text:
\section{Model-theoretic background}\label{sect:mt}

Let $\str M$ be an arbitrary (first-order) structure; its
theory is
\myboxed{\Th{\str M}}.
Let $T$ be an
arbitrary consistent (first-order) theory; its models compose the
class
\myboxed{\Mod T}.
Every class $\bm K$ of structures in some signature has a theory,
\myboxed{\Th{\bm K}}.  Then $\bm K\included\Mod{\Th{\bm K}}$; in case of
equality, $\bm K$ is \defn{elementary}.   Always, $\Th{\Mod T}=T$.

The structure $\str M$ has the universe $M$.  The structure denoted by
$\str M_M$ is the expansion of $\str M$ that has a name for every
element of $M$.
Then $\str M$ embeds in $\str N$ if and only if $\str M_M$ embeds in
an expansion of $\str N$.  However, although the class of
structures in
which $\str M$ embeds need not be elementary, the class of
structures in which $\str M_M$ embeds \emph{is} elementary.  The
theory of the latter class is the \defn{diagram}{} of $\str M$, or
\myboxed{\diag{\str M}}: it is
axiomatized by the quantifier-free sentences in $\Th{\str M_M}$
\cite[Thm 2.1.3, p.~24]{MR0153570}.
The class of structures in which $\str M_M$
embeds \emph{elementarily}
is also elementary, and its theory is just \myboxed{\Th{\str M_M}}.
The class of
substructures of models of $T$ is elementary, and its theory is
denoted by \myboxed{T_{\forall}}: this is axiomatized by the universal
sentences of $T$ \cite[Thm 3.3.2, p.~71]{MR0153570}.

By a \defn{system over}{} $\str M$, I mean a finite conjunction of
atomic and negated atomic formulas in the signature of $\str M_M$;
likewise, a system \defn{over}{} $T$ is in the signature
of $T$.  A structure $\str M$ \defn{solves}{} a system $\phi(\tuple x)$
if $\str M\models\Exists{\tuple x}\phi(\tuple x)$.  Note well here that
$\tuple x$, in boldface, is a \emph{tuple} of variables, perhaps
$(x^0,\dots,x^{n-1})$.  By an
\defn{extension}{} of a model of $T$, I mean another model of $T$ of
which the first is a substructure.  Two systems over a model $\str M$
of $T$ are \defn{equivalent}{} if they are soluble in the same extensions.

An \defn{existentially closed}{} model of $T$ is a model of $T$ that
solves every system over
itself that is soluble in some extension.  So a model $\str M$ of $T$ is
existentially closed if and only if
$T\cup\diag{\str M}\vdash\Th{\str M_M}_{\forall}$, that is, every
extension of a model of $T$ is a substructure of an elementary extension
  (\cite[\S~7]{MR0277372} or \cite[\S~2]{MR51:12518}).

A theory is \defn{model-complete}{} if its every
model is existentially closed.  An equivalent formulation explains the
name: $T$ is model-complete if and only if $T\cup\diag{\str M}$ is
complete whenever $\str M\models T$
\cite[Ch.~2]{MR0472504}.

Suppose every model of $T$ has an existentially closed extension.
Such is the case when $T$ is \defn{inductive}, that is, $\Mod{T}$ is
closed under unions of chains \cite[Thm~7.12]{MR0277372}: equivalently,
$T=T_{\forall\exists}$ \cite{MR0089813,MR0103812}.  Suppose further
that we have a uniform
first-order way to tell when systems over models of $T$ are soluble in
extensions: more precisely, suppose there is a function
\begin{equation}\label{eqn:fn}
\phi(\tuple x,\tuple y)\longmapsto
\robfn{\phi}(\tuple x,\tuple y),
\end{equation}
where
$\phi(\tuple x,\tuple y)$ ranges over the systems over $T$ (with
variables analyzed as shown),
such that, for every model $\str
M$ of $T$ and every tuple $\tuple a$ of parameters from $M$, the
system $\phi(\tuple x,\tuple a)$ is soluble in some extension of $\str
M$ just in case $\robfn{\phi}(\tuple x,\tuple a)$ is soluble in $\str M$.
Then the existentially closed
models of $T$ compose an elementary class, whose theory $T^*$ is
axiomatized by $T$ together with the sentences
\begin{equation}\label{eqn:sen}
  \Forall{\tuple y}(\Exists{\tuple x}\robfn{\phi}(\tuple x,\tuple
  y)\to
\Exists{\tuple x}\phi(\tuple x,\tuple y)).
\end{equation}
Immediately, $T^*$ is model-complete, so $T^*\cup\diag{\str M}$ is
complete when $\str M\models T^*$.  What is more,
$T^*\cup\diag{\str M}$ is complete whenever $\str M\models
T$ \cite[Thm 5.5.1]{MR0153570}.

In general, $T^*$ is a \defn{model-completion}{} of $T$ if
$T^*{}_{\forall}\included T\included T^*$
and
$T^*\cup\diag{\str M}$ is
complete whenever $\str M\models T$.  Model-completions are unique
\cite[(2.8)]{MR0091922}.   We have sketched the proof of part of the
following (the rest is \cite[(3.5)]{MR0091922}):

\begin{lemma}[Robinson's Criterion]
\label{lem:rob-crit}
  Let $T$ be inductive.  Then $T$ has a model-comple\-tion if and
  only if a function $\phi(\tuple x,\tuple
  y)\mapsto\robfn{\phi}(\tuple x,\tuple y)$ exists as
  in~\eqref{eqn:fn}.
In this case, the model-completion is axiomatized \emph{modulo} $T$ by
  the sentences in~\eqref{eqn:sen}.
\end{lemma}

If $T_{\forall}=T^*{}_{\forall}$ and $T^*$ is model-complete, then
$T^*$ is a \defn{model-companion}{} of~$T$
(\cite[\S~5]{MR0272613}; \emph{cf.}~\cite[\S~2]{MR0277372}).
Model-completions are model-companions, and model-companions are
unique \cite[Thm 5.3]{MR0272613}.
If $T$ has a
model-companion, then its models are just the existentially closed
models of $T$ \cite[Prop.~7.10]{MR0277372}.
Conversely, if $T$ is inductive,
and the class of existentially closed models
of $T$ is elementary, then the theory of this class is the
model-companion of $T$ \cite[Cor.~7.13]{MR0277372}.

\section{Fields with one derivation}\label{sect:one}

Let \myboxed{\DF}{} be the theory of fields with a derivation, and let
\myboxed{\DPF}{} be
the theory of models of $\DF$ that, for each prime $\ell$, satisfy also
\begin{equation*}
  \Forall x\Exists y(\underbrace{1+\dotsb+1}_{\ell}=0\land
  Dx=0\to\underbrace{y\dotsb y}_{\ell}=x).
\end{equation*}
So models of $\DPF$ are \defn{differentially perfect}.  A
subscript on the name of one of these theories will indicate a
required characteristic for the field.  In particular, we have $\DPF_0$,
which is the same as $\DF_0$.

Abraham Seidenberg
\cite{MR0082487} shows the existence of the function
in Lemma~\ref{lem:rob-crit} in case
$T$ is $\DPF_p$, where $p$ is prime or $0$.  Consequently:

\begin{theorem}[Robinson]
  $\DF_0$ has a model-completion, called \myboxed{\DCF_0}.
\end{theorem}

\begin{theorem}[Wood \cite{MR48:8227}]
  If $p$ is prime, then $\DF_p$ has a model-companion,
  \myboxed{\DCF_p}, which is the model-completion of $\DPF_p$.
\end{theorem}

\subsection{Single variables}\label{subsect:one-var}

Since it involves \emph{all} systems over a given theory,
Robinson's criterion yields the crudest possible axiomatization for a
model-completion.  By contrast, though the theory \myboxed{\ACF}{} of
algebraically closed fields is the model-completion of the theory of
fields, its axioms (\emph{modulo} the latter theory) can involve only
systems in one variable (indeed, single equations in one variable).
A generalization of this observation is the following, which can be
extracted from the proof of \cite[Thm~17.2, pp.~89--91]{MR0398817}
(see also \cite{MR0491149}):

\begin{lemma}[Blum's Criterion]\label{lem:blum}
  Say $T^*{}_{\forall}\included T\included T^*$.
  \begin{enumerate}
    \item
The theory $T^*$ is the model-completion of $T$ if and only if the
  commutative diagram
  \begin{equation*}
    \xymatrix{
\str M &\\
\str A \ar[u] \ar[r] & \str B \ar@{.>}[ul]
}
  \end{equation*}
of structures and embeddings
can be completed as indicated when
$\str A$ and $\str B$ are models of $T$ and $\str M$ is a
$\size{B}^+$-saturated model of $T^*$.
\item\label{item:blum2}
If $T=T_{\forall}$, it is enough to assume that $\str B$ is generated
over $\str A$
by a single element.
  \end{enumerate}
\end{lemma}

This allows a refinement of Lemma~\ref{lem:rob-crit} in a special
case:

\begin{lemma}
  Suppose $T=T_{\forall}$.  Then Lemma~\ref{lem:rob-crit} still holds
  when $\phi(\tuple x,\tuple y)$ is replaced with
  $\phi(x,\tuple y)$ (where $x$ is a single variable).
\end{lemma}

From Lemma~\ref{lem:blum},
Lenore Blum obtains Theorem~\ref{thm:dcf-1} below in characteristic
$0$, in which case the first two numbered conditions amount to
$K\models\ACF$ (\cite[pp.~298~ff.]{MR0398817} or~\cite{MR0491149}).
If $p>0$, then $\DF_p$ is not universal, so
part~\eqref{item:blum2} of Blum's criterion does
\emph{not} apply; Carol Wood instead uses a
primitive-element theorem of Seidenberg \cite{MR0049174} to obtain new
axioms for $\DCF_p$ \cite{MR50:9577}.  These can be combined with
Blum's axioms for $\DCF_0$ to yield the following.  (Here \myboxed{\SCF}{} is
the theory of separably closed fields.)

\begin{theorem}[Blum, Wood]\label{thm:dcf-1}
A model $(K,D)$ of $\DF$ is existentially closed if and only if
\begin{enumerate}
\item
$K\models\SCF$;
  \item
$(K,D)\models\DPF$;
\item
$(K,D)\models\Exists x(f(x,Dx,\dots,D^{n+1}x)=0\land
g(x,Dx,\dots,D^nx)\neq0)$ whenever $f$ and $g$ are ordinary
polynomials over $K$ in tuples $(x^0,\dots, x^{n+1})$ and
$(x^0,\dots,x^n)$ of variables
respectively such that $g\neq0$ and
$\partial f/\partial x^{n+1}\neq0$.
\end{enumerate}
Hence $\DF$ has a model-companion, \myboxed{\DCF}.
\end{theorem}

There is a similar characterization of the existentially closed
\emph{ordered} differential fields \cite{MR495120}.

\subsection{First derivatives}\label{sect:first-der}

Alternative simplified axioms for $\DCF$ are parallel to
those found for the model-companion \myboxed{\ACFA}{} of the theory of fields
with an automorphism \cite{MR99c:03046,MR2000f:03109}.  Suppose
$(K,D)\models\DPF$ and $K\models\SCF$.
Every system
over $(K,D)$ can be written as
\begin{equation}\label{eqn:sys}
  \bigwedge_ff(\tuple x,D\tuple x,\dots,D^n\tuple x)=0\land g(\tuple
  x,D\tuple x,\dots,D^n\tuple x)\neq0,
\end{equation}
where $g$ and the $f$ are ordinary polynomials over $K$.  This system is
equivalent to one that involves only first derivatives, namely
\begin{multline*}
  \bigwedge_ff(\tuple x_0,\dots,\tuple x_{n-1},\tuple y_{n-1})=0\land
g(\tuple x_0,\dots,\tuple x_{n-1},\tuple y_{n-1})\neq0\land\\
{}\land\bigwedge_{i+1<n}\tuple x_{i+1}=\tuple y_i\land
\bigwedge_{i<n}D\tuple x_i=\tuple y_i.
\end{multline*}
This system, in form, is a special case of the system
\begin{equation*}
  \bigwedge_ff(\tuple x,\tuple y)=0\land g(\tuple x,\tuple
  y)\neq0\land D\tuple x=\tuple y.
\end{equation*}
Suppose the latter system has the solution $(\tuple a,\tuple b)$.
Then $K(\tuple a,\tuple b)/K$ is separable
\cite[Lem.~1.5, p.~1328]{MR2114160}.  Let
$V$ and $W$ be the varieties over $K$ with generic points $\tuple a$
and $(\tuple a,\tuple b)$ respectively, let \myboxed{T_D(V)}{} be the
twisted tangent bundle of $V$, and let $U$ be the open subset of $W$
determined by the inequation $g\neq0$; then the situation can be
depicted thus:
\begin{equation*}
\SelectTips{cm}{}
\newdir{ >}{{}*!/-5pt/@{>}}
  \xymatrix{
U \ar@{ >->}[r] \ar@{-->>}[dr]
 & W \ar@{ >->}[r] \ar@{-->>}[d] & T_D(V) \ar@{-->>}[dl]^{(\tuple
  x,\tuple y)\mapsto\tuple x}\\
& V&
}
\end{equation*}
In characteristic $0$, the model $(K,D)$ of $\DF$ is existentially
closed if and only if, in every such geometric situation, $U$ contains
a $K$-rational point $(\tuple c,D\tuple c)$; this yields the so-called
geometric axioms for $\DCF_0$ found with Anand Pillay
\cite{MR99g:12006}.  In positive characteristic, it is still true
that, if $(\tuple a,\tuple b)$ is a generic point of $V$, then $D$
extends to $K(\tuple a)$ so that $D\tuple a=\tuple b$.  However, an
additional condition is needed to ensure that $D$ extends to all of
$K(\tuple a,\tuple b)$; it is enough to require that the projection of
$T_D(W)$ onto $T_D(V)$ contain a generic point of $W$; this yields
Piotr Kowalski's geometric axioms for $\DCF_p$ \cite{MR2119125}.

In an alternative geometric approach to $\DCF$, instead
of~\eqref{eqn:sys}, it is enough to look
at an arbitrary system of \emph{equations,}
$\bigwedge_ff(\tuple x,D\tuple x,\dots,D^n\tuple x)=0$.
We can take \emph{all} of the derivations out of the polynomials,
getting the equivalent system
\begin{equation*}
  \bigwedge_ff(\tuple x_0,\dots,\tuple
  x_n)=0\land\bigwedge_{i<n}D\tuple x_i=\tuple x_{i+1}.
\end{equation*}
This is a special case of
\begin{equation}\label{eqn:1-my}
  \bigwedge_ff(x^0,\dots,
  x^{n-1})=0\land\bigwedge_{i<k}Dx^i=g^i(x^0,\dots,x^{n-1}),
\end{equation}
where $k\leq n$ and the $g^i$ are rational functions over $K$.
Suppose this has solution $\tuple a$, which is a generic point of
$V$.  It is enough to assume that
$(a^0,\dots,a^{k-1})$ is a separating tran\-scendence-basis of $K(\tuple
a)/K$.  Then we have a dominant, separable rational map $\tuple
x\mapsto(x^0,\dots,x^{k-1})$ or $\phi$ from $V$ onto $\Aff^k$,
and another rational map $\tuple x\mapsto(g^0(\tuple x),\dots,g^{k-1}(\tuple
x))$ or $\psi$ from $V$ to $\Aff^k$.  So $(K,D)$ is existentially
closed if
and only if $V$ always has a $K$-rational point $P$ such that
$D(\phi(P))=\psi(P)$ \cite[Thm~1.6, p.~1328]{MR2114160}.

\section{Fields with several derivations}\label{sect:several}

Let \myboxed{\mDF}{} be the theory of fields with $m$ commuting derivations.
Tracey McGrail \cite{MR2001h:03066} axiomatizes the model-completion,
\myboxed{\mDCF_0}, of $\mDF_0$.  Alternative axiomatizations arise as
special cases in
work of Yoav Yaffe \cite{MR1807840} and Marcus Tressl
\cite{MR2159694}.  There is a common theme:  A differential ideal has
a generating set of a special form; in the terminology of Joseph Ritt
\cite[\S~I.5, p.~5]{MR0201431} (when $m=1$) and Ellis Kolchin
\cite[\S~I.10, pp.~81~ff.]{MR58:27929}, this is a \emph{characteristic set.}
There is a first-order way to tell, uniformly in the parameters,
whether a given set of differential
polynomials is a characteristic set of some differential ideal, and
then to tell, if it \emph{is}
a characteristic set, whether it has a root.  In short, the function
$\phi\mapsto\robfn{\phi}$ in Robinson's criterion
(Lemma~\ref{lem:rob-crit}) is defined for
sufficiently many systems $\phi$.  (Applying Blum's criterion, McGrail
and Yaffe consider only systems in one variable, so they must include
inequations in these systems; Tressl uses only equations, in
arbitrarily many variables.)

I do not give the definition of a characteristic set, as not all
ingredients of the definition are needed for the arguments presented
in \S~\ref{sect:resolution}.  However, some of the ingredients
\emph{are} needed; these are in~\ref{subsect:terms}.

\subsection{Differential forms}\label{subsect:fga}

In \cite{MR2000487} I attempted to apply the geometric approach
described in~\ref{sect:first-der} to $\mDF_0$.  I worked more
generally with $\DF^m_0$, where \myboxed{\DF^m}{} is the theory of structures
$(K,D_0,\dots,D_{m-1})$ such that $(K,D_i)\models\DF$ for each $i$, and
each bracket $[D_j,D_k]$ is a $K$-linear combination of the $D_i$.
(This is roughly what Yaffe did too.)
In
\cite[\S~2]{MR2114160} I made some minor corrections and otherwise adapted
the argument to arbitrary characteristic.  Nonetheless, in May, 2006,
Ehud Hrushovski showed me a counterexample to \cite[Thm~A,
  p.~926]{MR2000487}, a theorem that was an introductory formulation of
 \cite[Thm~5.7, p.~942]{MR2000487}.  Then I found an error at
 the end of the proof of the latter theorem.  Before offering a resolution of
 the problem, let me review the general situation.

Let $(K,D_0,\dots,D_{m-1})\models\DF^m$, and let $\lrv$ be the
$K$-linear span of the $D_i$.  Then $\lrv$ is a Lie-ring, as well as a
vector-space over $K$.  As a vector-space, $\lrv$ has a dual,
\myboxed{\lrv^*}; and there is a derivation \myboxed{\dee}{} from $K$ into
$\lrv^*$ given by
\begin{equation}\label{eqn:dee-def}
  D(\dee x)=Dx.
\end{equation}
Then $\lrv^*$ has a basis $(\dee t^i\colon i<\ell)$ for some $t^i$ in
$K$ and some $\ell$ no greater than $m$ \cite[Lem.~4.4,
  p.~932]{MR2000487}, and this basis is dual to a basis
$(\partial_i\colon i<\ell)$ of $\lrv$, where
$[\partial_i,\partial_j]=0$ in each case, and $\dee$ is then given by
\begin{equation}\label{eqn:dee}
  \dee x=\sum_{i<\ell}\dee t^i\cdot\partial_ix
\end{equation}
\cite[Lem.~4.7, p.~934]{MR2000487}.
We can use these ideas to find an extension of $(K,D_0,\dots,D_{m-1})$
in which the named derivations are linearly independent over the field.
Indeed, if $\ell<m$, then let $L=K(\alpha^{\ell},\dots,\alpha^{m-1})$,
where
$(\alpha^{\ell},\dots,\alpha^{m-1})$ is algebraically
independent over $K$.
Extend the original $\partial_i$ to $L$ so that they are $0$ at the $\alpha^j$;
then, if $\ell\leq k<m$, define $\partial_k$ to be $0$ on $K$ and to be
$\deltaup_k^j$ at $\alpha^j$.  Then $(\partial_i\colon i<m)$ is a
linearly independent $m$-tuple of commuting derivations on $L$; from this, we
obtain linearly independent extensions of the $D_i$ to $L$ such that
the brackets $[D_j,D_k]$ are the same linear combinations of the $D_i$
as before (\cite[Lem.~5.2, p.~937]{MR2000487}---or \cite[Lem.~2.1,
  p.~1930]{MR495120}, by a different method---in characteristic $0$;
generally,  \cite[Lem.~2.4, p.~1334]{MR2114160}).

Consequently, a model-companion $\DCF^m$ for $\DF^m$ can be obtained from a
model-com\-panion $\mDCF$ for $\mDF$ (\cite[Thm~5.3 and proof]{MR2000487} or
\cite[\S~3]{MR495120}).  This much stands, and differential forms
are convenient for establishing it.

But I tried also to obtain $\DCF^m$
independently as follows.  Suppose now we have
a separably closed field~$K$, along with a Lie-ring and
finite-dimensional space $\lrv$ of derivations
of $K$; as a space, $\lrv$ has a basis $(\partial_i\colon i<m)$, whose
dual is $(\dee t^i\colon i<m)$, so that the $\partial_i$ commute.  We
may assume  that
$(K,\partial_0,\dots,\partial_{m-1})$ is differentiably perfect
\cite[Lem.~2.4]{MR2114160}.
Every system over $(K,\partial_0,\dots,\partial_{m-1})$
 is equivalent to a system of the
form of~\eqref{eqn:1-my}, generalized to
\begin{equation}\label{eqn:m-my}
  \bigwedge_ff(x^0,\dots,x^{n-1})=0\land
  \bigwedge_{j<k}\bigwedge_{i<m}\partial_ix^j=g_i^j(\tuple x).
\end{equation}
By means of~\eqref{eqn:dee}, we can also write this as
\begin{equation}\label{eqn:many-my}
  \bigwedge_ff(x^0,\dots,x^{n-1})=0\land
  \bigwedge_{j<k}\dee x^j=\sum_{i<m}\dee t^i\cdot g_i^j(\tuple x).
\end{equation}
If $\tuple a$ is a solution (from some extension), it is enough to
assume that $(a^0,\dots,a^\ell)$ is a
separating transcendence-basis of $K(\tuple a)/K$ for some $\ell$ such
that $k\leq\ell<m$.  That we cannot generally assume $k=\ell$ is an important
difference from the case of one derivation; it is what causes the
difficulties in the case of several derivations.  The solution $\tuple a$
to~\eqref{eqn:many-my} can be
understood as follows.  First
we have the field $K(\tuple a)$, and then~\eqref{eqn:many-my} can be be
written as
\begin{equation}\label{eqn:step-1}
  \bigwedge_{j<k}\dee a^j=\sum_{i<m}\dee t^i\cdot g_i^j(\tuple a).
\end{equation}
A solution of this can be understood as a model
$(L,\tilde{\partial}_0,\dots,\tilde{\partial}_{m-1})$ of $\mDF$
extending $(K,\partial_0,\dots,\partial_{m-1})$ such that $K(\tuple
a)\included L$ and~\eqref{eqn:step-1} holds when $\dee
a^j=\sum_{i<m}\dee t^i\cdot\tilde{\partial}_ia^j$, that is,
\begin{equation}\label{eqn:jkim}
\bigwedge_{j<k}\bigwedge_{i<m}\tilde{\partial}_ia^j=g_i^j(\tuple a).
\end{equation}
Since the
$\tilde{\partial}_i$ commute, it is
necessary that
\begin{equation}\label{eqn:jkhim}
\bigwedge_{j<k}\bigwedge_{h<i<m}\tilde{\partial}_h(g_i^j(\tuple
a))=\tilde{\partial}_i(g_h^j(\tuple a))
\end{equation}
\cite[\S~1, p.~926]{MR2000487}.
Any derivative with respect to $\tilde{\partial}_i$ of an element of
$K(\tuple a)$ is a constant plus a linear combination
of the derivatives  $\tilde{\partial}_ia^j$, where $j<\ell$ (by
\cite[Fact~1.1 (0, 2)]{MR2114160}, for example); we know
what these derivatives $\tilde{\partial}_ia^j$ are when $j<k$,
by~\eqref{eqn:jkim};
so~\eqref{eqn:jkhim} becomes a linear system in the unknowns
$\tilde{\partial}_ia^j$ where $k\leq j<\ell$.

If $k=\ell$, then this linear system has no variables, so it is true
or false; its truth is a sufficient condition for~\eqref{eqn:step-1}
to have a solution.  If $k<\ell$, then the linear system is soluble or
not.  If it is soluble, then it is possible to extend the $\partial_i$
to derivations $\tilde{\partial}_i$ as required by~\eqref{eqn:jkim}
that commute on
$K(a^0,\dots,a^{k-1})$; but these derivations need not commute on all
of $K(\tuple a)$.  In \cite{MR2000487} I claimed that they could
commute, and that the solubility
of~\eqref{eqn:jkhim} was sufficient for solubility
of~\eqref{eqn:step-1} in the sense above.  I was wrong.

\subsection{A counterexample}
In the counterexample supplied by Hrushovski, the constants
$(m,k,\ell)$ of \S~\ref{subsect:fga} are $(2,2,3)$.
Let $(a,b,c)$ be an algebraically independent triple, and
consider the system
\begin{equation}\label{eqn:sys-ex}
  \dee a=\dee t^0\cdot c\cdot c+\dee t^1\cdot c,\qquad
\dee b=\dee t^0\cdot 2a+\dee t^1\cdot c;
\end{equation}
equivalently, by~\eqref{eqn:dee}, the system comprises the equations
  $\partial_0a= c\cdot c$, $\partial_1a=c$,
  $\partial_0b=2a$, and $\partial_1b=c$.
From these, we compute
\begin{equation*}
\partial_1\partial_0a=2c\cdot\partial_1c,\qquad
\partial_0\partial_1a=\partial_0c,\qquad
\partial_1\partial_0b=2\cdot\partial_1a=2c,\qquad
\partial_0\partial_1b=\partial_0c.
\end{equation*}
Equating $\partial_0\partial_1$ and $\partial_1\partial_0$ yields the
linear system
\begin{equation*}
  \partial_0c-2c\cdot\partial_1c=0,\qquad
  \partial_0c=2c,
\end{equation*}
which has the solution $\partial_0c=2c$, $\partial_1c=1$.  But then we
must have
$\partial_1\partial_0c=2\cdot\partial_1c=2$, while
$\partial_0\partial_1c=\partial_01=0$, which means~\eqref{eqn:sys-ex}
has no solution, contrary to my claim in~\cite{MR2000487}.

For the record, the mistake is at the end of the proof of
\cite[Thm~5.7, p.~942]{MR2000487} and can be seen as follows.
Write the system~\eqref{eqn:sys-ex} as
$\dee a=\alpha$, $\dee b=\beta$; then
\begin{gather}\label{eqn:dab}
\begin{aligned}
  \dee\alpha
&=\dee(\dee t^0\cdot c\cdot c+\dee t^1\cdot c)\\
&=\dee c\wedge\dee t^0\cdot 2c+\dee c\wedge\dee t^1\\
&=\dee c\wedge(\dee t^0\cdot 2c+\dee t^1),\\
\end{aligned}
\quad
\begin{aligned}
\dee\beta
&=\dee(\dee t^0\cdot 2a+\dee t^1\cdot c)\\
&=\dee a\wedge\dee t^0\cdot 2+\dee c\wedge\dee t^1\\
&=(\dee t^0\cdot
c\cdot c+\dee t^1\cdot c)
\wedge\dee t^0\cdot 2+\dee c\wedge\dee t^1\\
&=-\dee t^0\wedge\dee t^1\cdot 2c+\dee c\wedge\dee t^1\\
&=(\dee c-\dee t^0\cdot 2c)\wedge\dee t^1.
\end{aligned}
\end{gather}
Since also $\dee\beta=\dee^2b=0$,
\eqref{eqn:dab} imposes a condition on $\dee c\wedge\dee
t^1$, hence on $\partial_0c$; in particular, $\partial_0c=2c$, which
is what we found above.  But
there is no apparent
condition on $\partial_1c$, so I try introducing a new transcendental,
$d$, for this.  Then $\dee c=\dee t^0\cdot
2c+\dee t^1\cdot d$.  But this causes a problem, since it allows a
substitution in~\eqref{eqn:dab}, yielding
\begin{align*}
  \dee\alpha
&=(\dee t^0\cdot 2c+\dee t^1\cdot d)\wedge
(\dee t^0\cdot 2c+\dee t^1)\\
&=\dee t^0\wedge\dee t^1\cdot 2c(1-d),
\end{align*}
which means $d=1$, contrary to
assumption.  In short, the next to last sentence of the proof of
 \cite[Thm~5.7]{MR2000487} (beginning ``This ideal is linearly disjoint
 from'') is simply wrong.  (I had given no argument that it was
 correct.)

\section{Resolution}\label{sect:resolution}

To resolve the problem, it is better not to introduce differential
forms from the beginning, but to allow equations to involve any number
of applications of the derivations.  Examples (as
in~\ref{subsect:another}) may involve only one variable; but in
contrast to~\ref{subsect:one-var}, there does not seem to be an
advantage in restricting attention to this case.

\subsection{Another example}\label{subsect:another}

Over a
differential field $(K,\partial_0,\partial_1)$, consider the system
\begin{equation}\label{eqn:my-ex}
  \partial_0{}^n\partial_1{}^nx-x=0,\qquad
\partial_0{}^n\partial_1x-\partial_1{}^nx=0,
\end{equation}
where $n\geq2$.
Think of the derivatives $\partial_0{}^i\partial_1{}^jx$ as matrix
entries:
\begin{equation*}
  \begin{matrix}
    x&\partial_1x&\partial_1{}^2x&\dots\\
\partial_0x&\partial_0\partial_1x&\partial_0\partial_1{}^2x&\dots\\
\partial_0{}^2x&\partial_0{}^2\partial_1x&\partial_0{}^2\partial_1{}^2x&\dots\\
\hdotsfor{4}
  \end{matrix}
\end{equation*}
Treat each derivative  $\partial_0{}^i\partial_1{}^jx$ as a new
variable, $x^{(i,j)}$.  In case $n=3$, the system~\eqref{eqn:my-ex}
determines a $4\times4$ matrix $(a^{(i,j)})^{i<4}_{j<4}$ in which
$a^{(3,3)}=a^{(0,0)}$ and $a^{(3,1)}=a^{(0,3)}$; we can depict this as
follows:
\begin{equation}\label{eqn:matrix}
 \begin{matrix}
    a&*&*&b\\
    *&*&*&*\\
    *&*&*&*\\
    *&b&*&a
  \end{matrix}
\end{equation}
We first ask whether the derivations can be
extended to commuting derivations on the field
$K(a^{(i,j)}\colon(i,j)\leq(3,3))$, where $\leq$ is
the product order on $\vnn^2$, so that
$\partial_0a^{(i,j)}=a^{(i+1,j)}$ when $(i+1,j)\leq(3,3)$ and
$\partial_1a^{(i,j)}=a^{(i,j+1)}$ when $(i,j+1)\leq(3,3)$.
That is, writing $a^{(4,j)}$ for $\partial_0a^{(3,j)}$, and
$a^{(i,4)}$ for $\partial_1a^{(i,3)}$, we ask whether $\partial_0$ and
$\partial_1$ can map $K(a^{(i,j)}\colon(i,j)\leq(3,3))$ into
 $K(a^{(i,j)}\colon(i,j)\leq(4,3))$ and
 $K(a^{(i,j)}\colon(i,j)\leq(3,4))$ respectively, in the manner
suggested by the notation; and we ask further whether $\partial_0$ and
$\partial_1$ can still commute, which means there should be
$a^{(4,4)}$ so that
$\partial_0$ maps
$K(a^{(i,j)}\colon(i,j)\leq(3,4))$, and
$\partial_1$ maps
$K(a^{(i,j)}\colon(i,j)\leq(4,3))$,
into
$K(a^{(i,j)}\colon(i,j)\leq(4,4))$.
In short, can
the matrix~\eqref{eqn:matrix} be extended by one row and
column?  Since $a^{(3,3)}=a^{(0,0)}$, we must have
$a^{(4,3)}=a^{(1,0)}$, and so forth; none of this causes any problem,
and the new matrix can be depicted:
\begin{equation}\label{eqn:matrix1}
  \begin{array}{cccc|c}
    a     &d     &*     &b     &g     \\
    c     &e     &*     &f     &h     \\
    *     &*     &*     &*     &*     \\
    *     &b     &g     &a     &d     \\\cline{1-4}
    *     &f     &h     &\multicolumn{1}{c}{c}     &\multicolumn{1}{c}{e}
  \end{array}
\end{equation}
To extend the
  derivations further to
$K(a^{(i,j)}\colon(i,j)\leq(4,4))$, we require a new condition on
 the original matrix $(a^{(i,j)})^{i<4}_{j<4}$,
 namely, $a^{(3,0)}=a^{(0,2)}$; this comes out when we try to
  extend~\eqref{eqn:matrix1} by one column:
  \begin{equation}\label{eqn:matrix2}
 \begin{array}{cccc|cc}
    a     &d     &j     &b     &g     &a\\
    c     &e     &*     &f     &h     &c\\
    i     &*     &*     &*     &*     &i\\
    j     &b     &g     &a     &d     &j\\\cline{1-4}
    *     &f     &h
    &\multicolumn{1}{c}{c}&\multicolumn{1}{c}{e}&\multicolumn{1}{c}{*}
   \end{array}
  \end{equation}
Indeed, as $a^{(0,4)}=a^{(3,2)}$, so $a^{(0,5)}=a^{(3,3)}=a^{(0,0)}$,
so column $5$ must be the same as column $0$; also
$a^{(0,1)}=a^{(3,4)}$, whence $a^{(0,2)}=a^{(3,5)}=a^{(3,0)}$.
This does not mean that~\eqref{eqn:my-ex} is insoluble; it is.  But
the additional condition found
in~\eqref{eqn:matrix2} does mean that the associated non-homogeneous system
\begin{equation}\label{eqn:my-ex1}
  \partial_0{}^n\partial_1{}^nx-x=0,\qquad
\partial_0{}^n\partial_1x-\partial_1{}^nx=1
\end{equation}
is insoluble, although this is not clear from~\eqref{eqn:matrix}
or~\eqref{eqn:matrix1}.
Alternatively, we can work with the differential ideal, looking
for a characteristic set as mentioned at the beginning of this section:
\begin{equation*}
\begin{split}
&\phantom{{}={}}[\partial_0{}^n\partial_1{}^nx-x,\
\partial_0{}^n\partial_1x-\partial_1{}^nx-1]\\
&=[\partial_1{}^{2n-1}x-x,\
\partial_0{}^n\partial_1x-\partial_1{}^nx-1]\\
&=[\partial_1{}^{2n-1}x-x,\
\partial_0{}^n\partial_1x-\partial_1{}^nx-1,\
\partial_1{}^{3n-2}x-\partial_0{}^nx]\\
&=[\partial_1{}^{2n-1}x-x,\
\partial_0{}^n\partial_1x-\partial_1{}^nx-1,\
\partial_0{}^nx-\partial_1{}^{n-1}x]\\
&=[\partial_1{}^{2n-1}x-x,\
1,\
\partial_0{}^nx-\partial_1{}^{n-1}x]
=[1],
\end{split}
\end{equation*}
so~\eqref{eqn:my-ex1} is insoluble; but higher-order derivatives were
needed to discover this.

By the same computation, the differential ideal
$[\partial_0{}^n\partial_1{}^nx-x,\
\partial_0{}^n\partial_1x-\partial_1{}^nx]$ has the characteristic set
$\{\partial_1{}^{2n-1}x-x,
\partial_0{}^nx-\partial_1{}^{n-1}x\}$, which, in the manner described above,
determines the following matrix in case $n=2$:
\begin{equation}\label{eqn:matrix-n2}
\begin{matrix}
    a&b&*&a\\
    *&*&*&*\\
    b&*&*&*
  \end{matrix}
\end{equation}
Therefore~\eqref{eqn:matrix}
has a solution, simply because the ordinary ideal
$(x^{(0,2n-1)}-x,x^{(n,0)}-x^{(0,n-1)})$
has a zero.  One way of justifying this conclusion is
\cite[Lem.~3.1.2]{MR2001h:03066}; another way will be
Theorem~\ref{thm:second}, according to
which it is enough to
observe that the dependencies in~\eqref{eqn:matrix-n2} extend in a `nice'
way to a triangle:
\begin{equation}
\begin{array}{cccc|cc} \label{eqn:triangle}
  a&b&c&a&b&c\\
  d&e&f&d&e& \\
  b&c&a&b& &  \\\cline{1-4}
  e&f&d&\multicolumn{3}{c}{}  \\
  c&a& &\multicolumn{3}{c}{}  \\
  f& & &\multicolumn{3}{c}{}
\end{array}
\end{equation}

\subsection{Terminology}\label{subsect:terms}

I shall now avoid working
with differential polynomials as such, but shall work instead with the
algebraic dependencies that they determine.

Let $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$.  Higher-order
derivatives with respect to the $\partial_i$ can be indexed by
elements of $\vnn^m$:  so, for
$\partial_0{}^{\sigma(0)}\dotsb\partial_{m-1}{}^{\sigma(m-1)}x$, we
may write \myboxed{\partial^{\sigma}x}.
Let \myboxed{\leq}{} be the product ordering of $\vnn^m$.
Then the derivative $\partial^{\sigma}x$ is
\defn{below}{} $\partial^{\tau}x$ (and the latter is \defn{above}{} the
former) if $\sigma\leq\tau$.  If $n\in\vnn$, then $\leq$ may also be
the product ordering of $\vnn^m\times n$.

If
$\sigma\in\vnn^m$, let the sum $\sum_{i<m}\sigma(i)$ be denoted by
\myboxed{\size{\sigma}}: this is the \defn{height}{} of $\sigma$ or of
$\partial^{\sigma}x$.  (Kolchin \cite[\S~I.1, p.~59]{MR58:27929} uses
the word \emph{order}.)
If $n$ is a positive integer, let $\vnn^m\times n$ be (totally) ordered by
\myboxed{\tleq}, which is taken from
the left lexicographic ordering
of $\vnn^{m+1}$ by means of the embedding
\begin{equation*}
  (\xi,k)\longmapsto
(\size{\xi},k,\xi(0),\dots,\xi(m-2))
\end{equation*}
of $\vnn^m\times n$ in $\vnn^{m+1}$.
Then $(\vnn^m\times n,\tleq)$ is isomorphic to $(\vnn,\leq)$.
In case $n=1$, and $m$ is $2$ or $3$, the picture is thus:
\begin{center}
\hfill
  \begin{pspicture}(-1,-2.5)(3,0.5)
    \psset{radius=0.1}
    \Cnode(0,0){a}
    \uput[ul](0,0){$(0,0)$}
    \Cnode(1,0){b}
    \uput[u](1,0){$(0,1)$}
    \ncline{->}{a}{b}
    \Cnode(0,-1){c}
    \uput[l](0,-1){$(1,0)$}
    \ncline{->}{b}{c}
    \Cnode(2,0){d}
    \uput[ur](2,0){$(0,2)$}
    \ncline{->}{c}{d}
    \Cnode(1,-1){e}
    \uput[dr](1,-1){$(1,1)$}
    \ncline{->}{d}{e}
    \Cnode(0,-2){f}
    \uput[d](0,-2){$(2,0)$}
    \ncline{->}{e}{f}
  \end{pspicture}
\hfill
\begin{pspicture}(-2.5,-0.5)(2.5,3)
    \psset{radius=0.1}
    \Cnode(0,2.4){a}
    \uput[u](0,2.4){$(0,0,2)$}
    \Cnode(0.7,1.2){b}
    \uput[ur](0.7,1.2){$(0,1,1)$}
    \ncline{->}{a}{b}
    \Cnode(1.4,0){c}
    \uput[r](1.4,0){$(0,2,0)$}
    \ncline{->}{b}{c}
    \Cnode(-0.7,1.2){d}
    \uput[ul](-0.7,1.2){$(1,0,1)$}
    \ncline{->}{c}{d}
    \Cnode(0,0){e}
    \uput[d](0,0){$(1,1,0)$}
    \ncline{->}{d}{e}
    \Cnode(-1.4,0){f}
    \uput[l](-1.4,0){$(2,0,0)$}
    \ncline{->}{e}{f}
\end{pspicture}
\hfill
\mbox{}
\end{center}

We may
write $(\sigma,k)\tl\infty$ for all $(\sigma,k)$ in $\vnn^m\times n$.
Suppose $(x_h\colon h<n)$ is a tuple of indeterminates.  By ordering
the formal derivatives $\partial^{\sigma}x_k$ in terms of $(\sigma,k)$ and
$\tleq$, we have Kolchin's example of an \emph{orderly ranking} of
derivatives \cite[\S~I.8, p.~75]{MR58:27929}.  If
$(\sigma,k)\tl(\tau,\ell)$, I shall say that the derivative
$\partial^{\sigma}x_k$ is
\defn{less}{} than $\partial^{\tau}x_{\ell}$ or is a \defn{predecessor}{}
of $\partial^{\tau}x_{\ell}$, and
 $\partial^{\tau}x_{\ell}$ is \defn{greater}{} than
 $\partial^{\sigma}x_k$; likewise for the expressions $a_k^{\sigma}$
and $a_{\ell}^{\tau}$, introduced in~\eqref{eqn:L} below.  (So, the
terms just defined refer
to the total ordering $\tl$, while `below' and `above' refer to the
partial ordering $\leq$.)

Addition and subtraction on $\vnn$ induce corresponding operations on
$\vnn^m$.  Then
\begin{gather}\notag
\tau\leq\sigma+\tau,\\\notag
\partial^{\sigma}\partial^{\tau}x_k=
\partial^{\sigma+\tau}x_k,\\\label{eqn:sks}
(\sigma,k)\tleq(\sigma+\tau,k),\\\notag
(\sigma,k)\tleq(\tau,\ell)\iff
(\sigma+\rho,k)\tleq(\tau+\rho,\ell).
\end{gather}
If $i<m$, let \myboxed{\ichar}{} denote the
characteristic function of $\{i\}$ in $\vnn^m$, so that
$\partial^{\ichar}=\partial_i$, and more generally
$\partial_i\partial^{\sigma}=\partial^{\sigma+\ichar}$, and
$\partial_i\partial^{\sigma-\ichar}=\partial^{\sigma}$.

Let $L$ be an extension of $K$ with generators that are indexed by an initial
segment of $(\vnn^m\times n,\tleq)$; that is,
\begin{equation}\label{eqn:L}
  L=K(a_h^{\xi}\colon(\xi,h)\tl(\tau,\ell)),
\end{equation}
where $(\tau,\ell)\in\vnn^m\times n$, or possibly $(\tau,\ell)=\infty$,
in which case $L=K(a_h^{\xi}\colon(\xi,h)\in\vnn^m\times n)$.
By~\eqref{eqn:sks}, if $a_k^{\sigma+\ichar}$ is one of the generators
of $L/K$, then so is $a_k^{\sigma}$.  Let us say that $L$, with the
given generators, meets the \defn{differential condition}{} if there is
no obstacle to extending each derivation $\partial_i$ to a derivation
$D_i$ on $K(a_h^{\xi}\colon(\xi+\ichar,h)\tl(\tau,\ell))$ such that
\begin{equation}\label{eqn:Dia}
  D_ia_k^{\sigma}=a_k^{\sigma+\ichar}
\end{equation}
whenever $(\sigma+\ichar,k)\tl(\tau,\ell)$.
(So, if the right-hand member of~\eqref{eqn:Dia} is not defined,
then the left need not be defined.)  Formally, the
differential condition is that, if $f$ is a rational function over $K$
in variables $(x_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))$, where
$(\sigma+\ichar,k)\tl(\tau,\ell)$ for some $i$ in $m$, and if
\begin{equation}\label{eqn:fahx}
  f(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))=0,
\end{equation}
then we may apply $D_i$ to this, assuming~\eqref{eqn:Dia}, to get
\begin{equation}\label{eqn:sumeta}
  \sum_{(\eta,g)\tleq(\sigma,k)}
\frac{\partial f}{\partial
  x_g^{\eta}}(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))\cdot
a_g^{\eta+\ichar}+
f^{\partial_i}(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))=0.
\end{equation}
(Note well the assumption that
 $(\sigma+\ichar,k)\tl(\tau,\ell)$.  In~\eqref{eqn:sumeta}, each
of the $a_g^{\eta+\ichar}$ must exist, even though the
coefficient $(\partial f/\partial
 x_g^{\eta})(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))$ might be $0$ in
some cases when $\Char K>0$.)
So the differential condition is \emph{necessary} for the
extensibility of the $\partial_i$ as desired (see for example
\cite[Fact~1.1~(0)]{MR2114160}); sufficiency is part of
Lemma~\ref{lem:above} below.

An extension $(M,D_0,\dots,D_{m-1})$ of
$(K,\partial_0,\dots,\partial_{m-1})$ is \defn{compatible}{} with the
extension $L$ of $K$ given in~\eqref{eqn:L} if $L\included M$,
and~\eqref{eqn:Dia} holds whenever
$(\sigma+\ichar,k)\tl(\tau,\ell)$.

Borrowing some terminology
used for differential polynomials \cite[\S~IX.1, p.~163]{MR0201431},
let us say that a generator $a_k^{\sigma}$ of $L/K$ is a \defn{leader}{} if
it is algebraically dependent over $K$ on its predecessors, that is,
\begin{equation*}
a_k^{\sigma}\in\alg{K(a_h^{\xi}\colon(\xi,h)\tl(\sigma,k))}.
\end{equation*}
Then
$a_k^{\sigma}$ is a \defn{separable}{} leader if it is separably
algebraic over $K(a_h^{\xi}\colon(\xi,h)\tl(\sigma,k))$.  A separable
leader $a_k^{\sigma}$ is \defn{minimal}{} if
there is no separable leader strictly below it---no separable leader
$a_k^{\rho}$ such that $\rho<\sigma$.

For example, in the field $K(a^{(i,j)}\colon(i,j)\leq(3,3))$ depicted
in~\eqref{eqn:matrix} above, the generator
$a^{(3,3)}$ is a (non-minimal) separable leader, and
$a^{(3,1)}$ is a minimal separable leader.  But here we wanted $a^{(3,3)}$
ultimately to be a derivative of $a^{(3,1)}$, namely
 $\partial_1{}^2a^{(3,1)}$.  Passing to a larger field
in~\eqref{eqn:matrix2}, we found the
condition $a^{(3,0)}=a^{(0,2)}$; then $a^{(3,0)}$ became a new separable
leader, strictly below than the formerly minimal separable leader $a^{(3,1)}$.

\begin{lemma}\label{lem:above}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $L$ is
an extension
$K(a_h^{\xi}\colon(\xi,h)\tl(\tau,\ell))$ of $K$ meeting the
differential condition.  Then the derivations
$\partial_i$ extend to derivations $D_i$ from
$K(a_h^{\xi}\colon(\xi+\ichar,h)\tl(\tau,\ell))$ into $L$
 such that~\eqref{eqn:Dia} holds when $(\sigma+\ichar,k)\tl(\tau,\ell)$.
 If $a_k^{\sigma}$ is a
 separable leader, and $(\sigma+\ichar,k)\tl(\tau,\ell)$, then
 \begin{equation}\label{eqn:aksi}
   a_k^{\sigma+\ichar}\in K(a_h^{\xi}\colon (\xi,h)\tl(\sigma+\ichar,k))
 \end{equation}
(that is, $a_k^{\sigma+\ichar}$ is a rational function over $K$ of its
 predecessors);
in particular, $a_k^{\sigma+\ichar}$ is a separable leader.  Therefore
generators of $L/K$ that are above separable leaders are themselves
separable leaders.
\end{lemma}

\begin{proof}
Equation~\eqref{eqn:Dia} shows how to define $D_i$ at the
generators over $K$ of
$K(a_h^{\xi}\colon(\xi+\ichar,h)\tl(\tau,\ell))$.  Then $D_i$ extends
to the whole field by the rules of derivations: this extension is
unambiguous, hence well-defined, if the differential condition is met.

If $a_k^{\sigma}$ is a separable leader, then we may suppose
that~\eqref{eqn:fahx} expresses this; then~\eqref{eqn:sumeta} can be
solved for $a_k^{\sigma+\ichar}$, yielding~\eqref{eqn:aksi}.  (Or just
use, for example, \cite[Fact~1.1~(2)]{MR2114160}.)
In particular, if $a_k^{\rho-\ichar}$ is a separable leader, then so is
$a_k^{\rho}$.  Repeating this observation as needed completes the
proof.
\end{proof}

\subsection{An example}\label{subsect:example}

It is an exercise to
check that the
differential ideal
\begin{equation*}
    [\partial^{(1,1)}x-\partial^{(0,2)}x,
\partial^{(1,2)}x-\partial^{(2,0)}x],
\end{equation*}
with two generators,
is equal to $[f,g,h]$, where
\begin{equation*}
f=\partial^{(1,1)}x-\partial^{(0,2)}x,\qquad
g=\partial^{(0,3)}x-\partial^{(2,0)}x,\qquad
h=\partial^{(3,0)}x-\partial^{(2,1)}x.
\end{equation*}
The algebraic relations imposed by $f$, $g$, and $h$ are given in
the following triangle, where the minimal separable leaders in the
present sense are
underlined (though we have not yet checked the corresponding field
meets the differential condition):
\begin{equation}\label{eqn:triangle2}
  \begin{matrix}
    *&*&a&\underline b\\
    *&\underline a&*&\\
    b&c&&\\
    \underline c&&&
  \end{matrix}
\end{equation}
These leaders have common derivatives, which however impose no new
conditions. (In the terminology introduced by Azriel Rosenfeld \cite[\S~I.2,
  p.~397]{MR0107642} in characteristic $0$, the set $\{f,g,h\}$ is
\emph{coherent.})  For example,
\begin{align*}
  \partial^{(3,0)}g-
\partial^{(0,3)}h
&=\partial^{(3,0)}(\partial^{(0,3)}x-\partial^{(2,0)}x)-
\partial^{(0,3)}(\partial^{(3,0)}x-\partial^{(2,1)}x)\\
&=\partial^{(3,3)}x-\partial^{(5,0)}x-
\partial^{(3,3)}x+\partial^{(2,4)}x\\
&=\partial^{(2,4)}x-\partial^{(5,0)}x
\end{align*}
where we used derivatives as great as
$\partial^{(3,3)}x$; but we already had
\begin{align*}
  \partial^{(2,1)}g-\partial^{(2,0)}h
&=\partial^{(2,1)}(\partial^{(0,3)}x-\partial^{(2,0)}x)-
\partial^{(2,0)}(\partial^{(3,0)}x-\partial^{(2,1)}x)\\
&=\partial^{(2,4)}x-\partial^{(4,1)}x-
\partial^{(5,0)}x-\partial^{(4,1)}x\\
&=\partial^{(2,4)}-\partial^{(5,0)}x,
\end{align*}
where no derivative is as great as $\partial^{(3,3)}x$.  This checking
did require some derivatives that were not \emph{below}
$\partial^{(3,3)}x$; but they do all belong to the following larger
triangle, whose corresponding field meets the differential condition
(though note how the $c$ from~\eqref{eqn:triangle2} changes to $b$):
\begin{equation}\label{eqn:triangle3}
  \begin{array}{cccc|ccc}
    *&*&a&b&b&b&b\\
    *&a&b&b&b&b& \\
    b&b&b&b&b& & \\
    b&b&b&b& & & \\\cline{1-4}
    b&b&b&\multicolumn{4}{c}{} \\
    b&b& &\multicolumn{4}{c}{} \\
    b& & &\multicolumn{4}{c}{}
  \end{array}
\end{equation}

We shall see that extensibility to triangles as
in~\eqref{eqn:triangle3}, without introduction of new minimal
separable leaders, is sufficient to guarantee
solutions.

\subsection{A solubility condition}

If $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, then this model
has an extension whose underlying field is the separable closure of
$K$ (as by \cite[Lem.~3.4, p.~930]{MR2000487} and \cite[Lem.~2.4,
  p.~1334]{MR2114160}).  We shall need this in a more general form:

\begin{lemma}\label{lem:comm}
  Suppose a field $M$ has two subfields $L_0$ and $L_1$, which in turn
  have a common subfield $K$.  For each $i$ in $2$, suppose there is a
  derivation $D_i$ mapping $K$ into $L_i$ and $L_{1-i}$ into $M$.
  Then the bracket $[D_0,D_1]$ is a well-defined
  derivation on $K$.  Suppose it is the $0$-derivation, that is, the
  following diagram commutes.
  \begin{equation*}
    \begin{CD}
      K @>{D_1}>> L_0\\
@V{D_0}VV @VV{D_0}V\\
L_1 @>>{D_1}> M
    \end{CD}
  \end{equation*}
Suppose also that $a$ is an element of $M$ that is separably algebraic over
$K$.  Then each $D_i$ extends uniquely to $K(a)$, and $D_ia\in
L_{1-i}(a)$, so $D_{1-i}D_ia$ is well-defined.  Moreover, $[D_0,D_1]a=0$.
\end{lemma}

\begin{proof}
  The claim follows from standard facts, at least if $L_0=K=L_1$; but
  the proof is the same in the general case.  Indeed, though the
  derivations $D_0$ and $D_1$ are defined on $K$, their bracket
  $[D_0,D_1]$ need not be so, since the compositions $D_0D_1$ and
  $D_1D_0$ need not be so; but if they are, then $[D_0,D_1]$ is a
  \emph{derivation} on $K$.
  A derivation on $K$ extends uniquely to $\sep K$; if the derivation
  is $0$ on $K$, then it is $0$ on $\sep K$
  \cite[Fact~1.1~(2)]{MR2114160}.
In the present case, as $a\in \sep K$, so $D_ia\in L_{1-i}(a)$, and
  therefore $D_ia\in\sep{L_{1-i}{}}$; hence $D_{1-i}D_ia$ is defined.
Thus $[D_0,D_1]$ is
  defined on $K(a)$, where $a\in \sep K$; and if the bracket is $0$ on
  $K$, then is $0$ at $a$.
\end{proof}

Another example illustrates the following theorem in positive
characteristic.  Over a differential field
$(K,\partial_0,\partial_1)$, where $\Char K=p>0$,
the differential equation
\begin{equation*}
  (\partial_0x)^p+x=\partial_1x
\end{equation*}
determines an extension
$K(a^{\xi}\colon\size{\xi}\leq 2)$ of $K$ that meets the differential
condition: the generators form a triangle thus:
\begin{equation*}
  \begin{matrix}
    a&b^p+a&b^p+a\\
    b&b    &     \\
    c&     &
  \end{matrix}
\end{equation*}
where $(a,b,c)$ is algebraically independent over $K$.
In particular, $a^{(1,0)}$ (which has the value $b$) is an inseparable leader,
but none of the generators of height $2$ (namely, $a^{(0,2)}$,
$a^{(1,1)}$ and $a^{(2,0)}$) is an inseparable
leader.

\begin{theorem}\label{thm:first}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq2r\land
h<n)$ meeting the differential condition for some positive integers
$r$ and $n$.  Suppose further that, whenever $a_k^{\sigma}$ is a
minimal separable leader,
then $\size{\sigma}\leq r$.  Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension
$(M,D_0,\dots,D_{m-1})$
compatible with
$K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$.
\end{theorem}

\begin{proof}
  The claim can be compared to and perhaps derived from a
  differential-alge\-braic lemma of
  Rosenfeld \cite[\S~I.2]{MR0107642}, at least in
  characteristic~$0$.  Here I give an independent
  argument, for arbitrary characteristic.
We shall obtain $M$ recursively as
  $K(a_h^{\xi}\colon(\xi,h)\in\vnn^m\times n)$, at the same time
  proving inductively that the $\partial_i$ can be extended to $D_i$ so
  that~\eqref{eqn:Dia} holds in all cases.

Let $L=K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$; this is
$K(a_h^{\xi}\colon(\xi,h)\tleq((2r-1,0,\dots,0),n-1))$.  Then
by~\eqref{eqn:sumeta}, the differential
condition requires of the tuple
$(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$ only that it solve
some linear equations over $L$.  The hypothesis of our claim is that
there \emph{is} a solution, namely
 $(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$.  We may therefore
assume that this tuple is a
\emph{generic} solution of these equations.  In particular, no entry
of this tuple is an inseparable leader.  (Alternatively,
one might try choosing the entries of
$(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$ from the field $L$;
then this field
would be closed under the desired extensions $D_i$ of $\partial_i$,
and the derivations $D_i$ would commute on the subfield
$K(a_h^{\xi}\colon\size{\xi}+1<2r\land h<n)$; but they might not
commute on all of $L$.)

Now, as an inductive hypothesis, suppose we have the extension
$K(a_h^{\xi}\colon(\xi,h)\tl(\tau,{\ell}))$ of $K$ meeting the differential
condition, so that there are derivations $D_i$ as given by
Lemma~\ref{lem:above}; suppose also that
\begin{enumerate}
\item
if $a_k^{\sigma}$ is a minimal separable leader, then
$\size{\sigma}\leq r$;
\item
if $a_k^{\sigma}$ is an inseparable leader, then
$\size{\sigma}<2r$.
\end{enumerate}
We need
to choose $a_{\ell}^{\tau}$ in such a way that these conditions still
hold for $K(a_h^{\xi}\colon(\xi,h)\tleq(\tau,{\ell}))$.
The inductive
hypothesis is correct when $\size{\tau}\leq2r$; so we may
assume $\size{\tau}>2r$.
Hence, if $\tau(i)>0$, so that $\tau-\ichar$ is defined, then
$\size{\tau-\ichar}\geq2r$, so
$a_{\ell}^{\tau-\ichar}$ is not an inseparable leader.

If $a_{\ell}^{\tau-\ichar}$ is not a
leader at all, for any $i$ in $m$, then we may let $a_{\ell}^{\tau}$ be a
new transcendental, and we may define each derivative
$D_ia_{\ell}^{\tau-\ichar}$ as this \cite[Fact~1.1~(1)]{MR2114160}.

In the other case, $a_{\ell}^{\tau-\ichar}$ is a
separable leader for some $i$.  Then $D_ia_{\ell}^{\tau-\ichar}$ is determined
(Lemma~\ref{lem:above}).  We want to let $a_{\ell}^{\tau}$ be this
derivative.  However, possibly also $a_{\ell}^{\tau-\jchar}$ is a
separable leader, where $i\neq j$.  In this case, we must check that
\begin{equation}\label{eqn:comm}
  D_ja_{\ell}^{\tau-\jchar}=D_ia_{\ell}^{\tau-\ichar},
\end{equation}
that is, $[D_i,D_j]a_{\ell}^{\tau-\ichar-\jchar}=0$.

There are minimal separable leaders $a_{\ell}^{\pi}$ and $a_{\ell}^{\rho}$
below $a_{\ell}^{\tau-\ichar}$ and
$a_{\ell}^{\tau-\jchar}$ respectively.
Let $\nu$ be $\pi\wedge\rho$,
the least upper bound of $\{\pi,\rho\}$ with respect to $\leq$.
Then $\nu\leq\tau$.  But
$\size{\nu}\leq\size{\pi}+\size{\rho}\leq2r<\size{\tau}$; so $\nu<\tau$.
Hence $\nu\leq\tau-\kchar$ for some $k$ in $m$, which means
$a_{\ell}^{\nu}$ is below $a_{\ell}^{\tau-\kchar}$.
Consequently,
\begin{enumerate}
  \item
$a_{\ell}^{\pi}$ is below both $a_{\ell}^{\tau-\ichar}$ and
    $a_{\ell}^{\tau-\kchar}$;
\item
$a_{\ell}^{\rho}$ is below both $a_{\ell}^{\tau-\jchar}$ and
    $a_{\ell}^{\tau-\kchar}$.
\end{enumerate}
If $k=j$, then $a_{\ell}^{\pi}$ is below
$a_{\ell}^{\tau-\ichar-\jchar}$, so this is a separable leader.  As
$D_i$ and $D_j$ commute on
$K(a_h^{\xi}\colon (\xi,h)\tl(\tau-\ichar-\jchar,{\ell}))$ by the
differential condition, they must
commute also at $a_{\ell}^{\tau-\ichar-\jchar}$ (Lemma~\ref{lem:comm}),
so~\eqref{eqn:comm} is established.  The argument is the same if
$k=i$.  If $k$ is different from $i$ and $j$, then again the same
argument yields
 $D_ja_{\ell}^{\tau-\jchar}=D_ka_{\ell}^{\tau-\kchar}$ and
 $D_ka_{\ell}^{\tau-\kchar}=D_ia_{\ell}^{\tau-\ichar}$,
so~\eqref{eqn:comm} holds.

In no case did we introduce a new minimal separable leader or an
  inseparable leader.  This completes the induction and the
  proof.
\end{proof}

The claim at the end of the last subsection (\ref{subsect:example}) is
now justified.

In terms of differential polynomials and ideals, the theorem can be
understood as follows.  Given the hypothesis of the theorem, let $S$
be the set of differential polynomials
$f(\partial^{\xi}x_h\colon\size{\xi}<2r\land h<n)$, where $f$
is an ordinary polynomial over $K$ such that
$f(a^{\xi}_h\colon\size{\xi}<2r\land h<n)=0$.  Then $S$
includes a characteristic set for the differential ideal that it
generates.

We can now characterize the existentially closed models of $\mDF$ by
means of the following lemma.  The lemma follows from unproved
statements in \cite[\S~0.17, p.~49]{MR58:27929}; let's just prove it here.

\begin{lemma}\label{lem:fin}
For every $m$ in $\vnn$ and positive integer $n$,
  every antichain of $(\vnn^m\times n,\leq)$ is finite.
\end{lemma}

\begin{proof}
The general case
follows from the case when $n=1$, since if $S$ is an antichain of
$(\vnn^m\times n,\leq)$, then
\begin{equation*}
S=\bigcup_{j<n}\{(\xi,h)\in S\colon h=j\},
\end{equation*}
and each component of the union is in bijection with
an antichain of $(\vnn^m,\leq)$.  As an inductive hypothesis,
suppose every antichain of
$(\vnn^{\ell},\leq)$ is finite; but suppose also, if possible, that
there is an infinite
antichain $S$ of $(\vnn^{\ell+1},\leq)$.  Then $S$ contains some
$\sigma$.  By inductive hypothesis, the set
\begin{equation*}
  \bigcup_{j\leq\ell}\bigcup_{i\leq\sigma(j)}\{\xi\in S\colon\xi(j)=i\}
\end{equation*}
is a finite union of finite sets, so its complement with respect to
$S$ has infinitely many elements $\tau$; but then $\sigma<\tau$, so
$S$ was not an antichain.
\end{proof}

\begin{theorem}\label{thm:dcf}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$.  Then the
following are equivalent:
\begin{enumerate}
  \item\label{item:ec}
The model
$(K,\partial_0,\dots,\partial_{m-1})$ of $\mDF$ is existentially closed.
\item\label{item:<}
For all positive integers $r$ and $n$, if $K$ has an extension
$K(a_h^{\xi}\colon\size{\xi}\leq2r\land h<n)$ meeting the
differential condition such that $\size{\sigma}\leq r$ whenever
$a_k^{\sigma}$ is a minimal separable leader, then the tuple
$(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$ has a
specialization $(\partial^{\xi}b_h\colon\size{\xi}<2r\land
h<n)$ for some tuple $(b_h\colon h<n)$ of elements of $K$.
\end{enumerate}
\end{theorem}

\begin{proof}
Assume~\eqref{item:ec} and the hypothesis of~\eqref{item:<}.  Let $S$ be a
  (finite) generating set of the ideal of
 $(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$ over $K$.  By
  Theorem~\ref{thm:first}, the system
  \begin{equation*}
    \bigwedge_{f\in S}f(\partial^{\xi}x_h\colon\size{\xi}<2r\land h<n)=0
  \end{equation*}
has a solution in some extension, hence it has a solution
in $K$ itself, which means the conclusion of~\eqref{item:<} holds.
So~\eqref{item:<} is necessary for~\eqref{item:ec}.

Every system over $(K,\partial_0,\dots,\partial_{m-1})$ is equivalent to a
system of equations.  Suppose such a system has a
solution $(a_h\colon h<n)$ in some extension.
Then the extension $K(\partial^{\xi}a_h\colon(\xi,h)\in(\vnn^m\times
n))$ has a \emph{finite} set of minimal separable leaders, by
Lemma~\ref{lem:fin}, since this
set is indexed by an antichain of $(\vnn^m\times n,\leq)$.
Hence there is $r$ large enough that all of these minimal separable
leaders are also generators of
$K(a_h^{\xi}\colon\size{\xi}\leq r\land h<n)$.  We may assume also
that $r$ is large enough that $\size{\sigma}\leq r$ for every
derivative $\partial^{\sigma}x_k$ that appears in the original
system.  The hypothesis of~\eqref{item:<} is now satisfied when each
$a^{\sigma}_k$ is taken as $\partial^{\sigma}a_k$.  If the conclusion
of~\eqref{item:<} follows, then $(b_h\colon h<n)$ is a solution of the
original system.
Thus,~\eqref{item:<} is sufficient
for~\eqref{item:ec}.
\end{proof}

\begin{corollary}\label{cor:dcf}
  The theory $\mDF$ has a model-companion, $\mDCF$.
\end{corollary}

\begin{proof}
We consider all possible situations in which
the hypothesis of~\eqref{item:<} in the theorem is satisfied.
In each case, there is a finite tuple $\tuple c$ of parameters from
$K$, there is a universal formula $\phi(\tuple y)$, and there is a system
$\psi(\tuple x,\tuple y)$ such that:
\begin{enumerate}
  \item
$(K,\partial_0,\dots,\partial_{m-1})\models\phi(\tuple c)$;
\item
$(K,\partial_0,\dots,\partial_{m-1})\models\psi(\tuple b,\tuple c)$ if
  and only if $\tuple b$ is a tuple $(b_h\colon h<n)$ as in the
  conclusion of~\eqref{item:<};
\item
for all models $\str M$ of $\mDF$ and all $\tuple d$ from $M$, if
$\str M\models\phi(\tuple d)$, then the system $\psi(\tuple x,\tuple
d)$ is soluble in some extension of $\str M$.
\end{enumerate}
Then $\mDCF$ is axiomatized \emph{modulo} $\mDF$ by all of the sentences
\begin{equation*}
  \Forall{\tuple x}(\phi(\tuple x)\to\Exists{\tuple y}\psi(\tuple
  x,\tuple y))
\end{equation*}
that can arise in this way.  To check this in detail, we note that
$\phi(\tuple c)$ and $\psi(\tuple x,\tuple c)$ can be such that:
\begin{enumerate}
  \item
the system $\psi(\tuple x,\tuple c)$ includes equations
\begin{equation}\label{eqn:ideal}
f(\partial^{\xi}x_h\colon\size{\xi}< 2r\land h<n)=0,
\end{equation}
where
    the (ordinary) polynomials $f$ generate the ideal of
    $(a_h^{\xi}\colon\size{\xi}< 2r\land h<n)$ over $K$;
\item
for each leader of $K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$, there
is an equation~\eqref{eqn:ideal} in $\psi(\tuple x,\tuple c)$ such
that the variables appearing in $f$ correspond only to that leader and
its predecessors that are \emph{not} leaders, so that (since these
non-leading predecessors are algebraically independent) $f$ is in
effect a constant multiple of the minimal polynomial of the leader
over (the extension of $K$ generated by) those predecessors;
\item
then $\phi(\tuple c)$ says that those $f$ are irreducible (this is why
$\phi$ is universal);
\item
also, $\phi(\tuple c)$ shows explicitly how nothing new results when
the equations~\eqref{eqn:ideal} are differentiated: if this
differentiation does not introduce a new derivative (namely, some
$\partial^{\sigma}x_k$ where $\size{\sigma}=2r$), then the new
equation is an explicit linear combination of the original equations;
in the cases where new derivatives are introduced, the resulting
linear system has an explicit solution.
\end{enumerate}
This can all have been done so that $\tuple c$ can be extracted to yield
$\phi(\tuple y)$ and $\psi(\tuple x,\tuple y)$ as desired.
\end{proof}

\subsection{Differential forms again}

The condition in Theorem~\ref{thm:first} can be adjusted to yield the
following:

\begin{theorem}\label{thm:second}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq\size{\mu}\land
h<n)$ meeting the differential condition for some $\mu$ in $\vnn^m$
and some positive integer $n$.  Suppose further that, if $a_k^{\sigma}$ is a
minimal separable leader,
then $\sigma\leq\mu$.  Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension
compatible with
$K(a_h^{\xi}\colon\size{\xi}<\size{\mu}\land h<n)$.
\end{theorem}

\begin{proof}
  The proof is as for Theorem~\ref{thm:first}, \emph{mutatis
  mutandis.}  What needs adjusting is the choosing of
  $a_{\ell}^{\tau}$ in case both $a_{\ell}^{\tau-\ichar}$ and
  $a_{\ell}^{\tau-\jchar}$ are separable leaders.  Again we have
  minimal separable leaders
 $a_{\ell}^{\pi}$ and $a_{\ell}^{\rho}$
below $a_{\ell}^{\tau-\ichar}$ and
$a_{\ell}^{\tau-\jchar}$ respectively.  Since
  $\size{\mu}<\size{\tau}$, there is some $k$ in $m$
  such that $\mu(k)<\tau(k)$.  If $k=j$, then
  $\pi(j)\leq\mu(j)<\tau(j)$, so
  $\pi(j)\leq(\tau-\jchar)(j)=(\tau-\ichar-\jchar)(j)$.  Then
  $\pi\leq\tau-\ichar-\jchar$, so $a_{\ell}^{\pi}$ is below
  $a_{\ell}^{\tau-\ichar-\jchar}$.  Now we can proceed as before.
\end{proof}

As Theorem~\ref{thm:first} yields Theorem~\ref{thm:dcf}, so
Theorem~\ref{thm:second} yields a characterization of the existentially
closed models of $\mDF$.  Moreover,
Theorems~\ref{thm:first} and~\ref{thm:second} can be combined in the
following way:

\begin{theorem}\label{thm:third}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq2r\land
h<n)$ meeting the differential condition for some positive integers $n$
and $r$.  Suppose further that, for each $k$ in $m$, either
 $\sigma\leq r$ whenever $a_k^{\sigma}$ is a
minimal separable leader, or else there is some $\tau$ in $\vnn^m$
such that $\size{\tau}=2r$, and
 $\size{\sigma}\leq\size{\tau}$ whenever $a_k^{\sigma}$ is a
minimal separable leader.
Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension
compatible with
$K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$.
\end{theorem}

\begin{proof}
  Combine the proofs of
Theorems~\ref{thm:first} and~\ref{thm:second}.
\end{proof}

Again there is a corresponding characterization of the models of
$\mDCF$.

Suppose $K(a_h^{\xi}\colon\size{\xi}\leq\size{\mu}\land h<n)$ is as in
the hypothesis of Theorem~\ref{thm:second}, and $\size{\mu}>2$.  If
$\size{\tau}\leq2$ and $\size{\sigma+\tau}\leq\size{\mu}$, let
$a_{(\sigma,k)}^{\tau}$ denote $a_k^{\sigma+\tau}$.  Then the
extension of $K$ can be written as
\begin{equation}\label{eqn:ext}
  K(a_{(\xi,h)}^{\eta}\colon
  \size{\eta}\leq2\land(\xi,h)\tl((0,\dots,0,\size{\mu}-1),0)),
\end{equation}
but this no longer need satisfy the conditions in
Theorem~\ref{thm:third}.  For example, in case $m=2$ and $n=1$ and
$\mu=(3,0)$, the original extension might determine the following
triangle, with the single leader underlined:
\begin{equation}\label{eqn:a-k}
  \begin{matrix}
    a&b&d&g\\
    c&e&h& \\
    f&k& & \\
    \underline a& & &
  \end{matrix}
\end{equation}
Then the derived extension in~\eqref{eqn:ext} determines three
triangles thus, with minimal separable leaders underlined:
\begin{equation*}
  \begin{matrix}
    a&\underline b&d\\
    \underline c&e& \\
    f           & &
  \end{matrix}
\qquad\qquad
\begin{matrix}
  b&d&g\\
  e&h& \\
  k& &
\end{matrix}
\qquad\qquad
\begin{matrix}
  c&\underline e&h\\
  f&k&            \\
  \underline a& &
\end{matrix}
\end{equation*}
We know that the two minimal separable leaders in the last triangle
will not cause a problem; but we don't know this directly from the
tree small triangles.  In the present context, the
system~\eqref{eqn:step-1} is
\begin{align*}
  \dee a&=\dee t^0\cdot c+\dee t^1\cdot b,&
  \dee b&=\dee t^0\cdot e+\dee t^1\cdot d,&
  \dee c&=\dee t^0\cdot f+\dee t^1\cdot e,\\
  \dee d&=\dee t^0\cdot h+\dee t^1\cdot g,&
  \dee e&=\dee t^0\cdot k+\dee t^1\cdot h,&
  \dee f&=\dee t^0\cdot a+\dee t^1\cdot k,
\end{align*}
and then~\eqref{eqn:jkim} consists of $\partial_0g=\partial_1h$,
$\partial_0h=\partial_1k$, and
$\partial_0k=b$.  As linear equations, these are soluble, but they
don't carry the information in~\eqref{eqn:a-k} that lets us know that
the corresponding differential system is soluble.

\subsection{Another sufficient condition}

If $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and
$K(a_h^{\xi}\colon\size{\xi}\leq\size{\pi}\land h<n)$ is an extension
of $K$ meeting the differential condition, this by itself is not enough
to ensure that $(K,\partial_0,\dots,\partial_{m-1})$
has an extension compatible with
$K(a_h^{\xi}\colon\size{\xi}<\size{\pi}\land h<n)$.  However, if such
an extension does exist, then this can be shown by means of the
Theorem~\ref{thm:first}, provided $\size{\pi}$ can be made large
enough:  this is
Theorem~\ref{thm:s} below, which relies on the existence of
bounds as in the following.

\begin{lemma}
  For all positive integers $m$ and $n$, for all sequences $(a_i\colon
  i\in\vnn)$ of positive integers, there is a
  bound on the length of strictly increasing chains
  \begin{equation}\label{eqn:chain}
    S_0\pincluded S_1\pincluded S_2\pincluded\dotsb
  \end{equation}
of antichains $S_k$ of $(\vnn^m\times n,\leq)$, where also
$S_k\included\{(\xi,h)\colon\size{\xi}\leq a_k\}$.
\end{lemma}

\begin{proof}
Divide and conquer.
  First reduce to the case when $n=1$.  Indeed, suppose the claim
  does hold in this case.  Suppose also, as an inductive hypothesis,
  that the claim holds when $n=\ell$.  Now fix $m$ and the sequence
  $(a_i\colon i\in\vnn)$ or rather  $(a(i)\colon i\in\vnn)$, and
  consider arbitrary chains as
  in~\eqref{eqn:chain}, where $n=\ell+1$.  Analyze each $S_k$ as
  $S_k'\cup S_k''$, where
  \begin{align*}
    S_k' &=\{(\xi,h)\in S_k\colon h<\ell\},\\
    S_k''&=\{(\xi,h)\in S_k\colon h=\ell\}.
  \end{align*}
For each $k$ such that $S_{k+1}$ exists, at least one of the
inclusions $S_k'\included S_{k+1}'$ and $S_k''\included S_{k+1}''$
is strict; also, by our assumption, there is an upper bound
$f(k)$ on those $r$ such that
\begin{equation}\label{eqn:sk''}
  S_k''\pincluded S_{k+1}''\pincluded\dotsb\pincluded S_{r-1}''.
\end{equation}
The function $f$ depends only on
$m$, $n$, and $(a_i\colon i\in\vnn)$), not on the choice
of chain in~\eqref{eqn:chain}.

Let $k(0)=0$, and if $k(i)$ has been chosen, let $k(i+1)$ be the least
$r$, if it exists, such that $S_{k(i)}'\pincluded S_r'$.
Here $k$ does depend on the chain.
But if $r$ is maximal in~\eqref{eqn:sk''}, and $S_r'$ exists,
then $S_k'\pincluded S_r'$.
Hence
$k(i+1)\leq f(k(i))$.  Since the function $f$ is not necessarily
increasing, we derive from it the increasing function $g$, where
$g(k)=\max_{i\leq k}f(i)$.  Then
\begin{equation}\label{eqn:krg}
  k(r)\leq f(k(r-1))\leq g(k(r-1))\leq g\circ g(k(r-2))\leq\dotsb
\leq
\overbrace{g\circ\dotsb\circ g}^r(0)=g^r(0).
\end{equation}
In particular,
$S_{k(r)}\included\{(\xi,h)\colon\size{\xi}\leq
a(g^r(0))\}$.  The sequence $(a(g^i(0))\colon i\in\vnn)$ does not
depend on the original chain.
Hence the inductive hypothesis applies to the chain
\begin{equation}\label{eqn:chain2}
  S_{k(0)}'\pincluded S_{k(1)}'\pincluded\dotsb,
\end{equation}
showing
that there is $s$ (independent of the original chain) such that
$k(s)$ is defined, and $r\leq s$ for all entries $S_{k(r)}'$
in~\eqref{eqn:chain2}.
Hence also, by~\eqref{eqn:krg}, if
$S_r$ is an entry in~\eqref{eqn:chain2}, then $r\leq k(s)\leq
g^s(0)$.

Now suppose $S_r'$ is the final entry in~\eqref{eqn:chain2}.  Then
$S_r''\pincluded S_{r+1}''\pincluded\dotsb$; but if $S_t''$ is an
entry of this chain, then $t<f(r)\leq g(r)\leq g(g^s(0))=g^{s+1}(0)$.

Therefore the
original chain in~\eqref{eqn:chain} has a final entry $S_t$, where
$t<g^{s+1}(0)$.  Thus the claim holds when $n=\ell+1$.  By induction, the
claim holds for all positive $n$, provided it holds when $n=1$.

It remains to show that, for all positive $m$, for all sequences
$(a_i\colon i\in\vnn)$, there is a bound on the length of chains
\begin{equation}\label{eqn:chain3}
  S_0\pincluded S_1\pincluded S_2\pincluded\dotsb
\end{equation}
of antichains $S_k$ of $(\vnn^m,\leq)$,
where $S_k\included\{\xi\colon\size{\xi}\leq a_k\}$.  The claim is
trivially true when $m=1$.  Suppose it is true when $m=\ell$.  Now let
$m=\ell+1$, and suppose we have a chain as in~\eqref{eqn:chain3}.  We
may assume that $S_0$ contains some $\sigma$.  If $i<m$ and
$j\in\vnn$, let
\begin{equation*}
  S_k^{i,\,j}=\{\xi\in S_k\colon\xi(i)=j\}.
\end{equation*}
Then the inductive hypothesis applies to chains of the form
\begin{equation*}
  S_{k(0)}^{i,\,j}\pincluded
  S_{k(1)}^{i,\,j}\pincluded
  S_{k(2)}^{i,\,j}\pincluded\dotsb.
\end{equation*}
Moreover, if $\tau\in S_k$, then $\tau(i)\leq\sigma(i)$ for some $i$
in $m$ (since
$\sigma$ is also in $S_k$, and this is an antichain).  Hence
\begin{equation*}
  S_k=\bigcup_{i<m}\bigcup_{j\leq\sigma(i)}S_k^{i,\,j},
\end{equation*}
a union of no more than $\size{\sigma}$-many sets, hence no more than
$a_0$-many
sets.  So the proof can proceed as in the reduction to $n=1$.
\end{proof}

\begin{theorem}\label{thm:s}
  Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $r$
  and $n$ are positive integers.  Then there is a positive integer $s$
  such that $r\leq s$, and if $K(a_h^{\xi}\colon\size{\xi}\leq s\land
  h<n)$ meets the differential condition, then
  $(K,\partial_0,\dots,\partial_{m-1})$ has an extension that is
  compatible with $K(a_h^{\xi}\colon\size{\xi}\leq
  r\land h<n)$.
\end{theorem}

\begin{proof}
Suppose $K(a_h^{\xi}\colon\size{\xi}\leq 2^tr\land h<n)$
meets the differential condition for some~$t$.  When $u\leq t$, let
  $K_u=K(a_h^{\xi}\colon\size{\xi}\leq 2^ur\land h<n)$, and let $S_u$
  be the set of minimal separable leaders of $K_u$.  Then we have an
  increasing chain $S_0\included S_1\included\dots\included S_t$.
  By the preceding lemma, there is a value of $t$, depending only on
  $m$, $r$, and $n$, large
  enough that this chain cannot be strictly increasing.  Then
  $S_u=S_{u+1}$ for some $u$ less than this~$t$.
  Then $K_{u+1}$ satisfies the hypothesis of Theorem~\ref{thm:first}.
So $(K,\partial_0,\dots,\partial_{m-1})$ has an extension compatible
  with $K(a_h^{\xi}\colon\size{\xi}<2^{u+1}r\land h<n)$, and \emph{a
  fortiori} with
 $K(a_h^{\xi}\colon\size{\xi}\leq r\land h<n)$.  In short, the desired
  $s$ is $2^tr$.
\end{proof}

This theorem yields another yet characterization of the models of
$\mDCF$.

redpajama_set_name: RedPajamaArXiv


Unlawful to engage in any game of chance or make any wager by which any money of other article of value may be lost or won. Unlawful to ride or drive any animal of the horse kind in the streets or alley faster than a moderate trot except in cases of necessity. Unlawful to make or allow any loud noise disorder or tumult to disturb peace and quiet of town or inhabitants. Unlawful to allow animals of horse kind standing in the streets or alley without hitching. Unlawful to be found in state of intoxication within city limits.
Prohibited cattle from running at large on certain streets and alleys in town from Nov. 1st until April 1st between 8 a.m. and 4 p.m. on following streets: Main or First street, Second Street between Front and Sycamore, and on Front, Walnut, Market, Mill and Sycamore Streets between First and Second. Also alleys within these bounds.
unclean, filthy or unhealthy substances whenever notified by Marshall. For each violation a fine of not more than $5.00 nor less than $1.00.
Unlawful for owners of hogs, pigs, or shoats to allow them to run at large in streets, alleys, or public ground unless with rings in the nose to prevent them from rooting. Duty of Marshall to take up all animals found at large and hold them for payment of penalties. Must show proof of ownership and pay 40 cents per head plus cost of feed. Marshall shall keep 25 cents per head. In case animals held three days and not claimed they may be sold after proper notice given.
Making it necessary for downtown signs to be at least seven and one half feet above the sidewalk. subject to fine of $2.00 - $5.00. Same fine for not providing a light in the cellar ways of those businesses below the sidewalk. Unlawful to block the sidewalk with vehicle, boxes, barrels, furniture, other material or by hitching or tying any animal on any sidewalk except for loading and unloading purposes. Four feet of space next to building may be used for display purposes.
To Prohibit the depositing of coal or wood ashes, decayed vegetation or garbage in street or alleys or to obstruct streets or alleys with logs, lumber, wood boxes or material of any kind.
Prohibits the throwing of clubs, stone or other missiles along or across any street, alley or unenclosed lands within the town.


text:
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From the old Sea King to the graceful Prince, from the loving sisters to the perfidious Sea Witch and her Pacts: in the world of Dark Tales, new characters appear...
Fairyland is a game of grand and fantastical ideas inspired by books and tales like The Adventures of Peter Pan, Alice in Wonderland, and the Blue Book of Fairy Tales (along with the Red, Green, Yellow, Grey Books, and so...
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redpajama_set_name: RedPajamaCommonCrawl


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Have you been drawn by the theory of finance, or are you more amazed by the practice of finance? The scholastic finance thesis will appeal highly to theoreticians, and must this follow the accepted format of a service thesis: a problem/introduction declaration followed by a literature evaluation, method, argument, conclusion, and recommendations. If your useful finance thesis is a research study of how one business raised loan for its brand-new item advancement cycle, the point is to focus in firmly on that business's finance practices, not overwhelm the reader with unneeded allusions to scholastic literature, or to stay on a complicated method.
As in economics, accounting, and other topics, it pays to be imaginative when looking into, structuring, and composing your finance thesis. Family finance– the normal budget plan– is a kind of finance that all of us practice. If your goal is to compose a finance thesis with optimal appeal to non-insiders in finance, pursue a thesis to which a big number of readers can relate. The finest location to discover finance argumentation examples is on the Web. This is the only method to successfully utilize a finance example that you have actually discovered online and pass it off as your own. As soon as you have actually discovered a strong finance argumentation example to work with you'll desire to compose up an overview design template to work from. Utilizing the example you can fill in the blanks composing in point form the primary points utilized in the example paper.
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DatingAdvice Forum | How does he really feel? HELP!!!!!!
How does he really feel? HELP!!!!!!
This guy says he likes me but he always hangs out with this other girl in class and just waves and stares at me in class. They have known each other for 2 years and we have known each other for about 8 months. He hangs out with me at lunch but that's fixing too change and if he spends the time with her the only time we have to talk too each other does he really want to be with me? Ever since he told me he liked me she hasn't been at school so I don't know how it's gonna be. I didn't tell him I like him but I'm planning too, but if she's always there how am I supposed too? I thought about showing her how close we are but I'm not one of those girls that brags. I have horrible relationship anxiety. When he was with me the last time we saw her leave and I could tell he wanted to go talk to her but he stayed with me and didn't. He has met my dad and my whole family I have talked to his little brother and his dad. I'm so confused and I cry about this every night. I NEED HELP!
It's hard to tell from the information you've given if he likes you or her, but my guess is that he likes both of you. Perhaps he hasn't decided which of you he likes best. If he knew this other girl before you, perhaps he's just trying not to be rude to her by continuing to talk to her, even when he is with you.
It can be difficult for guys because the onus is always on us to make the first move, and if he's shy or unsure if you like him that way, then he might be anxious about making the first move.
Yes he does like you. Just make it official and hint you would like to be go to the movies or somehting fun together. Sometimes males don't get hints so try to make it obvious as you can.
You're good. You'll be great together.
He probably likes both of you, so if I were you I'd just move on. If he really likes you he'll find his way to you.
Reply To: How does he really feel? HELP!!!!!!


text:
Q: Is getNumPartitions an RDD action or transformation? Is RDD's getNumPartitions a transformation or an action?

A: Is it an action?
NO, If you do getNumPartition() on any RDD, It just fetches the meta(partitions.length) of an RDD and no tasks will be submitted.
Is it a transformation?
My understanding about transformation is "It should return an RDD when that function applied on RDD". But getNumPartition() return integer.

Check the definition of function in RDD.scala
  @Since("1.6.0")
  final def getNumPartitions: Int = partitions.length

redpajama_set_name: RedPajamaStackExchange


For many people sunlight is a source of well-being. Especially after the long winter we like to stay outside. Sunlight provides positive effects such as heat, stimulation of vitamin D synthesis and release of "feel-good-hormones" such as serotonin - a neurotransmitter of happiness. Many people still underestimate the risk of the sun and its potential to cause skin cancer. Meanwhile, skin cancer is the most common cancer of the Caucasian population, with around 240.000 people diagnosed yearly. This is roughly the population of Kiel in Germany. The incidence of skin cancer is rising rapidly.
The ozone layer protecting the earth from the dangerous ultraviolet rays becomes thinner. According to the World Meteorological Organization (WMO) in Geneva, the ozone layer over the Artic was destroyed by about 40 percent this winter. This means an increasing health risk for the people. Therfore a good sun protection is extremely important.


text:
Set Backs & Milestones: The Rizes & Falls
Part 17: Life of Dreams Vol.3, Is this Real? Mental Fitness?
I think I must have went to court 3 or 4 times in the first couple weeks, with no results... Mainly to set bail and what not. After four attempts to set bail unsuccessfully, as well as not having anybody willing to bail me out my first lawyer set up some time for me to be evaluated psychiatrically.... So, I went back to holding, to wait some more...
The first month was so active and disappointing, that it could have broke just about anybody... There was a lot of waiting, not knowing, I was given the run around by just about everybody I knew, something was not right...
The Psych assessments were supposed to take a month, my understanding was that I was going to moved to a new facility, what really happened was that there was about three days worth of tests, spread out over a months time. This was very defeating, two days were mainly testing, and sitting in a bare cell for the rest of the day.... every question and answer was analyzed and scrutinized, it made it very hard to speak openly without being judged...
The final day of the assessments were spent waiting in that empty cell, while all the doctors involved discussed results, I guess... Then I was lead into an office, where in front of me sat about 9 doctors and others taking notes, like court reporters or something... how many times can someone be ask the same question "Do you hear voices?" That question til this day, from a doctor makes me want to strangle somebody.... basically, if you don't hear voices, there is nothing wrong with you? Three days and what seemed like a thousand tests.... for that?
I must have went back to court shortly before Christmas, but I don't think it was for anything other than to find out that I was mentally fit to stand trial... after that day, I never saw a court room for about 3 months... we now had to wait for disclosure, bail was never set, so I never really had a bail hearing, but I also never had anybody to bail me out any way.... so why get my hopes up now?
When I got back to the jail house, we were all shuffled around and I had a new cell mate at the end of the day... with Christmas just around the corner and no knowledge of my future, I was stuck, my life in limbo, but calmer for some reason, I started reading and lifting weights as much as I could, to keep my mind occupied.... The dreams had not stopped, in fact, they had gotten increasingly violent and disturbing, but now not always reoccurring in the same manner...
The was a strike looming for the union workers that ran the jail... tensions were high all the time, but this made things even worse...
It was late January 1996, when our range got word that there had been a huge riot down the hall, and more were expected... and the strike was inevitable... the guards were sick of doing their jobs, or wanted more money... some bullshit that affected our lives directly...
About 2 weeks before the strike actually started, a new guy came in and had smuggled something, he was a known crack head to many of the guys and ended up getting jumped by about 12 or so of them.... this lead to us all getting locked down for the rest of the month leading up to the strike...
When the strike eventually kicked in, they shuffled everybody again... it took hours to get in & out of the building for court dates, so many inmates were stuck sitting in court vans that whole time... I didn't have to experience this until sometime in March.
We basically did everything from our cells, we were let out in pairs to make calls or shower, we ate in the cells, etc... this went on from Feb. until Mid-April...
I was still in a state of disbelief, I would wake up some nights think that I was still in a dream, sometimes, it put me in such deep thought, that I believed I was stuck in a coma somewhere and all of my surroundings were part of something bigger, inside my mind... I was losing my grip on reality, and I had by this point been completely sober for months... It made me wonder how I could be feeling this way and still have been found competent...
Disclosure came in sometime in March, and my lawyer showed up to go through it with me. I now had a lawyer, that was being paid through legal aid, but working with my parents... She didn't really believe anything I said because of what she had talked to my parents about, behind my back... it was a tough night, what I saw in the disclosure...
I found out who were my friends, and who weren't. What people really thought about me, and the things they said about me regarding the charges... I couldn't believe what I saw, it took a couple hours to go through, it was a thick as a dictionary, but lacked any evidence, photos, etc... it was all statements from people in the neighbourhood.
When all you have is time, to think, and time to yourself... it really makes you dwell on things... before the end of the night I was placed on suicide watch. Not really suicidal, just so they could keep an eye on me... I wasn't even able to speak, it was the equivalent of having a whole neighbourhood, rat you out, for lack of a better way to say it...
Having spent a great portion of my time before ending up in jail trying, sincerely, to turn my life around, not that anybody even took notice... I held on to what really cleared my thoughts and kept me in good spirits as much as it could... Hip Hop music was all that kept me together....
Now I was torn, between trying to change my life now, at that point, and plotting some kind of revenge, something I wanted so bad some days...
My Pre-trial was set, and reset about three times before my lawyer decided to plea it out rather than take the case any further... Revenge and anger fuel me through my workouts helping me vent those feelings...
Hip Hop fueled my positivity, believe it or not... in a way it never had before... it both kept me going, and helped keep my mind off the things that sent my thoughts and dreams into sequences of rage...
The pink walls and enclosed cells, was taking my mental state into the unknown... a place I didn't want to be, but a place that made me survive without even trying very hard, it was a natural feeling, but an unnatural state of mind... I needed balance... my last court date was coming...


Ryan Schofield's impressive performances for Huddersfield Town have seen him called up to the England Under-20s squad.
Ryan has one cap for the England Under-18 side, keeping a clean sheet in a 4-0 win over host nation Qatar in March 2017.
More recently, in his first ever loan spell in senior football, he made six starts for Vanarama National League North club FC United of Manchester, helping the team win the Manchester Premier Cup in the process.
During the 2016/17 season, Ryan – who joined Town at Under-9 level from Lepton Highlanders - became the Under-18s number one playing in 25 out of a possible 29 matches.
He also signed a new professional contract at the Club, committing his future to Town until 2019.
CLICK HERE to purchase a shirt with Schofield on the back!
HOGG & SCHOFIELD RETURN TO TRAINING!


text:
Massachusetts Leads With First Major U.S. Contact Tracing Program
Health authorities are breaking out the old Yellow Pages to fight COVID-19
ByAlex Hickey
Health authorities are breaking out the Yellow Pages to fight COVID-19. Last Friday, Massachusetts Gov. Charlie Baker detailed a "powerful tool" to stop the coronavirus's spread: contact tracing.
How it works: When a patient tests positive, you make a list of everyone they came in close contact with. Then, you find those people and make sure they self-isolate before infecting others.
That sounds straightforward, but contact tracing a new patient typically takes three days, according to Stat, which is "an insurmountable hurdle in the U.S., with its low numbers of public health workers and tens of thousands of new cases every day."
It's worked elsewhere
South Korea used high-tech contact tracing to tame its outbreak. The government compiled GPS data, credit card swipes, and other info into a public log showing where COVID-19 patients had traveled.
Some countries (including the U.S.) are trying other methods, including looking at smartphone location data and developing Bluetooth systems that provide warnings if you've crossed paths with an infected person.
The hangup: Despite its widespread use in places like Singapore, contact tracing has raised concerns about privacy and governments following citizens' whereabouts. Still, it's a term you'll be hearing a lot more of in the coming weeks.


The province is establishing a justice centre in Kenora as part of its $16.4 million guns and gangs strategy, with the funds earmarked for crime prevention, enforcement, prosecution and alternative justice programs meant to steer people away from the court and prison system — but at this point, the centre won't be a physical building.
The goal of the centres is to coordinate the police and the judicial system with health and social service agencies, the province said.
Although there are no firm plans to put shovels in the ground for this program, Kenora-Rainy River MPP Greg Rickford told reporters that he's talked to Grand Council Treaty Three Grand Chief Francis Kavanaugh about developing a physical location for it.
"We don't need physical infrastructure for this to work but I think, I personally believe, it might be appropriate here," Rickford said.
"These centres will integrate justice facilities with prevention and intervention supports to hold individuals accountable while connecting them with services that prevent crime and break the cycle of offending," the province said in a statement.
Rickford noted Kenora has a "serious requirement for an alternative way to look at justice," pointing to Indigenous over-representation in the justice system and Kenora's overburdened jail.
Among the programs the province is planning are a youth violence prevention program meant to strengthen social connections to reduce risk factors associated with gang violence; an Indigenous youth prevention and intervention program fund aimed at supporting Indigenous communities and First Nations police services to keep Indigenous kids from being recruited into gangs; a gang intervention and exit program specifically meant to prevent Indigenous women and girls being exploited, recruited or victimized by gangs; and programs with certain school boards aimed at addressing overrepresentation of some groups in suspensions and expulsions.
"It has some program features to it, obviously it deals with violence, it deals with alternative decision making within the justice system," Rickford said.
When asked whether any new alternative justice programming is being created in Kenora, Rickford responded by saying the province will roll it out in the coming months.
The province's recent funding announcements for upgrades at the Kenora Jail and the new supportive housing facility introduced early in March are part of a chaptered book, Rickford said, a three-pronged approach to addressing issues of homelessness and justice in Kenora.
The $16.4 million fund for the overall gun and gang crime reduction plan comes in part from $11 million in federal funds Ottawa gave the province earlier in March.


1. Was there anything from the sermon, the big idea, or Galatians 1:1-5 that stood out to you, challenged you, or confused you?
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own terms; rather, he is speaking exactly what Jesus told him face-to-face.
2. When do you find it hardest to accept apostolic New Testament teaching? Why?
Paul reminds them (and us!) that Jesus + Nothing = Everything. What Jesus did on the cross is totally sufficient to make us acceptable before God.
5. Which of these four characteristics of the cross most stand out to you and why? Why is it crucial to realize that Paul mentions absolutely nothing that we do in verses 1-5 when he summarizes the gospel?
The gospel keeps getting better and better! Jesus didn't just die to stop the death sentence that was hanging over us; he died also to rescue us from the prison of our sin. So we would have new life in Him, forever. Joy to us, glory to God.


text: As a quasi-historical, quasi-legendary figure of consistently great popularity, King Arthur has been subject to an extraordinary amount of reinvention and rewriting: as a Christian hero and war-leader; as an ineffective king and pathetic cuckold; and as a tragic figure of noble but doomed intentions. As we trace Arthur's evolution and that of principal knights, we will ask what underlies the appeal of this figure whose consistent reappearance in western culture has performed the medieval prophecy that he would be rex quondam et futurus: the once and future king. Readings will include early Latin and Welsh texts, the great Old French romances of Chretien de Troyes (Yvain, Lancelot, Perceval), and the extraordinary Morte d'Arthur of Sir Thomas Malory.


text:
Il Fyen Rundt è una corsa in linea maschile di ciclismo su strada che si disputa nei dintorni di Odense, in Danimarca. Dal 2008 è stata inserita nel circuito UCI Europe Tour classe 1.2.

Albo d'oro
Aggiornato all'edizione 2022

Note

Collegamenti esterni

redpajama_set_name: RedPajamaWikipedia


Summer is the best time to go on adventurous and fun date nights. Below I have some affordable & unique date ideas that inspire you & help make date night that much more special!
A Summer bucket list is great because you will never have to think about another date idea once you have both created the bucket list. Ideally, this should be made at the start of Summer, s you have more time to complete the list. But, it is never too late to jump in an start crossing off adventures.
To create the bucket list, you should both write down everything & anything you want to do over the Summer. When it comes to date night, or if you just want to get out, you both decide on one of the items to complete and cross off. This keeps your date nights fresh and fun while removing the stress factor of figuring out what to do.
I really love Kelle Hamptons Summer Bucket list. Her list is full of amazing ideas and I love the way that she decorated the list & displays it in her home. Check out her list over at Kellehamptons.com to get loads of ideas for your Summer bucket list.
Movie nights are a date night go-to staple. But, why not mix it up and host a backyard movie screening. The Summer heat makes the backyard the perfect escape when the sun falls and the house is still hot from the day. Plus, you get to lay under the stars with your cutie – What is more romantic than that?
Don't know what you would need to get your theatre set up? Todaywedate.com has all the information on how to set-up your backyard movie theatre. From the seating to the projector and even movie suggestions, everything you need is listed in the blog post.
As someone with a SERIOUS sweet tooth, when it comes to movies there better be some candy laying around. I love Celebrationsathomeblog.com's idea for a fondue dessert platter. She combines cut up fruit with mini donuts and lots of other sweet treats. All centered around a pot of delicious melted chocolate for dipping.
Another way to get out into nature and spend some quaily time together is with a nice picnic. If the weather is nice, why not take a hike to a more secluded location and set-up a picnic. This is a great opportunity to get some cardio in & spend some time alone.
When it comes to a picnic menu, I love having lots of things to nibble on. Which is why I love Jojotastic.com's charcuterie platter idea. This idea is super simple and requires almost no prep work – the meat counter will do all of the work for you. Pick out a few types of meat and cheeses, crackers, fruit, and anything else that you would like to combine and snack on.
If there is one thing that I love it is doing things by theme – Every part should be a theme party as far as I am concerned. So why not have a themed date? Love everything Parisian? why not throw on your beret and make some French food together. After dinner, you can throw on a French movie to really transport you to Paris.
I love thedatingdivas.com's idea for an England themed date night. They take this themed date to the next level with activities, meals, and even the invitations. I love how detailed her post is and she even provides printouts to make this date a sinch. I really want to do this date just to try out her recipe for shepherds pie!
Summer is the perfect time to grab your bae and head to a show. Almost every band is touring during the summer making the options endless. Concert tickets can be kinda pricey after the ticket price and all the added fees. If you are on a budget then look to see if there are any free events going on. I always find cool little festivals or free shows during the summer month.
Get together all your favorite couples and have a potluck group date. This is fun because you get to see all your friends skills in the kitchen. This also makes having a party with friends less stressful because each guest will be coming with food – leaving you more time to enjoy your company.
I love Mylifeasamrs.com's perfect potluck receipt for mushroom, garlic, and parmesan flatbread. Everyone loves pizza & mushrooms are typically a pretty party friendly veggie. This recipe looks so easy & delicious – it will definitely be a crowd pleaser.
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Nov. 16, 2018 3:40 p.m.
Asher Trousdell was surprised when water dripped through the bouquet of flowers he was holding Friday morning and onto his clothes.
It's safe to say the nine-year-old is new to greenhouses. Most nine-year-olds are.
The flowers Asher and his twin brother Nolan held were in the 21 degrees Eurosa Farms greenhouse in Central Saanich where Help Fill a Dream, Country Grocer, Eurosa Farms and other partners launched their eighth annual Holiday Campaign. Eurosa, a family business started by owner Ryan Worsfold's Dutch immigrant parents, ships its flowers as far south as Portland and as far east as Winnipeg.
For the campaign, Eurosa sells bouquets of flowers at a discount to Country Grocer who in turn sells them as a holiday fundraiser to support Help Fill a Dream. All seven Country Grocer stores will sell the bouquets as well as Christmas trees as part of the campaign.
The money goes towards things like Help Fill a Dream's Bear Essentials program that provided the Trousdell family with tie-downs to secure Asher's wheelchair in the family van. Help Fill a Dream is also known for flying families to Disney World, which is exactly where the Trousdell's are going later this year.
"And Universal Studios," Nolan said.
Credit Universal Orlando's Wizarding World of Harry Potter for that draw.
Born fourteen weeks early, Asher and Nolan faced challenges their whole life. Asher survived an early brain hemorrhage and was later diagnosed with cerebral palsy at as a one-year-old. Trousdell's suspicion that Nolan, who still wasn't walking as a two-year-old, also had cerebral palsy was confirmed when he was three.
The two are happy Grade 4 students at View Royal Elementary.
Craig Smith, Help Fill A Dream executive director, said this year's Holiday Campaign will break the $100,000 mark in total funds raised through the campaign.
"We are consistently humbled by the commitment and passion of the Country Grocer employees who make this campaign such a success every year," Smith said. "They make it personal. The thoughtfulness and kindness of Country Grocer's customers and suppliers exceed our expectations year after year.
Disney is the most popular vacation but there are other requests, Smith said, including a family of eight who was flown to the 2018 World Cup.