\chapter{Meet Continuous Congruence
Identities}\label{strong_rectangulation2}

The class of all congruence lattices of algebras in a
variety $\vr V$ is usually too complicated to describe in
any nontrivial way.  There is more hope for a description of
the class 
$\vr L(\vr V)$\index{$\vr L_{MC}(\vr V)$@$\vr L(\vr V)$ (the congruence prevariety of~$\vr V$)}\index{LAAMC(V)@$\vr L(\vr V)$ (the congruence prevariety of~$\vr V$)} 
of lattices embeddable in
congruence lattices of algebras in $\vr V$.  It is evident
that $\vr L(\vr V)$ is closed under the formation of
isomorphic lattices and sublattices. Moreover, the map
\begin{equation}\label{embed_product} \prod_{i\in I}
\Conb(\m a_i) \to \Conb\Bigl(\prod_{i\in I} \m
a_i\Bigr)\colon (\theta_i)_{i\in I}\mapsto \Theta\,,
\end{equation} where $\wec{a}\wrel{\Theta}\wec{b}$ if
$a_i\wrel{\theta_i} b_i$ for all $i$ and $a_j = b_j$ for all
but finitely many $j$, is an embedding, so $\vr L(\vr V)$ is
closed under the formation of products. This makes $\vr
L(\vr V)$ a prevariety, which we call the {\bf congruence
prevariety}\index{congruence prevariety} of~$\vr V$.

Any prevariety is axiomatizable by a class of formulas of
the form
\begin{equation}\label{Horn} \bigwedge_{i\in I}
\varepsilon_i(\wec{x}) \to \varepsilon(\wec{x})
\end{equation} where $\varepsilon(\wec{x})$ and each
$\varepsilon_i(\wec{x})$ is an identity and $\wec{x}$ and
$I$ may be infinite (see Theorem~9.2.2 of \cite{hodges}).
Therefore, in studying the shapes of congruence lattices in
varieties, it is natural to start by considering the
satisfaction of sentences of this form.  The most obvious
problems are still unsolved.

\begin{prb}\label{1rst-order_ax}
  Which congruence prevarieties are first-order
  axiomatizable?
\end{prb}

In other words, for which varieties $\vr V$ is it true that
$\vr L(\vr V)$ can be axiomatized by sentences like
(\ref{Horn}) where $\wec{x}$ and $I$ are finite?
Equivalently, when is $\vr L(\vr V)$ a quasi-variety?  It is
known that if $\vr V$ is congruence $n$-permutable for some
$n$, then $\vr L(\vr V)$ is first-order axiomatizable,
\cite{czedli2, herrmann-poguntke1, herrmann-poguntke2}.  It
is also known that if $\vr V$ has a Taylor term but does not
satisfy a nontrivial congruence identity, then $\vr L(\vr
V)$ is not first-order axiomatizable, \cite{k-nation}. The
full answer to Problem~\ref{1rst-order_ax} is not yet
known. We do not even know the answer to the following
special case.

\begin{prb}\label{1rst-order_ax_2}
  Is the congruence prevariety of a congruence modular
  variety first-order axiomatizable?
\end{prb}

The congruence prevariety is first-order axiomatizable if
and only if it is a quasi-variety, so it is a natural
problem to consider the satisfaction of congruence
quasi-identities.

\begin{prb}\label{cong_quasi_maltsev}
  Is the class of varieties satisfying a given set of
  congruence quasi-identities definable by (idempotent,
  linear) Maltsev conditions?
\end{prb}

Problem~\ref{cong_quasi_maltsev} is known to have a positive
answer when posed for identities in place of
quasi-identities.  
A.~Pixley~\cite{pixley2}\index{Pixley, A.}  
and R.~Wille~\cite{wille}\index{Wille, R.} each
described an algorithm to produce a family of idempotent
linear Maltsev conditions defining the class of varieties
satisfying a given congruence identity.  By suitably
modifying the Pixley--Wille algorithm, partial results have
been obtained for special types of quasi-identities (cf.\
\cite{czedli, czedli1, czedli2, czedli-day, jonsson3}).  The
strongest of these is the result we have recorded as
Theorem~\ref{czedli-day-W}, which states that
Problem~\ref{cong_quasi_maltsev} has an affirmative answer
for quasi-identities satisfying $\W$.  The most that one
could hope for is that if $\vr U$ and $\vr V$ satisfy the
same idempotent Maltsev conditions, then they have the same
congruence prevariety.  The most attractive special case of
this is the case where $\vr U = \vr S\vr E\vr T$, since it
follows from Whitman's Theorem\index{Whitman's Theorem}
(that every lattice is embeddable into a lattice of
equivalence relations, cf.\ Section VII.8 of
\cite{birkhoff}) that $\vr L(\vr S\vr E\vr T)$ is the class
$\vr L$ of all lattices.

\begin{prb}\label{same_prevariety}
  Suppose that a variety $\vr V$ satisfies no nontrivial
  idempotent Maltsev condition.  Is it true that $\vr L(\vr
  V) = \vr L$?  Is it true at least that $\vr L(\vr V)$
  contains the lattice $\Pi_4$ of all equivalence relations
  on a $4$-element~set?
\end{prb}

\noindent We expect a negative answer to both questions
asked in Problem~\ref{same_prevariety}, but see
Theorem~\ref{czedli-day_thm} for a positive partial result.

The purpose of this chapter is to explain how the presence
or absence of a Taylor term affects the shapes of congruence
lattices in a variety.  The Pixley--Wille algorithm shows
that any variety that satisfies a nontrivial congruence
identity in the language of lattices satisfies a nontrivial
idempotent linear Maltsev condition and therefore has a
Taylor term.  The converse is not true; congruence
identities are not flexible enough to say anything
nontrivial about varieties satisfying very weak idempotent
Maltsev conditions.  In order to make nontrivial statements
about the shapes of congruence lattices in arbitrary
varieties with a Taylor term, we need to consider sentences
other than lattice identities.  In this chapter we will
describe congruence lattice shapes in three different ways:
by using quasi-identities in the language of ordinary
lattices, by using identities in the language of meet
continuous lattices, and by identifying omitted sublattices.
We will exhibit a nontrivial quasi-identity $\sigma$ in the
language of lattice theory such that $\vr V$ has a Taylor
term if and only if $\vr L(\vr V)\models \sigma$
(Theorem~\ref{typ1condcor}~(2)).  We will prove that there
is a nontrivial identity $\tau$ in the language of meet
continuous lattice theory such that $\vr V$ has a Taylor
term if and only if $\vr L(\vr V)\models \tau$
(Theorem~\ref{typ1condcor}~(4)).  Finally, we will exhibit a
finite lattice~$\m l$ such that $\vr V$ has a Taylor term if
and only if $\m l \notin \vr L(\vr V)$
(Theorem~\ref{omit_d1_sub}).

\section{Maltsev Conditions From Congruence
Identities}\label{4.1}\index{Pixley--Wille algorithm, the|(}

If $\vr V$ is a variety, let $\vr L_{MC}(\vr V)$ be the
class of all meet continuous lattices $\m l$ for which there
is a meet continuous embedding of $\m l$ into $\Conb(\m a)$
for some $\m a\in \vr V$. The class $\vr L_{MC}(\vr V)$ is
the meet continuous analogue of the congruence prevariety of
$\vr V$.  It is also a prevariety, since the mapping
described in (\ref{embed_product}) preserves complete joins.
It seems appropriate to us to identify the study of shapes
of congruence lattices in varieties with the study of $\vr
L_{MC}(\vr V)$.  In this short monograph we only begin to
scratch the surface. In this chapter we will show that a
variety has a Taylor term if and only if $\vr L_{MC}(\vr V)$
satisfies a nontrivial meet continuous identity.

In this section we will show that the Pixley--Wille
algorithm can be extended to meet continuous identities,
which will show that the class of varieties satisfying a
given meet continuous congruence identity is definable by
idempotent linear Maltsev conditions.  The proof is
straightforward, but is included because there is a
compactness condition associated to Maltsev conditions that
we establish by proving the somewhat nonobvious fact that
any meet continuous identity is equivalent to a set of
identities involving only finitely many variables~apiece.

Using the same arguments as those used for ordinary lattice
identities (cf.\ the third paragraph of
Section~\ref{lattice_theory}), it can be shown that any
family of meet continuous lattice identities is equivalent
to a family of inclusions of the form $P\leq Q$ where $P$
and $Q$ are meet continuous lattice words. Modulo the
identities defining meet continuous lattices, we may
identify $P$ and $Q$ with elements from some free meet
continuous lattice, hence with ideals from some free
lattice.  Now, an inclusion $P\leq Q$ between ideals is
equivalent to a family of inclusions $(p]\leq Q$, $p\in P$,
where the included ideal is principal, therefore any family
of meet continuous lattice identities is equivalent to a
family of inclusions of the form $p\leq Q$ where $p$ and $Q$
are meet continuous lattice words with no infinitary join
occurring in $p$.  We refine this in the following~lemma.

\begin{lm}\label{bound_variables}
  Any family of meet continuous lattice identities is
  equivalent, modulo the identities defining the variety of
  meet continuous lattices, to a family of inclusions of the
  form $p\leq Q$ where both $p$ and~$Q$ are meet continuous
  lattice words in finitely many variables, and no
  infinitary join occurs in~$p$.
\end{lm}

\begin{proof}
  We work modulo the identities defining the class of meet
  continuous lattices, so we may assume that $p$ and $Q$ are
  elements in the free meet continuous lattice $\m f$
  generated by some set $\{x_1, x_2, x_3, \ldots\}$.  A~meet
  continuous lattice $\m l$ satisfies the inclusion $p\leq
  Q$ if for every meet continuous homomorphism
  $\varphi\colon \m f\to \m l$ it is the case that
  $\varphi(p)\leq \varphi(Q)$.  Assume that this holds for
  $\m l$.  Since $p$ has no infinitary joins, we may assume
  that the variables that appear in $p$ are among $\{x_1,
  \ldots, x_n\}$.  Let $\varepsilon$ be the meet continuous
  endomorphism of $\m f$ defined on the generators by
  $x_i\mapsto x_i$ if $1\leq i\leq n$ and $x_i\mapsto 0$
  otherwise.  Then $\varphi\circ\varepsilon\colon \m f\to \m
  l$ is a meet continuous homomorphism, so
  \[
  \varphi\circ\varepsilon(p)\leq \varphi\circ\varepsilon(Q)
  \]
  by the assumption on $\m l$.  But $\varepsilon$ is a
  retraction of $\m f$ onto its meet continuous sublattice
  generated by $\{x_1, \ldots, x_n\}$, and this sublattice
  contains $p$, so $\varepsilon(p) = p$. Thus
  $\varphi(p)\leq \varphi(Q')$ where $Q' := \lb
  \varepsilon(Q)\rb$.  Since $\varphi$ was arbitrary, this
  shows that if $p\leq Q$ holds in $\m l$, then $p\leq Q'$
  holds in $\m l$.  The converse is obvious: $\varepsilon$
  is decreasing on generators, so $Q' = \varepsilon(Q) \leq
  Q$ in $\m f$, and so $p\leq Q'$ implies $p\leq Q$.
  Therefore, a meet continuous lattice satisfies $p\leq Q$
  if and only if it satisfies $p\leq Q'$. The inclusion
  $p\leq Q'$ involves elements of the (free) sublattice of
  $\m f$ generated by $\{x_1,\ldots,x_n\}$, so our original
  inclusion $p\leq Q$ is equivalent to one in a finite set
  of variables.
\end{proof}

Lemma~\ref{bound_variables} implies that there are at most
$2^{2^{\aleph_0}}$ varieties of meet continuous lattices,
since there are at most $2^{\aleph_0}$ inequivalent
inclusions of the form $p\leq Q$ in finitely many variables.
We do not know if this upper bound is attained, but it is
not hard to see that there are at least $2^{\aleph_0}$
varieties of meet continuous lattices. (Argument: If $\vr V$
is a variety of lattices, then the subclass of meet
continuous lattices in $\vr V$ generates~$\vr V$ since it
contains the ideal lattices of members of $\vr V$.  This
subclass is a variety in the language whose nonlogical
symbols are $\{\bigvee, \wedge\}$, since it is closed under
$\Ho, \Su$ and $\Pd$.  This proves that there are at least
as many meet continuous lattice varieties as there are
ordinary lattice varieties, and it is well known that are
$2^{\aleph_0}$ ordinary lattice varieties.)

\begin{prb}\label{meet-continuous_prob}
  Determine the number of meet continuous lattice varieties.
\end{prb}

\begin{thm}\label{pixley-wille_extension}
  The class of varieties satisfying a given set of meet
  continuous congruence identities is definable by a family
  of idempotent linear Maltsev conditions.
\end{thm}

\begin{proof}
  It suffices to prove that the class of varieties whose
  congruence lattices satisfy a single inclusion $p\leq Q$,
  where $p$ is an ordinary lattice word and $Q$ is a meet
  continuous lattice word in the same set of variables, is
  definable by a family of idempotent linear Maltsev
  conditions.

  If $p = p(x_1,\ldots,x_n)$ and $\theta_1,\ldots,\theta_n$
  are congruences on an algebra, then
  \[
  p(\theta_1,\ldots, \theta_n) = \bigcup_{i\in \omega}
  p_i(\theta_1,\ldots, \theta_n)
  \]
  where $p_i$ is the term in the signature $\{\circ,
  \wedge\}$ obtained from~$p$ by replacing each occurrence of
  binary join with the $i$-fold relational product and
  keeping each occurrence of $\wedge$.  For example, if $p =
  x_1\wedge(x_2\vee x_3)$, then $p_1 = x_1\wedge x_2$,\; $p_2
  = x_1\wedge (x_2\circ x_3)$,\; $p_3 = x_1\wedge (x_2\circ
  x_3\circ x_2)$, etc.  Now, the inclusion
  $p(\theta_1,\ldots, \theta_n) = \bigcup
  p_i(\theta_1,\ldots, \theta_n) \leq Q(\theta_1,\ldots,
  \theta_n)$ holds in some congruence lattice if and only if
  $p_i(\theta_1,\ldots, \theta_n)\subseteq
  Q(\theta_1,\ldots, \theta_n)$ holds for every $i$. We will
  show that for any single $i$ the inclusion $p_i \subseteq
  Q$ can be characterized by a single Maltsev condition.

  As noted in Corollary~\ref{freese_thm}, there is a natural
  identification of~$Q$ with an ideal in the free lattice
  generated by $\{x_1,\ldots,x_n\}$.  This free lattice is
  countable, so the ideal associated to $Q$ has a cofinal
  sequence $q^0\leq q^1\leq q^2\leq \cdots$ of lattice
  words.  Let $q_j^k$ be the word in $\{\circ,\wedge\}$
  obtained from $q^k$ by replacing each join with $j$-fold
  alternating $\circ$-composition and keeping each
  occurrence of $\wedge$.  For example, if $q^k = (x_1\wedge
  x_2)\vee (x_1\wedge x_3)$, then $q^k_3 = (x_1\wedge x_2)
  \circ (x_1\wedge x_3)\circ (x_1\wedge x_2)$.  Let $r_j =
  q_j^0\circ q_j^1\circ\cdots\circ q_j^j$.  If
  $\theta_1,\ldots,\theta_n$ are congruences on some
  algebra, $\circ$~represents relational product, and
  $\wedge$ represents intersection of relations, then
  $q_j^k(\theta_1,\ldots, \theta_n) \subseteq
  r_j(\theta_1,\ldots, \theta_n)\subseteq
  q^j(\theta_1,\ldots, \theta_n)$ for all~$k\leq j$,~so
  \[
  Q(\theta_1,\ldots, \theta_n) = \bigcup_{j\in \omega}
  r_j(\theta_1,\ldots, \theta_n)\,.
  \]
  We will argue that
  \begin{enumerate}\label{ab}
  \item[(a)] any variety satisfying the congruence inclusion
    $p_i\subseteq Q = \bigcup r_j$ must satisfy a congruence
    inclusion of the form $p_i\subseteq r_j$ for some $j$,
    and that
    \item[(b)] the class of varieties satisfying any
    particular inclusion $p_i\subseteq r_j$ is definable by
    an idempotent linear strong Maltsev condition.
  \end{enumerate}
  If this strong Maltsev condition is denoted $\sigma_j$,
  then $\sigma_j\vdash \sigma_{j+1}$ (since
  $r_j(\theta_1,\ldots, \theta_n) \subseteq
  r_{j+1}(\theta_1,\ldots, \theta_n)$ for any choice of the
  $\theta$'s).  Therefore, (a) and~(b) will show that the
  class of varieties satisfying $p_i\subseteq Q$ is
  definable by the Maltsev condition $\bigvee \sigma_j$.

  Next we concern ourselves with the problem of constructing
  an idempotent linear strong Maltsev condition associated
  to a congruence inclusion $p\subseteq r$ where $p$ and $r$
  are terms in the signature $\{\circ, \wedge\}$.  The
  construction is a straightforward generalization of
  Maltsev's construction of a strong Maltsev condition
  defining congruence permutability, which many readers will
  already know. We will illustrate the construction by
  deriving the Maltsev condition associated with the
  congruence inclusion
  \begin{equation}\label{cong_incl}
    \theta_1\wedge(\theta_2\circ\theta_3)\subseteq
    (\theta_1\wedge\theta_2)\circ (\theta_1\wedge\theta_3)\,,
  \end{equation} which is $p\subseteq r$ for the
  $\{\circ,\wedge\}$-terms $p(x_1,x_2,x_3) =
  x_1\wedge(x_2\circ x_3)$ and $r(x_1,x_2,x_3) = (x_1\wedge
  x_2)\circ (x_1\wedge x_3)$.

  We begin the construction by building a directed labeled
  graph associated to a $\{\circ,\wedge\}$-term $p$.  Start
  with a graph $\vr G_1(p)$ having two vertices $y_1$ and
  $y_2$ connected by an edge $(y_1, y_2)$ directed from left
  to right and labeled with $p$, as depicted in
  Figure~\ref{fig5}.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(20,5) \thicklines \put(0,0){\circle*{1.6}}
\put(20,0){\circle*{1.6}} \put(0,0){\line(1,0){20}}
\put(11,0){\vector(1,0){1}} \put(-2,2){$y_1$} \put(9,2){$p$}
\put(19,2){$y_2$}
\end{picture}
\end{center}
\caption{ \ }\label{fig5}
\end{figure} From a partially completed graph $\vr G_i(p)$
the construction continues by selecting an edge directed
from left to right, labeled by a term $w$, and connecting
vertices $y_i$ and $y_j$.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(70,5) \thicklines \put(0,0){\circle*{1.6}}
\put(20,0){\circle*{1.6}} \put(0,0){\line(1,0){20}}
\put(11,0){\vector(1,0){1}} \put(-2,2){$y_i$}
\put(6,2){$u\wedge v$} \put(19,2){$y_j$}
%
\put(33,0){$\Rightarrow$}
%
\put(50,0){\circle*{1.6}} \put(70,0){\circle*{1.6}}
\bezier{200}(50,0),(60,5),(70,0)
\bezier{200}(50,0),(60,-5),(70,0)
\put(61,2.45){\vector(1,0){1}}
\put(61,-2.45){\vector(1,0){1}} \put(48,2){$y_i$}
\put(59,4){$u$} \put(59,-6.5){$v$} \put(69,2){$y_j$}
\end{picture}
\end{center}
\caption{ \ }\label{fig6}
\end{figure} If $w = u\wedge v$, then $\vr G_{i+1}(p)$ is the
graph obtained from $\vr G_i(p)$ by replacing the edge
labeled $w$ with two edges directed from left to right,
labeled $u$ and $v$ respectively, connecting the same
vertices in parallel, as in Figure~\ref{fig6}.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(90,5) \thicklines \put(0,0){\circle*{1.6}}
\put(20,0){\circle*{1.6}} \put(0,0){\line(1,0){20}}
\put(11,0){\vector(1,0){1}} \put(-2,2){$y_i$}
\put(6,2){$u\circ v$} \put(19,2){$y_j$}
%
\put(33,0){$\Rightarrow$}
%
\put(50,0){\circle*{1.6}} \put(70,0){\circle*{1.6}}
\put(90,0){\circle*{1.6}} \put(50,0){\line(1,0){40}}
\put(61,0){\vector(1,0){1}} \put(81,0){\vector(1,0){1}}
\put(48,2){$y_i$} \put(59,2){$u$} \put(68,2){$y_k$}
\put(79,2){$v$} \put(89,2){$y_j$}
\end{picture}
\end{center}
\caption{ \ }\label{fig7}
\end{figure} If instead $w = u\circ v$, then construct $\vr
G_{i+1}(p)$ by replacing the edge labeled $w$ with two edges
directed from left to right, labeled $u$ and $v$, connecting
the same vertices in serial through a new vertex, as in
Figure~\ref{fig7}.  If neither of the above steps can be
performed, then all edges are labeled by variables and the
construction is complete.  The construction will eventually
end since each step alters the complexity of exactly one
edge label, and reduces the complexity of that label without
introducing any more labels of the same or larger
complexity.

In our illustrative example the term $p$ equals
$x_1\wedge(x_2\circ x_3)$ and the graphs that arise in the
three steps of the construction are depicted in
Figure~\ref{fig7.5}.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(110,22) \thicklines
\put(0,12){\circle*{1.6}} \put(20,12){\circle*{1.6}}
\put(0,12){\line(1,0){20}} \put(11,12){\vector(1,0){1}}
\put(-2,14){$y_1$} \put(9,14){$p$} \put(6.5,-2){$\vr
G_{1}(p)$} \put(19,14){$y_2$}
%
\put(28,12){$\Rightarrow$}
%
\put(40,12){\circle*{1.6}} \put(60,12){\circle*{1.6}}
\bezier{200}(40,12),(50,17),(60,12)
\bezier{200}(40,12),(50,7),(60,12)
\put(51,14.45){\vector(1,0){1}}
\put(51,9.55){\vector(1,0){1}} \put(38,14){$y_1$}
\put(48.5,17.5){$x_1$} \put(45,5.5){$x_2\circ x_3$}
\put(46.5,-2){$\vr G_{2}(p)$} \put(59,14){$y_2$}
%
\put(68,12){$\Rightarrow$}
%
\put(80,12){\circle*{1.6}} \put(95,12){\circle*{1.6}}
\put(110,12){\circle*{1.6}} \put(80,12){\line(1,0){30}}
\bezier{200}(80,12),(95,27),(110,12)
\put(88,12){\vector(1,0){1}} \put(103,12){\vector(1,0){1}}
\put(95.35,19.4){\vector(1,0){1}} \put(77.5,14){$y_1$}
\put(93,14){$y_3$} \put(110,14){$y_2$} \put(93,22){$x_1$}
\put(86,7.5){$x_2$} \put(101,7.5){$x_3$} \put(91.5,-2){$\vr
G_{3}(p)$}
\end{picture}
\end{center}
\caption{The construction of $\vr G(p)$ for
$p=x_1\wedge(x_2\circ x_3)$}\label{fig7.5}
\end{figure}

Call the final graph in the construction $\vr G(p)$.  We may
assume that the vertices are $Y = \{y_1,\ldots, y_m\}$ and
that the edge labels are the variables occurring in $p$,
which are among the variables $x_1,\ldots, x_n$.  The key
features of $\vr G(p)$ are stated in Claim~\ref{clmG(p)}.

\begin{clm}\label{clmG(p)}
  Let\/ $\m a$ be an algebra and $\alpha_i\in\Con(\m a)$ for
  $1\leq i\leq n$.
  \begin{enumerate}
  \item[(1)] If\/ $Y\to A\colon y_s\mapsto a_s$ is an
    assignment with the property that $(a_i,a_j)\in\alpha_k$
    whenever $(y_i,y_j)$ is an $x_k$-labeled edge of\/~$\vr
    G(p)$, then $(a_1,a_2)\in p(\alpha_1,\ldots,\alpha_n)$.
  \item[(2)] Conversely, given any $(a_1,a_2)\in
    p(\alpha_1,\ldots,\alpha_n)$, there is an assignment\/
    $Y\to A\colon y_s\to a_s$ extending $y_1\mapsto a_1$,
    $y_2\mapsto a_2$ such that $(a_i,a_j)\in \alpha_k$
    whenever $(y_i,y_j)$ is an $x_k$-labeled edge of\/~$\vr
    G(p)$.
  \end{enumerate}
\end{clm}


To prove item~(1), choose and fix assignments $Y\to A\colon
y_s\mapsto a_s$ of vertex labels of $\vr G(p)$ to elements
of $A$ and $X\to \Con(\m a)\colon x_s\mapsto \alpha_s$ of
edge labels to congruences on $\m a$.  With respect to these
choices, consider the following property of some $\vr
G_{\ell}(p)$: ``For each $w$-labeled edge $(y_i,y_j)$ the
pair $(a_i,a_j)$ is in
$w(\overline{\alpha}):=w(\alpha_1,\ldots,\alpha_n)$.''
Item~(1) asserts that if this property holds for $\vr G(p)$,
then it holds for $\vr G_1(p)$.  Therefore, it is enough to
show that if the property holds for $\vr G_{\ell+1}(p)$,
then it holds for $\vr G_{\ell}(p)$. For a construction step
of the `parallel type' (Figure~\ref{fig6}) this is a
consequence of the fact that if $(a_i,a_j)\in
u(\overline{\alpha})$ and $(a_i,a_j)\in
v(\overline{\alpha})$, then $(a_i,a_j)\in u\wedge
v(\overline{\alpha})$.  For a construction step of the
`serial type' (Figure~\ref{fig7}) this is a consequence of
the fact that if $(a_i,a_k)\in u(\overline{\alpha})$ and
$(a_k,a_j)\in v(\overline{\alpha})$, then $(a_i,a_j)\in
u\circ v(\overline{\alpha})$.


For (2), choose and fix an assignment $X\to \Con(\m a)\colon
x_s\mapsto \alpha_s$.  Consider this property of an
assignment $Y\to A\colon y_s\mapsto a_s$ of vertex labels of
$\vr G_{\ell}(p)$ to elements of $A$: ``For each $w$-labeled
edge $(y_i,y_j)$ the pair $(a_i,a_j)$ is in
$w(\overline{\alpha})$.''  Item (2) asserts that any
assignment $y_1\mapsto a_1$, $y_2\mapsto a_2$ of $\vr
G_1(p)$ which satisfies this property can be extended to an
assignment of the vertex labels of $\vr G(p)$ which
satisfies this property.  It is enough to prove that if an
assignment of the vertex labels of $\vr G_{\ell}(p)$ has the
property, then this assignment can be extended to an
assignment of the vertex labels of $\vr G_{\ell+1}(p)$ that
has the property.  This is trivial for a construction step
of the parallel type (Figure~\ref{fig6}), since no new
vertex labels are introduced and the assignment $y_s\mapsto
a_s$ that witnesses the satisfaction of the property for
$\vr G_{\ell}(p)$ also witnesses it for $\vr G_{\ell+1}(p)$.
For a construction step of the serial type
(Figure~\ref{fig7}), a new vertex with label $y_k$ is
introduced to replace an edge $(y_i,y_j)$ labeled $u\circ v$
with edges $(y_1,y_k)$ and $(y_k,y_j)$ labeled $u$ and $v$
respectively.  In this case, extend the witnessing
assignment $y_s\mapsto a_s$ for $\vr G_i(p)$ by defining
$y_k\mapsto a_k$ where $a_k$~is chosen so that $(a_i,a_k)\in
u(\overline{\alpha})$ and $(a_k,a_j)\in
v(\overline{\alpha})$.  It is possible to make such a
choice, since $u\circ v(\overline{\alpha}) =
u(\overline{\alpha}) \circ v(\overline{\alpha})$, and such a
choice witnesses the desired property for~$\vr
G_{\ell+1}(p)$. Thus the claim is proved.

\medskip

Now construct a directed graph $\vr G(r)$ for $r$ in exactly
the same way, except use $z$'s instead of $y$'s for the
vertices.  See Figure~\ref{fig7.6} for~$\vr G(r)$ when $r$
equals $(x_1\wedge x_2)\circ (x_1\wedge x_3)$.

\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(40,12) \thicklines \put(0,6){\circle*{1.6}}
\put(20,6){\circle*{1.6}} \bezier{200}(0,6),(10,11),(20,6)
\bezier{200}(0,6),(10,1),(20,6)
\put(11,8.45){\vector(1,0){1}}
\put(11,3.55){\vector(1,0){1}} \put(-2.5,8.5){$z_1$}
\put(8.5,11.5){$x_1$} \put(8.5,-0.5){$x_2$}
\put(19,8.5){$z_3$} \put(20,6){\circle*{1.6}}
\put(40,6){\circle*{1.6}} \bezier{200}(20,6),(30,11),(40,6)
\bezier{200}(20,6),(30,1),(40,6)
\put(31,8.45){\vector(1,0){1}}
\put(31,3.55){\vector(1,0){1}} \put(28.5,11.5){$x_1$}
\put(28,-0.5){$x_3$} \put(39.5,8.5){$z_2$}
\end{picture}
\end{center}
\caption{$\vr G(r)$ for $r=(x_1\circ x_2)\wedge(x_1\circ
x_3)$}\label{fig7.6}
\end{figure}

If the set of vertex labels for $\vr G(r)$ is $\{z_1,
\ldots, z_{\ell}\}$, then consider the following strong
Maltsev condition associated to $p\subseteq r$: There exist
$m$-ary terms $F_1, \ldots, F_{\ell}$ for which the
following identities are satisfied:
\begin{enumerate}
\item[(i)] $F_1(y_1,\ldots, y_m)\approx y_1$,
\item[(ii)] $F_2(y_1,\ldots, y_m)\approx y_2$, and
\item[(iii)] an identity of the form $F_i(\mbox{\rm
    variables})\approx F_j(\mbox{\rm variables})$ for each
  edge $(z_i, z_j)$ in $\vr G(r)$.  The pattern of the
  variables is determined as follows. Given a variable
  $x_k$, let $E_k$ be the equivalence relation on the set
  $Y$ of vertex labels of $\vr G(p)$ defined by
  $y_i\wrel{E_k} y_j$ if $y_i$ and $y_j$ belong to the same
  ``$x_k$-component'', by which we mean that there is an
  undirected path from~$y_i$ to~$y_j$ in~$\vr G(p)$
  consisting of $x_k$-labeled edges.  Let $\tau_k\colon Y\to
  Y$ be a transversal for $E_k$.  Now, if $(z_i,z_j)$ is
  labeled by $x_k$, then the identity associated with $(z_i,
  z_j)$ is
  \[
  F_i\big(\tau_k(y_1), \ldots, \tau_k(y_m)\big)\approx
  F_j\big(\tau_k(y_1), \ldots, \tau_k(y_m)\big)\,.
  \]
\end{enumerate} This is a linear strong Maltsev condition.
The idempotence of $F_1$, i.e., $F_1(x,\ldots, x)\approx x$,
is a consequence of identity (i) from above.  There is a
directed path in $\vr G(r)$ from the vertex labeled $z_1$ to
any other vertex, and for any $(z_i, z_j)$ in such a path
the identity from (iii) above has the consequence
$F_i(x,\ldots, x)\approx F_j(x,\ldots, x)$.  Thus, the
idempotence of each $F_k$ follows from the identities (i)
and (iii).

In our illustrative example, (\ref{cong_incl}), we have $p =
x_1\wedge (x_2\circ x_3)$ and $r = (x_1\wedge x_2)\circ
(x_1\wedge x_3)$, and $\vr G(p)$ and $\vr G(r)$ are depicted
in Figures~\ref{fig7.5} and~\ref{fig7.6}.  The associated
Maltsev condition involves three ternary terms $F_1$, $F_2$
and $F_3$, corresponding to the $z$'s in
Figure~\ref{fig7.6}.  The equivalence relation $E_k$ on the
vertex labels of $\vr G(p)$ associated to the
$x_k$-component can be read off Figure~\ref{fig7.5}; we
represent them by their partition of $\{y_1, y_2, y_3\}$:
$E_1\leftrightarrow y_1y_2/y_3$, $E_2\leftrightarrow
y_1y_3/y_2$, and $E_3\leftrightarrow y_1/y_2y_3$. The
transversals we choose for these equivalence relations are
the functions~$\tau_k$ which assign to each element of $Y$
the $E_k$-related element whose subscript is smallest.  With
these choices, the Maltsev condition asserts that the
following identities hold.
\begin{itemize}
\item $F_1(y_1,y_2,y_3)\approx y_1$,
\item $F_2(y_1,y_2,y_3)\approx y_2$,
\item and corresponding to $\tau_1$, $\tau_2$, and $\tau_3$:
  \begin{itemize}
  \item $F_1(y_1,y_1,y_3)\approx F_2(y_1,y_1,y_3)\approx
    F_3(y_1,y_1,y_3)$,
  \item $F_1(y_1,y_2,y_1)\approx F_3(y_1,y_2,y_1)$, and
  \item $F_2(y_1,y_2,y_2)\approx F_3(y_1,y_2,y_2)$.
  \end{itemize}
\end{itemize}

Next we argue that the congruence lattices of algebras in a
variety~$\vr V$ satisfy $p\subseteq r$ if and only if $\vr
V$ satisfies the strong Maltsev condition defined above in
(i)---(iii).  We begin by saying what we will mean by a
``generic'' pair $(u,v)\in p$.  Let $\m f = \m f_{\vr V}(Y)$
be the free algebra in~$\vr V$ generated by $Y =
\{y_1,\ldots,y_m\}$.  For each edge label $x_k$ in $\vr
G(p)$ let $\theta_k$~be the congruence on $\m f$ generated
by all pairs $(y_s, y_t)$ that are edges in~$\vr G(p)$ with
label~$x_k$, i.e., $\theta_k = \Cg f(E_k)$ where $E_k$ is
defined as above. This is the least congruence on $\m f$
that relates generators $y_i$ and $y_j$ if $y_i$~is
connected to $y_j$ in $\vr G(p)$ by edges labeled $x_k$. By
Lemma~\ref{T_theta}~(1), $\theta_k$~is the kernel of the
retraction $\widehat{\tau_k}\colon \m f\to \m f$ induced by
the function $\tau_k\colon Y\to Y$ defined in (iii) above.
It is the pair $(y_1,y_2)=:(u,v)$ that will be called a {\bf
  generic}\index{generic pair} pair $(u,v)\in p$.  The fact
that a ``generic pair in~$p$'' actually is in
$p(\theta_1,\ldots,\theta_n)$ follows from
Claim~\ref{clmG(p)}~(1).

If $\vr V$ satisfies the congruence inclusion $p\subseteq
r$, then the generic pair in~$p$ must belong to $r$, i.e.,
$(y_1,y_2)\in r(\theta_1,\ldots,\theta_n)$ in $\m f$. Using
only the fact that $(y_1,y_2)\in
r(\theta_1,\ldots,\theta_n)$ we will prove that $\vr V$
satisfies the Maltsev condition described above.  By
Claim~\ref{clmG(p)}~(2), the fact that $(y_1,y_2)\in
r(\theta_1,\ldots,\theta_n)$ implies that there exist $f_1 =
y_1$, $f_2 = y_2$, $f_3, \ldots, f_{\ell}\in F$ such that
the assignment $z_s\mapsto f_s$ has the property that
$(f_i,f_j)\in\theta_k$ whenever $(z_i,z_j)$ is an
$x_k$-labeled edge of $\vr G(r)$. Since $\m f$ is generated
by $Y = \{y_1,\ldots,y_m\}$, there exist terms $F_i$ such
that $f_i = F_i(y_1,\ldots,y_m)$ for all $i$.  Since
$F_1(y_1,\ldots,y_m) = y_1$ holds as a relation among the
free generators of $\m f$, it follows that
$F_1(y_1,\ldots,y_m)\approx y_1$ holds as an identity
in~$\vr V$. Similarly, $F_2(y_1,\ldots,y_m)\approx y_2$
holds in~$\vr V$.  For each $x_k$-labeled edge $(z_i,z_j)$
we have
\[
\begin{array}{r@{}l} F_i\big(\tau_k(y_1), \ldots,
\tau_k(y_m)\big) &{} = \widehat{\tau_k}\big(F_i(y_1, \ldots,
y_m)\big)\\ &{} =\widehat{\tau_k}(f_i)\\ &{} =
\widehat{\tau_k}(f_j)\\ &{} =F_j\big(\tau_k(y_1), \ldots,
\tau_k(y_m)\big)\,,
\end{array}
\]
because $(f_i,f_j)\in\theta_k = \ker(\widehat{\tau_k})$.
Since $F_i\big(\tau_k(\wec{y})\big) =
F_j\big(\tau_k(\wec{y})\big)$ holds as a relation among free
generators, $F_i\big(\tau_k(\wec{y})\big) \approx
F_j\big(\tau_k(\wec{y})\big)$ holds as an identity in $\vr
V$.  Altogether this shows that $\vr V$ satisfies the
Maltsev condition described in (i)---(iii) above.

Conversely assume that $\vr V$ satisfies the Maltsev
condition we have described.  Choose any $\m a\in {\vr V}$
and congruences $\alpha_1,\ldots,\alpha_n$. To show that
$p(\alpha_1,\ldots,\alpha_n)\subseteq
r(\alpha_1,\ldots,\alpha_n)$, choose $(a_1,a_2)\in
p(\alpha_1,\ldots,\alpha_n)$.  According to
Claim~\ref{clmG(p)}~(2), there is an assignment $v\colon
Y\to A\colon y_s\mapsto a_s$ of the variables of $\vr G(p)$
in $A$ such that $(a_i,a_j)\in\alpha_k$ if $(y_i,y_j)$ is
labeled by~$x_k$.  If $\widehat{v}\colon \m f_{\vr V}(Y)\to
\m a$ is the extension of $v$ to a homomorphism, and
$\theta_k$~is the congruence generated by all pairs
$(y_i,y_j)$ whose edge label in~$\vr G(p)$ is~$x_k$, then
$\widehat{v}(\theta_k)\subseteq \alpha_k$ since
$\widehat{v}\big((y_i,y_j)\big)=(a_i,a_j)\in\alpha_k$ for
any generating pair $(y_i,y_j)$ of $\theta_k$.

Now consider the assignment $Z\to F\colon z_s\mapsto
F_s(y_1,\ldots,y_m)$ where $Z = \{z_1,\ldots,z_{\ell}\}$.
The identities (i)---(iii) of our Maltsev condition
guarantee respectively that
\begin{enumerate}
\item[(i)] $z_1\mapsto F_1(\wec{y})=y_1$,
\item[(ii)] $z_2\mapsto F_2(\wec{y})=y_2$, and
\item[(iii)] if $(z_i,z_j)$ is an $x_k$-labeled edge of $\vr
  G(r)$, then
  $\big(F_i(\wec{y}),F_j(\wec{y})\big)\!\in\theta_k$.
\end{enumerate} By Claim~\ref{clmG(p)}~(1), these conditions
yield $(y_1,y_2)\in r(\theta_1,\ldots,\theta_n)$, hence
\[
(a_1,a_2) = \widehat{v}\big((y_1,y_2)\big) \in
\widehat{v}\big(r(\theta_1,\ldots,\theta_n)\big) \subseteq
r(\alpha_1,\ldots,\alpha_n)\,.
\]
Since the pair $(a_1,a_2)\in p(\alpha_1,\ldots,\alpha_n)$
was arbitrary, we derive that $p\subseteq r$ holds
throughout~$\vr V$.

Recall that we reduced the proof of
Theorem~\ref{pixley-wille_extension} to two claims: (a) any
variety satisfying the congruence inclusion $p_i\subseteq Q
= \bigcup r_j$ must satisfy a congruence inclusion of the
form $p_i\subseteq r_j$ for some~$j$, and that (b) the class
of varieties satisfying any particular inclusion
$p_i\subseteq r_j$ is definable by an idempotent linear
strong Maltsev condition (see page~\pageref{ab}). We have
just completed the proof of a strong form of~(b): the class
of varieties satisfying $p_i\subseteq r_j$ is definable by
an idempotent linear strong Maltsev condition {\em which
  expresses the fact that $(u,v)\in r_j$ for a generic pair
  $(u,v)$ of\/ $p_i$}.  Thus, we obtain~(a) as well: a
variety~$\vr V$ that satisfies $p_i\subseteq Q = \bigcup
r_j$ must satisfy $(u,v)\in r_j$ for some $j$ and some
generic pair $(u,v)$ of $p_i$, and for this $i$ and~$j$ we
have~$p_i\subseteq r_j$.
\end{proof}

For future reference we point out that this proof shows that
the class of varieties satisfying a congruence inclusion of
the form $p\subseteq r$, where $p$ and $r$~are
$\{\circ,\wedge\}$-terms, is definable by a strong Maltsev
condition, and that the class of varieties satisfying a
congruence inclusion of the form $p\subseteq
\bigcup_{i\in\omega} r_i$, where $p$ and $r_i$ are
$\{\circ,\wedge\}$-terms is definable by a Maltsev condition,
provided the $r_i$'s are increasing.

\begin{cor}\label{pixley-wille_cor} If a variety satisfies a
nontrivial meet continuous congruence identity, then it
satisfies a nontrivial idempotent Maltsev condition.
\end{cor}\index{Pixley--Wille algorithm, the|)}

\section{Congruence Identities From Maltsev
Conditions}\label{4.2}

We prove the converse to Corollary~\ref{pixley-wille_cor} in
Theorem~\ref{typ1condcor} of this section.

\begin{thm}\label{typ1cond}
  Let\/ $\vr V$ be a variety that satisfies a nontrivial
  idempotent Maltsev condition. If\/ $\m a\in {\vr V}$ has
  congruences $\alpha, \beta, \gamma, \delta$
  satisfying
  \begin{enumerate}
  \item[(1)] $\alpha\cap (\beta\circ \gamma) \cap
    (\gamma\circ \beta) \subseteq \delta \subseteq
    \alpha\subseteq \beta\vee\gamma$ and
  \item[(2)] $\C c(\beta \vee \gamma, \alpha; \delta)$,
  \end{enumerate} then $\delta = \alpha$.
\end{thm}

\begin{proof}
  Let $f(x_1,\ldots,x_n)$ be a Taylor term for the
  variety~$\vr V$, and let $N = \{1,\ldots, n\}$ be the set
  of subscripts of variables in this term.  The $i$-th
  Taylor identity has the form $f(x_1,\ldots, x_n) \approx
  f(y_1,\ldots, y_n)$ where $(x_j,y_j)\in \{(x,x), (x,y),
  (y,x), (y,y)\}$ for all $j$, and $(x_i,y_i)=(x,y)$.  As in
  the proof of Theorem~\ref{abel_tol=abel_cong}, partition
  $N$ into blocks $\{B_1, \ldots, B_5\}$:
  \begin{enumerate}
  \item[(1)] $B_1$ consists of those $j\in N$ for which
    $(x_j,y_j)= (x,x)$,
  \item[(2)] $B_2$ consists of those $j\neq i$ for which
    $(x_j,y_j)= (x,y)$,
  \item[(3)] $B_3=\{i\}$,
  \item[(4)] $B_4$ consists of those $j\in N$ for which
    $(x_j,y_j)= (y,x)$, and
  \item[(5)] $B_5$ consists of those $j$ for which
    $(x_j,y_j)= (y,y)$.
  \end{enumerate}
  Let $f_i(z_1,z_2,z_3,z_4,z_5)$ be the $5$-ary term
  obtained from $f$ by substituting the variable $z_1$ for
  the variable $x_j$ of $f(x_1,\ldots,x_n)$ if $j\in B_1$,
  $z_2$ if $j\in B_2$, and so on.  The $i$-th Taylor
  identity can be written as $f_i(x,x,x,y,y)\approx
  f_i(x,y,y,x,y)$.

  Now suppose that $\m a\in {\vr V}$ has congruences
  $\alpha$, $\beta$, $\gamma$, $\delta$ satisfying all the
  hypotheses of the theorem, and also satisfying $\alpha
  \neq \delta$.  Then, since $\delta<
  \alpha\leq\beta\vee\gamma$, there is a least $k$ such
  that $\alpha\cap (\beta\circ_k\gamma)\not\subseteq
  \delta$.\footnote{Recall that $\beta\circ_k\gamma =
    \beta\circ\gamma\circ \beta\cdots$, with $k-1$
    occurrences of $\circ$.}  This $k$ is at least $2$ since
  \[
  (\alpha\cap\beta)\cup(\alpha\cap\gamma)\subseteq
  \alpha\cap (\beta\circ \gamma) \cap (\gamma\circ
  \beta) \subseteq \delta\,.
  \]
  By interchanging the roles of $\beta$ and $\gamma$ if
  necessary we may assume that $\alpha\cap
  (\beta\circ_{k-1}\gamma)\subseteq \delta$ and
  $\alpha\cap (\gamma\circ_{k-1}\beta)\subseteq \delta$.

  Choose $(u,v)\in \alpha\cap(\beta\circ_k\gamma) -
  \delta$, and suppose that
  \[
  u = u_0 \wrel{\beta}u_1\wrel{\gamma}u_2\wrel{\cdots} u_k=v
  \]
  is a $\beta\circ_k\gamma$-chain connecting $u$ to $v$.  If
  $p(x,y,z) = f_i(u_0,x,y,z,u_1)$, then for $k>2$ the
  sequence
  \[
  \begin{array}{r@{}l}
    r=p(u_1,u_1,u_0)&{} \wrel{\gamma}\/
    p(u_1,u_2,u_0)\wrel{\beta}\/ p(u_0,u_3,u_1)\\
    &{} \wrel{\gamma}\/ p(u_0,u_4,u_1) \wrel{\beta}
    \cdots\; p(u_0,u_k,u_1) = s
  \end{array}
  \]
  is a $\gamma\circ_{k-1}\beta$-chain. Moreover, the $i$-th
  Taylor identity implies that
  \begin{align*}
    p(u_0,u_0,u_1) 
    &{} = f_i(u_0, u_0, u_0, u_1 ,u_1)\\
    &{} = f_i(u_0, u_1, u_1, u_0 ,u_1) = p(u_1, u_1, u_0)\,,
  \end{align*}
  so $r = p(u_0,u_0,u_1)$ is $\alpha$-related to $s = p(u_0,
  u_k, u_1)$ (since $(u_0,u_k)\in\alpha$).  Thus
  $(r,s)\in\alpha\cap(\gamma\circ_{k-1}\beta)\subseteq
  \delta$ when $k>2$.  If $k=2$, then the same conclusion
  (that $(r,s)\in\delta$) may be reached as follows.
  Starting with the same $\beta\circ_k\gamma$-chain $u =
  u_0\wrel{\beta}u_1 \wrel{\gamma}u_2 = v$, and the same
  polynomial $p(x,y,z) = f_i(u_0,x,y,z,u_1)$, we obtain a
  chain
  \[
  r = p(u_1,u_1,u_0)\wrel{\gamma} p(u_1,u_2,u_0)\wrel{\beta}
  p(u_0,u_2,u_1) = s\,.
  \]
  This chain is not shorter (it has length $k=2$ also), but
  it begins with $\gamma$ instead of $\beta$. All other
  conclusions from the $k>2$ case remain valid, in
  particular we have
  \[
  r = p(u_1,u_1,u_0)=p(u_0,u_0,u_1)\wrel{\alpha}
  p(u_0,u_k,u_1) = s\,.
  \]
  This shows that $(r,s)\in \alpha\cap(\gamma\circ\beta)$.
  But since $(r,s)$ is an image of $(u_0,u_2) = (u,v)$ under
  the polynomial $p(u_0,x,u_1)$, and
  $(u,v)\in\beta\circ\gamma$, we even have $(r,s)\in
  \alpha\cap(\beta\circ\gamma)\cap
  (\gamma\circ\beta)\subseteq \delta$.

  We now have
  \[
  f_i(u_0,u_0,u_0,u_1,u_1) = p(u_0,u_0,u_1) = r
  \]
  and
  \[
  f_i(u_0,u_0,u_k,u_1,u_1) = p(u_0,u_k,u_1) = s\,,
  \]
  so from $(r,s)\in\delta$ we get
  \[
  f_i(u_0,u_0,u_0,\underline{u_1},\underline{u_1})
  \wrel{{\delta}}
  f_i(u_0,u_0,u_k,\underline{u_1},\underline{u_1})\,.
  \]
  Since $\C c(\beta\vee\gamma,\alpha;\delta)$ holds, we may
  change underlined values from $u_1$ to $u_0$ to obtain
  \[
  f_i(u_0,u_0,u_0,\underline{u_0},\underline{u_0})
  \wrel{{\delta}}
  f_i(u_0,u_0,u_k,\underline{u_0},\underline{u_0})\,,
  \]
  which we write as
  \[
  f_i(u,u,u,u,u) \wrel{{\delta}} f_i(u,u,v,u,u)\,.
  \]
  This implies that for our original Taylor term $f$ we have
  \[
  f(u,\ldots, u, v, u, \ldots, u) \wrel{\delta} f(u, u,
  \ldots, u) = u
  \]
  where the $v$ occurs in the $i$-th argument of~$f$. This
  holds for every~$i$.  We now argue that the relation
  \[
  f(v,\dots,v,u,\dots,u)\wrel{\delta} u\,,
  \]
  where the last $v$ occurs in the $i$-th argument of~$f$,
  also holds for every~$i$.

  The facts that $(u,v)\in\alpha\leq \beta\vee\gamma$ and
  $\C c(\beta\vee\gamma,\alpha;\delta)$ imply that we may
  change the underlined $u$ to $v$ in
  \[
  f(v,\dots, v, \underline{u}, u,\dots,u)\wrel{\delta} u =
  f(u,\dots, u, \underline{u}, u,\dots, u)
  \]
  to obtain
  \[
  f(v,\dots, v, \underline{v}, u,\dots,u)\wrel{\delta} u =
  f(u,\dots, u, \underline{v}, u,\dots, u)\,.
  \]
  By our earlier conclusions, the value $f(u,\dots, u, v,
  u,\dots, u)$ on the right side is $\delta$-related to $u$.
  By induction and the idempotence of~$f$ we get that $v =
  f(v, v, \ldots, v)\wrel{\delta} u$, which is a
  contradiction.
\end{proof}


The hypotheses of the previous theorem are consequences of
purely order-theoretic conditions. For the statement of the
next lemma, let $(p, q, r)$ be a triple of lattice
variables, and define lattice words
\begin{gather*}\label{triangle-terms}
p_{[0]}:=p\,,\qquad q_{[0]}:=q\,,\qquad r_{[0]}:=r\,,\\
p_{[n+1]}:= p_{[n]}\wedge (q_{[n]}\vee r_{[n]})\,,\qquad
q_{[n+1]}:= q_{[n]}\wedge (p_{[n]}\vee r_{[n]})\,,\\
r_{[n+1]}:= r_{[n]}\wedge (p_{[n]}\vee q_{[n]})\,.
\end{gather*}
If $(\alpha, \beta, \gamma)$ is a triple of congruences on
$\m a$, then let
\[
\alpha_{[n]}:=p_{[n]}(\alpha,\beta,\gamma)\,,\quad
\beta_{[n]}:=q_{[n]}(\alpha,\beta,\gamma)\,,\quad
\gamma_{[n]}:=r_{[n]}(\alpha,\beta,\gamma)\,.
\]

\begin{lm}\label{typ1condlm} 
  Let $(\alpha, \beta, \gamma)$ be a triple of congruences
  on $\m a$, and let\/ $\delta$ be a congruence on $\m a$.
  The following two conditions imply the corresponding
  conditions of Theorem~\ref{typ1cond}.
  \begin{enumerate}
  \item[(1)]
    $\alpha_{[n]}\leq\delta\leq\alpha\leq\beta\vee\gamma$
    for some $n$.
  \item[(2)] $\alpha\wedge\beta = \alpha\wedge\gamma =
    \delta$.
  \end{enumerate}
\end{lm}

\begin{proof} 
  Choose $(u, w)\in \alpha\cap(\beta\circ\gamma)$, then
  there is a $v$ such that $u\wrel{\beta}v\wrel{\gamma}w$.
  For $k=0$, these three elements are related as in the
  following triangle:
\begin{figure}[h]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(18,22)
\put(0,3.5){\circle*{1.2}}
\put(18,3.5){\circle*{1.2}}
\put(9,18.5){\circle*{1.2}}
\thicklines
\put(0,3.5){\line(1,0){18}}
\put(0,3.5){\line(3,5){9}}
\put(18,3.5){\line(-3,5){9}}
%
\put(-2,0){$u$}
\put(17.5,0){$w$}
\put(8,21){$v$}
%
\put(6,0.5){$\alpha_{[k]}$}
\put(-3,12.5){$\beta_{[k]}$}
\put(14,12.5){$\gamma_{[k]}$}
\end{picture}
\end{center}
\caption{ \ }\label{fig8}
\end{figure}

If $u$, $v$ and $w$ are related in this way for some $k\geq
0$, then we have $(u,w)\in
\alpha_{[k]}\cap(\beta_{[k]}\vee\gamma_{[k]})
=\alpha_{[k+1]}$ and similarly $(u,v)\in\beta_{[k+1]}$ and
$(v,w)\in\gamma_{[k+1]}$. This shows that
$\alpha\cap(\beta\circ\gamma)\subseteq \alpha_{[k]}$ for all
$k$.  If $\alpha_{[n]}\leq\delta$ holds for some $n$, then
\[
\alpha\cap (\beta\circ \gamma) \cap (\gamma\circ
\beta) \subseteq \alpha\cap (\beta\circ \gamma) \subseteq
\alpha_{[n]}\subseteq \delta\,,
\]
so the first condition of this lemma implies the first
condition of Theorem~\ref{typ1cond}.

Now if $\alpha\wedge\beta = \alpha\wedge\gamma = \delta$, then
we have $\C c(\beta,\alpha;\delta)$ and $\C
c(\gamma,\alpha;\delta)$ according to item (8) of
Theorem~\ref{basic_centrality}.  By the semidistributivity
of the centralizer in its first variable
(Theorem~\ref{basic_centrality}~(5)) we derive that $\C
c(\beta\vee\gamma,\alpha;\delta)$ holds.  This shows that
the conditions of this lemma imply those of
Theorem~\ref{typ1cond}.
\end{proof}

Now we are prepared to prove the main result of this
section.

\begin{thm}\label{typ1condcor}
  The following conditions on a variety\/ $\vr V$ are
  equivalent.
  \begin{enumerate}
  \item[(1)] $\vr V$ satisfies an idempotent Maltsev
    condition that fails in the variety of sets.
  \item[(2)] The quasi-identity
    \begin{equation}\label{weak_W}
      \big((p\wedge q)\approx s\big)
      \wrel{\&} \big((p\wedge r)\approx s\big)
      \wrel{\&} (p_{[2]}\approx s)
      \wrel{\to} \big((p\wedge (q\vee r))\approx s\big)
    \end{equation} holds in the congruence lattices of algebras
    in $\vr v$ (the lattice word\/~$p_{[2]}$ is defined
    on page~\pageref{triangle-terms}). 
  \item[(3)] Quasi-identity (\ref{weak_W}) holds in the meet
    continuous congruence variety of\/ $\vr V$.
  \item[(4)] $\vr V$ satisfies a nontrivial meet continuous
    congruence identity.
  \item[(5)] $\vr V$ satisfies a congruence inclusion of the
    form
    \[
    \alpha\cap(\beta\circ_{4}\gamma)\subseteq
    w(\alpha, \beta, \gamma)
    \]
    for some lattice word $w(p,q,r)$ such that
    $w(p,q,r)<p\wedge (q\vee r)$ in the free lattice $\m
    f_{\vr L}(\{p,q,r\})$.
  \item[(6)] For some $k\geq 4$, $\vr V$ satisfies a
    congruence inclusion of the form
    \[
    \alpha\cap(\beta\circ_{k}\gamma)\subseteq w(\alpha,
    \beta, \gamma)\,,
    \]
    where $w(p,q,r) < p\wedge (q\vee r)$ in\/ $\m f_{\vr
      L}(\{p,q,r\})$.
  \end{enumerate}
\end{thm}

A seventh equivalent condition is described in the remarks
following the proof.

\begin{proof}
  To see that (1)$\mimplies$(2), choose an assignment of
  variables $p\mapsto \alpha$, $q\mapsto \beta$, $r\mapsto
  \gamma$, $s\mapsto \delta$ in $\Conb(\m a)$ for some $\m
  a\in {\vr V}$ for which the premises of quasi-identity
  (\ref{weak_W}) are satisfied.  Then the congruences
  $\alpha' := \alpha\wedge(\beta\vee\gamma)$, $\beta$,
  $\gamma$ and $\delta$ satisfy the conditions of
  Lemma~\ref{typ1condlm}.  (This uses the fact that
  $\alpha'\leq \alpha$, so $\alpha'_{[2]}\leq \alpha_{[2]} =
  \delta$.)  Hence by Theorem~\ref{typ1cond} we have
  $\alpha\wedge(\beta\vee\gamma) = \alpha' = \delta$, which
  shows that the conclusion of quasi-identity (\ref{weak_W})
  is satisfied by this assignment.

  To show that (2)$\miff$(3) we argue that quasi-identity
  (\ref{weak_W}) is equivalent to a meet continuous lattice
  identity.  This will follow from Theorem~\ref{strange-thm}
  if we verify that the lattice words $w_1(p,q,r):=p\wedge
  q$, $w_2(p,q,r):=p\wedge r$, and $w_3(p,q,r):=p_{[2]}$
  satisfy the hypotheses of Theorem~\ref{strange-thm}.  For
  this we must check the comparabilities $w_1\leq q$,
  $w_2\leq r$ and $w_i\leq p\wedge (q\vee r)$ in the free
  lattice over $\{p,q,r\}$. The only nontrivial thing to
  verify is that $p_{[2]}\leq p\wedge (q\vee r)$.  This
  follows from the definition $p_{[2]} := p_{[1]}\wedge
  (q_{[1]}\vee r_{[1]})\leq p_{[1]} := p\wedge (q\vee r)$.

  The implication (3)$\mimplies$(4) follows from the fact
  that quasi-identity (\ref{weak_W}) fails in some meet
  continuous lattice, e.g. the lattice $\m d_1$ defined in
  Theorem~\ref{not_jonsson_rival}.  The implication
  (4)$\mimplies$(1) is Corollary~\ref{pixley-wille_cor}, so
  we have proved that the first four statements of this
  theorem are equivalent.

  By Remark~\ref{sdm_rem}, quasi-identity (\ref{weak_W}) is
  equivalent to the meet continuous identity
  \begin{equation}\label{identity_in_(3)}
    \delta^{\wec{w}}(p,q,r)\approx p\wedge(q\vee r)
  \end{equation} for some meet continuous lattice word
  $\delta^{\wec{w}}$, computed from the premises of
  (\ref{weak_W}), which satisfies $\delta^{\wec{w}}(p,q,r)\leq
  p\wedge(q\vee r)$ in $\m f_{\vr L_{MC}}(\{p,q,r\})$.  Since
  this identity is nontrivial, $\delta^{\wec{w}}(p,q,r)$ is in
  fact {strictly} smaller than $p\wedge(q\vee r)$ in $\m
  f_{\vr L_{MC}}(\{p,q,r\})$.  Because of the equivalence of
  the first four statements of this theorem, we may assume
  that the meet continuous congruence identity referred to in
  statement (4) is identity~(\ref{identity_in_(3)}).

  To prove (4)$\mimplies$(5) we need to show that $\vr V$
  satisfies the congruence inclusion
  $\alpha\cap(\beta\circ_{4}\gamma)\subseteq w(\alpha,
  \beta, \gamma)$ for some word~$w$ satisfying $w(p,q,r) <
  p\wedge(q\vee r)$. We begin with a generic occurrence of
  the situation ``$(u,v)\in \alpha\cap(\beta\circ_4
  \gamma)$'' in $\vr V$.  That is, we let $\m f=\m f_{\vr
    V}(u,v,x,y,z)$ be the free $\vr V$-algebra on five
  generators, and define $\alpha = \Cg a(u,v)$, $\beta = \Cg
  a\big((u,x),(y,z)\big)$, and $\gamma = \Cg a\big((x,y),
  (z,v)\big)$. The free generators are related as shown in
  Figure~\ref{fig9}.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(60,16)
\put(0,4.5){\circle*{1.2}}
\put(15,4.5){\circle*{1.2}}
\put(30,4.5){\circle*{1.2}}
\put(45,4.5){\circle*{1.2}}
\put(60,4.5){\circle*{1.2}}
\thicklines
\put(0,4.5){\line(1,0){60}}
{\bezier{250}(0,4.5)(7.5,14.5)(30,14.5)}
{\bezier{250}(30,14.5)(52.5,14.5)(60,4.5)}
\put(6,0){$\beta$}
\put(21,0.5){$\gamma$}
\put(36,0){$\beta$}
\put(51,0.5){$\gamma$}
\put(29,16.5){$\alpha$}
\put(-1,0.5){$u$}
\put(14,0.5){$x$}
\put(29,0.5){$y$}
\put(44,0.5){$z$}
\put(59,0.5){$v$}
\end{picture}
\end{center}
\caption{ \ }\label{fig9}
\end{figure}
Since $\alpha\cap(\beta\circ_4 \gamma)\subseteq
\alpha\cap(\beta\vee \gamma)$ and $\vr V$ satisfies
(\ref{identity_in_(3)}) as a congruence identity we get that
\[
(u,v)\in \alpha\cap(\beta\circ_4 \gamma) \subseteq
\alpha\cap(\beta\vee \gamma) =
\delta^{\wec{w}}(\alpha,\beta,\gamma)\,.
\]
Under the identification of $\m f_{\vr L_{MC}}(\{p,q,r\})$
with the ideal lattice of $\m f_{\vr L}(\{p,q,r\})$ the word
$\delta^{\wec{w}}(p,q,r)$ corresponds to an ideal $D$ of
ordinary lattice words $w(p,q,r)$, each of which is strictly
less than $p\wedge (q\vee r)$ (since $w(p,q,r)\leq
\delta^{\wec{w}}(p,q,r) < p\wedge (q\vee r)$).  The
assignment $p\mapsto \alpha$, $q\mapsto \beta$, $r\mapsto
\gamma$ extends to a homomorphism $\varphi\colon \m f_{\vr
  L_{MC}}(\{p,q,r\})\to \Conb(\m f)$ into the congruence
lattice of the 5-generated free algebra of $\vr V$.  This
homomorphism assigns to each $w\in D$ the congruence
$w(\alpha,\beta,\gamma)$, and assigns to $\delta^{\wec{w}}$
the union $\bigcup_{w\in D} w(\alpha,\beta,\gamma) =
\delta^{\wec{w}}(\alpha,\beta,\gamma)$.  Since $(u,v)\in
\delta^{\wec{w}}(\alpha,\beta,\gamma)$, we must have
$(u,v)\in w(\alpha,\beta,\gamma)$ for some $w\in D$.  Since
$(u,v)$ is a generic element of $\alpha\cap
(\beta\circ_4\gamma)$ the inclusion
$\alpha\cap(\beta\circ_4\gamma)\subseteq
w(\alpha,\beta,\gamma)$ holds throughout $\vr V$.  This
completes the proof that (4)$\mimplies$(5).


The implication (5)$\mimplies$(6) is trivial; we now prove
that (6)$\mimplies$(1).
Item (6) asserts that $\vr V$ satisfies a certain congruence
inclusion in the signature $\{\vee, \wedge, \circ\}$.  The
class of varieties satisfying any such inclusion is
definable by idempotent Maltsev conditions.  To prove that
at least one of the defining Maltsev conditions satisfied by
$\vr V$ is nontrivial it suffices to prove that the
congruence inclusion in (6) fails in the variety of sets.
To show this, let $\m a = \lb\{a_0, a_1, \ldots,
a_k\};\emptyset\rb$ be a $(k+1)$-element set, and define
$\alpha = \Cg a(a_0,a_k)$, $\beta = \Cg
a(\{(a_{i},a_{i+1})\mid \mbox{\rm $i$ even}\})$ and $\gamma
= \Cg a(\{(a_{i},a_{i+1})\mid \mbox{\rm $i$ odd}\})$.  Then
$\alpha$, $\beta$ and $\gamma$ relate the elements of~$\m a$
as indicated in Figure~\ref{fig10}. 
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(75,16)
\put(0,4.5){\circle*{1.2}}
\put(15,4.5){\circle*{1.2}}
\put(30,4.5){\circle*{1.2}}
\put(45,4.5){\circle*{1.2}}
\put(60,4.5){\circle*{1.2}}
\put(75,4.5){\circle*{1.2}}
%
\put(65,4.5){\circle*{.7}}
\put(67.5,4.5){\circle*{.7}}
\put(70,4.5){\circle*{.7}} \thicklines
\put(0,4.5){\line(1,0){60}}
{\bezier{250}(0,4.5)(7.5,14.5)(37.5,14.5)}
{\bezier{250}(37.5,14.5)(67.5,14.5)(75,4.5)}
%
\put(6,0){$\beta$}
\put(21,0.5){$\gamma$}
\put(36,0){$\beta$}
\put(51,0.5){$\gamma$}
\put(36,16.5){$\alpha$}
\put(-2,0.5){$a_0$}
\put(13,0.5){$a_1$}
\put(28,0.5){$a_2$}
\put(43,0.5){$a_3$}
\put(58,0.5){$a_4$}
\put(73,0.5){$a_k$}
\end{picture}
\end{center}
\caption{ \ }\label{fig10}
\end{figure}
These congruences
generate a sublattice of $\Conb(\m a) = \Eqb(A)$ isomorphic
to $\m d_1$, labeled as in Figure~\ref{fig11}.
Let $\varphi\colon \m f_{\vr L}(\{p,q,r\})\to \Conb(\m a)$
be the lattice homomorphism determined by $p\mapsto \alpha$,
$q\mapsto \beta$, $r\mapsto \gamma$.  Since
$\im(\varphi)\cong \m d_1$ is a finite lower bounded lattice,
for each $a\in \im(\varphi)$ there is a least element $b\in
\m f_{\vr L}(\{p,q,r\})$ such that $\varphi(b)\geq a$.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(30,40)
\put(0,15){\circle*{1.2}}
\put(0,25){\circle*{1.2}}
\put(15,5){\circle*{1.2}}
\put(15,15){\circle*{1.2}}
\put(15,35){\circle*{1.2}}
\put(30,15){\circle*{1.2}}
\put(30,25){\circle*{1.2}}
\put(0,15){\line(3,-2){15}}
\put(0,15){\line(0,1){10}}
\put(15,15){\line(-3,2){15}}
\put(15,15){\line(3,2){15}}
\put(15,15){\line(0,-1){10}}
\put(15,35){\line(-3,-2){15}}
\put(15,35){\line(3,-2){15}}
\put(30,15){\line(-3,-2){15}}
\put(30,15){\line(0,1){10}}
\put(11,13){$\alpha$}
\put(-4.5,13){$\beta$}
\put(32,13){$\gamma$}
\put(14,0){$0$}
\put(14,37){$1$}
\end{picture}
\end{center}
\caption{ \ }\label{fig11}
\end{figure}
There is a procedure described in Lemma~2.7
of~\cite{freese-jezek-nation} for calculating this least
$b$.  Applying it to our situation shows that the least
$b\in \m f_{\vr L}(\{p,q,r\})$ such that $\varphi(b)\geq
\alpha$ is $b = p\wedge(q\vee r)$. Since we assume in (4)
that $w(p,q,r) < p\wedge(q\vee r)$, we get that
\[
w(\alpha,\beta,\gamma)= \varphi\big(w(p,q,r)\big)
\not\geq \alpha = \Cg a(u,v)\,.
\]
Hence $(u,v)\in \alpha\cap(\beta\circ_4 \gamma) -
w(\alpha,\beta,\gamma)$, which proves that the inclusion in
(5) fails in the variety of sets.
\end{proof}

\begin{rem}
  Quasi-identity~(\ref{weak_W}) satisfies $\W$.  (Recall
  from Section~\ref{lattice_theory} that to check that a
  quasi-identity $Q$ satisfies $\W$ it suffices to check
  that $\W$ holds in the partial lattice $\m p(Q)$ of the
  $Q$-configuration.  For (\ref{weak_W}) this partial
  lattice has only twelve distinct elements, so the
  verification is not difficult.  In fact, if one first uses
  the axioms of lattice theory to simplify
  $p_{[2]}=p_{[1]}\wedge (q_{[1]}\vee r_{[1]})$ to
  \[
  p\wedge (q_{[1]}\vee r_{[1]}) = 
  p\wedge \big[\big(q\wedge (p\vee r)\big)
  \vee \big(r\wedge (p\vee q)\big)\big]\,,
  \]
  then the associated partial lattice only has nine
  elements.)  The theorem just proved, together with
  Theorem~\ref{czedli-day-W}, shows that
  quasi-identity (\ref{weak_W}) is the weakest congruence
  quasi-identity in the collection of nontrivial congruence
  quasi-identities satisfying $\W$.  That is, if $Q$ is any
  nontrivial quasi-identity satisfying $\W$ and $Q$ holds in
  all congruence lattices of algebras in some variety $\vr
  V$, then $\vr V$ satisfies a nontrivial idempotent Maltsev
  condition by Theorem~\ref{czedli-day-W}.  Now, from
  (1)$\mimplies$(2) of Theorem~\ref{typ1condcor} we get that
  quasi-identity (\ref{weak_W}) is also satisfied by all
  congruence lattices of algebras in $\vr V$. Thus any
  variety satisfying $Q$ as a congruence quasi-identity also
  satisfies (\ref{weak_W}).
\end{rem}

\begin{prb}\label{weakest_prb}
  Is there a weakest congruence quasi-identity (not
  necessarily satisfying $\W$)?
\end{prb}

Problem~\ref{weakest_prb} is related to
Problems~\ref{cong_quasi_maltsev} and \ref{same_prevariety}.

\begin{rem}
  It is natural to be curious about the effect of replacing
  $p_{[2]}$ with some other $p_{[n]}$ in item (2) of
  Theorem~\ref{typ1condcor}.  The quasi-identities
  \[
  Q_n:\; \big((p\wedge q)\approx s\big)
  \wrel{\&} \big((p\wedge r)\approx s\big)
  \wrel{\&} (p_{[n]}\approx s) \wrel{\to}
  \big((p\wedge (q\vee r))\approx s\big)
  \]
  become formally stronger as $n$ increases, since
  \[
  \big((p\wedge q)\approx s\big)
  \wrel{\&} \big((p\wedge r)\approx s\big)
  \wrel{\&} (p_{[n]}\approx s) 
  \wrel{\to} (p_{[n+1]}\approx s)
  \]
  holds in all lattices.  But $Q_1$ is a tautology, and for
  $n\geq 2$ all $Q_n$ are equivalent.  For, if some
  assignment $p\mapsto a$, $q\mapsto b$, $r\mapsto c$,
  $s\mapsto d$ witnesses a failure of $Q_n$ but not
  $Q_{n-1}$, then $p\mapsto a_{[n-2]}$, $q\mapsto
  b_{[n-2]}$, $r\mapsto c_{[n-2]}$, $s\mapsto d$ is an
  assignment witnessing a failure of $Q_2$.

  It is also natural to be curious about the effect of
  replacing the numbers $4$ and $k\; (\geq 4)$ in items (5)
  and (6) of Theorem~\ref{typ1condcor} with smaller
  numbers. It happens that there is a slightly more
  complicated condition for $k=3$ that is equivalent to the
  other conditions of Theorem~\ref{typ1condcor}, but there
  is no similar condition for~$k=2$.

  To state the condition for $k=3$, recursively define
  ternary lattice words as follows: 
  \begin{gather*}\label{prep_4_7}
  p^0 := p\,, \qquad r^0 := r\,,\\
  p^{n+1} := p\wedge(q\vee r^n)\,,
  \qquad\mbox{and}\qquad
  r^{n+1} := r\wedge(q\vee p^n)\,.
  \end{gather*}
  The equivalent condition for $k=3$ is:
  \begin{enumerate}
  \item[(7)] $\vr V$ satisfies a congruence inclusion of the
    form
    \[
    \alpha\cap(\beta\circ_{3}\gamma)\subseteq w(\alpha,
    \beta, \gamma)\,,
    \]
    where $w(p,q,r)$ is a lattice word such that
    $w(p,q,r)\not\geq p^{n}$ for any $n$ in the free lattice
    $\m f_{\vr L}(\{p,q,r\})$.
  \end{enumerate}

  To see why it is necessary to phrase item (7) in this more
  complicated way, suppose that $(u,v)\in
  \alpha\cap(\beta\circ\gamma\circ\beta)$.  Then, for
  $k=0$, there are elements $x$ and $y$ related to $u$ and
  $v$ as in Figure~\ref{fig12}. But it is easy to see
\begin{figure}[h]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(20,25)
\put(0,5){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(20,5){\circle*{1.2}}
\put(20,20){\circle*{1.2}} \thicklines
\put(0,5){\line(1,0){20}}
\put(0,5){\line(0,1){15}}
\put(20,20){\line(-1,0){20}}
\put(20,20){\line(0,-1){15}}
\put(-5,11){$\beta$}
\put(7,22){$\gamma^k$}
\put(22,11){$\beta$}
\put(7,0){$\alpha^k$}
\put(-1,1){$u$}
\put(-1,22){$x$}
\put(19,22){$y$}
\put(19,1){$v$}
\end{picture}
\end{center}
\caption{ \ }\label{fig12}
\end{figure}
that if $u$, $v$, $x$ and $y$ are related in this way for
some $k$, then $(u,v)\in \alpha^k\cap(\beta\vee \gamma^k)
= \alpha^{k+1}$ and $(x,y)\in \gamma^k\cap(\beta\vee
\alpha^k) = \gamma^{k+1}$. By induction
$\alpha\cap(\beta\circ_3\gamma) \subseteq\alpha^k$ for all
$k$.  This shows that if a lattice word $w(p,q,r)$ is above
$p^n$ for some $n$, then
$\alpha\cap(\beta\circ_3\gamma)\subseteq \alpha^n
\subseteq w(\alpha,\beta,\gamma)$ holds in any lattice of
equivalence relations.  Thus, for an inclusion
$\alpha\cap(\beta\circ_3\gamma) \subseteq
w(\alpha,\beta,\gamma)$ to be nontrivial, we must have
$w(p,q,r)\not\geq p^n$ for every $n$.  On the other hand, if
$k\geq 4$, then inclusions of the form
$\alpha\cap(\beta\circ_k\gamma) \subseteq
w(\alpha,\beta,\gamma)$ are nontrivial as soon as the single
inequality $w(p,q,r)\not\geq p^1 = p\wedge(q\vee r)$ is
satisfied.

Fortunately, that is the only modification of the condition
on~$w$ that is needed.  To prove (4)$\mimplies$(7), observe
that the assignment of variables in $\m e_1$ defined by
$p\mapsto \alpha$, $q\mapsto \beta$, $r\mapsto\gamma$,
$s\mapsto 0$ is one that satisfies the premises but not the
conclusion of quasi-identity (\ref{weak_W}).
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(20,35)
\put(0,20){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,15){\circle*{1.2}}
\put(10,35){\circle*{1.2}}
\put(20,15){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
%
\put(10,5){\line(-2,3){10}}
\put(10,5){\line(0,1){10}}
\put(10,5){\line(1,1){10}}
\put(10,35){\line(-2,-3){10}}
\put(10,35){\line(1,-1){10}}
\put(20,25){\line(-1,-1){10}}
\put(20,25){\line(0,-1){10}}
%
\put(9,0){$0$}
\put(6,14){$\alpha$}
\put(-4.5,18){$\beta$}
\put(22,14){$\gamma$}
\end{picture}
\end{center}
\caption{The lattice $\m e_1$}\label{fig13}
\end{figure}

An examination of the third paragraph of the proof of
Theorem~\ref{strange-thm} reveals that the same assignment
must therefore witness a failure of the equivalent identity
(\ref{identity_in_(3)}).  Thus, if $\varphi\colon \m f_{\vr
  L_{MC}}(\{p,q,r\})\to \m e_1$ is the extension of the
assignment $p\mapsto \alpha$, $q\mapsto \beta$,
$r\mapsto\gamma$ to a homomorphism, then
$\varphi\big(\delta^{\wec{w}}(p,q,r)\big) <
\varphi\big(p\wedge(q\vee r)\big) = \alpha$.  On the other
hand, an easy induction shows that $\varphi(p^n)=\alpha$ for
all $n$.  Hence $\delta^{\wec{w}}(p,q,r)\not\geq p_n$ for
all $n$ in $\m f_{\vr L_{MC}}(\{p,q,r\})$.  Now, as in the
proof of Theorem~\ref{typ1condcor} (4)$\mimplies$(5), the
word $w(p,q,r)$ lies below $\delta^{\wec{w}}(p,q,r)$ in $\m
f_{\vr L_{MC}}(\{p,q,r\})$, hence satisfies
$w(p,q,r)\not\geq p^n$ for all $n$ in $\m f_{\vr
  L}(\{p,q,r\})$.  This proves (4)$\mimplies$(7).

One can prove that (7)$\mimplies$(1) in essentially the same
way we proved (6)$\mimplies$(1) using the number $3$ in
place of $k$.  This time, define $\m a =
\lb\{a_0,a_1,a_2,a_3\};\emptyset\rb$, $\alpha = \Cg
a(a_0,a_3)$, $\beta = \Cg a\big((a_0,a_1),(a_2,a_3)\big)$
and $\gamma = \Cg a(a_1,a_2)$. These congruences on~$\m a$
generate a sublattice of~$\Conb(\m a)$ isomorphic to the
lattice $\m e_1$, therefore $\m e_1$ should be used in the
argument in place of $\m d_1$.  Unlike $\m d_1$, the lattice
$\m e_1$ is not lower bounded.  However it can be shown
(using Theorems~2.2 and~2.3 of~\cite{freese-jezek-nation})
that if $\varphi\colon \m f_{\vr L}(\{p,q,r\})\to \Conb(\m
a)$ is the lattice homomorphism determined by $p\mapsto
\alpha$, $q\mapsto \beta$, $r\mapsto \gamma$, then the set
of all $b\in \m f_{\vr L}(\{p,q,r\})$ such that
$\varphi(b)\geq \alpha$ is the filter $\{t\mid t(p,q,r)\geq
p^n, \;\mbox{\rm for some $n$}\}$.  If~$w$ satisfies the
conditions given in item (7), then $w$ is not in this
filter, so $\varphi(w)\not\geq\alpha$.  As in the proof we
gave for (6)$\mimplies$(1), this implies that the congruence
inclusion $\alpha\cap(\beta\circ_3\gamma)\subseteq
w(\alpha,\beta,\gamma)$ fails in $\m a$, hence fails in the
variety of~sets.

There is no analogous condition for $k=2$ that is equivalent
to the other conditions of Theorem~\ref{typ1condcor}.  The
reason for this is that the condition that
$\alpha\cap(\beta\circ\gamma)\subseteq
w(\alpha,\beta,\gamma)$ is trivial if $w(p,q,r)\geq p_{[n]}$
for some~$n$, as we showed in the first paragraph of the
proof of Lemma~\ref{typ1condlm}.  On the other hand,
Theorems~2.2 and 2.3 of \cite{freese-jezek-nation} can be
used to show that if $w(p,q,r)\not\geq p_{[n]}$ for all $n$,
then $\varphi(w)\not\geq \alpha$ for the homomorphism $h: \m
f_{\vr L}(\{p,q,r\})\rightarrow \m m_3: p\mapsto\alpha$,
$q\mapsto \beta$, $r\mapsto \gamma$, where $\m m_3$ is
labeled as in Figure~\ref{fig14}.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(20,25)
\put(0,10){\circle*{1.2}}
\put(10,0){\circle*{1.2}}
\put(10,10){\circle*{1.2}}
\put(10,20){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(10,0){\line(-1,1){10}}
\put(10,0){\line(0,1){10}}
\put(10,0){\line(1,1){10}}
\put(10,20){\line(-1,-1){10}}
\put(10,20){\line(0,-1){10}}
\put(10,20){\line(1,-1){10}}
\put(-4,8.5){$\alpha$}
\put(6,8){$\beta$}
\put(21.5,8.5){$\gamma$}
\end{picture}
\end{center}
\caption{The lattice $\m m_3$}\label{fig14}
\end{figure}
Therefore, if the congruence inclusion
\hbox{$\alpha\cap(\beta\circ\gamma)\subseteq
  w(\alpha,\beta,\gamma)$} is nontrivial and holds
throughout $\vr V$, then whenever $\alpha$, $\beta$ and
$\gamma$ are congruences on some $\m a\in {\vr V}$ that
generate a copy of~$\m m_3$ it must be that
\hbox{$\alpha\cap(\beta\circ\gamma)\subseteq \alpha\cap
  w(\alpha,\beta,\gamma) \lneqq \alpha$.}  In particular,
the idempotent Maltsev condition associated with the
inclusion \hbox{$\alpha\cap(\beta\circ\gamma)\subseteq
  w(\alpha,\beta,\gamma)$} precludes the possibility that
$\alpha$, $\beta$ and $\gamma$ generate a copy of $\m m_3$
where $\beta\circ\gamma = \beta\vee\gamma$.  Such an
idempotent Maltsev condition must fail in any nontrivial
variety of modules, since any nontrivial variety of modules
contains members whose congruence lattices contain
sublattices of permuting congruences isomorphic to $\m m_3$.
But we will prove in Theorem~\ref{SD-meet-thm}
(10)$\mimplies$(1) that any idempotent Maltsev condition
that fails in every nontrivial variety of modules implies
congruence meet semidistributivity.  This shows that a
congruence inclusion of the type
$\alpha\cap(\beta\circ\gamma)\subseteq
w(\alpha,\beta,\gamma)$ for some lattice word $w$ is trivial
if $w(p,q,r)\geq p_{[n]}$ for some $n$, and implies
congruence meet semidistributivity if $w(p,q,r)\not\geq
p_{[n]}$ for all $n$.  Therefore any such inclusion is
either too weak or too strong to be equivalent to the
existence of a Taylor term.

We emphasize, however, that this conclusion was reached
under the assumption that $w$ is a lattice word. There do
exist expressions $W(p,q,r)$ in the symbols $\vee, \wedge$
{\it and} $\circ$ such that the congruence inclusion
$\alpha\cap(\beta\circ\gamma)\subseteq
W(\alpha,\beta,\gamma)$ is nontrivial but does not imply
congruence meet semidistributivity for a variety. (E.g.,
Theorem~\ref{cong_id2}~(2) of this monograph, or Theorem~4.8
(2) and (3) of \cite{k-szendrei2}.)
\end{rem}


\section{Omitted Sublattices}\label{4.3}

In this section we are going to show that if a variety $\vr
V$ has a Taylor term, then certain lattices cannot be
embedded in congruence lattices of algebras in $\vr V$.

\begin{thm}\label{taylor_forbidden} Let $\vr V$ be a
  variety that satisfies a nontrivial idempotent Maltsev
  condition and let\/~$\m a$ be a member of\/~$\vr V$.
  \begin{enumerate}
  \item[(1)] The lattices in Figure~\ref{fig15} do not occur
    in the meet continuous congruence variety of\/~$\vr V$.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(110,45)
\put(0,20){\circle*{1.2}}
\put(0,30){\circle*{1.2}}
\put(15,10){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(15,40){\circle*{1.2}}
\put(30,20){\circle*{1.2}}
\put(30,30){\circle*{1.2}}
%
\put(45,25){\circle*{1.2}}
\put(55,10){\circle*{1.2}}
\put(55,20){\circle*{1.2}}
\put(55,40){\circle*{1.2}}
\put(65,20){\circle*{1.2}}
\put(65,30){\circle*{1.2}}
%
\put(80,20){\circle*{1.2}}
\put(90,10){\circle*{1.2}}
\put(90,30){\circle*{1.2}}
\put(95,25){\circle*{1.2}}
\put(100,20){\circle*{1.2}}
\put(100,40){\circle*{1.2}}
\put(110,30){\circle*{1.2}}
%
\put(0,20){\line(3,-2){15}}
\put(0,20){\line(0,1){10}}
\put(15,20){\line(-3,2){15}}
\put(15,20){\line(3,2){15}}
\put(15,20){\line(0,-1){10}}
\put(15,40){\line(-3,-2){15}}
\put(15,40){\line(3,-2){15}}
\put(30,20){\line(-3,-2){15}}
\put(30,20){\line(0,1){10}}
%
\put(55,10){\line(-2,3){10}}
\put(55,10){\line(0,1){10}}
\put(55,10){\line(1,1){10}}
\put(55,40){\line(-2,-3){10}}
\put(55,40){\line(1,-1){10}}
\put(65,30){\line(-1,-1){10}}
\put(65,30){\line(0,-1){10}}
%
\put(80,20){\line(1,1){20}}
\put(80,20){\line(1,-1){10}}
\put(90,30){\line(1,-1){10}}
\put(110,30){\line(-1,1){10}}
\put(110,30){\line(-1,-1){20}}
%
\put(14,5){$\delta$}
\put(54,5){$\delta$}
\put(89,5){$\delta$}
%
\put(11,18){$\alpha$}
\put(51,19){$\alpha$}
\put(75.5,19){$\alpha$}
%
\put(-4.5,18){$\beta$}
\put(40.5,23){$\beta$}
\put(91,22.5){$\gamma$}
%
\put(32,18){$\gamma$}
\put(67,19){$\gamma$}
\put(112,29){$\beta$}
%
\put(13,0){$\m d_1$}
\put(53,0){$\m e_1$}
\put(94,0){$\m g$}
\end{picture}
\end{center}
\caption{ \ }\label{fig15}
\end{figure}

\newpage

\item[(2)] The lattices in Figure~\ref{fig16} with the given
  centralities do not occur as sublattices of $\Conb(\m a)$.

\begin{figure}[ht]
\begin{center}\setlength{\unitlength}{1mm}
\begin{picture}(80,48)
\put(0,23){\circle*{1.2}}
\put(10,8){\circle*{1.2}}
\put(10,38){\circle*{1.2}}
\put(20,18){\circle*{1.2}}
\put(20,28){\circle*{1.2}}
%
\put(50,18){\circle*{1.2}}
\put(50,28){\circle*{1.2}}
\put(65,8){\circle*{1.2}}
\put(65,28){\circle*{1.2}}
\put(65,38){\circle*{1.2}}
\put(80,18){\circle*{1.2}}
\put(80,28){\circle*{1.2}}
%
\put(10,8){\line(-2,3){10}}
\put(10,8){\line(1,1){10}}
\put(10,38){\line(-2,-3){10}}
\put(10,38){\line(1,-1){10}}
\put(20,28){\line(0,-1){10}}
%
\put(50,18){\line(3,-2){15}}
\put(50,18){\line(3,2){15}}
\put(50,18){\line(0,1){10}}
\put(65,38){\line(-3,-2){15}}
\put(65,38){\line(0,-1){10}}
\put(65,38){\line(3,-2){15}}
\put(80,18){\line(-3,-2){15}}
\put(80,18){\line(-3,2){15}}
\put(80,18){\line(0,1){10}}
%
\put(-4.5,21){$\beta$}
\put(22,17){$\gamma=\delta$}
\put(22,27){$\alpha$}
%
\put(45.5,27){$\beta$}
\put(82,17){$\gamma=\delta$}
\put(82,27){$\alpha$}
\put(64,23){$\sigma$}
\put(64,40){$\tau$}
%
\put(-13,45){\mbox{\rm $\C c(\beta,\alpha;\delta)$ or $\C
c(\beta,\beta;\beta\wedge\delta)$}}
\put(55,45){$\C
c(\tau,\tau;\sigma)$}
%
\put(7.5,0){$\m n_5$}
\put(62.5,0){$\m d_2$}
\end{picture}
\end{center}
\caption{ \ }\label{fig16}
\end{figure}
\item[(3)] The lattices in Figure~\ref{fig17} do not occur
as sublattices of\/  \linebreak 
$\Conb(\m a)$.
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(70,44)
\put(0,23){\circle*{1.2}}
\put(10,8){\circle*{1.2}}
\put(10,28){\circle*{1.2}}
\put(10,38){\circle*{1.2}}
\put(20,18){\circle*{1.2}}
\put(20,28){\circle*{1.2}}
%
\put(50,23){\circle*{1.2}}
\put(60,8){\circle*{1.2}}
\put(60,23){\circle*{1.2}}
\put(60,38){\circle*{1.2}}
\put(70,18){\circle*{1.2}}
\put(70,28){\circle*{1.2}}
%
\put(10,8){\line(-2,3){10}}
\put(10,8){\line(1,1){10}}
\put(10,28){\line(0,1){10}}
\put(10,28){\line(1,-1){10}}
\put(10,38){\line(-2,-3){10}}
\put(10,38){\line(1,-1){10}}
\put(20,28){\line(0,-1){10}}
%
\put(60,8){\line(-2,3){10}}
\put(60,8){\line(1,1){10}}
\put(60,8){\line(0,1){30}}
\put(60,38){\line(-2,-3){10}}
\put(60,38){\line(1,-1){10}}
\put(70,28){\line(0,-1){10}}
%
\put(-4.5,21){$\beta$}
\put(12,27){$\alpha$}
\put(22,27){$\gamma$}
\put(22,17){$\delta$}
%
\put(7.5,0){$\m e_2$}
%
\put(45.5,21){$\beta$}
\put(62,21){$\alpha$}
\put(72,27){$\gamma$}
\put(72,17){$\delta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig17}
\end{figure}
\end{enumerate}
\end{thm}

\begin{proof}
  For the first claim of this theorem, the fact that $\vr V$
  satisfies a nontrivial idempotent Maltsev condition
  implies that the congruence lattices of algebras in $\vr
  V$ satisfy quasi-identity (\ref{weak_W}) of
  Theorem~\ref{typ1condcor}~(2).  But in the proof of
  Theorem~\ref{typ1condcor} we explained why that
  quasi-identity is equivalent to a meet continuous
  identity, so in fact all members of the meet continuous
  congruence variety of $\vr V$ satisfy quasi-identity
  (\ref{weak_W}).  Item (1) of this theorem follows from the
  fact that $\m d_1, \m e_1$ and $\m g$ all fail
  quasi-identity (\ref{weak_W}) of
  Theorem~\ref{typ1condcor}. (In each case the assignment
  $p\mapsto \alpha$, $q\mapsto \beta$, $r\mapsto \gamma$,
  $s\mapsto \delta$ witnesses a failure.)

  For the second claim, observe that $\alpha_{[2]}\leq
  \delta \leq \alpha\leq \beta\vee\gamma$ holds for the
  indicated congruences in both $\m n_5$ and $\m d_2$.  By
  Lemma~\ref{typ1condlm}, this is enough to establish the
  first condition of Theorem~\ref{typ1cond}.  To show that
  the two configurations are forbidden it suffices to show
  that the second condition of Theorem~\ref{typ1cond} is
  satisfied, since $\delta\neq\alpha$.  The second condition
  of Theorem~\ref{typ1cond}, that $\C
  c(\beta\vee\gamma,\alpha;\delta)$ holds, is equivalent to
  the conjunction of $\C c(\beta,\alpha;\delta)$ and $\C
  c(\gamma,\alpha;\delta)$ since the centralizer relation is
  monotone and meet semidistributive in its first variable.
  In the two examples $\C c(\gamma,\alpha;\delta)$ holds
  simply because $\gamma = \delta$.  In the first assertion
  concerning $\m n_5$ we have $\C c(\beta,\alpha;\delta)$
  because we have assumed it. The second assertion
  concerning $\m n_5$ can be reduced to the first, as
  follows.  Our assumption that $\C C(\beta,
  \beta;\beta\wedge\delta)$ holds implies $\C
  C(\beta\vee\delta, \beta\vee\delta;\delta)$ using
  Theorem~\ref{perspective_abelian_taylor}, and then $\C
  C(\beta, \alpha;\delta)$ follows from the monotonicity of
  the centralizer in its first two variables.

  For $\m d_2$ we have assumed only that $\C
  c(\tau,\tau;\sigma)$ holds.  From this and $\alpha, \beta
  \leq \tau$ we get that $\C c(\beta,\alpha;\sigma)$ holds.
  We have $\C c(\beta,\alpha;\alpha)$ trivially, so by item
  (6) of Theorem~\ref{basic_centrality} and the fact that
  $\delta = \alpha\wedge\sigma$ we get that $\C
  c(\beta,\alpha;\delta)$ holds.  Since $\alpha$, $\beta$
  and $\delta$ generate a sublattice isomorphic to $\m n_5$,
  we get a contradiction from the argument in the previous
  paragraph.

  For the part of the third claim concerning $\m e_2$, $\C
  C(\alpha,\beta;\beta\wedge\delta)$ and $\C
  C(\gamma,\beta;\beta\wedge\delta)$ hold by
  Theorem~\ref{basic_centrality}~(8).  We can use
  Theorem~\ref{basic_centrality}~(5) and (1) to conclude
  first that $\C
  C(\alpha\vee\gamma,\beta;\beta\wedge\delta)$ holds, and
  then that $\C C(\beta,\beta;\beta\wedge\delta)$ holds. By
  applying item~(2) of this theorem to the sublattice
  generated by $\beta, \gamma$ and $\delta$ we obtain a
  contradiction.  One can handle the second lattice in the
  third claim by the same arguments.
\end{proof}

\begin{rem}
  The combination of Theorem~\ref{not_jonsson_rival}~(1) and
  Theorem~\ref{taylor_forbidden} shows that if $\m l$ is a
  finite sublattice of $\Conb(\m a)$ where $\m a$ has a
  Taylor term, and $\m m_3$ is not a sublattice of $\m l$,
  then $\m l$ is meet semidistributive.  This is the
  strongest conclusion about $\m l$ that can be derived from
  these hypotheses, since the variety $\vr S$ of
  semilattices has a Taylor term, $\m m_3$ is not embeddable
  in $\Conb(\m a)$ for any $\m a\in {\vr S}$, yet every
  finite meet semidistributive lattice is embeddable in the
  congruence lattice of some semilattice.  (To show this,
  combine Theorem~4.4.12 of \cite{gorbunov} with the main
  result of \cite{fajtlowicz-schmidt}.)
\end{rem}

\section{Admitted Sublattices}\label{4.3.5}

Theorem~\ref{taylor_forbidden} shows that if $\vr V$ has a
Taylor term, then certain lattices cannot be embedded in
$\Conb(\m a)$ for any $\m a\in \vr V$.  If, on the other
hand, $\vr V$ does not have a Taylor term, then many
lattices must appear as sublattices of congruence lattices
of algebras in $\vr V$.  The strongest result of this sort
that we know is the following.

\begin{thm}\label{czedli-day_thm}
  Let $\vr V$ be a variety that satisfies no nontrivial
  idempotent Maltsev condition.  If\/ $\m l$ is a finitely
  presentable lattice satisfying Whitman's condition $\W$
  (Definition~\ref{(W)_df}), then $\m l\in \vr L(\vr V)$.
\end{thm}

\begin{proof}
  In order to prove that $\m l$ is a member of $\vr L(\vr
  V)$ it suffices to prove that it is possible to separate
  the points of $\m l$ by homomorphisms into members of $\vr
  L(\vr V)$, since $\vr L(\vr V)$ is a prevariety.  So,
  choose elements $p\neq q$ in $\m l$ which are to be
  separated.

  Suppose that $\m l$ is presented by $\lb G \mid R\rb$
  where $G = \{g_1,\ldots, g_n\}$, $\wec{g} =
  (g_1,\ldots,g_n)$ and $R = \{p_1(\wec{g}) =
  q_1(\wec{g}),\ldots, p_m(\wec{g}) = q_m(\wec{g})\}$.
  There exist $n$-ary terms $p(\wec{x})$ and $q(\wec{x})$
  such that $p = p(\wec{g})$ and $q = q(\wec{g})$.  The
  quasi-identity
  \begin{equation}\label{quasi-identity}
    \bigwedge_{i=1}^m
    \big(p_i(\wec{x}) \approx q_i(\wec{x})\big)\rightarrow
    \big(p(\wec{x}) \approx q(\wec{x})\big)
  \end{equation}
  is nontrivial because it fails in $\m l$ when $\wec{x} =
  \wec{g}$.  The quasi-identity~(\ref{quasi-identity})
  satisfies $\W$ (Definition~\ref{quasi_sat_W}), since $\m
  l$ does, so by Theorem~\ref{czedli-day-W} the class of
  varieties satisfying (\ref{quasi-identity}) is definable
  by idempotent Maltsev conditions.

  What this shows is that if $\vr V$ satisfies no nontrivial
  idempotent Maltsev condition, then $\vr V$ has an algebra
  $\m a$ such that $\Conb(\m a)$ fails
  quasi-identity~(\ref{quasi-identity}).  If
  $(\theta_1,\ldots,\theta_n)$ is a witness of the failure,
  then the function $g_i\mapsto \theta_i$ extends to a
  homomorphism $h\colon \m l\to \Conb(\m a)$ for which
  $h(p)\neq h(q)$.  This proves the theorem.
\end{proof}

In the case where $\vr V$ is locally finite and $\m l$ is
finite, Theorem~\ref{czedli-day_thm} is part of Theorem~6.22
of \cite{hobby-mckenzie}.

For later reference, and to illustrate
Theorem~\ref{czedli-day_thm}, we next describe a class of
lattices satisfying $\W$.

\begin{df}\label{graph_lattice}
  A {\bf graph} is a pair $G = \lb V; E\rb$ where $V$ is
  a~set, $E$~is a set of $2$-element subsets of~$V$, and
  $V\cap E = \emptyset$. Elements of $V$ are called {\bf
    vertices} and elements of $E$ are called {\bf edges}.
  $G$ is {\bf discrete} if~$E = \emptyset$.

  If $G$ is a graph, then the {\bf lattice of $G$} (or {\bf
    graph lattice})\index{graph lattice} is the partially
  ordered set $\m L[G]$\index{$\m L[G]$ (graph lattice)}
  obtained from $\lb V\cup E; \in\rb$ by adding a new top
  element $1$ and a new bottom element $0$.
\end{df}

$\m l[G]$ has height $\leq 3$ and, when $V\neq \emptyset$,
has $V$ as its set of atoms and has each element of $E$ as a
coatom of height $2$. If $G$ has at least one edge, then the
height of $\m [G]$ is exactly $3$.

\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(15,43)
\put(15,5){\circle*{1.2}}
\put(15,35){\circle*{1.2}}
%
\put(0,16){\line(0,1){8}}
\put(15,16){\line(0,1){8}}
\put(15,5){\line(-3,2){13}}
\put(15,5){\line(3,2){13}}
\put(15,35){\line(-3,-2){13}}
\put(15,35){\line(3,-4){14}}
\put(15,35){\line(0,-1){9}}
%
\put(15,15){\oval(32,2)}
\put(7.5,25){\oval(17,2)}
%
\put(-6,14){$V$}
\put(-6,24){$E$}
\put(14,0){$0$}
\put(14,37){$1$}
\end{picture}
\end{center}
\caption{The shape of $\m l[G]$}\label{fig18}
\end{figure}
$\m l[G]$ is indeed a lattice. For, $\lb V\cup E; \in\rb$
is a poset of height $1$ that does not have elements ordered
as in Figure~\ref{fig19},
since we have defined graphs to exclude multiple edges $e,
f$ between distinct vertices $a, b$.  It is easy to see that
if one adds new top and bottom elements to any poset of
height $1$ with this property one always obtains a lattice.
(Any poset of height $1$ omitting the poset of
Figure~\ref{fig19} has the property that any two distinct
elements have at most one upper bound and one lower bound,
so adding top and bottom elements results in a poset where
every pair of distinct elements has a unique least upper
bound and greatest lower bound.)

\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(20,20)
\put(0,4){\circle*{1.2}}
\put(0,14){\circle*{1.2}}
\put(20,4){\circle*{1.2}}
\put(20,14){\circle*{1.2}}
\put(0,4){\line(2,1){20}}
\put(0,4){\line(0,1){10}}
\put(20,4){\line(-2,1){20}}
\put(20,4){\line(0,1){10}}
\put(-1,0){$a$}
\put(19,0){$b$}
\put(-1,16){$e$}
\put(19,16){$f$}
\end{picture}
\end{center}
\caption{ \ }\label{fig19}
\end{figure}

\begin{lm}\label{W_char}
  A lattice $\m l$ of height $\leq 3$ satisfies $\W$ if and
  only if\/ $\m l$ has no sublattice isomorphic to $\m
  m_{3,3}$ (Figure~\ref{fig19.5}).
\begin{figure}[ht]
\begin{center} \setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(0,10){\circle*{1.2}}
\put(10,0){\circle*{1.2}}
\put(10,10){\circle*{1.2}}
\put(10,20){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(20,20){\circle*{1.2}}
\put(20,30){\circle*{1.2}}
\put(30,20){\circle*{1.2}}
\put(10,0){\line(-1,1){10}}
\put(10,0){\line(0,1){10}}
\put(10,0){\line(1,1){10}}
\put(20,10){\line(-1,1){10}}
\put(20,10){\line(0,1){10}}
\put(20,10){\line(1,1){10}}
\put(20,30){\line(-1,-1){10}}
\put(20,30){\line(0,-1){10}}
\put(20,30){\line(1,-1){10}}
\put(10,20){\line(-1,-1){10}}
\put(10,20){\line(0,-1){10}}
%
\put(6,19){$v$}
\put(22,19){$a$}
\put(32,19){$b$}
\put(-4,9){$c$}
\put(6,9){$d$}
\put(22,9){$u$}
\end{picture}
\end{center}
\caption{$\m m_{3,3}$}\label{fig19.5}\index{$\m m_{3,3}$}
\end{figure}

\noindent
In particular, any graph lattice or dual of a graph lattice
satisfies~$\W$.
\end{lm}

\begin{proof}
  If $\m l$ fails $\W$, then there exist elements $a, b, c,
  d\in L$ such that $a\wedge b = u\leq v = c\vee d$ and
  $\{a,b,c,d\}\cap I[u,v] = \emptyset$.  These conditions
  force $0< u< x < 1$, where $x\in \{a,b\}$, and $0 < y < v
  < 1$, where $y\in \{c,d\}$.  Since the height of $\m l$ is
  $\leq 3$, it must be that $u$, $c$ and $d$ are atoms,
  while $v$, $a$ and $b$ are coatoms (of height $2$).  In
  particular, this gives us that $u<v$.  The interval
  $I[u,1]$ has height $2$, so it is isomorphic to~$\m m_m$
  for some $m$, and $a$, $b$ and $v$ are distinct coatoms in
  this interval.  Similarly, $I[0,v]$ is isomorphic to~$\m
  m_n$ for some $n\geq 3$, and $c$, $d$ and $u$ are distinct
  atoms in this interval. This, is enough to show that $a$,
  $b$, $c$ and $d$ generate a sublattice of $\m l$
  isomorphic to the one in Figure~\ref{fig19.5}.

  Conversely, if $\m l$ satisfies $\W$, then every
  sublattice of~$\m l$ must satisfy~$\W$.  Since $\m
  m_{3,3}$ does not, it cannot be a sublattice of $\m l$.

  To see that any graph lattice omits $\m m_{3,3}$ as a
  sublattice, it is enough to observe that the element $v$
  in Figure~\ref{fig19.5} has height $2$ and dominates at
  least three distinct atoms.  If this happened in a
  sublattice of a graph lattice, then $v$ would represent an
  edge incident to at least three distinct vertices, which
  is impossible.  A dual argument works for duals of graph
  lattices.
\end{proof}

\begin{lm}\label{graph_lm}
  If\/ $G$ is a nondiscrete graph with at least $4$
  vertices, then $\m l[G]$ is a nonmodular lattice.  If, in
  addition, $G$ is not isomorphic to the complete bipartite
  graph $K_{1,m}$ for some $m$, then $\m l[G]$ is a simple
  lattice.
\end{lm}

(The complete bipartite graph $K_{1,m}$ has vertex set that
is a disjoint union of sets $V_1$ and $V_m$ where $|V_1|=1$
and $|V_m|=m$, and edge set consisting of all $\{x,y\}$ with
$(x,y)\in V_1\times V_m$. The reader can check that $\m
l[K_{1,m}]$ is not simple.)

\begin{proof}
  For the first statement, suppose that $e=\{a,b\}$ is an
  edge of $G$ and that $c\notin \{a,b\}$ is a vertex of $G$
  not adjacent to $a$.  Then $\{0, 1, a, c, e\}$ is a
  sublattice of $G$ isomorphic to the nonmodular lattice~$\N
  5$.  It follows from this that if $G$ has at least one
  edge and $\m l[G]$ is modular, then $G$ is a complete
  graph.  If $G$ has at least four distinct vertices, say
  $a$, $b$, $c$ and $d$, then the edges $\{a,b\}$ and
  $\{c,d\}$ are coatoms of $\m l[G]$ that meet to $0$. This
  cannot happen in a modular lattice of height~$3$.  Hence,
  if $G$ is nondiscrete with at least $4$ vertices, then $\m
  l[G]$ is nonmodular.

  Now assume in addition that $G$ is not isomorphic to
  $K_{1,m}$ for any~$m$.  Let $\theta$ be a nontrivial
  congruence on $\m l[G]$, and let $u<v$ be distinct,
  comparable, $\theta$-related elements. The hypotheses
  guarantee that there is an atom $a\leq v$ for which
  $a\not\leq u$.  (If $v$ is an atom, choose $a=v$; if~$v$
  is an edge, choose $a$ to be any incident vertex not equal
  to $u$; if $v$ is the top, choose $a$ to be one of the
  four atoms not below $u$.)  Hence $(0,a) = (a\wedge
  u,a\wedge v)\in\theta$.  If there is a vertex~$b$ not
  adjacent to~$a$, then $(b,1) = (0\vee b, a\vee
  b)\in\theta$.  Now for each vertex $c\neq b$ we have
  $(0,c) = (c\wedge b, c\wedge 1)\in\theta$.  By hypothesis,
  $G$ has at least $3$ distinct vertices other than $b$, so
  $\m l[G]$ has at least $3$ distinct atoms $\theta$-related
  to~$0$.  But then the join of these atoms is
  $\theta$-related to $0$, and the join of any $3$ atoms in
  a graph lattice is the top.  Thus, $\theta$ is the total
  relation in the case that there exists a vertex $b$ not
  adjacent to $a$.

  We may assume that $a$ is adjacent to all other vertices.
  Since $G$~is not isomorphic to $K_{1,m}$ there exists an
  edge $e=\{p,q\}$ with $a\notin \{p,q\}$. In this case,
  $(e,1) = (0\vee e,a\vee e)\in\theta$.  Therefore, for
  every $r\notin \{p,q\}$, we have $(0,r)=(r\wedge e,r\wedge
  1)\in\theta$.  Using the same arguments we used for~$a$,
  we conclude that $r$ is adjacent to every other vertex in
  $G$.  Since $p$~is adjacent to $q$, and every $r\notin
  \{p,q\}$ is adjacent to every vertex, we conclude that $G$
  is a complete graph. Moreover, arguing by symmetry, we get
  that $(0,r)\in\theta$ for every vertex $r$ that fails to
  lie on some edge (every vertex has this property!). Hence
  $0$ is $\theta$-related to the join of the atoms of $\m
  l[G]$, which is $1$.  This proves that $\m l[G]$ is
  simple.
\end{proof}

From Theorem~\ref{czedli-day_thm} and Lemmas~\ref{W_char}
and \ref{graph_lm} we derive the following.

\begin{thm}\label{graph_lattice_thm}
  If\/ $\vr V$ has no Taylor term, then any graph lattice is
  embeddable in the congruence lattice of some member
  of\/~$\vr V$.  Hence, there exist arbitrarily large
  finite, simple, nonmodular lattices that are embeddable in
  congruence lattices of members of\/~$\vr V$.
\end{thm}

If we had defined graphs to allow loops on vertices, then we
would have found that the corresponding graph lattices still
satisfy $\W$, so Theorem~\ref{graph_lattice_thm} also
applies to these lattices.  Lattices of graphs with loops
are still usually nonmodular, but they need not be
simple. For example, this type of graph lattice is not
simple if the underlying graph has at least one isolated
vertex with a loop.

\bigskip

We have shown that certain lattices are not embeddable into
congruence lattices of algebras in $\vr V$ if $\vr V$ has a
Taylor term, while certain lattices must be embeddable if
$\vr V$ does not have a Taylor term. The two classes that we
have described overlap, allowing us to characterize the
existence of a Taylor term according of the shapes of
congruence lattices of algebras in $\vr V$.

\begin{thm}\label{omit_d1_sub}
  Let\/ ${\vr V}$ be a variety and let\/ $\m l$ be a
  finitely presented lattice satisfying $\W$ but failing
  quasi-identity (\ref{weak_W}). (For example, $\m l$~could
  be\/ $\m d_1$, $\m e_1$ or\/~$\m g$.)  The following
  conditions are equivalent.
  \begin{enumerate}
  \item[(1)] $\vr V$ satisfies an idempotent Maltsev
    condition that fails in the variety of sets.
  \item[(2)] $\vr V$ satisfies a nontrivial meet continuous
    congruence identity.
  \item[(3)] $\m l$ does not appear as a sublattice of\/
    $\Conb(\m a)$ for any $\m a\in {\vr V}$.
  \item[(4)] $\m l$ is not in the meet continuous congruence
    variety of $\vr V$.
\end{enumerate}
\end{thm}

\begin{proof}
  The equivalence of (1) and (2) follows from
  Theorems~\ref{wtaylor_term} and~\ref{typ1condcor}.  That
  (1)$\mimplies$(4) follows from the argument we used in
  Theorem~\ref{taylor_forbidden}~(1), which only required
  that the lattice in question failed quasi-identity
  (\ref{weak_W}).  The implication (4)$\mimplies$(3) is
  trivial.  Finally, the contrapositive of (3)$\mimplies$(1)
  follows from Theorem~\ref{czedli-day_thm}.
\end{proof}


\begin{rem}
  As we stated at the end of
  Section~\ref{mc_lattice_theory}, the lattice~$\m d_1$ is
  subdirectly irreducible and projective in ${\vr L}_{MC}$.
  Therefore the equivalence of items (3) and (4) of
  Theorem~\ref{omit_d1_sub} for $\m d_1$ is to be expected.
  But $\m e_1$ and $\m g$ are not projective in ${\vr
    L}_{MC}$, which makes the equivalence of (3) and (4) for
  these lattices somewhat surprising.
\end{rem}

