\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 is 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 
%meet continuous 
sublattices. Moreover, the map
\begin{equation}\label{embed_product}
\prod_{i\in I} \Conb(\m a_i) \to 
\Conb\biggl(\prod_{i\in I} \m a_i\biggr)\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 
%meet continuous 
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
(cf.\ 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}

%\noindent
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 quasivariety?
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}
If 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 quasivariety.
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.}
%independently 
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 $\vr V$ 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.
% on this problem.

%\bigskip

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;
% It turns out that 
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
%(which satisfies $\W$)
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 for which there
is a meet continuous embedding of $\m l$
into $\Conb(\m a)$ for some $\m a\in \vr V$.
$\vr L_{MC}(\vr V)$ 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 
%each.
apiece.
%In addition to this, it seems useful to include
%the details of the Pixley--Wille algorithm
%here in order to fix language for
%later arguments.
%%so that it can be 
%%%used without detail
%%referred to in 
%%later arguments.
%%%in this monograph.

Using the same arguments as those used for ordinary lattice
identities (cf.\ the second 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
%This shows that
%%modulo the identities defining the variety
%%of all meet continuous lattices,  
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 occuring 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 continous lattice $\m f$
generated by some 
%(possibly infinite) 
set 
$\{x_1, x_2, x_3, \ldots\}$.
A meet continuous lattice 
$\m l$ satisfies the inclusion
$p\leq Q$ if 
%and only 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}

%We note in passing that 
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.
%as one sees by identifying $p$ and $Q$ with
%an element and ideal of a finitely generated free lattice.
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
$\{\circ, \wedge\}$ that is obtained from $p$
by replacing each occurrence of
binary join with the $i$-fold relational product
and keeping each occurrence of $\meet$.
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 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 (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. 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 
%associated to the inclusion
%$x_1\circ x_2\subseteq x_2\circ x_1$ 
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 
$p(x_1,x_2,x_3) = x_1\wedge(x_2\circ x_3)$
and
$r(x_1,x_2,x_3) = (x_1\wedge x_2)\circ (x_1\wedge x_3)$.


We begin 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 $\vr G(p)$. 
%$\vr G(p)$ is connected. 
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 let
$\alpha_i\in\Con(\m a)$ for $1\leq i\leq n$.
\begin{enumerate}
\item[(1)]
If $Y\to A\colon y_s\mapsto a_s$ is an assignment
such 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}


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 
%$\vr G_{\ell}(p)$ that there exists 
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.

\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
%(not necessarily directed)
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
\[
F_i(\tau_k(y_1), \ldots, \tau_k(y_m))\approx 
F_j(\tau_k(y_1), \ldots, \tau_k(y_m)).
\]
\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 $p\subseteq r$, 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 
\[
\begin{array}{rl}
F_i(\tau_k(y_1), \ldots, \tau_k(y_m)) &= 
\widehat{\tau_k}(F_i(y_1, \ldots, y_m))\\
&=\widehat{\tau_k}(f_i)\\
&= \widehat{\tau_k}(f_k)\\
&=F_j(\tau_k(y_1), \ldots, \tau_k(y_m)),
\end{array}
\]
since $(f_i,f_j)\in\theta_k = \ker(\widehat{\tau_k})$.
Since 
$F_i(\tau_k(\wec{y})) = F_j(\tau_k(\wec{y}))$
%$F_i(\tau_k(y_1), \ldots, \tau_k(y_m)) =
%F_j(\tau_k(y_1), \ldots, \tau_k(y_m))$ 
holds as a relation
among free generators, 
$F_i(\tau_k(\wec{y})) \approx F_j(\tau_k(\wec{y}))$
%$F_i(\tau_k(y_1), \ldots, \tau_k(y_m))\approx 
%F_j(\tau_k(y_1), \ldots, \tau_k(y_m))$
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 $\widehat{v}((y_i,y_j))=(a_i,a_j)\in\alpha_k$
for any generating pair $(y_i,y_j)$
of $\theta_k$.
%According to our definitions, $(y_1,y_2)$
%is a generic pair of $p$. 

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 
$(F_i(\wec{y}),F_j(\wec{y}))\in\theta_k$.
\end{enumerate}
By Claim~\ref{clmG(p)}~(1), these conditions yield
$(y_1,y_2)\in r(\theta_1,\ldots,\theta_n)$,
hence 
\[
(a_1,a_2) = \widehat{v}((y_1,y_2)) \in 
\widehat{v}(r(\theta_1,\ldots,\theta_n))
\subseteq r(\alpha_1,\ldots,\alpha_n).
\]
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 in the third paragraph of this proof we 
reduced the proof of this theorem 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. 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 $\vr V$, 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 $\{B_1, \ldots, B_5\}$
where 
$B_1$ consists of those $j\in N$ for which
$(x_j,y_j)= (x,x)$, $B_2$ consists of
those $j\neq i$ for which 
$(x_j,y_j)= (x,y)$, $B_3=\{i\}$,
$B_4$ consists of
those $j\in N$ for which
$(x_j,y_j)= (y,x)$, and $B_5$
consists of those $j$ for which
$(x_j,y_j)= (y,y)$.
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 is now
$f_i(x,x,x,y,y)\approx f_i(x,y,y,x,y)$.

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 $\delta\leq \alpha\leq\beta\vee\gamma$,
there is a least $k$ such that 
$\alpha\cap 
(\beta\circ_k\gamma)\not\subseteq \delta$.
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 
$$
u = u_0
\wrel{\beta}u_1\wrel{\gamma}u_2\cdots u_k=v
$$
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
$$
\begin{array}{rl}
r=p(u_1,u_1,u_0)&\gamma\;\;
p(u_1,u_2,u_0)\wrel{\;\beta\;}
p(u_0,u_3,u_1)\\
&\gamma\;\;
p(u_0,u_4,u_1) \wrel{\;\beta\;}
%p(u_0,u_5,u_1) \wrel{\;\gamma\;}
\cdots
p(u_0,u_k,u_1)=s
\end{array}
$$
is a $\gamma\circ_{k-1}\beta$-chain. Moreover, the 
$i$-th Taylor identity implies that 
$$
p(u_0,u_0,u_1) = f_i(u_0, u_0, u_0, u_1 ,u_1)
= f_i(u_0, u_1, u_1, u_0 ,u_1) = p(u_1, u_1, u_0),
$$
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
$$
f_i(u_0,u_0,u_0,\underline{u_1},\underline{u_1}) 
%\wrel{\equiv_{\delta}}
\wrel{{\delta}}
f_i(u_0,u_0,u_k,\underline{u_1},\underline{u_1}).
$$
Since $\C c(\beta\vee\gamma,\alpha;\delta)$ holds,
we may change underlined values from $u_1$ to $u_0$
to obtain
$$
f_i(u_0,u_0,u_0,\underline{u_0},\underline{u_0}) 
%\wrel{\equiv_{\delta}} 
\wrel{{\delta}} 
f_i(u_0,u_0,u_k,\underline{u_0},\underline{u_0}) ,
$$
which we write as
$$
f_i(u,u,u,u,u) 
%\wrel{\equiv_{\delta}} 
\wrel{{\delta}} 
f_i(u,u,v,u,u).
$$
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~$i$. 
%Then for $i=n$ we get from the
%idempotence of~$f$ that $v\wrel{\delta}u$, 
%the contradiction that finishes the proof. 

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 idempotence we get
$v = f(v, v, \ldots, v)\wrel{\delta} u$,
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
$p_{[0]}:=p$, $q_{[0]}:=q$, $r_{[0]}:=r$,
and 
$p_{[n+1]}:= p_{[n]}\wedge (q_{[n]}\vee r_{[n]})$,
$q_{[n+1]}:= q_{[n]}\wedge (p_{[n]}\vee r_{[n]})$,
$r_{[n+1]}:= r_{[n]}\wedge (p_{[n]}\vee q_{[n]})$.
If $(\alpha, \beta, \gamma)$ is a triple of congruences
on $\m a$, 
then $\alpha_{[n]}:=p_{[n]}(\alpha,\beta,\gamma),
\beta_{[n]}:=q_{[n]}(\alpha,\beta,\gamma),
\gamma_{[n]}:=r_{[n]}(\alpha,\beta,\gamma)\in\Con(\m a)$.

\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 $(u, w)\in \alpha\cap(\beta\circ\gamma)$,
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}[ht]
\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 $(u,w)\in
\alpha_{[k]}\cap(\beta_{[k]}\vee\gamma_{[k]})
=\alpha_{[k+1]}$ 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 we have
$$
\alpha\cap (\beta\circ \gamma) 
\cap (\gamma\circ \beta) 
\subseteq \alpha\cap (\beta\circ \gamma) 
\subseteq \alpha_{[n]}\subseteq \delta,
$$
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}
((p\wedge q)\approx s) \wrel{\&}
((p\wedge r)\approx s) \wrel{\&}
(p_{[2]}\approx s) \wrel{\to} 
((p\wedge (q\vee r))\approx s)
\end{equation}
holds in the congruence lattices of
algebras in $\vr v$.
\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 $w(p,q,r) < p\wedge (q\vee r)$
%such that 
%$$w(p,q,r)\not\geq p\wedge(q\vee r)$$
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 $(1)\Longrightarrow(2)$, 
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 show that the 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 $\m d_1$.
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}
%$\delta^{\wec{w}}(p,q,r)\approx p\wedge (q\vee r)$
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)}).
%$\delta^{\wec{w}}(p,q,r)\approx p\wedge(q\vee r)$.

To prove that $\vr V$ satisfies 
$\alpha\cap(\beta\circ_{4}\gamma)\subseteq
w(\alpha, \beta, \gamma)$ 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
let $\alpha = \Cg a(u,v), \beta = \Cg a((u,x),(y,z))$, 
and $\gamma = \Cg a((x,y), (z,v))$. The
free generators are related as indicated
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 
$w(p,q,r)\leq \delta^{\wec{w}}(p,q,r) < p\wedge (q\vee r)$).
%If $\alpha, \beta$ and $\gamma$ are the congruences
%in the generic occurrence of
%the situation ``$(u,v)\in \alpha\cap(\beta\circ_4 \gamma)$'',
%then t
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 $\vr V$.
%Since $w\in D$ we have $w(p,q,r) < p\wedge (q\vee r)$,
%so we certainly also have $w(p,q,r)\not\geq 
%p\wedge (q\vee r)$. 
This completes the proof
that (4)$\mimplies$(5).


The implication (5)$\mimplies$(6)
is trivial, so we now prove that (6)$\mimplies$(1).
%The assertion that $\vr V$
%satisfies a congruence inclusion of the
%form $\alpha\cap (\beta\circ_k\circ \gamma)\subseteq 
%w(\alpha,\beta,\gamma)$
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 let
$\alpha = \Cg a(a_0,a_k), 
\beta = \Cg a(\{(a_{i},a_{i+1})\;|\;\mbox{\rm $i$ even}\}), 
\gamma = \Cg a(\{(a_{i},a_{i+1})\;|\;\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}.
\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}
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 
$\varphi(b)\geq a$.
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 
$$
w(\alpha,\beta,\gamma)=
\varphi(w(p,q,r)) \not\geq 
%\varphi(p\wedge(q\vee r)) =
\alpha = \Cg a(u,v)\,.
$$
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 
$$
p\wedge (q_{[1]}\vee r_{[1]})=p\wedge ((q\wedge (p\vee r))\vee (r\wedge (p\vee q))),
$$
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
$$Q_n:\,\,
((p\wedge q)\approx s) \wrel{\&}
((p\wedge r)\approx s) \wrel{\&}
(p_{[n]}\approx s) \wrel{\to} 
((p\wedge (q\vee r))\approx s)
$$
become formally stronger as $n$ increases,
since 
$$
((p\wedge q)\approx s) \wrel{\&}
((p\wedge r)\approx s) \wrel{\&}
(p_{[n]}\approx s) \wrel{\to} 
(p_{[n+1]}\approx s)
$$
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:
$p^0 = p, r^0 = r,
p^{n+1} = p\wedge(q\vee r^n)$, and
$r^{n+1} = r\wedge(q\vee p^n)$.
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}.
\begin{figure}[ht]
\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}
But it is easy to see 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 all $n$.
This is different from inclusions of
the form $\alpha\cap(\beta\circ_k\gamma)
\subseteq w(\alpha,\beta,\gamma)$
for $k\geq 4$, since these inclusions are nontrivial
as soon as $w(p,q,r)\not\geq p^1 = p\wedge(q\vee r)$.

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){$\delta$}
\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 
$\varphi(\delta^{\wec{w}}(p,q,r)) < \varphi(p\wedge(q\vee r)) = \alpha$.
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
that (6)$\mimplies$(1) using
the number $3$ in place of $k$.
This time, 
define $\m a = \lb\{a_0,a_1,a_2,a_3\};\emptyset\rb$,
$\alpha = \Cg a(a_0,a_3)$, 
$\beta = \Cg a((a_0,a_1),(a_2,a_3))$
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\;|\;t(p,q,r)\geq p^n, \;\mbox{\rm for some $n$}\}$.
If $w$ is as 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}(20,20)
\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(7,-7){$\m m_3$}
\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 
$\alpha\cap(\beta\circ\gamma)\subseteq w(\alpha,\beta,\gamma)$
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 $\alpha\cap(\beta\circ\gamma)\subseteq 
\alpha\cap w(\alpha,\beta,\gamma) \lneqq \alpha$.
In particular, the idempotent Maltsev condition
associated with $\alpha\cap(\beta\circ\gamma)\subseteq 
w(\alpha,\beta,\gamma)$ 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)
%But it follows from the results of 
%\cite{k-szendrei2} 
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 show that if $\vr V$
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}(120,45)
\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}

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

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(80,53)
\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 \hfill\linebreak
$\Conb(\m a)$.

\bigskip   %%placed here to get figure in correct position!

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(70,38)
\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){$\m e_2$}
\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.
We start with our assumption that
$\C C(\beta, \beta;\beta\wedge\delta)$ holds,
derive $\C C(\beta\vee\delta, \beta\vee\delta;\delta)$
using Theorem~\ref{perspective_abelian_taylor},
and then derive $\C C(\beta, \alpha;\delta)$
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 $\alpha, \beta \leq \tau$
%, by monotonicity, 
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).
%Since $\alpha\wedge\beta = \alpha\wedge\gamma = \alpha\wedge\delta$,
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 
$\alpha, \beta$ and $\delta$
we obtain a contradiction.
One can handle the second lattice
in the third claim by similar arguments.
%This completes the proof.
\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 
%(as explained to us by J.~B.~Nation) 
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 
$\lb G \wrel{|} R\rb$ 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 (p_i(\wec{x}) = q_i(\wec{x}))\rightarrow
(p(\wec{x}) = q(\wec{x})).
\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 
%the partially ordered set 
$\lb V\cup E; \in\rb$ 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 $2$.

\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){$1$}
%\put(-25,30){$\m l[G]$}
\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}, 
\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}
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.

\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}

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 $0< u< x < 1$, $x\in \{a,b\}$, and 
$0 < y < v < 1$, $y\in \{c,d\}$.
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 any 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 simple nonmodular lattice
of height $3$.
\end{lm}

\begin{proof}
Since $G$ has at least one edge, the height
of $G$ is exactly $3$. If $G$ is modular,
then the join of $2$ distinct elements of height $1$
in $\m l[G]$ (atoms, or vertices)
is an element of height $2$ (a coatom, or edge).
This means that any two vertices of $G$ are
adjacent, hence $G$ is a complete graph.
Now suppose that $a, b, c$ and $d$ are
$4$ distinct vertices of $G$. Then 
$e:=\{a,b\}$ and $f:=\{c,d\}$ are 
disjoint edges of $G$, so they are
coatoms that meet to zero in $\m l[G]$;
i.e., they are elements of coheight $1$
whose meet has coheight $> 2$.
This is impossible in a modular lattice.

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 $a$.

In the remaining case $G$ is a complete graph on 
at least $4$ vertices. If $\ell$ is any edge
not incident to $a$, then $(\ell,1) = (0\vee\ell,a\vee\ell)\in\theta$.
From this it follows that the meet of all edges $\ell$
not incident to the vertex $a$ is also $\theta$-related to $1$.
These are the edges from the induced complete graph
on the vertex set $V-\{a\}$.
Since $V-\{a\}$ has at least $3$ elements, 
and a complete graph on $\geq 3$ vertices
has no vertex
incident to all edges, the meet of all edges
not incident to $a$ is zero. Again, $\theta$ is the total relation.
\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 
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.

%The next theorem combines and strengthens the assertions made
%about $\m d_1$ above and in Theorem~\ref{taylor_forbidden}.

\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)] $\vr V$ 
%has a Taylor term.
satisfies an 
%nontrivial idempotent Maltsev condition.
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},
$\m d_1$ 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}

