution of $[\nu\times\bs0]$ for $[\bs b]$, we should
convert the two terms with $\bs0^\dag b\bs0^\dag$ on the right, to $-\bs0^\dag b^\dag\bs0^\dag$ first.

Step 5: Generating blocks. We now show precisely how, given a block of basis elements characterised
by a given composition component $w$ and parameterised by the corresponding $2^{|w|}$ basis elements $\bs\beta$ of $\mathbb R\la\mathbb B\ra$,
we can use the three standard actions to enumerate all the monomial generators that produce the basis elements
of that block at leading order, and also establish the corresponding signature coefficient associated with
each such basis element. Let us examine the three standard actions given in
Corollaries <ref> and <ref> more closely.
If we examine the right-hand side of $[\bs a]\,[\bs b]\,\bigl([cw]\times\bs\beta\bigr)$
in Corollary <ref>, then we observe that in terms of descent order, the first
composition term `$[(a+1)(b+1)cw]$' on the right is highest, and thus we retain that term only.
In Corollary <ref>, at leading order, the action $[\bs a]\,[\bs0]\,\bigl([cw]\times\bs\beta\bigr)$ in unique, while
the right-hand side of $[\bs0]\,[\bs b]\,\bigl([cw]\times\bs\beta\bigr)$ contains two terms, the
first of which is higher in terms of descent order, which is the one we retain.
Thus at leading order the three standard actions on $[cw]\times\bs\beta$ are:

\begin{align*}
[\bs a]\,[\bs b]\,(\cdot)&\!=\!\chi(cw)\!\cdot\![(a+1)(b+1)cw]\!\times\!\bigl((1,0,1,0)\lhd\bs\beta+(-1)^{|w|}\,(0,1,0,1)\lhd\bs\beta^\dag\bigr)+\cdots,\\
[\bs a]\,[\bs0]\,(\cdot)&\!=\!\chi(cw)\!\cdot\![(a+1)(c+1)w]\!\times\!\bigl((1,-1)\lhd\bs\beta+(-1)^{|w|}(1,-1)\lhd\bs\beta^\dag\bigr)+\cdots,\\
[\bs0]\,[\bs b]\,(\cdot)&\!=\!\chi(cw)\!\cdot\![(b+2)cw]\!\times\!\bigl((2,0)\lhd\bs\beta+(-1)^{|w|}(0,2)\lhd\bs\beta^\dag\bigr)+\cdots.
\end{align*}

Here we have used the homomorphic properties of $\chi$, in particular that $\chi(a\ot b\ot cw)=\chi(a\ot 0\ot cw)=\chi(0\ot b\ot cw)=\chi(cw)$.
Consider the following respective replacements in each of the three actions above:
(i) $a\to a-1$, $b\to b-1$, $c\to\nu$; (ii) $a\to a-1$, $c\to c-1$ and (iii) $b\to b-2$.
With these three choices, each of the actions generates the same composition $acw$—in the first case we relabel $b$ as $c$ and
in the third case we relabel $b$ as $a$. Recall from our coarse-grain overview in Step 3 that to enumerate the generators
corresponding to basis elements with composition components with $k\geqslant2$ parts, we apply the first action to the generators
at level $k-2$, and the two special actions to the generators at level $k-1$, taking into account the nilpotent action outlined
just below in Lemma <ref>.
We note that, for any sequence $\hat u\in\mathbb R\la\mathbb B\ra$, with $|\hat u|=2^{k-2}$, we have,

\begin{align*}
(1,0,1,0)\lhd\hat u&=(1,1)\lhd(1,0)\lhd\hat u=(1,1)\lhd(\hat u,0),\\
(0,1,0,1)\lhd\hat u^\dag&=(1,1)\lhd(0,1)\lhd\hat u^\dag=(1,1)\lhd(0,{\hat u}^\dag),
\end{align*}

where $(\hat u,0)$ and $(0,\hat u^\dag)$ are of length $2^{k-1}$.
Putting these observations together, we have thus established the following lemma.

At leading order, with the choices mentioned above, the following three standard actions generate the same composition with the
respective $\mathbb R\la\mathbb B\ra$ components indicated,

\begin{align*}
[\bs{a-1}]\,[\bs{c-1}]\,\bigl([w]\!\times\!\bs\beta\bigr)&\!=\!\chi(w)\cdot[acw]\times\bigl((1,1)\lhd\bigl((\hat u,0)
-(-1)^{|w|}(\hat u,0)^\dag\bigr)\bigr),\\
[\bs{a-1}]\,[\bs0]\,\bigl([(c-1)w]\!\times\!\bs\beta\bigr)&\!=\!\chi((c-1)w)\cdot[acw]\times\bigl((1,-1)\lhd\bigl((\hat a,\hat b)
+(-1)^{|w|}(\hat a,\hat b)^\dag\bigr)\bigr),\\
[\bs0]\,[\bs{a-2}]\,\bigl([cw]\!\times\!\bs\beta\bigr)&\!=\!\chi(cw)\cdot[acw]\times\bigl((2,0)\lhd(\hat a,\hat b)
+(-1)^{|w|}(0,2)\lhd(\hat a,\hat b)^\dag\bigr).
\end{align*}

Here, in the first case $\bs\beta=(\hat u,0)\in\mathbb R\la\mathbb B\ra$ with $\hat u$ arbitrary, and in the second and third cases
$\bs\beta=(\hat a,\hat b)\in\mathbb R\la\mathbb B\ra$ is arbitrary. Each such $\bs\beta$ is of length $2^{|acw|-1}$,
and $\hat a$ and $\hat b$ have the same length—matching that of $\hat u$.

Note, in the statement of Lemma <ref>, the case of the first action which corresponds to
the action $[\bs a]\,[\bs b]\,(\cdot)$ applied to $[cw]\times\bs\beta$ in the discussion preceding the Lemma.
In that discussion, when we set $c\to\nu$, we equivalently replaced $cw$ by $w$. This means that
we should effectively consider the length of $w$ to be one less than would otherwise be the
case. This explains why the sign in front of the term with the factor $(-1)^{|w|}$ in the first action case is negative
in the statement of the Lemma.

Some further clarifications on the statement of Lemma <ref> are required. Note that,

\begin{equation*}
\end{equation*}

where $\nu$ is the empty composition and $(1)$ is the element of $\mathbb R\la\mathbb B\ra$ corresponding to
compositions of zero parts. The special action $[\bs0]\,[\bs{a-2}]\,(\cdot)$  in Lemma <ref>
still applies when the argument $[cw]\times\bs\beta=[\nu]\times(1)$ and thus when $(\hat a,\hat b)=(1)$.
The result is that at leading order we have,

\begin{equation*}
\end{equation*}

Here, by convention, we take $\chi(\nu)\coloneqq1$. Since we have taken $cw\to\nu$,
we can think of the number of parts of $w$ to be `$-1$', explaining the sign
in front of the $\mathbb R\la\mathbb B\ra$-element $(0,2)$. This is consistent with
just computing $[\bs0]\,[\bs{a-2}]\,[\bs0]$. Further, the first two actions in Lemma <ref>
don't make sense when $cw\to\nu$, though if $w\to\nu$, the special action $[\bs{a-1}]\,[\bs0]\,(\cdot)$
applies with the appropriate adaptations. And of course we can compute

Finally, we now also observe the following (aforementioned) nilpotency property.

At leading order, if we first apply the action $[\bs0]\,[\bs b]\,(\cdot)$ to an arbitrary $\mathbb R\la\mathbb B\ra$ component,
then apply the action $[\bs a]\,[\bs 0]\,(\cdot)$ to the result, this generates the zero $\mathbb R\la\mathbb B\ra$ component.
In other words at leading order we have,

\begin{equation*}
[\bs a]\,[\bs 0]\,\bigl([\bs0]\,[\bs b]\,(\cdot)\bigr)=0,
\end{equation*}

where the `$\,0$' on the right-hand side represents the zero $\mathbb R\la\mathbb B\ra$ component.

We focus on the effect of the actions on the $\mathbb R\la\mathbb B\ra$ components only.
The third (special) action applied to the input $(a,b)$ generates $2\cdot(a,b,\pm b^\dag,\pm a^\dag)$.
Set $A,B\in\mathbb R\la\mathbb B\ra$ to be the sub-components $A\coloneqq(a,b)$ and $B\coloneqq\pm(b^\dag,a^\dag)$.
With these identifications we note that
$B=\pm A^\dag$. Ignoring the real factor $2$, apply the second (special) action to the input $(A,B)$.
This is (note the sign of the second term of the action changes), $(1,-1)\lhd\bigl((A,B)\mp(A,B)^\dag\bigr)$,
which equals, $(A\mp B^\dag,B\mp A^\dag,-A\pm B^\dag,-B\pm A^\dag)$.
Since $B=\pm A^\dag$, this result is the zero $\mathbb R\la\mathbb B\ra$ component.

We now explore, through a series of examples, how to construct the generators and coefficient blocks
associated with any given composition. In particular we consider the cases of compositions with
$1$, $2$ and $3$ parts, before exploring the case of any given composition. Compositions
containing a `$1$' need to be singled out, as explained below.

[One-part compositions]
We observe that there are two basis elements corresponding to the one-part composition $w=a$, namely,
$[a]\times(1,0)$ and $[a]\times(0,1)$. We assume $n=a\geqslant2$. At leading we know from the corresponding
signature expansion $[\bs a]=[a]\times(1,0)+\cdots$. From our discussion succeeding Lemma <ref>,
we know the first two actions do not make sense when $cw\to\nu$, while the final special action
does make sense. As we saw directly, at leading order we have $[\bs0]\,[\bs{a-2}]\,[\bs0]=[a]\times(2,-2)$.
We have thus enumerated the generators corresponding to $[a]\times(1,0)$ and $[a]\times(0,1)$ and that
the signature coefficient matrix is,

\begin{equation*}
A_0=\begin{pmatrix} 1 & 2\\ 0 & -2 \end{pmatrix}.
\end{equation*}

[Two-part compositions]
Consider the basis elements with a two-part composition component $a_1a_2$, i.e. basis
elements of the form $[a_1a_2]\times\bs\beta_i$, where the $\bs\beta_i$ are the
four basis elements of length $4$, which are zero apart from a `$1$' in the $i$th position.
For the moment assume neither $a_1$ nor $a_2$ equal $1$; we consider each of these two special
cases separately below. Using Lemma <ref>, noting that for
each of the standard actions our goal is to obtain the composition component $[a_1a_2]$
on the right-hand side, we observe the following. For the first action, setting $w=\nu$,
$a=a_1$ and $c=a_2$, we find that at leading order, we get,

\begin{equation*}
\end{equation*}

The first special action in Lemma <ref>, with the same identifications gives,

\begin{equation*}
[\bs{a_1-1}]\,[\bs0]\,\bigl([a_2-1]\times(\hat a,\hat b)\bigr)=[a_1a_2]\times\bigl((1,-1)\lhd\bigl((\hat a,\hat b)+(\hat b,\hat a)\bigr)\bigr).
\end{equation*}

We saw in Example <ref>, the basis element $[a_2-1]\times(\hat a,\hat b)$
can be generated both by the corresponding signature expansion $[\bs{a_2-1}]=[a_2-1]\times(1,0)+\cdots$,
and by the generator $[\bs0]\,[\bs{a-2}]\,[\bs0]$. We discount the latter case due to the nilpotent action property.
Hence using this expression for $[\bs{a_2-1}]$ and inserting $(\hat a,\hat b)=(0,1)$ into the expression above, we deduce,

\begin{equation*}
\end{equation*}

to leading order. Now consider the second special action in Lemma <ref>.
Again with the same identifications for $a$, $c$ and $w$, we observe that to leading order,

\begin{equation*}
[\bs0]\,[\bs{a_1-2}]\,\bigl([a_2]\times(\hat a,\hat b)\bigr)=[a_1a_2]\times\bigl((2,0)\lhd(\hat a,\hat b)+(0,2)\lhd(\hat b,\hat a)\bigr).
\end{equation*}

We know from  Example <ref>, the basis element $[a_2]\times(\hat a,\hat b)$
can be generated either by the signature expansion $[\bs{a_2}]=[a_2]\times(1,0)+\cdots$,
or by the generator $[\bs0]\,[\bs{a_2-2}]\,[\bs0]=[a_2]\times(2,-2)+\cdots$. Respectively substituting the expressions
$[a_2]\times(1,0)$ and $[a_2]\times(2,-2)$ for $[a_2]\times(\hat a,\hat b)$ in the relation just above, we observe that to leading order,

\begin{align*}
\end{align*}

We have thus enumerated the four generators corresponding to the four basis elements $[a_1a_2]\times(1,0,0,0)$,
$[a_1a_2]\times(0,1,0,0)$, $[a_1a_2]\times(0,0,1,0)$ and $[a_1a_2]\times(0,0,0,1)$. They are
$[\bs{a_1-1}]\,[\bs{a_2-1}]\,[\bs0]$, $[\bs{a_1-1}]\,[\bs0]\,[\bs{a_2-1}]$, $[\bs0]\,[\bs{a_1-2}]\,[\bs{a_2}]$
and the quintic generator $[\bs0]\,[\bs{a_1-2}]\,[\bs0]\,[\bs{a_2-2}]\,[\bs0]$. The corresponding signature coefficient matrix is,

\begin{equation*}
1 & 1 & 2 & 4 \\
-1 & 1 & 0 & -4 \\
1 & -1& 0 & -4 \\
-1 & -1 & 2& 4
\end{pmatrix}
\end{equation*}

which is the subsystem coefficient matrix $A_2$ in Examples <ref> and <ref>
respectively concerning the quartic and quintic non-commutative nonlinear Schrödinger equations.

Let us now consider the case when $a_2=1$. If we substitute this value for $a_2$ into the generators above,
we see that the first two generators coincide and are given by $[\bs{a_1-1}]\,[\bs0]\,[\bs{0}]=[a_11]\times(1,-1,1,-1)+\cdots$
and $[\bs{a_1-1}]\,[\bs0]\,[\bs{0}]=[a_11]\times(1,1,-1,-1)+\cdots$. Since we can add them together under
the same coefficient $c_{(a_1-1)00}$, in this case we have the single generator,
The third generator above becomes, $[\bs0]\,[\bs{a_1-2}]\,[\bs1]=[a_11]\times(2,0,0,2)+\cdots$.
The final quintic generator cannot be a generator in this case if we insist on only including signature expansions
corresponding to non-negative integers. There are thus only two independent generators. Hence this this case,
the corresponding signature coefficient matrix, ignoring the middle two rows, is

\begin{equation*}
A_1\coloneqq\begin{pmatrix} 2 & 2\\ -2 & 2 \end{pmatrix}.
\end{equation*}

See Examples <ref> and <ref> and the equations
for the coefficients $c_{(n-2)00}$ and $c_{0(n-3)1}$ in those cases for when $w=(n-1)1$, as
well as with the coefficients in Tables <ref>–<ref>.
Note, when $a_1=a_2=1$, there is only one generator, $[\bs0]^3$, as we saw in Example <ref>.
We treat the more general case when $a_1=1$ at the end of this step..

[Three-part compositions]
Consider basis elements with a three-part composition component $a_1a_2a_3$, i.e. basis
elements of the form $[a_1a_2a_3]\times\bs\beta_i$, where the $\bs\beta_i$ for $i=1,\ldots,8$,
contain `$1$' in the $i$th position and zeros in the remaining seven positions.
For the moment assume neither $a_1$ nor $a_2$ nor $a_3$ are unity.
Using Lemma <ref>, the standard actions, setting $a=a_1$, $c=a_2$ and $w=a_3$ give to leading order,

\begin{align*}
[\bs{a_1-1}]\,[\bs{a_2-1}]\,\bigl([a_3]\times(\hat u,0)\bigr)&=[a_1a_2a_3]\times\bigl((1,1)\lhd(\hat u,\hat u^\dag)\bigr),\\
[\bs{a_1-1}]\,[\bs0]\,\bigl([(a_2-1)a_3]\times(\hat a,\hat b)\bigr)&=\chi((a_2-1)a_3)\cdot[a_1a_2a_3]\\
&\qquad\qquad\qquad\quad\times\bigl((1,-1)\lhd\bigl((\hat a,\hat b)
-(\hat a,\hat b)^\dag\bigr)\bigr),\\
[\bs0]\,[\bs{a_1-2}]\,\bigl([a_2a_3]\times(\hat a,\hat b)\bigr)&=\chi(a_2a_3)\cdot[a_1a_2a_3]\times(2\hat a,2\hat b,-2\hat b^\dag,-2\hat a^\dag).
\end{align*}

We observe, with these three relations, the task of finding the generators for any basis element
with a three-part composition component, becomes the task of finding the generators for the basis element with the one-part
composition component `$[a_3]$' in the first case, and then the generators for basis elements with the two-part
components `$[(a_2-1)a_3]$' and `$[a_2a_3]$' in the second and third cases. We can construct the
generators in these cases via Examples <ref> and <ref> just above.
In the first case, from Example <ref>, the two generators for $[a_3]\times(1,0)$ and $[a_3]\times(0,1)$
are $[\bs a_3]=[a_3]\times(1,0)+\cdots$ and $[\bs0]\,[\bs{a_3-2}]\,[\bs0]=[a_3]\times(2,-2)+\cdots$.
Hence if we substitute these expressions into the first case above, respectively setting $\hat u=(1,0)$
and then $\hat u=(2,-2)$, we find,

\begin{align*}
\end{align*}

For the second case above with composition component `$[(a_2-1)a_3]$', we know from Example <ref>,
there are four possible generators. However once we observe the nilpotent action property, we are
left with two, namely, $[\bs{a_2-2}]\,[\bs{a_3-1}]\,[\bs0]=[(a_2-1)a_3]\times(1,-1,1,-1)+\cdots$ and
$[\bs{a_2-2}]\,[\bs0]\,[\bs{a_3-1}]=[(a_2-1)a_3]\times(1,1,-1,-1)+\cdots$. Substituting these expressions
into the second case above, respectively setting $(\hat a,\hat b)=(1,-1,1,-1)$ and then $(\hat a,\hat b)=(1,1,-1,-1)$, we find,

\begin{align*}
\end{align*}

For the third case above with composition component `$[a_2a_3]$', again, we know from Example <ref>,
there are four possible generators. These are all four of the generators shown in Example <ref>
once we replace $a_1$ and $a_2$ therein respectively by $a_2$ and $a_3$. If we substitute the corresponding
four expressions with the replacements mentioned into the third case above, respectively setting
$(\hat a,\hat b)=(1,-1,1,-1)$, $(\hat a,\hat b)=(1,1,-1,-1)$, $(\hat a,\hat b)=(2,0,0,2)$ and then $(\hat a,\hat b)=(4,-4,-4,4)$,
we find at leading order,

\begin{align*}
\end{align*}

and finally,

\begin{equation*}
\!=\!\chi(a_2a_3)\cdot[a_1a_2a_3]\times(8,-8,-8,8,-8,8,8,-8).
\end{equation*}

Hence, for the eight basis elements $[a_1a_2a_3]\times\bs\beta_i$, $i=1,\ldots,8$,
with the columns corresponding to the generators above in descent order following by degree,
the corresponding signature coefficient matrix, is the full rank matrix,

\begin{equation*}
1 &  2 &  2 & 2  & 2  & 2  & 4  & 8 \\
0 & -2 & -2 & 2  & -2 & 2  & 0  & -8 \\
0 & -2 &  2 & -2 & 2  & -2 & 0  & -8 \\
1 & 2  & -2 & -2 & -2 & -2 & 4  & 8 \\
1 & 2  & -2 & -2 &  2 & 2  & -4 & -8 \\
0 & -2 &  2 & -2 & -2 & 2  & 0  & 8 \\
0 & -2 & -2 &  2 &  2 & -2 & 0  & 8 \\
1 & 2  & 2  & 2  & -2 & -2 & -4 & -8
\end{pmatrix},
\end{equation*}

where columns $3$ and $4$ should involve the factor $\chi((a_2-1)a_3)$, while
columns $5$ through to $8$ should involve the factor $\chi(a_2a_3)$. The factors
are omitted in $A_3$ for clarity.

Examples <ref>–<ref> precisely reflect the analysis we outlined in Step 3.

We can now discern the pattern. Suppose we wish to construct all the generators corresponding to a full set of $2^k$
basis elements associated with a given composition component $[a_1a_2\cdots a_k]$. We preclude
for the moment, that any of $a_1$ through to $a_k$ are equal to `$1$'. Using the standard actions
in Lemma <ref>, we find,

\begin{align*}
[\bs{a_1-1}]\,[\bs{a_2-1}]\,\bigl([a_3\cdots a_k]\times(\hat u,0)\bigr)&=[a_1\cdots a_k]\\
&\quad\times\bigl((1,1)\lhd\bigl((\hat u,0)-(-1)^{k-2}(0,\hat u^\dag)\bigr)\bigr),\\
[\bs{a_1-1}]\,[\bs0]\,\bigl([(a_2-1)a_3\cdots a_k]\times(\hat a,\hat b)\bigr)&=\chi((a_2-1)a_3)\cdot[a_1\cdots a_k]\\
&\quad\times\bigl((1,-1)\lhd\bigl((\hat a,\hat b)+(-1)^{k-2}(\hat a,\hat b)^\dag\bigr)\bigr),\\
[\bs0]\,[\bs{a_1-2}]\,\bigl([a_2\cdots a_k]\times(\hat a,\hat b)\bigr)&=\chi(a_2a_3)\cdot[a_1\cdots a_k]\\
&\quad\times\bigl((2,0)\lhd(\hat a,\hat b)+(-1)^{k-2}(0,2)\lhd(\hat a,\hat b)^\dag\bigr),
\end{align*}

to leading order. We then act iteratively to substitute for the generators corresponding to the composition:
(i) $[a_3\cdots a_k]$ with $(k-2)$-parts; (ii) $[(a_2-1)a_3\cdots a_k]$ with $(k-1)$-parts, taking into account
the nilpotency action property; and (iii) $[a_2\cdots a_k]$ with $(k-1)$-parts. We know from Step 3,
for a given composition $a_1\cdots a_k$, we can construct $2^k$ unique generators in this way.
In Step 6 we demonstrate that the corresponding signature coefficient matrix generated in this way has full rank.
However, at this stage, we note that we have the following straightforward result.

If we use the procedure above to construct the $2^k$ generators associated with the $2^k$ basis elements
with a given composition component $a_1\cdots a_k$, then each such set of generators is unique to the given composition $a_1\cdots a_k$,
i.e. each such set of generators corresponding to a given composition $a_1\cdots a_k$, does not appear elsewhere
in generator sets for other compositions.

Lastly, apart from the case of the composition `$a_11$' in Example <ref>, including the case `$11$',
our analysis above has precluded compositions $a_1\cdots a_k$ containing `$1$' in the composition sequence.
We saw at the end of Example <ref>, that provided $a_1\neq1$ then the set of generators
for the basis elements with composition components $a_11$ reduces to two generators only, however, both generators
only generate the basis elements $[a_11]\times(1,0,0,0)$ and $[a_11]\times(0,0,0,1)$.
Consider the case of the composition $a_1a_21$, with both $a_1\neq1$ and $a_2\neq1$.
Using arguments analogous to those for the case `$a_11$' at the end of Example <ref>,
if we examine the eight generators listed in Example <ref>,
we observe that the second and last cannot be generators in the case when $a_3=1$,
while the third and fourth generators combine, and the fifth and sixth generators combine,
in much the same way as for the case of `$a_11$' in Example <ref>.
The latter two correspond to adding the third and fourth, and also the fifth and sixth, columns
in the $8\times8$-matrix $A_3$ above. Thus for the case of the composition $a_1a_21$, the resulting coefficient matrix, ignoring the
second, third, sixth and seventh rows which are zero, corresponds to the coefficient matrix $A_3^\prime$
in Example <ref>. With these last two examples in hand, we deduce that for any composition
of the form $a_1a_2\cdots a_{k-1}1$, where we preclude any of $a_1$ through to $a_{k-1}$ to be `$1$',
we have a unique set of generators in the sense of Lemma <ref>, albeit with
a signature coefficient matrix of size $2^{k-1}\times2^{k-1}$. Now consider the case when the
composition component is $[1a_2\cdots a_k]$, assume for the moment none of $a_2$ throught to $a_k$ equal `$1$'.
Looking at the standard actions in Lemma <ref>, we observe that for basis elements with
such a composition component, the two valid actions are,

\begin{align*}
[\bs0]\,[\bs{a_2-1}]\,\bigl([a_3\cdots a_k]\times(\hat u,0)\bigr)&=\chi(a_3\cdots a_k)\cdot[a_1\cdots a_k]\times(\star,\star),\\
[\bs0]\,[\bs0]\,\bigl([(a_2-1)a_3\cdots a_k]\times(\hat a,\hat b)\bigr)&=\chi((a_2-1)a_3\cdots a_k)\cdot[a_1\cdots a_k]\times(\star,\star),
\end{align*}

where the two expressions $(\star,\star)$ are proxies for the appropriate $\mathbb R\la\mathbb B\ra$-components whose exact form is
not important at this stage. However, we now observe that if we apply the final action in Lemma <ref> respectively
for the cases of the compositions $[(a_2+1)a_3\cdots a_k]$ and $[2(a_2-1)a_3\cdots a_k]$, we find,

\begin{align*}
[\bs0]\,[\bs{a_2-1}]\,\bigl([a_3\cdots a_k]\times(\hat u,0)\bigr)&=\chi(a_3\cdots a_k)\cdot[(a_2+1)\cdots a_k]\times(\star,\star),\\
[\bs0]\,[\bs0]\,\bigl([(a_2-1)a_3\cdots a_k]\times(\hat a,\hat b)\bigr)&=\chi((a_2-1)a_3\cdots a_k)\cdot[2(a_2-1)a_3\cdots a_k]\times(\star,\star).
\end{align*}

We observe that the first two respective factors of the generators and their arguments match the two cases corresponding to
the composition $[1a_2\cdots a_k]$. However the latter two cases generate basis elements with the respective composition components
$[(a_2+1)a_3\cdots a_k]$ and $[2(a_2-1)a_3\cdots a_k]$, both of which occur before the composition $[1a_2\cdots a_k]$ in descent order.
Thus these generators will not be new. A similar scenario could occur if one or more letters $a_2$ through to $a_k$ are equal to $1$.
In our main proof below in Step 8, we are able to discount any compositions $a_1\cdots a_k$ in which any one of the letters $a_1$ through to $a_{k-1}$
is equal to `$1$'.

Step 6: Full rank blocks.
Our goal in this step is to show that for a given composition of length $k$, for which in general there are $2^k$ different possible
$\mathbb R\la\mathbb B\ra$ components, there are $2^k$ independent generators, generated by the first action
acting on generators at level $k-2$ and the second and third special actions on generators at level $k-1$.
The following results establish that this is indeed the case.

We have the following, at leading order:

(i) Given an independent set of input $\mathbb R\la\mathbb B\ra$ components of length $2^{k-1}$,
of the form $(u,0)$ for the first action, or of the form $(a,b)$ for the second and third actions,
each individual action produces an independent set of $\mathbb R\la\mathbb B\ra$ components of length $2^k$;

(ii) Given any arbitrary length $2^{k-1}$ non-zero inputs, of the form $(u,0)$ for the first action or of the form $(a,b)$
for the second and third actions, the set of three actions generate independent $\mathbb R\la\mathbb B\ra$ components of length $2^k$.

We focus on the effect of the actions on the $\mathbb R\la\mathbb B\ra$ components only.
In order, consider (i). It is sufficient to prove the result for two independent inputs as the general case
follows suit. Consider the first action and suppose $u$ and $\hat u$ are two non-trivial independent
$\mathbb R\la\mathbb B\ra$ components. Consider an arbitrary linear combination, with scalar coefficients $\kappa_1$ and $\kappa_2$,
of the first action applied to the input $(u,0)$ and the first action applied to the input $(\hat u,0)$.
Set the linear combination to zero. This gives,
$\kappa_1\cdot(u,\pm u^\dag,u,\pm u^\dag)+\kappa_2\cdot(\hat u,\pm\hat u^\dag,\hat u,\pm\hat u^\dag)=0$,
where the right-hand side represents the zero $\mathbb R\la\mathbb B\ra$ component of the appropriate length.
Pairing up, we observe the equation above is equivalent to $\kappa_1\cdot u+\kappa_2\cdot\hat u=0$,
with the other pairings generating the same equation. Since by assumption $u$ and $\hat u$ are two independent
$\mathbb R\la\mathbb B\ra$ components, the result follows. Now consider the second action.
Suppose $(a,b)$ and $(\hat a,\hat b)$ are two non-trivial independent $\mathbb R\la\mathbb B\ra$ components.
As above, we construct the arbitrary linear combination,
$\kappa_1\cdot(a\pm b^\dag,b\pm a^\dag,-a\mp b^\dag,-b\mp a^\dag)
+\kappa_2\cdot(\hat a\pm\hat b^\dag,\hat b\pm\hat a^\dag,-\hat a\mp\hat b^\dag,-\hat b\mp\hat a^\dag)=0$,
for arbitrary scalar coefficients $\kappa_1$ and $\kappa_2$. We assume $a\neq \pm b^\dag$ and
$\hat a\neq\pm\hat b^\dag$—we observe from our proof of Lemma<ref> that the
second action is trivial if and only if $a=\pm b^\dag$.
Since in the last equation the final two components generate the same equation as the first two, the last equation is equivalent to,
$\kappa_1\cdot(a\pm b^\dag,b\pm a^\dag)+\kappa_2\cdot(\hat a\pm b^\dag,\hat b\pm a^\dag)=0$.
This reduces to $\kappa_1\cdot(a,b)+\kappa_2\cdot(\hat a,\hat b)=0$.
Hence by our assumption on $(a,b)$ and $(\hat a,\hat b)$, the result follows.
We now consider the third action.
Suppose $(a,b)$ and $(\hat a,\hat b)$ are two non-trivial independent $\mathbb R\la\mathbb B\ra$ components,
As above, we construct the linear combination,
$\kappa_1\cdot(a,b,\pm b^\dag,\pm a^\dag)+\kappa_2\cdot(\hat a,\hat b,\pm\hat b^\dag,\pm\hat a^\dag)=0$,
for arbitrary scalar coefficients $\kappa_1$ and $\kappa_2$. This last equation is equivalent to
$\kappa_1\cdot(a,b)+\kappa_2\cdot(\hat a,\hat b)=0$—the final two components
generate the same equation as the first two. By our independence assumption on $(a,b)$ and $(\hat a,\hat b)$,
the result follows.

We now consider (ii). For arbitrary $(u,0)$ and $(a,b)$ in $\mathbb R\la\mathbb B\ra$,
consider the following linear combination of the three actions, set equal to the zero $\mathbb R\la\mathbb B\ra$ component, namely:
where $\kappa_1$, $\kappa_2$ and $\kappa_3$ are arbitrary scalar coefficients.
Note we assume $u$ and $(a,b)$ are non-trivial.
If $\kappa_1\neq0$ and $\kappa_2=\kappa_3=0$ then we observe that necessarily $u$
must be the zero $\mathbb R\la\mathbb B\ra$ component, which contradicts our assumptions.
Similarly if $\kappa_3\neq0$ and $\kappa_1=\kappa_2=0$ then necessarily $(a,b)$ is
the zero $\mathbb R\la\mathbb B\ra$ component, which again contradicts our assumptions.
If $\kappa_1\neq0$, $\kappa_2\neq0$ and $\kappa_3=0$, the first and second components above
reveal that necessarily $\kappa_1\cdot u+\kappa_2\cdot(a-b^\dag)=0$ and $\kappa_1\cdot u^\dag+\kappa_2\cdot(b-a^\dag)=0$.
Taking the adjoint of the second equation and adding the result to the first equation, implies $u=0$.
The third and fourth components generate the same information. Thus we have a contradiction.
Analogously, in the cases $\kappa_1\neq0$, $\kappa_3\neq0$ and $\kappa_2=0$, as well as
$\kappa_2\neq0$, $\kappa_3\neq0$ and $\kappa_1=0$, it is straightforward to show that
a necessary consequence is that $(a,b)$ is the zero $\mathbb R\la\mathbb B\ra$ component,
and we have a contradiction. Now consider the case when all of $\kappa_1$, $\kappa_2$, $\kappa_3$
are non-zero. Pairing up the first component from the linear combination above with the
adjoint of the fourth component reveals that necessarily $a$ is the zero $\mathbb R\la\mathbb B\ra$ component.
Pairing the second component and the adjoint of the third component reveals that necessarily $b$
is the zero $\mathbb R\la\mathbb B\ra$ component. We thus reach another contradiction.
The final case we have not considered is the case $\kappa_1=\kappa_3=0$ and $\kappa_2\neq0$. In this case we necessarily
deduce $a=b^\dag$. As we have seen above, this is precisely the condition we need to rule
out for the input when we apply the second action. The proof is complete.

Putting the results of this and the previous steps together, we observe the following.

Suppose we are given a composition $a_1\cdots a_k\in\mathcal C$ of $k$-parts.
Assume that all of $a_1$ through to $a_{k-1}$ are not equal to `$1$'.
Then associated with the $2^k$ set of basis elements with composition component $a_1\cdots a_k$,
are a unique set of $2^k$ generators, and the signature coefficient matrix has full rank.
If $a_k=1$, the statement still holds but instead with $2^{k-1}$ basis elements and $2^{k-1}$ generators.

Step 7: Composition and generator counts.
In light of Proposition <ref>, we are interested in the following counts.
Given $n\in\mathbb N$, what are the total numbers of: (i) Generators; (ii) Basis elements
with composition components avoiding `$1$', i.e. compositions $a_1\cdots a_k$ for which
none of the letters $a_1$ through to $a_k$ are `$1$';
(iii) Basis elements with compositions ending in `$1$' with rest of the composition avoiding `$1$',
i.e. compositions of the form $a_1\cdots a_{k-1}1$ for which $a_1\cdots a_{k-1}$ avoids `$1$'.
Such information will help us keep track of size of the linear system of equations for the unknown coeffcients $\{c_\star\}$ we solve
as part of the proof of Theorem <ref> below in Step 8.

Let us begin with the total number of generators, i.e. item (i). Note that all monomial generators
are of odd-degree. For a given $n\in\mathbb N$, there is one generator of degree $1$, namely $[\bs{n}]$.
We then need to enumerate all the possible degree $3$ generators. Each Pöppe product corresponds to
increasing the order of the compositions they generate by $1$, and there are two Pöppe products in any
degree $3$ generator. Hence all the degree $3$ generators consist of all the possible compositions
of $(n-2)$ with $1$, $2$ and $3$ parts and all the possible ways to assort them into three factors
which can include “packing factors” of $0$. So for example, the only $1$-part compositions of $(n-2)$
assorted in this way are $[\bs{n-2}]\,[\bs0]\,[\bs0]$, $[\bs0]\,[\bs{n-2}]\,[\bs0]$ and $[\bs0]\,[\bs0]\,[\bs{n-2}]$.
The first set of generators of this form associated with the $2$-part compositions of $(n-2)$ are
$[\bs{n-3}]\,[\bs1]\,[\bs0]$, $[\bs{n-3}]\,[\bs0]\,[\bs1]$ and $[\bs0]\,[\bs{n-3}]\,[\bs1]$, and so forth.
These are just the weak compositions of $(n-2)$ into three parts.
The next set of generators are those of degree $5$, and all the generators of this degree would
consist of the weak compositions of $(n-4)$ with $5$ parts, and so forth.
The number of weak compositions of $m$ into $k$ parts is $m+k-1$ choose $k-1$.
We can also think of this as the number of ways of distributing $m$ balls into $k$ slots, allowing empty slots.
From our discussion above, when $n$ is odd, we see that we are interested in, for $k=1,3,5,\ldots,n$,
the number of ways of distributing $m=n-k+1$ balls into $k$ slots, or in other words,

\begin{equation*}
\sum_{k=1\text{($k$ odd)}}^{n} \begin{pmatrix} n\\ k-1\end{pmatrix}=\sum_{\ell=0}^{\frac12(n-1)}\begin{pmatrix} n\\ 2\ell\end{pmatrix}=2^{n-1},
\end{equation*}

where we use the substitution $k=2\ell+1$ for the second sum. That the sum total on the right equals $2^{n-1}$ follows, with some care,
from the corresponding result in Lemma <ref>. Similarly, when $n$ is even,
we are interested in, for $k=1,3,5,\ldots,n+1$, the number of ways of distributing $m=n-k+1$ balls into $k$ slots, or in other words,

\begin{equation*}
1+\sum_{k=1\text{($k$ odd)}}^{n-1} \begin{pmatrix} n\\ k-1\end{pmatrix}=1+\sum_{\ell=0}^{\frac12(n-2)}\begin{pmatrix} n\\ 2\ell\end{pmatrix}=2^{n-1}.
\end{equation*}

Again we used the substitution $k=2\ell+1$ for the second sum. Note that the initial `$1$' in the sum corresponds to the
case $k=n+1$, i.e. corresponding to the monomial generator $[\bs0]^{n+1}$. That the sum total is $2^{n-1}$ again follows
from the first result in Lemma <ref>. We have thus established the following.

Given $n\in\mathbb N$, the total number of monomial generators is equal to $2^{n-1}$.

The number of compositions of $n$ with $k$-parts is $n-1$ choose $k-1$, and accumulating
these coefficients from $k=1$ to $k=n$, the total number of compositions of $n$ is $2^{n-1}$.
Of course associated with each composition, there are one or more basis elements.
For example for a composition $a_1\cdots a_k$ with $k$-parts which avoids $1$, there are $2^{k}$ associated basis elements.
The number of compositions of $n$ into $k$ parts avoiding $1$ is given by,

\begin{equation*}
\begin{pmatrix} n-k-1\\k-1\end{pmatrix}.
\end{equation*}

To see this we observe the following—see Axenovich and Ueckerdt <cit.> or Beck and Robbins [13].
There is a bijection between: (a) the arrangements of $n$ balls into $k$ slots with each slot containing two or more balls;
and (b) the arrangements of $n-k$ balls into $k$ slots with no empty slots. For the map from (a) to (b),
we simply remove one ball from each slot. For the map from (b) to (a) we just add one ball to each slot.
The count for (b) is $n-k-1$ choose $k-1$, giving the result above. Thus, given that for a given composition with $k$ parts that
avoids `$1$' there are $2^k$ corresponding basis elements, the total number of basis elements
with composition components that avoid `$1$' is given, with $\lambda=2$, respectively when $n$ is odd and then when $n$ is even, by,

\begin{equation*}
p(n;\lambda)\coloneqq\sum_{k=1}^{(n-1)/2}\begin{pmatrix} n-k-1\\k-1\end{pmatrix}\lambda^k
\qquad\text{and}\qquad
p(n;\lambda)\coloneqq\sum_{k=1}^{n/2}\begin{pmatrix} n-k-1\\k-1\end{pmatrix}\lambda^k.
\end{equation*}

By direct enumeration, as well as from Examples <ref>–<ref>,
we know $p(2;2)=p(3;2)=2$, $p(4;2)=6$ and $p(5;2)=10$.
In fact, in general, we have the following result.

Given an integer $n\geqslant2$ and a real number $\lambda>0$,
the weighted sum $p=p(n;\lambda)$ of the total number of basis elements
with composition components which avoid `$1$' satisfies the weighted Fibonacci sequence satisfying,

\begin{equation*}
\end{equation*}

In particular, when $\lambda=2$, $p(2;2)=p(3;2)=2$ and $p(n;2)=\frac23(2^{n-1}+(-1)^n)$.

By direct computation, for $n$ odd, we observe, $p(n-1;\lambda)+\lambda\,p(n-2;\lambda)$ equals,

\begin{align*}
\sum_{k=1}^{(n-1)/2}&\begin{pmatrix} n-k-2\\k-1\end{pmatrix}\lambda^k+\sum_{k=1}^{(n-3)/2}\begin{pmatrix} n-k-3\\k-1\end{pmatrix}\lambda^{k+1}\\
&=\begin{pmatrix} n-3\\0\end{pmatrix}\lambda+\sum_{k=2}^{(n-1)/2}\Biggl(\begin{pmatrix} n-k-2\\k-1\end{pmatrix}
+\begin{pmatrix} n-k-2\\k-2\end{pmatrix}\Biggr)\lambda^k,
\end{align*}

which equals $p(n;\lambda)$ once we combine the two terms in the coefficient of $\lambda^k$ shown and observe that
the coefficient of the $\lambda$ term is one. Note that in the
first step we made the change of variables $\ell=k+1$ in the second sum, before relabelling $\ell$ as $k$.
When $n$ is even, we similarly observe that $p(n-1;\lambda)+\lambda\,p(n-2;\lambda)$ equals,

\begin{align*}
\sum_{k=1}^{(n-2)/2}&\begin{pmatrix} n-k-2\\k-1\end{pmatrix}\lambda^k+\sum_{k=1}^{(n-2)/2}\begin{pmatrix} n-k-3\\k-1\end{pmatrix}\lambda^{k+1}\\
&=\begin{pmatrix} n-3\\0\end{pmatrix}\lambda+\sum_{k=2}^{(n-2)/2}\Biggl(\begin{pmatrix} n-k-2\\k-1\end{pmatrix}
+\begin{pmatrix} n-k-2\\k-2\end{pmatrix}\Biggr)\lambda^k+\begin{pmatrix} n/2-2\\ n/2-2\end{pmatrix}\lambda^{n/2},
\end{align*}

which equals $p(n;\lambda)$ once we combine the terms in the coefficient of $\lambda^k$, and
observe that the coefficients of the $\lambda$ and $\lambda^{n/2}$ terms are one.
We also used the same change of variables in the first step.
The final statement specific to $\lambda=2$, follows directly by solving the difference
equation for $p=p(n;2)$ for the initial conditions indicated.

The result of Lemma <ref> provides an answer to item (ii), stated at the beginning of this step.
Item (iii) is now straightforward. The total number of compositions with $k$-parts of the form $a_1\cdots a_{k-1}1$
for which $a_1\cdots a_{k-1}$ avoids `$1$' is simply $n-k-1$ choose $k-2$.
This is because here we require $n-1$ balls to fit into $k-1$ slots with each slot containing two or more balls.
Hence the total number of basis elements associated with composition components which end in `$1$', but avoid `$1$' elsewhere,
when $n$ is odd so $n-1$ is even, is given by,

\begin{equation*}
\sum_{k=2}^{(n+1)/2}\begin{pmatrix} n-k-1\\k-2\end{pmatrix}\lambda^{k-1}=\sum_{k=1}^{(n-1)/2}\begin{pmatrix} n-k-2\\k-1\end{pmatrix}\lambda^{k},
\end{equation*}

which equals $p(n-1;\lambda)$. When $n$ is even and thus $n-1$ is odd, the same count is,

\begin{equation*}
\sum_{k=2}^{n/2}\begin{pmatrix} n-k-1\\k-2\end{pmatrix}\lambda^{k-1}=\sum_{k=1}^{(n-2)/2}\begin{pmatrix} n-k-2\\k-1\end{pmatrix}\lambda^{k},
\end{equation*}

which equals $p(n-1;\lambda)$. Finally we observe the following.

The total number of basis elements with composition components which avoid `$1$', or end in `$1$' and avoid `$1$' elsewhere,
equals $2^{n-1}$.

The count in question is $p(n;2)+p(n-1;2)$. Using the explicit soluton for $p(n;2)$ given in Lemma <ref>,
the result follows.

Combining the results of Lemmas <ref> and <ref>, we deduce the
rather remarkable fact:

For any given $n\in\mathbb N$, the total number of basis elements whose composition component avoids `$1$', or ends in `$1$' but avoids `$1$' elsewhere,
exactly equals the total number of generators.

Naturally this result is important in our proof of Theorem <ref> just below. We remark that the inclusion of
basis elements whose composition components end in `$1$' but avoid `$1$' elsewhere, rather than composition components
containing `$1$' at some other position with avoidance elsewhere, is merely a consequence of the ordering we have imposed,
namely the descent order.

Step 8: Proof of Theorem <ref>.
In this last step we provide the overall proof of our main theorem. We combine together the knowledge
we gained in Steps 1–7. This final stage of the argument, though significantly adapted, is analogous to that outlined for
the non-commutative potential Korteweg–de Vries hierarchy in Malham [68].

To complete the proof, we essentially construct a table of signature coefficients for the arbitrary order $n$ case,
much like Tables  <ref>–<ref> for the $n=4$ and $n=5$ cases.
Indeed we refer to these example tables to demonstrate examples of the general procedure.
We already know from Steps 1-7 that we can construct a linear algebraic equations of the form $AC=B$,
where the vector $C$ of length $2^{n-1}$ lists the unknown coefficients of the generator monomials
in the Pöppe polynomial $\pi_n=\pi_n([\bs0],[\bs1],\ldots,[\bs n])$. The vector $B$, whose length exceeds $2^{n-1}$,
is a vector of zeros apart from a single non-zero value `$1$' in the first position when $n$ is odd and
in the second position when $n$ is even. This is due to the ordering we impose which we outline briefly now,
and in some more detail just below.
The coefficients of $C$, are ordered according to the descent order and blocks as outlined in Steps 3 and 5.
The signature coefficient matrix $A$ has a lower triangular block form.
It has $2^{n-1}$ columns and its total number of rows exceeds $2^{n-1}$, though equals the length of $B$.
The columns of $A$ are parametrised by the blocks of monomial generators mentioned, or equivalently the order of the coeffcients in $C$.
The rows of $A$ are parametrised by the basis elements, characterised by the composition components of the basis elements listed in descent order,
and within individual composition $w$, the basis elements are listed according to the binary order of the $\mathbb R\la\mathbb B\ra$-components.
The number of such $\mathbb R\la\mathbb B\ra$-components is $2^{|w|-\upsilon(w)}$, where $\upsilon(w)$ counts the number of $1$'s in the
composition $w$.

Let us outline the forms of $C$ and $A$ in some more detail. We can be brief as much of the procedure has already been outlined in Steps 1–7.
Corresponding to the pair of basis elements with a one-part composition component, namely $[n]\times(1,0)$ and $[n]\times(0,1)$,
the first pair of coefficients in $C$ are $c_n$ and $c_{0(n-2)0}$, corresponding to the generators $[\bs n]$ and $[\bs0]\,[\bs{n-2}]\,[\bs0]$.
The corresponding $2\times2$ top-left block in $A$ is the matrix $A_0$. All the remaining entries in the first two rows of $A$ to the right
of this block are zero. We move onto basis elements whose composition components have $k=2$ parts. In descent order the first set of
basis elements consists of $[(n-1)1]\times(1,0,0,0)$ and $[(n-1)1]\times(0,0,0,1)$. We know from Example <ref> in Step 5,
these are the only two relevant basis elements for corresponding to compositions ending in `$1$'. The corresponding coefficients in $C$
are $c_{(n-2)00}$ and $c_{0(n-3)1}$ associated with the generators $[\bs{n-2}]\,[\bs0]\,[\bs0]$ and $[\bs0]\,[\bs{n-3}]\,[\bs1]$.
The corresponding coefficient matrix is the block $A_1$ taking up rows and columns $3$ and $4$ in $A$, with the remaining entries
in rows $3$ and $4$ to the right of this block being zero. This first set of blocks is important once we start addressing the issue
of the uniqueness of the solution $C$ to the linear system, or equivalently the consistency of the overall linear system.
Next we consider the blocks of basis elements parametrised by two-part composition components of the form $a_1a_2$
in descent order, with neither $a_1$ nor $a_2$ equal to zero. Associated with any such composition, there are four
basis elements $[a_1a_2]\times\bs\beta_i$, $i=1,2,3,4$, and the corresponding coefficients in $C$
are $c_{(a_1-1)(a_2-1)0}$, $c_{(a_1-1)0(a_2-1)}$, $c_{0(a_1-2)a_2}$ and $c_{0(a_1-2)0(a_2-2)0}$,
corresponding to the monomial generators outlined in Example <ref>.
For each block of four basis elements with such a composition component, the corresponding signature coefficient matrix is $A_2$.
As we run through the compositions $a_1a_2$ which avoid `$1$' in descent order,
the coefficient matrix $A_2$ occupies rows and columns $4(n-a_1-1)+i$ for $i=1,2,3,4$ in the signature coefficient matrix $A$.
All the entries in $A$ in the these rows to the right of these blocks are zero.
We know from Step 5, that the basis elements with the two-part composition component $1(n-1)$ does not occupy any new
columns in $A$, but instead, the two rows corresponding to the two basis elements concerned only contain
non-zero entries in the columns parametrised by $4(n-a_1-1)+i$ for $i=1,2,3,4$ for $a_1\neq1$.

We move onto blocks of basis elements corresponding to composition components with $k=3$ parts. We know from
our arguments in Step 5 that we can focus on composition components of the form $a_1a_2a_3$ which avoid `$1$'
or end in `$1$' and avoid `$1$' elsewhere. Basis elements with composition components lying in the complement of
this set do not generate “new” columns, or equivalently, the non-zero entries in the rows in $A$ corresponding
to these basis elements only occupy columns we have already encountered/parametrised. Let us examine the
blocks we generate as we run through the compositions $a_1a_2a_3$ which avoid `$1$' or end in `$1$' and avoid `$1$' elsewhere.
We know from Step 5, that for each of the former compositions we generate the block matrix $A_3$,
modified to include the column factors mentioned in Step 5, associated with the $8$ corresponding basis elements
and the $8$ “new” monomial generator columns. For each of the latter compositions we generate a block matrix $A_3^\prime$
associated with the $4$ corresponding basis elements and the $4$ “new” monomial generator columns. The block matrices
$A_3$ and $A_3^\prime$ are diagonal blocks in $A$, with all entries in the corresponding rows they occupy to the right
of these blocks equal to zero.

We know from our analysis in Step 5 and in particular Lemma <ref> and the discussion immediately following this
lemma, that we have the following. For every composition $w$ of $n$ that avoids `$1$', or ends in `$1$' but avoids `$1$' elsewhere, we can construct
a $2^{|w|}\times2^{|w|}$ block matrix $A_{w}$ in the former case, or a $2^{|w|-1}\times2^{|w|-1}$ block matrix $A_w^\prime$ in the latter case,
which occupies a distinct diagonal block in $A$. Here by “distinct”, we mean that its rows and columns do not coincide with the rows
and columns of any of the analogous block matrices corresponding to basis elements with such composition components. All the entries
in $A$ in the rows occupied by these blocks and to the right of them, are zero. Thus for general $n$, the signature coefficient matrix $A$
is indeed a lower block triangular matrix. Further, from our results in Step 6, we know that each of the block matrices $A_w$ and $A_w^\prime$
has full rank. Even further, from our results in Step 7, we know that the total number of basis elements corresponding to such
composition components and generating the rows of such diagonal blocks, exactly equals the total number of monomial generators
generating the columns of such diagonal blocks. This means that if we ignore the basis elements with composition components
in the complementary set for the moment, then we can proceed block by block, starting with $A_0$, and solve for the corresponding
set of Pöppe polynomial coefficients in $C$, until we precisely exhaust the blocks and uniquely recover $C$.

The top left block entries in the signature coefficient matrix $A$
at any order $n$, depending on whether $n$ is odd (top) or $n$ is even (bottom).
The coefficients are the $\chi$-images of the signature entries shown.
The forms shown for the case of when $n$ is odd or even, are used to prove the
consistency of the overdetermined linear system of algebraic equations for
the Pöppe polynomial coefficients. The first rows are not shown.

$n$ odd $\phantom{\biggl|}$                              $[\bs n]$     $[\bs{n-2}]\,[\bs0]^2$                   $[\bs0]\,[\bs{n-3}]\,[\bs1]$         $\cdots$

$[\bs0(n-1)\bs0 1\bs0]$ $\phantom{\Bigl|}$                $(n-1)1$      $2\cdot\bigl((n-2)\ot0\ot0\bigr)$       $2\cdot\bigl(0\ot(n-3)\ot1\bigr)$

$[\bs0(n-1)\bs0^\dag 1\bs0^\dag]$ $\phantom{\Bigl|}$                     $-2\cdot\bigl((n-2)\ot0\ot0\bigr)$      $2\cdot\bigl(0\ot(n-3)\ot1\bigr)$

$~~\vdots$ $\phantom{\Bigl|}$

$n$ even  $\phantom{\biggl|}$        $[\bs n]$     $[\bs0]\,[\bs{n-2}]\,[\bs0]$        $[\bs{n-2}]\,[\bs0]^2$     $[\bs0]\,[\bs{n-3}]\,[\bs1]$

$[\bs0 n\bs0]$  $\phantom{\Bigl|}$        $n$           $2\cdot\bigl(0\ot(n-2)\ot0\bigr)$

$[\bs0(n-1)\bs0 1\bs0]$  $\phantom{\Bigl|}$      $(n-1)1$     $2\cdot\bigl(0\ot(n-2)\ot0\bigr)$     $2\cdot\bigl((n-2)\ot0\ot0\bigr)$     $2\cdot\bigl(0\ot(n-3)\ot1\bigr)$

$[\bs0(n-1)\bs0^\dag 1\bs0^\dag]$ $\phantom{\Bigl|}$         $2\cdot\bigl(0\ot(n-2)\ot0\bigr)$     $-2\cdot\bigl((n-2)\ot0\ot0\bigr)$    $2\cdot\bigl(0\ot(n-3)\ot1\bigr)$

The rest of the proof is now concerned with demonstrating the consistency of the remaining rows/linear equations for the coefficients
in $C$ associated with those basis elements with composition components that contain a `$1$', not including the instances of
compositions ending in `$1$', but avoiding it elsewhere. We heavily rely on the fact that the linear system of algebraic equations for $C$
is almost homogeneous apart from the single unit entry in the first position when $n$ is odd and in the second position when $n$ is even.
The first phase of this section of the proof focuses on the blocks associated with basis elements with $1$ and $2$-part composition components.
Assume for the moment that $n$ is odd. In this instance the first two linear equations for the coefficients $c_n$ and $c_{0(n-2)0}$
are $c_n+2\cdot c_{0(n-2)0}=1$ and $-2\cdot c_{0(n-2)0}=0$. Thus when $n$ is odd, we always have $c_{0(n-2)0}=0$. This means that all
the entries in the second column of the signature coefficient matrix $A$ are not relevant to the linear system of equations
for $C$ and we thus eliminate this column in $A$ when $n$ is odd. Implementing this, and ignoring the first row in $A$ corresponding
to the basis element $[\bs0 n\bs0]$, the top left corner of $A$ has the form shown in the top part of Table <ref>.
Now assume that $n$ is even. The first two linear for the coefficients $c_n$ and $c_{0(n-2)0}$ in this instance are
are $c_n+2\cdot c_{0(n-2)0}=0$ and $-2\cdot c_{0(n-2)0}=1$. When $n$ is even we swap over the first two rows in the
signature coefficient matrix $A$ which is equivalent to swapping the order of the first two linear equations shown.
If we ignore the new first row of $A$ corresponding to the nonhomogeneous equation $-2\cdot c_{0(n-2)0}=1$,
then the top left corner of $A$ has the form shown in the bottom part of Table <ref>.
All the remaining rows and columns in the signature coefficient matrix $A$, whether $n$ is odd or even, remain
the same. However, with the first rows ignored, the system of linear equations that remains is homogeneous.
And, of course, the top left corners in the case that $n$ is odd or even have the forms shown in Table <ref>.
In both cases in Table <ref> we trace a diagonal starting from the top left non-zero
coefficient, which is $\chi\bigl((n-1)1\bigr)=n$ when $n$ is odd, and $\chi(n)=1$ when $n$ is even.
When $n$ is even the next two entries along this diagonal are $\chi\bigl(2\cdot(0\ot(n-2)\ot0)\bigr)=2$ and
$\chi\bigl(-2\cdot((n-2)\ot0\ot0)\bigr)=-2$. When $n$ is odd the next diagonal entry is $\chi\bigl(-2\cdot((n-2)\ot0\ot0)\bigr)=-2$.
Thereafter the diagonal entries for the cases of $n$ being even or odd are the same.
In Table <ref>, when $n$ is even, after swapping the first two rows, though not ignoring the new first row yet,
we can view the diagonal we have identified as the diagonal just below the leading diagonal. Similarly in Tables <ref>
and <ref>, when $n$ is odd, after eliminating the second column but still retaining the first row, we can again view
the diagonal we have identified as that just below the leading diagonal in those Tables.
In either Table <ref> or in Tables <ref> and <ref>, let us call the diagonal just
below the leading diagonal the `sub-diagonal'. Consider for example Tables <ref> and <ref>.
If we follow the sub-diagonal with the view of retaining non-zero terms along it, we observe we meet a obstruction
in the first $4\times4$ block with matrix $A_2$ characterised by the composition component $32$. The problem is
that while the sub-diagonal of $A_2$ has non-zero entries, the next term along the diagonal that lies in the last column
of that $A_2$ block, but beneath the entire block, and in fact the row corresponding to the basis element $[23]\times(1,0,0,0)$
is zero. However there is a quick fix to this obstruction. That is to simply swap the two columns in the signature coefficient matrix
$A$ corresponding to the final two columns of the $A_2$ block characterised by the composition component $23$.
Such a swap simply corresponds to changing the order of the monomial generators.
We can see from Tables <ref> and <ref> that this column-swap procedure would guarantee
that the next entry in the sub-diagonal would be non-zero. We can then continue to consider the sub-diagonal
in the $A_2$ block corresponding to the basis elements $[23]\times\bs\beta_i$ for $i=1,2,3,4$. However we observe
a similar obstruction necessitating an anologous swap of the columns of $A$ corresponding to the final two
columns of this second $A_4$ block. This is again enacted to ensure that the term in the final column
immediately below this $A_4$ block is non-zero. The term in question corresponds to the signature coefficient
$\chi(0\ob0\ot3)$ in the row corresponding to $[14]\times(1,0,0,0)$. We have thus established, for all the
basis elements with $1$ and $2$-part composition components, a complete diagonal with all entries non-zero, and which can act as pivots.
Since all the linear equations corresponding to the rows we are considering (we are ignoring the top row) are homogeneous
we can use Gaussian elimination to render all the entries in the columns below the sub-diagonal to be zero.

The procedure for the case of general $n$ proceeds in exactly the same manner as the $n=5$ case we have just
outlined, except that now we need to establish that when we swap the columns over as just outlined, we are
guaranteed a non-zero entry in the corresponding sub-diagonal entry. We also need to guarantee that the
entry immediately below the final column of the $2\times2$ block $A_1$ corresponding to the rows $[(n-1)1]\times(1,0,0,0)$
and $[(n-1)1]\times(0,0,0,1)$ is also non-zero. This case corresponds to the column given by $[\bs0]\,[\bs{n-3}]\,[\bs{1}]$.
In the other cases of the $A_4$ blocks corresponding to the basis elements $[(n-m)m]\times\bs\beta_i$ for $i=1,2,3,4$,
the column in question is the third column in the $A_4$ block corresponding to the column given by $[\bs0]\,[\bs{n-m-2}]\,[\bs{m}]$.
In particular this means we can treat the $m=1$ and $m=2,\ldots,n-2$ cases simultaneously. Indeed, using Corollary <ref>
in Step 4, we observe that at leading order we have,

\begin{equation*}
\end{equation*}

where we have used that at leading order $[\bs{m}]=[m]\times(1,0)+\cdots$. The first term on the
right is the term we expect at leading order for this generator, while the second term on the right
is the column corresponding to the rows in the next block down—we see that $(n-m-1)(m+1)$ is obtained
from $(n-m)m$ by replacing $m$ by $m+1$, which is one composition further down in descent order.
Thus indeed we are guaranteed that the next entry in the sub-diagonal is non-zero when we enact the column swap.
Further we observe that in the case of the final $A_4$ block corresponding to the composition $2(n-2)$
for which $m=n-2$, the corresponding generator is $[\bs0]\,[\bs0]\,[\bs{n-2}]$ while the corresponding
row containing the sub-diagonal entry of interest is $[1(n-1)]\times(1,0,0,0)$. At leading order we have,

\begin{equation*}
\end{equation*}

which thus guarantees a final sub-diagonal non-zero entry. We are thus in the exact same situation
as described for the $n=5$ case just above, and we can use the sub-diagonal entries as pivots
to render all the entries in $A$, in all the sub-diagonal columns, below the sub-diagonal to be zero.

The second phase of this section of the proof now focuses on all the blocks associated with
basis elements with composition components with $k$-parts with $k\geqslant3$. This phase is
more straightforward. Let us focus on the $3$-part composition cases to begin with.
The first $3$-part composition in descent order is $(n-2)11$, and we know from Step 5 that there
are no “new” generators associated with with any such composition that lies in the set of
compositions complementary to those avoiding `$1$' or ending in `$1$', but avoiding it elsehwere.
Hence we can use Gaussian elimination, using the pivots from the
sub-diagonal outlined for the $1$ and $2$-part composition cases just outlined,
to render the entries in for the two rows/basis elements concerned here,
$[(n-2)11]\times\bs\beta_1$ and $[(n-2)11]\times\bs\beta_8$, equal to zero.
Next we consider the block of rows/basis elements corresponding to the composition $(n-3)21$.
As outlined in Step 5 this block is associated with $4$ generators and the block matrix $A_3^\prime$.
We know this has full rank and we can thus use the leading diagonal as pivots to render
all entries in the corresponding columns of $A$ below this diagonal to be zero.
The next block is associated with the composition $(n-3)12$, which with a `$1$' in the
middle is not associated with any new generators, and from our Gaussian elimination processes
thus far has all row entries rendered zero. The next blocks are associated with the compositions $(n-4)31$,
$(n-4)22$, $(n-4)14$. The first two of these compositions are associated with separate copies of the $8\times8$ matrix $A_3$ (with
the columns mentioned in Step 5 suitably scaled) and a total of $16$ generators (one set of $8$ each). The leading diagonals
of both copies of $A_3$ can again be used as pivots to render all the entries, below this diagonal in the columns of $A$ associated
with these two copies, equal to zero. The entries in the rows associated with the third composition $(n-4)13$ will have been
rendered zero in the Gaussian elimination process just outlined for the other two compositions. And so forth, we can see
that we can proceed in descent order through the blocks associated with $3$-part compositions, either in the case of
compositions that avoid `$1$' or end in `$1$' but avoid it elsewhere, using the diagonals of the blocks associated
with $A_3$ or $A_3^\prime$, to render the corresponding entries in $A$ below these diagonals to be zero, or recognising
for the blocks associated with the complementary set of compositions, the entries in the rows of those block will
already be rendered zero. The procedure for all further blocks associated with compositions of $4$ or more parts
proceeds exactly analogously. Naturally that the corresponding diagonals with non-zero entries exist for all
blocks associated with compositions that avoid `$1$' or end in `$1$' but avoid it elsewhere, is guaranteed
by the results in Step 6, in particular Proposition <ref>.

Hence we have thus rendered all the entries in all the rows corresponding to basis elements with
composition components which lie in the set complementary to those that avoid `$1$' or end in `$1$' but avoid it elsewhere,
equal to zero. Briefly returning to the rows/blocks associated with the $1$ and $2$ part compositions, still
ignoring the top row as indicated in Table <ref>. A quick count reveals that when $n$ is even,
we have $3+4(n-1)$ rows, i.e. homogneous linear equations, in $4+4(n-1)$ unknowns, while when $n$ is odd,
we have $2+4(n-1)$ homogeneous linear equations, in $3+4(n-1)$ unknowns. In either case when $n$ is even or odd,
proceeding through all the other blocks associated with compositions of three or more parts, the remaining
number of homogeneous linear equations equals the remaining number of unknowns (as outlined in the first section of this proof).
Hence, in either case when $n$ is even or odd, we can solve the entire system of linear homogeneous equations,
with one less equation then the total number of unknowns, to find expressions for all the unknowns in terms of only one of them.
In the case that $n$ is odd, we solve for all of them in terms of $c_n$. In the case that $n$ is even, we solve for
all of them in terms of $c_{0(n-2)0}$.
We now re-introduce the very first row we ignored at the beginning of this second, “consistency”, section of the proof.
When $n$ is odd, that first equation is $c_n+2\cdot c_{0(n-2)0}=1$. Since we have an expression for $c_{0(n-2)0}$ in terms of $c_n$
from the homogeneous set of linear equations, we can substitute that expression into this non-homogeneous linear equation
and determine $c_n$. When $n$ is even, the first equation is $-2\cdot c_{0(n-2)0}=1$ or equivalently $c_{0(n-2)0}=-1/2$.
Since in this case we have expressions for all the other unknowns in terms of $c_{0(n-2)0}$, this fixes the values of
all the other unknowns. In either case, whether $n$ is odd or even, we have established a unique solution $C$, and
the proof is complete.

As mentioned, the overall proof in Step 8 just above is analogous to that for
the non-commutative potential Korteweg–de Vries hierarchy in Malham [68].
Therein we proceed by considering compositions with $k=1$, $k=2$, and so forth, parts.
In that case there are no blocks as there are no $\mathbb R\la\mathbb B\ra$ components.
The basis elements are just compositions of $n$, there are no skew forms.
Also, as it is the potential equation, the generators are just monomials of
the signature expansions $\bs n$, with $n\in\mathbb N$—in particular there are
no generators corresponding to $[\bs0]$. Further, since there are no skew forms,
the generators can be of even or odd degree. See in particular Section 6 in Malham [68].
We observed at the end of Step 7 above that the total number of generators equalled
the total number of basis elements with composition components that avoided `$1$' together
with those that ended in `$1$' but avoided it elsewhere. It would be natural to
wonder whether a similar situation occurs for the case of the non-commutative potential Korteweg–de Vries hierarchy,
and indeed retrospectively, we can establish the exact same result for that hierarchy.
In fact we can show, since there are no blocks and no generators akin to `$[\bs0]$', that
for each set of compositions with $k$ parts, the number of monomial generators with $k$ factors
equals the number of compositions (the basis elements here) that avoid `$1$' together
with those that ended in `$1$' but avoided it elsewhere. Again, that we single out those
compositions ending in `$1$' is just an artefact of the descent order we impose.
To see this fact, we observe from Section 6 in Malham [68],
that the number of generators with $k$ factors, say of the form $(\bs n_1)(\bs n_2)\cdots(\bs n_k)$
with the Pöppe product, is given by $n-k$ choose $k-1$. This is because each Pöppe product
adds a `$1$' to one of the composition parts in the eventual expansion in compositions.
The complete set of such monomial generators is exhausted by those with $k=1,2,\ldots,\frac12(n+1)$ parts.
We already know from Step 7 above, that the number of compositions of $n$ with $k$ parts that
avoid `$1$' equals $n-k-1$ choose $k-1$ for $k=1,2,\ldots,\frac12(n-1)$, and the number of
compositions that end in `$1$' but avoid it elsewhere equals $n-k-1$ choose $k-2$ for $k=2,\ldots,\frac12(n+1)$.
If we restrict ourselves to $k=2,\ldots,\frac12(n-1)$, we observe that the number of compositions
satisfying either property is given by,

\begin{equation*}
\begin{pmatrix} n-k-1\\ k-1\end{pmatrix}+\begin{pmatrix} n-k-1\\ k-2\end{pmatrix}=\begin{pmatrix} n-k\\ k-1\end{pmatrix},
\end{equation*}

with equality following by simply adding the two relevant fractions on the left.
The cases $k=1$ and $k=\frac12(n+1)$, for which the number of such compositions and generators is singular,
just follows by inspection.

§ CONCLUSION

There are many open directions of research we intend to pursue based on the combinatorial algebraic approach we introduced herein.
One direction we have not directly addressed herein is that of alternative formulations of the modified Korteweg–de Vries hierarchy members
of orders $3$, $5$ and higher. See for example Liu and Athorne [65], Olver and Sokolov [81], Oevel and Rogers [80] and Gerdjikov [47].
For example, the alternative non-commutative modified Korteweg–De Vries equation has the form,

\begin{equation*}
\pa_t g=\pa^3g+3\bigl(g(\pa^2g)-(\pa^2g)g\bigr)-6g(\pa g)g.
\end{equation*}

Note that the polynomial partial differential field includes even degree terms. Such alternative forms can be obtained
from non-commutative modified Korteweg–De Vries equation via a suitable gauge transformation as, for example,
outlined in detail in Carillo and Schiebold [23]. The combinatorial algebraic structure we have developed
would seem a natural context to investigate such alternative hierarchy forms further. Closely related is the
Miura transformation. This is particularly simple in our context. Assuming the order $n=2m+1$ with $m\in\mathbb N$
is odd, then since $\mathcal I^2=\id$, the base dispersion equation for $P$ is,

\begin{equation*}
\pa_t P=(-1)^{m+1}\pa^{2m+1}P.
\end{equation*}

We can assume this to be the base equation for the non-commutative potential Korteweg–de Vries hierarchy considered in Malham [68].
In that case the solution $G^{\mathrm{pKdV}}$ is given by $G^{\mathrm{pKdV}}=P(\id-P)^{-1}$.
For the non-commutative modified Korteweg–de Vries hierarchy, we observe that when $n$ is odd we can
assume the solution $G^{\mathrm{mKdV}}$ to have the form $G^{\mathrm{mKdV}}=2P(\id+P)^{-1}(\id-P)^{-1}$,
i.e. replacing the `$\mathrm{i}P$' everywhere simply by $P$, and all our results in Section <ref> and thereafter follow through.
This is because we carried through the quantity `$\mathrm{i}P$' throughout our computations in Section <ref>
and, in particular, into our abstract encoding. For example, our computation for $\pa_t[V]$ preceding Definition <ref>
carries through with this replacement with $V\coloneqq(\id-P)^{-1}$, $P^\dag=-P$ and $V^\dag=(\id+P)^{-1}$.
For convenience we set $U^{\mathrm{pKdV}}\coloneqq(\id-P)^{-1}$ and $U^{\mathrm{mKdV}}\coloneqq(\id+P)^{-1}(\id-P)^{-1}$.
Note that by operator partial fractions we have $U^{\mathrm{pKdV}}=\id+PU^{\mathrm{pKdV}}$ so $\pa G^{\mathrm{pKdV}}=\pa U^{\mathrm{pKdV}}$.
Then as in Doikou et al. <cit.> we observe that since $U^{\mathrm{pKdV}}=(\id+P)U^{\mathrm{mKdV}}$ we have,

\begin{align*}
&&\pa U^{\mathrm{pKdV}}&=\pa\bigl(PU^{\mathrm{mKdV}}\bigr)+\pa U^{\mathrm{mKdV}}\\
\Leftrightarrow&&\pa U^{\mathrm{pKdV}}&=\pa\bigl(PU^{\mathrm{mKdV}}\bigr)+U^{\mathrm{mKdV}}\pa(P^2)U^{\mathrm{mKdV}}\\
\Rightarrow    &&\pa\lb G^{\mathrm{pKdV}}\rb&=\pa\lb G^{\mathrm{mKdV}}\rb+\lb G^{\mathrm{mKdV}}\rb^2.
\end{align*}

In the last step we used the Pöppe product rule. This represents the Miura transformation giving the connection
between the non-commutative potential, and modified, Korteweg–de Vries hierarchies. A natural question is what
the translation (likely non-trivial) of this result is at even orders?

A natural formulation for Hankel and Toeplitz operators is the $L^2$ Hardy spaces $\mathbb H_{\pm}$, corresponding to the
upper and lower half complex plane; see for example Peller [83]. This can be thought of as the Fourier transform
representation of the formulation we gave in Section <ref>. Recently this context has been used to
prove interesting integrability results/connections for the cubic Szegö equation, see Pocovnicu [84],
Grellier and Gerard [50] and Gerard and Pushnitski [48],
and to extend regularity results for the Korteweg–de Vries equation, see Grudsky and Rybkin [51, 52, 53].
There is a natural decomposition $L^2(\R)=\mathbb H_+\oplus \mathbb H_-$ and thus an immediate direction to pursue
would be to consider our Marchenko equation and Fredholm Grassmannian flow in this context and establish a
connection to the results of, for example, Grellier and Gerard [50] and Grudsky and Rybkin [53].

At the abstract algebra level, for the skew-Pöppe algebra
$\mathbb C[\mathbb Z_{\bs0}]\cong\mathbb C[\mathcal C]\times\mathbb R\la\mathbb B\ra$,
there are many open questions as follows:
(i) The skew-Pöppe algebra $\mathbb C[\mathbb Z_{\bs0}]$, endowed with the triple product in Lemma <ref>,
constitutes a triple system or ternary algebra. See for example Meyberg <cit.> or Ricciardo <cit.>.
Exploring this context is very much of interest.
(ii) The Pöppe products in Lemma <ref> are quasi-Leibniz products in which the `quasi' label refers to the term additional to the
two expected Leibniz terms which essentially involves inserting a `$1$' between the two terms in the product (as well as a factor `$2$').
A natural question is, is it possible to establish an isomorphism between between this skew-Pöppe algebra and the corresponding
skew-Pöppe algebra endowed with the triple product based on the Pöppe products in Lemma <ref>
without the `quasi' terms? This will necessarily require a fix of the non-quasi product for low order terms, for example those
involving products with `$[\bs0]$' and so forth.
The analogy is the isomorphism between the shuffle algebra and the quasi-shuffle algebra proved by Hoffman [60].
Establishing such an isomorphism would significantly simplify the proofs of the results herein and would help to
establish (iii) and (iv) just below.
(iii) We observe that in our main result we sought Pöppe polynomial expansions $\pi_n=\pi_n([\bs0],[\bs1],\ldots,[\bs n])$
for the endomorphisms $[\bs 0n\bs0]$ when $n$ is odd, and $[\bs0 n\bs0^\dag]$ when $n$ is even.
However more generally we might ask the question of whether there exists Pöppe polynomial expansions for any of the basis elements
in $\mathbb C[\mathbb Z_{\bs0}]$? In other words can we express any basis element in $\mathbb C[\mathbb Z_{\bs0}]$ as a linear combination
of monomials of the form $[\bs{n_1}]\,[\bs{n_2}]\,\cdots\,[\bs{n_k}]$?
(iv) A directly related broader question then is, does there exist an isomorphism between the algebra of
odd-degree monomial forms $[\bs{n_1}]\,[\bs{n_2}]\,\cdots\,[\bs{n_k}]$ with $n_i\in\mathbb N\cup\{0\}$ endowed with the concatenation product,
and the skew-Pöppe algebra? The connection is provided by the signature expansions. The odd-degree monomial form parametrising factors $n_1n_2\cdots n_k$
are actually weak compositions.
(v) Can we establish a co-algebra associated with the skew-Pöppe algebra $\mathbb C[\mathcal C]\times\mathbb R\la\mathbb B\ra$?
This was achieved for Pöppe algebra in Malham <cit.>.
Here we have to deal with the $\mathbb R\la\mathbb B\ra$-components. Indeed, we have already started in this direction.
(vi) Establishing such a co-algebra, or at least a refined de-Pöppe co-product $\Delta_n$ associated with $\mathbb C[\mathcal C]\times\mathbb R\la\mathbb B\ra$,
would be useful. Consider the odd-degree monomial $[\bs{n_1}]\,[\bs{n_2}]\,\cdots\,[\bs{n_k}]$ with $n_1n_2\cdots n_k$ a weak composition of $n$.
Using the signature expansions for each of the factors, $[\bs{n_1}]\,[\bs{n_2}]\,\cdots\,[\bs{n_k}]$ can be expressed in the form,

\begin{equation*}
\sum \chi(w_1\ot w_2\ot\cdots\ot w_k)\cdot\bigl([w_1]\times\bs\beta_1(|w_1|)\bigr)\,\bigl([w_2]\times\bs\beta_1(|w_2|)\bigr)\,\cdots\,\bigl([w_k]\times\bs\beta_1(|w_k|)\bigr),
\end{equation*}

where the sum is over all basis elements $[w_i]\times\bs\beta_1(|w_i|)$ for $i=1,\ldots,k$, with $w_i\in\mathcal C(n_i)$ and
$\bs\beta_1(|w_i|)\in\mathbb R\la\mathbb B\ra$ of length $2^{|w_i|}$ with the first component equal to `$1$' as the only non-zero component.
If we compute all the odd-degree Pöppe products on the right, we generate the following form,

\begin{equation*}
\sum_{w\in\mathcal C(n)}\sum_{i=1}^{2^{|w|}}\chi_{\bs\beta_i}\Bigl(\bigl(\Delta_k([w]\times\bs\beta_i\bigr)\Bigr)\cdot\bigl([w]\times\bs\beta_i\bigr).
\end{equation*}

In this expression, the $\bs\beta_i$ are the natural basis elements of $\mathbb R\la\mathbb B\ra$ of length $2^{|w|}$,
containing a `$1$' in the $i$th position and zeros elsewhere. The combined pair of sums correspond to a sum over all possible basis elements,
for example over all the left-most column elemnts in Tables <ref> and <ref>.
The co-product $\Delta_k$ generates all forms $w_1\ot w_2\ot\cdots\ot w_k$ such that the odd-degree Pöppe product
$\bigl([w_1]\times\bs\beta_1(|w_1|)\bigr)\,\cdots\,\bigl([w_k]\times\bs\beta_1(|w_k|)\bigr)$ generates $[w]\times\bs\beta_i$.
The homomorphic map $\chi_{\bs\beta_i}$ records the signature coefficient $\chi(w_1\ot w_2\ot\cdots\ot w_k)$ together
with the factor in the $2^{|w|}\times2^{|w|}$ block associated with the composition $[w]$ as outlined in Section <ref>.
See for example the matrix $A_3$ from that section. It records the factor associated with the $\bs\beta_i$ row and the
$[\bs{\sigma(w_1)}]\,[\bs{\sigma(w_1)}]\,\cdots\,[\bs{\sigma(w_1)}]$ column, where $\sigma(w_i)$ represents the sum of
all the factors in the composition $w_i$. Any terms resulting from the `quasi' term in the Pöppe product are included.
Recall we can systematically generate all the blocks and all such factors
using the appropriate three standard actions—see Step 5 in Section <ref>.
Let $\mathcal C^{\ast}(n)$ denote the set of all odd-length weak compositions $v$ of $n$ such that $\sigma(v)+|v|-1=n$.
Assuming we have established such a co-product $\Delta_k$, then we observe, we can express any Pöppe polynomial,
or even an arbitrary sum of Pöppe polynomials, in the form,

\begin{align*}
\sum_{n\geqslant1}\pi_n&=\sum_{w\in\mathcal C}\Pi\bigl([w]\bigr)\cdot\bigl([w]\times\bs\beta_i\bigr),
\intertext{where}
\Pi\bigl([w]\bigr)&=\sum_{i=1}^{2^{|w|}}\sum_{v\in\mathcal C^\ast(\sigma(w))} c_v\,\chi_{\bs\beta_i}\bigl(\Delta_{|v|}([w]\times\bs\beta_i)\bigr).
\end{align*}

Thus, in principle, we can express the whole hierarchy as the co-algebra sum $\Pi$.

SJAM would like to thank the EPSRC for the Mathematical Sciences Small Grant EP/X018784/1.
It is also a pleasure to acknowledge very interesting discussions with Alexander Pushnitski and Alexei Rybkin in connection with our work herein.

§ DECLARATIONS

§.§ Funding and/or Conflicts of interests/Competing interests

SJAM received funding from the EPSRC for the Mathematical Sciences Small Grant EP/X018784/1.
There are no conflicts of interests or competing interests.

§.§ Data availability statement

No data was used in this work.

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§ COARSE-GRAINED CO-ALGEBRA STRUCTURE

Recall Remark <ref> and Examples <ref> and <ref> in Section <ref>,
as well as Examples <ref>–<ref> in Section <ref>. At each order $n\geqslant0$,
the Pöppe polynomials $\pi_n=\pi_n([\bs0],[\bs1],\ldots,[\bs n])$ consist of linear combinations of
multi-factor products of signature expansions of the form,

\begin{equation*}
[\bs n_1]\,[\bs n_2]\,\cdots\,[\bs n_k]
=\sum\chi(w_1\ot w_2\ot\cdots\ot w_k)\cdot[w_1\times\bs\varphi_1]\,[w_2\times\bs\varphi_2]\,\cdots\,[w_k\times\bs\varphi_k],
\end{equation*}

where the sum is over all words $w_1\times\bs\varphi_1$ with $w_1\in\mathcal C(n_1)$, $w_2\times\bs\varphi_2$ with $w_2\in\mathcal C(n_2)$, and so forth.
Here the $n_i$ for $i=1,\ldots,k$ are non-negative integers. Note we have utilised the natural homomorphic property of $\chi$ we invoked in Section <ref>.
In general, as we outlined in Remark <ref>, the arguments $w_1\ot\cdots\ot w_k$ of $\chi$ shown, might also involve quasi-product tensors `$\hat\ot$',
which invoke scalar multiples of `$2$'.
The forms $[w_1\times\bs\varphi_1]\,[w_2\times\bs\varphi_2]\,\cdots\,[w_k\times\bs\varphi_k]$ on the right generate many different
skew or symmetric forms $[w\times\bs\varphi]$ or $\{w\times\bs\varphi\}$ in $\mathbb C\la\mathbb Z_{\bs 0}\ra$ via the Pöppe product,
as we saw in Examples <ref>–<ref>.
In Tables <ref> and <ref> in Section <ref> we observe the basis elements generated by
products of the form $[\bs n_1]\,[\bs n_2]\,\cdots\,[\bs n_k]$.
Indeed, we record therein in each column, the arguments of $\chi$, generated by such products, associated with those basis elements.
Hence, for example in Table <ref>, the multi-factor product $[\bs0]\,[\bs1]^2$ generates the basis element $[\bs02\bs02\bs0^\dag]$
with $\chi$-factor argument `$0\ot1\ot1$'. It is thus natural to ask the following question. Given a basis element, can we enumerate
all the possible multi-factor products of signature expansions that generated it? We have already achieved this “at leading order”
via the standard actions in Lemma <ref> in Step 5 in Section <ref>.
However our goal here is to enumerate all the possible terms.

We address this question using a coarse-grained approach, sufficient for our illustrative purposes here, as follows.
We consider a simpler Pöppe algebra in which all the skew and symmetric forms have the form
$[\bs0 a_1\bs0 a_2\bs0\cdots\bs0 a_k\bs0]$ or $\{\bs0 a_1\bs0 a_2\bs0\cdots\bs0 a_k\bs0\}$,
i.e. there are no adjoint components $\bs0^\dag$, or even adjoint letters in the composition words $w$. Further we assume
the resulting Pöppe products have the form (recall Remark <ref>),

\begin{align*}
\end{align*}

We assume products with $[\bs0]=[\nu\times\bs0]$ correspond to those in Remark <ref>.
Suppose we are given a basis element $[\bs0 a_1\bs0 a_2\bs0\cdots\bs0 a_k\bs0]$ in $\mathbb C[\mathbb Z_{\bs 0}]\subset\mathbb C\la\mathbb Z_{\bs 0}\ra$,
with a given composition component $w=a_1a_2\cdots a_k\in\mathcal C$.
We seek to enumerate all possible multi-factor products $[\bs n_1]\,[\bs n_2]\,\cdots\,[\bs n_k]$, where the products are the coarse-grained
forms above, that generate $[\bs0 a_1\bs0 a_2\bs0\cdots\bs0 a_k\bs0]$. Note, the $n_i\geqslant0$ for $i=1,\ldots,k$,
so the skew-form $[\bs0]$ is included as a possible factor.

The coarse-grained algebra we have outlined is similar to the Pöppe algebra
outlined in Malham <cit.> for the non-commutative Korteweg–de Vries hierarchy.
Let us nominate the latter, the KdV-Pöppe algebra. The overall product rule is essentially the same.
And, indeed, for our coarse-grained algebra, as for the KdV-Pöppe algebra,
we could simply not record the intervening $\bs0$'s and assume they are implicitly there.
There are however two differences between the coarse-grained algebra here and the KdV-Pöppe algebra.
The first is we distinguish skew and symmetric forms; more on this presently.
The second is that we allow for a stand-alone $[\bs0]=[\nu\times\bs0]$ term,
that has the properties outlined in Remark <ref>. In particular, any single factor $[\bs0]$ in a monomial,
as a result of the Pöppe product outlined above, adds one to the beginning/end of the composition component of its nearest
neighbour without lengthening the composition, and also adds a quasi-product term.

To formalise the process of enumerating all the possible multi-factor products that generate a given basis element $[\bs0 a_1\bs0 a_2\bs0\cdots\bs0 a_k\bs0]$,
we define a set of de-Pöppe co-products as follows. Note hereafter we indeed drop the intervening $\bs0$'s in our basis element encoding
and simply use $[w]$ and $\{w\}$ to denote the basis elements and $[\nu]$ to denote $[\bs0]=[\nu\times\bs0]$.

For any $n\in\mathbb N$ and composition $w\in\mathcal C$, we define the set of de-Pöppe co-products $\Delta_n(w)$ of $w$ to be,

\begin{equation*}
\Delta_n(w)=\begin{cases}
\sum\bigl\langle[u_1]\,[u_2]\,\cdots\,[u_{n+1}],\{w\}\bigr\rangle_{\mathbb C\la\mathcal C\ra}\,u_1\ot u_2\ot\cdots\ot u_{n+1} & \text{when $n$ is odd},\\
\sum\bigl\langle[u_1]\,[u_2]\,\cdots\,[u_{n+1}],\,[w]\,\bigr\rangle_{\mathbb C\la\mathcal C\ra}\,u_1\ot u_2\ot\cdots\ot u_{n+1} & \text{when $n$ is even}.
\end{cases}
\end{equation*}

Here, the sums are over all $u_i\in\mathcal C$, $i=1\ldots,n+1$, and $\la\,\cdot\,,\,\cdot\,\ra_{\mathbb C\la\mathcal C\ra}$
is the inner product on $\mathbb C\la\mathcal C\ra$ defined for any $u,v\in\mathcal C$ by,

\begin{equation*}
\la u,v\ra_{\mathbb C\la\mathcal C\ra}\coloneqq\begin{cases} 1, & \text{if}~u=v,\\ 0, & \text{if}~u\neq v.\end{cases}
\end{equation*}

The idea is that $\Delta_n(w)$ enumerates all the possible degree $(n+1)$ Pöppe products of the form $[u_1]\,[u_2]\,\cdots\,[u_{n+1}]$ that
might generate $[w]$ or $\{w\}$, depending on whether $n$ is even or odd, respectively. For more details on co-products,
see for example Klausner [64]. Note we denote $\Delta\coloneqq\Delta_1$.
Naturally it would be desirable to have an explicit formula for $\Delta_n(w)$. In order to represent such a formula, we need
the map $\theta\colon\mathcal C\cup\{0\}\times\mathcal C\cup\{0\}\to\mathcal C\times\mathcal C$.
For any compositions $u,v\in\mathcal C$, we define,

\begin{equation*}
\theta\colon u\ot v\mapsto u\ot v,\quad\theta\colon u0\ot v\mapsto u\ot v
\quad\text{and}\quad\theta\colon u\ot0v\mapsto u\ot v.
\end{equation*}

In the first case, $u$ or $v$ could be the empty composition/word.
Hereafter we use `$0$' to denote the empty composition/word $\nu$ in this context;
the reason for this will become apparent presently.
An explicit formula for the de-Pöppe co-product is the following.

The co-product $\Delta\coloneqq\Delta_1$ can be characterised as follows. For any composition $w=a_1\cdots a_n\in\mathcal C$,
we have $\Delta(w)$ is given by, $\Delta(w)=\theta\circ\Delta^\prime(w)$, where $\Delta^\prime(w)$ is given by,

\begin{align*}
\Delta^\prime(a_1&\;\cdots a_n)=0\ot(a_1-1)a_2\cdots a_n\\
&\;+\sum_{k=1}^{n-1}\bigl(a_1\cdots a_{k-1}(a_k-1)\ot a_{k+1}\cdots a_n+a_1\cdots a_k\ot(a_{k+1}-1)a_{k+2}\cdots a_n\bigr)\\
&\;+a_1\cdots a_{n-1}(a_n-1)\ot0.
\end{align*}

We denote $\Delta^\prime$ as the primitive de-Pöppe co-product.
Since $w=a_1\cdots a_n\in\mathcal C$ the letters $(a_k-1)$ or $(a_{k+1}-1)$ in $\Delta^\prime(w)$ could be the letter `$\,0$', for any $k=1,\ldots,n-1$.

In the primitive de-Pöppe co-product term $\Delta^\prime(a_1\cdots a_n)$, the first and last terms account for the fact
that $w=a_1\cdots a_n$ could be generated, respectively, by the Pöppe products $[\bs0]\{(a_1-1)a_2\cdots a_n\}$
and $\{a_1\cdots a_{n-1}(a_n-1)\}\,[\bs0]$. Then for each $k=1$ through to $k=n-1$,
the pair of terms in the sum in $\Delta^\prime(w)$, account for the fact that $w=a_1\cdots a_n$ is generated
by either of the Pöppe products, $[a_1\cdots a_{k-1}(a_k-1)]\{a_{k+1}\cdots a_n\}$ or $[a_1\cdots a_k]\{(a_{k+1}-1)a_{k+2}\cdots a_n\}$.
The role of the map $\theta$ is to detect if any of the $(a_1-1)$ through to $(a_n-1)$ are in fact
zero, which will occur if any of the $a_1$ through to $a_n$ equal $1$ and any such compositions $w=a_1\cdots a_n$
are generated through the quasi-product term in the Pöppe product.

[Basic co-products]
For any one-part composition $n\in\mathcal C$ we observe that we have $\Delta(n)=0\ot(n-1)+(n-1)\ot0$.
For two-part compositions, say $w=a_1a_2$, we have,

\begin{equation*}
\Delta^\prime(a_1a_2)=0\ot(a_1-1)a_2+(a_1-1)\ot a_2+a_1\ot(a_2-1)+a_1(a_2-1)\ot0,
\end{equation*}

for the primitive co-product. Some more specific examples, with the map $\theta$ applied, are,

\begin{align*}
\Delta(11)&=0\ot 1+0\ot1+1\ot0+1\ot0,\\
\Delta(21)&=0\ot11+1\ot 1+2\ot0+2\ot0,\\
\Delta(32)&=0\ot22+2\ot 2+3\ot1+31\ot0,\\
\Delta(321)&=0\ot221+2\ot21+3\ot11+31\ot1+32\ot0+32\ot0,\\
\Delta(312)&=0\ot212+2\ot12+3\ot2+3\ot2+31\ot1+311\ot0,\\
\Delta(322)&=0\ot222+2\ot22+3\ot12+31\ot2+32\ot1+321\ot0.
\end{align*}

Herein we impose that $\Delta^\prime(0)=\emptyset$, i.e. the empty set.
This is because un Section <ref> we are concerned
with partial differential fields involving the solution field and its derivatives,
i.e. in Pöppe polynomials of the form $\pi_n=\pi_n([\bs0],[\bs1],\ldots,[\bs n])$.
However, as we saw in Section <ref>, we can in principle also consider
Pöppe polynomials involving integrals of the solution. The natural extension of
the co-product in that situation, from the one-part composition case in Example <ref>,
would be $\Delta(0)=0\ot(-1)+(-1)\ot0$.

Let us now consider how to compute the co-product $\Delta_2(w)$, which is supposed to enumerate all the
possible triple Pöppe products that might generate $[w]$. The associativity of the Póppe product plays
a crucial role here. Naturally we would like $\Delta_2$ to be generated by either of the equivalent expressions
$(\id\ot\Delta)\circ\Delta$ or $(\Delta\ot\id)\circ\Delta$. However some care and modification is required.
To this end let us examine the result of the expressions $(\id\ot\Delta^\prime)$ and $(\Delta^\prime\ot\id)$
on a generic composition. Here we highlight these issues by considering a generic composition $w=a_1a_2$
with two parts only. We treat the general case $w=a_1\cdots a_n$ as part of the proof of Lemma <ref> below.
Hence consider applying $(\id\ot\Delta^\prime)$ and $(\Delta^\prime\ot\id)$ to the expression for $\Delta^\prime(a_1a_2)$
in Example <ref>. We observe that,

\begin{align*}
(\id\ot\Delta^\prime)\circ\Delta^\prime(a_1a_2)=&\;0\ot0\ot(a_1-2)a_2+0\ot(a_1-2)\ot a_2+0\ot(a_1-1)\ot(a_2-1)\\
\intertext{and}
(\Delta^\prime\ot\id)\circ\Delta^\prime(a_1a_2)=&\;0\ot(a_1-2)\ot a_2+(a_1-2)\ot0\ot a_2+0\ot(a_1-1)\ot(a_2-1)\\
\end{align*}

We notice that the two terms $0\ot0\ot(a_1-2)a_2$ and $a_1\ot0\ot(a_2-2)$ in the $(\id\ot\Delta^\prime)$ computation
do not appear in the $(\Delta^\prime\ot\id)$ computation, and vice-versa for the two terms $(a_1-2)\ot0\ot a_2$ and $a_1(a_2-2)\ot0\ot0$.
All four terms cannot ultimately contribute to $\Delta_2(a_1a_2)$. First, consider the term $0\ot0\ot(a_1-2)a_2$.
By the associativity of the Pöppe product, it does not matter in which order we compute $[0]\,[0]\,[(a_1-2)a_2]$.
If we compute the pair $[0][0]=2\cdot\{1\}$ first, then we see that the final triple product will generate a composition
with three parts. The same argument applies to $a_1(a_2-2)\ot0\ot0$. We use the map $\Theta_{\mathrm{L}}\colon{\mathcal C}^{\times3}\to\R$
to project out such terms (their image is the empty set), i.e. to project out terms that correspond to a triple
product that generates a composition whose length is too long. The `$\mathrm{L}$' label is for `length'.
Second, consider the term $a_1\ot0\ot(a_2-2)$. Again, by associativity of the Pöppe product, if we compute
the pair $[a_1]\,[0]$ first, then we see that the corresponding triple product generates a composition with the letter $(a_1+1)$,
which is the wrong original composition, or it generates a composition with the first two letters $a_11$ meaning that
the composition has too many parts. The same problem occurs with $(a_1-2)\ot0\ot a_2$. We use the map
$\Theta_{\mathrm{A}}\colon{\mathcal C}^{\times3}\to\R$ to project such terms to the empty set.
The `$\mathrm{A}$' label is for `associativity' as it originates from the associativity property of the
underlying Pöppe product and we wish to distingiush such terms from those projected out by $\Theta_{\mathrm{L}}$.
Under these two projections the two forms above coincide, both with themselves
and the definition for $\Delta_2$ in Definition <ref>. We summarise the result of this
discussion in the following lemma, whose full proof is given at the end of this section.

The co-product $\Delta_2$ is given by,

\begin{equation*}
\Delta_2\equiv(\Theta_{\mathrm{A}}\circ\Theta_{\mathrm{L}})\circ\bigl((\id\ot\Delta^\prime)\circ\Delta^\prime\bigr)
\equiv(\Theta_{\mathrm{A}}\circ\Theta_{\mathrm{L}})\circ\bigl((\Delta^\prime\ot\id)\circ\Delta^\prime\bigr).
\end{equation*}

We are now in a position where we can compute all the higher co-products $\Delta_n$ by successively
applying the formulae in Lemmas <ref> and <ref>, and in principle,
to outline a modified co-algebra structure. The former task is our goal here in order to elucidate the
which monomials of signature expansions in any Pöppe polynomial contribute to a given basis element.
In the following, in a suitable sense, we successively apply
to given compositions. We thereby retain the terms that might otherwise be projected out by $\Theta_{\mathrm{A}}$.
We choose to keep these for bookkeeping purposes only, which is useful as we see presently.
We can apply the projector $\Theta_{\mathrm{A}}$ a-posteriori if we choose; indeed we remark on this
at various junctures.

The remainder of this section is devoted to computing the co-products, modulo $\Theta_{\mathrm{A}}$,
of compositions of of lengths $1$, $2$ and $3$, via three examples. The insights gained corroborate the analysis
involving the leading order terms in the proof of our main result Theorem <ref> in Section <ref>.
For covenience, we use the notation $\Delta_0$ to simply denote the identity endomorphism.

[One-part compositions]
In Example <ref>, we saw that $\Delta(n)=0\ot(n-1)+(n-1)\ot0$.
Hence we observe, $\Delta_0(n)=n$ and $\Delta_2(n)=0\ot(n-2)\ot0$.
Thus, the only odd-degree generators of the one-part composition $n\in\mathcal C$ are
$[n]$ and $[0]\,[n-2]\,[0]$.

[Two-part compositions]
Consider a generic two-part composition of the form `$(n-m)m$'.
Naturally we have $\Delta_0\bigl((n-m)m\bigr)=(n-m)m$. Assume that $m\geqslant2$.
From our discussion preceding Lemma <ref>, we observe,

\begin{align*}
\Delta_2^\prime\bigl((n&-m)m\bigr)\\
&\;+(n-m-1)\ot(m-1)\ot0+(n-m-1)\ot0\ot(m-1)+0\ot(n-m-2)\ot m\\
\end{align*}

Since $(n-m-1)(m-1)$ is a composition of $n-2$, the term $0\ot(n-m-1)(m-1)\ot0$ corresponds to
a term generated by $[0]\,[n-2]\,[0]$, from Example <ref>, here involving the corresponding two-part composition.
See for example the third column in Table <ref>.
Since our main goal presently is to enumerate possible generators of composition components of the form `$(n-m)m$',
we ignore this term in $\Delta_2^\prime\bigl((n-m)m\bigr)$ hereafter. It corresponds to a
generator we have already come across, i.e. it does not correspond to a “new” generator.
Further we observe that on the right, the first line can be obtained from the second line by the replacement $m\to m-1$.
In other words, for a given $m$, the second line indicates the generators
$[n-m-1]\,[m-1]\,[0]$, $[n-m-1]\,[0]\,[m-1]$ and $[0]\,[n-m-2]\,[m]$, while the
generators $[n-m]\,[m-2]\,[0]$, $[n-m]\,[0]\,[m-2]$ and $[0]\,[n-m-1]\,[m-1]$
indicated by the first line, would have already been used/appeared at level $m-1$—we would have already come across them.
Note that the term $(n-m)\ot0\ot(m-2)$ would actually be projected out by $\Theta_{\mathrm{A}}$.
We thus retain only the “new” terms, i.e. those involving “new” generators, namely those on the second line and set,

\begin{equation*}
\Delta_2^\prime\bigl((n-m)m\bigr)_{\mathrm{new}}\!\coloneqq(n-m-1)\ot(m-1)\ot0+(n-m-1)\ot0\ot(m-1)+0\ot(n-m-2)\ot m.
\end{equation*}

Note that this also predicts the generators for the case when $m=1$ for which it collapses to $2\cdot (n-2)\ot0\ot0+0\ot(n-3)\ot1$.
Our coarse-grained approach captures the appropriate generators in the context of the algebra $\mathbb C[\mathbb Z_{\bs0}]$.
Indeed, comparison with all the $m\geqslant1$ cases in Tables <ref>-<ref>, and more generally with
Example <ref> in Step 5 in Section <ref>, confirms that all the degree three generators
corresponding to the basis element $[n(n-m)\times\bs\varphi]$ are parameterised by the forms in $\Delta_2^\prime\bigl((n-m)m\bigr)_{\mathrm{new}}$.
Next, applying $\Theta_{\mathrm{L}}\circ(\id^{\ot2}\ot\Delta^\prime)$ to $\Delta_2^\prime\bigl((n-m)m\bigr)_{\mathrm{new}}$ generates,

\begin{align*}
\Delta_3^\prime\bigl((n-m)m\bigr)_{\mathrm{new}}\coloneqq&\;(n-m)\ot0\ot(m-3)\ot0+0\ot(n-m-1)\ot(m-2)\ot0\\
\end{align*}

Lastly, applying $\Theta_{\mathrm{L}}\circ(\id^{\ot3}\ot\Delta^\prime)$ to this expression
generates the pair of terms given by the form
Since the first term is obtained from the second term by the replacement $m\to m-1$, we set,

\begin{equation*}
\Delta_4^\prime\bigl((n-m)m\bigr)_{\mathrm{new}}\coloneqq0\ot(n-m-2)\ot0\ot(m-2)\ot0.
\end{equation*}

Hence the only remaining possible “new” generator of the basis element with composition $(n-m)m$ is $[0]\,[n-m-2]\,[0]\,[m-2]\,[0]$.
Correspondingly, in the context of the algebra $\mathbb C[\mathbb Z_{\bs0}]$, we observe in Tables <ref>-<ref>,
and more generally in Example <ref>, the final degree five generator of the  basis element $[n(n-m)\times\bs\varphi]$ is represented by

[Three-part compositions]
For the three-part composition `$(n-m)(m-\ell)\ell$', by first computing $\Delta^\prime\bigl((n-m)(m-\ell)\ell\bigr)$
and then applying $(\id\ot\Delta^\prime)$ to the result, we observe,

\begin{align*}
\Delta_2^\prime\bigl((n-m)(m-\ell)\ell\bigr)
\end{align*}

We observe that the terms in the first through third, and sixth through to the eight, lines,
are not “new”. They involve terms which factors are just compositions of the corresponding
terms in $\Delta_2^\prime\bigl((n-m)m\bigr)$ in Example <ref>.
In the remaining third and fourth lines, the third line can be obtained from the fourth
by the replacement $m\to m-1$, so the terms in the third line are not “new”.
Then in the the fourth line, the first term is obtained from the second by the replacement
$\ell\to\ell-1$. So only the second term $(n-m-1)\ot(m-\ell-1)\ot\ell$ is “new” which
corresponds to the generator $[n-m-1]\,[m-\ell-1]\,[\ell]$. This corresponds to the
degree three generator of the basis element $[(n-m)(m-\ell)\ell\times\bs\varphi]$
in Tables <ref> and <ref>, and
Example <ref> in Step 5 in Section <ref>.
Briefly, we can proceed further as follows. Computing $\Delta_4^\prime\bigl((n-m)(m-\ell)\ell\bigr)$
as in Example  <ref> and focusing on the “new” terms, we generate the six quintic generators
for basis elements of the form $[(n-m)(m-\ell)\ell\times\bs\varphi]$. These correspond
to the six quintic generators indicated in Example <ref>.
Then further, by applying $(\id^{\ot4}\times\Delta^\prime)$ to the $\Delta_4^\prime$ computation just mentioned,
and then applying $(\id^{\ot5}\times\Delta^\prime)$ to that, we generate $\Delta_6^\prime\bigl((n-m)(m-\ell)\ell\bigr)$.
Focusing on the “new” terms gives $0\ot(n-m-2)\ot0\ot(m-\ell-2)\ot0\ot(\ell-2)\ot0$.
This corresponds to the septic generator in Example <ref>.

In our conclusions Section <ref>, we indicate how we can develop and utilise a more refined version
of this co-algebra structure further. We finish this section with a proof of Lemma <ref>.

Consider the application of $(\id\ot\Delta^\prime)$ and $(\Delta^\prime\ot\id)$ to the expression
for $\Delta^\prime(w)$ in Lemma <ref>. We observe that,

\begin{align*}
(\id&\ot\Delta^\prime)\circ\Delta^\prime(a_1\cdots a_n)\\
=&\;0\ot\biggl(0\ot(a_1-2)a_2\cdots a_n+\sum_{k=1}^{n-1}\Bigl((a_1-1)a_2\cdots a_{k-1}(a_k-1)\ot a_{k+1}\cdots a_n\\
&\;\qquad\qquad+(a_1-1)a_2\cdots a_k\ot(a_{k+1}-1)a_{k+2}\cdots a_n\Bigr)+(a_1-1)a_2\cdots a_{n-1}(a_n-1)\ot0\biggr)\\
&\;+\sum_{k=1}^{n-1}a_1\cdots a_{k-1}(a_k-1)\ot\biggl(0\ot(a_{k+1}-1)a_{k+2}\cdots a_n\\
&\;\qquad\qquad+\sum_{\ell=k+1}^{n-1}\Bigl(a_{k+1}\cdots a_{\ell-1}(a_\ell-1)\ot a_{\ell+1}\cdots a_n\\
&\;\qquad\qquad\qquad\qquad+a_{k+1}\cdots a_\ell\ot(a_{\ell+1}-1)a_{\ell+2}\cdots a_n\Bigr)+a_{k+1}\cdots a_{n-1}(a_n-1)\ot0\biggr)\\
&\;+\sum_{k=1}^{n-1}a_1\cdots a_k\ot\biggl(0\ot(a_{k+1}-2)a_{k+2}\cdots a_n\\
&\;\qquad+\sum_{\ell=k+1}^{n-1}\Bigl((a_{k+1}-1)a_{k+2}\cdots a_{\ell-1}(a_\ell-1)\ot a_{\ell+1}\cdots a_n\\
&\;\qquad\qquad\qquad\qquad\qquad+(a_{k+1}-1)a_{k+2}\cdots a_\ell\ot(a_{\ell+1}-1)a_{\ell+2}\cdots a_n\Bigr)\\
&\;\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+(a_{k+1}-1)a_{k+2}\cdots a_{n-1}(a_n-1)\ot0\biggr).
\end{align*}

In the expression above, we use the convention that when $k=1$ in the first line on the right,
the term $(a_1-1)a_2\cdots a_{k-1}(a_k-1)\ot a_{k+1}\cdots a_n$ is to be interpreted as $(a_1-2)\ot a_2\cdots a_n$,
and for $\ell=k+1$ the terms $(a_{k+1}-1)a_{k+2}\cdots a_{\ell-1}(a_\ell-1)\ot a_{\ell+1}\cdots a_n$ are to be interpreted
as $(a_{k+1}-2)\ot a_{k+2}\cdots a_n$. We observe that if we apply the projectors $\Theta_{\mathrm{L}}$
and $\Theta_{\mathrm{A}}$ to the expression above, then the terms $0\ot0\ot(a_1-2)a_2\cdots a_n$
and $a_1\cdots a_k\ot0\ot(a_{k+1}-2)a_{k+2}\cdots a_n$ for $k=1,\cdots,n-1$, would be projected out.
Next we observe that,

\begin{align*}
(\Delta^\prime&\ot\id)\circ\Delta^\prime(a_1\cdots a_n)\\
=&\;\sum_{k=1}^{n-1}\biggl(0\ot(a_1-1)a_2\cdots a_{k-1}(a_k-1)+\sum_{\ell=1}^{k-1}\Bigl(a_1\cdots a_{\ell-1}(a_\ell-1)\ot a_{\ell+1}\cdots a_{k-1}(a_k-1)\\
&\;\quad+a_1\cdots a_\ell\ot(a_{\ell+1}-1)a_{\ell+2}\cdots a_{k-1}(a_k-1)\Bigr)+a_1\cdots a_{k-1}(a_k-2)\ot 0\biggr)\ot a_{k+1}\cdots a_n\\
&\;+\sum_{k=1}^{n-1}\biggl(0\ot(a_1-1)a_2\cdots a_k+\sum_{\ell=1}^{k-1}\Bigl(a_1\cdots a_{\ell-1}(a_\ell-1)\ot a_{\ell+1}\cdots a_k\\
&\;\quad+a_1\cdots a_\ell\ot(a_{\ell+1}-1)a_{\ell+2}\cdots a_k\Bigr)+a_1\cdots a_{k-1}(a_k-1)\ot0\biggr)\ot(a_{k+1}-1)a_{k+2}\cdots a_n\\
&\;+\biggl(0\ot(a_1-1)a_2\cdots a_{n-1}(a_n-1)+\sum_{k=1}^{n-1}\Bigl(a_1\cdots a_{k-1}(a_k-1)\ot a_{k+1}\cdots a_{n-1}(a_n-1)\\
&\;\qquad\qquad+a_1\cdots a_k\ot(a_{k+1}-1)a_{k+2}\cdots a_{n-1}(a_n-1)\Bigr)+a_1\cdots a_{n-1}(a_n-2)\ot0\biggr)\ot0.
\end{align*}

We use the convention that when $\ell=k-1$ the terms $a_1\cdots a_\ell\ot(a_{\ell+1}-1)a_{\ell+2}\cdots a_{k-1}(a_k-1)$ are to be interpreted as
$a_1\cdots a_{k-1}\ot(a_k-2)$ and when $k=n-1$ the term $a_1\cdots a_k\ot(a_{k+1}-1)a_{k+2}\cdots a_{n-1}(a_n-1)$ is to be interpreted as $a_1\cdots a_{n-1}\ot(a_n-2)$.
Further, if we apply the projectors $\Theta_{\mathrm{L}}$ and $\Theta_{\mathrm{A}}$ to the expression just above, then the terms
$a_1\cdots a_{k-1}(a_k-2)\ot 0\ot a_{k+1}\cdots a_n$ for $k=1,\cdots,n-1$ and $a_1\cdots a_{n-1}(a_n-2)\ot0\ot0$ are projected out.
Now it comes down to comparing the expressions for $(\id\ot\Delta^\prime)\circ\Delta^\prime(a_1\cdots a_n)$
and $(\Delta^\prime\ot\id)\circ\Delta^\prime(a_1\cdots a_n)$ above. It is easy to check all the terms.
Indeed we see that all the terms starting with a `$0$' factor in the latter expression correspond exactly
to the terms in the first two lines of the former expression. The terms in both expressions involving a `$0$' factor
in the middle are straightforwardly matched up, as are all the terms ending in a `$0$' factor. All that remains
are the double sum terms in each expression. By considering the latter expression and reversing the order of the
summations (for $k$ and $\ell$) and then exchanging the dummy variable labels $k$ and $\ell$, the double sum
expressions in $(\Delta^\prime\ot\id)\circ\Delta^\prime(a_1\cdots a_n)$ are straightforwardly matched to those in
$(\id\ot\Delta^\prime)\circ\Delta^\prime(a_1\cdots a_n)$. The proof is complete.