\chapter{Congruence Meet and Join Semidistributivity}\label{congruence_sdj}

In this chapter we will characterize the satisfaction
of a nontrivial congruence identity in several new ways
(Theorems~\ref{main2}, \ref{SD-join/Mod},
\ref{omit_d2_sub},
\ref{typ1condcor_dual}, and \ref{cong_id2}). 
%These results are 
%%mainly focused on the problem of understanding 
Our goal is to understand
congruence join semidistributivity
and its relationship to the satisfaction
of a nontrivial congruence identity.
Although most of
the results of this chapter concern 
varieties satisfying congruence identities, 
we start the chapter by recalling known results
about congruence meet semidistributivity and
we end the chapter
with new results (Theorem~\ref{sdj-thm}) 
that provide several characterizations
of congruence join semidistributivity for varieties.
These results 
generalize Theorems~9.8, 9.10 and 9.11 of \cite{hobby-mckenzie}.

As a byproduct, we solve an old problem
of B.~J\'{o}nsson 
\index{J\'{o}nsson, B.}
(Problem 2.18 of \cite{jonsson3})
by showing that
the class of congruence join semidistributive varieties
is definable by a Maltsev condition.

It should be noted that any congruence join semidistributive
variety is also congruence meet semidistributive
(see Exercise 7.14~(10) of \cite{hobby-mckenzie}
or (1)$\mimplies$(6) of Theorem~\ref{sdj-thm}
of this chapter).
Therefore {\em congruence join semidistributivity}
and {\em congruence semidistributivity} are 
equivalent properties for varieties.

\section{Congruence Meet Semidistributivity}\label{sdm_sect}

Using straightforward arguments, it follows from a 
combination of the statements in
Theorem~\ref{basic_centrality}
%\footnote{Specifically, 
%from items (1), (8) and (10) of Theorem~\ref{basic_centrality}.}
that if $\alpha$ and $\beta$ are congruences
on an algebra $\m a$, then 
$[\alpha,\beta] \leq \alpha\wedge\beta$ and 
$(\alpha\wedge\beta)/[\alpha,\beta]$
is an abelian congruence on $\m a/[\alpha,\beta]$.
Therefore, if $\vr V$ is a variety that omits abelian
congruences,  it follows that 
$[\alpha,\beta] = \alpha\wedge\beta$ for any 
$\alpha, \beta\in \Con(\m a)$ for any $\m a\in \vr V$.
Theorem~\ref{basic_centrality}~(5) implies
that the commutator operation is
semidistributive 
in its first variable
i.e., 
\[
[\alpha,\gamma] = \delta
\wrel{\&} 
[\beta,\gamma] = \delta
\wrel{\to}
[\alpha\vee\beta,\gamma] = \delta\,.
\]
Hence if $\vr V$ omits abelian congruences,
then the congruence lattices of members
of $\vr V$ must satisfy the ordinary 
meet semidistributive law:
\[
\alpha\wedge\gamma = \delta
\wrel{\&} 
\beta\wedge\gamma = \delta
\wrel{\to}
(\alpha\vee\beta)\wedge\gamma = \delta\,.
\]
The converse of this statement is true,
but is much more difficult to prove.
It was established for locally
finite varieties in Theorem~9.10
of \cite{hobby-mckenzie} using tame congruence theory,
and in full generality in 
Corollary~4.7 of \cite{k-szendrei2}
using combinatorial arguments.
This result links many different properties,
as we will show in the next theorem.

%***Compare this statement with Theorem~\ref{sdj-thm}~(7).***

Recall the type of meet continuous 
lattice words defined 
before Corollary~\ref{strange-cor}:
if $x, y, z$ are variables let
$y_0 = y$, $z_0 = z$,
$y_{n+1} = y\vee(x\wedge z_n)$, and
$z_{n+1} = z\vee(x\wedge y_n)$. 
Let $y_{\omega} = \bigvee_{n<\omega} y_n$
and let 
$z_{\omega} = \bigvee_{n<\omega} z_n$.
When it is necessary to indicate the variables
involved and their order we will write 
$y_{\omega}(x,y,z)$ for $y_{\omega}$ and 
$z_{\omega}(x,y,z)$ for $z_{\omega}$.

\begin{thm}\label{SD-meet-thm}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
%for $\vr V$, any
%$\m a\in {\vr V}$, and any $\alpha,\beta,\gamma\in\Con(\m a)$.
\begin{enumerate}
\item[(1)] $\vr V$ is congruence meet semidistributive.
\item[(2)] There is a positive integer $n$ such that 
if $\m a\in {\vr V}$ and 
$\alpha, \beta, \gamma\in \Con(\m a)$, then 
\[\alpha\cap(\beta\circ\gamma)
\subseteq \beta_n\,,\] 
where $\beta_n := y_n(\alpha,\beta,\gamma)$.
\item[(3)] $[\alpha,\beta] = \alpha\wedge\beta$. 
for all 
$\alpha, \beta\in \Con(\m a)$ and all $\m a\in {\vr V}$.
\item[(4)] No member of $\vr V$ has a nontrivial 
abelian congruence.
\item[(5)] No member of $\vr V$ has a nontrivial 
abelian tolerance.
\item[(6)] 
$\C C(\alpha,\beta;\delta)\miff \beta\wedge(\alpha\vee(\beta\wedge\delta))\leq
\delta$
for all $\m a\in {\vr V}$ and all 
$\alpha, \beta, \delta\in \Con(\m a)$.
\item[(7)] 
$\vr V$ satisfies the meet continuous congruence
identity 
\[
x\wedge(y\vee z) \approx x\wedge y_{\omega}\,,
%\vee(x\wedge z_{\omega})\,, 
%(x,y,z)
%\alpha\wedge(\beta\vee\gamma) = \alpha\wedge \beta_{\omega}\,.
\]
where $y_{\omega} = y_{\omega}(x,y,z)$.
% and 
%$z_{\omega} = z_{\omega}(x,y,z)$.
\item[(8)] The meet continuous congruence variety
of $\vr V$ is meet semidistributive.
% variety
%of meet continuous lattices.
\item[(9)] $\m m_3$ is not embeddable 
in $\Conb(\m a)$ for any $\m a\in {\vr V}$.
\item[(10)] ${\vr V}$ satisfies a family of idempotent Maltsev
conditions that, taken together, fail in any nontrivial variety of modules.
(Equivalently, $\vr V$ satisfies a {\em single} 
idempotent Maltsev condition that fails 
in any nontrivial variety of modules.)
\end{enumerate}
\end{thm}

\begin{proof}
The equivalence of items (1), (2) and (3) 
is Corollary~4.7 of \cite{k-szendrei2}.
%(The proof that (3)$\mimplies$(1)
%is outlined in the first paragraph
%of this section.)

If (3) holds, and $\m a\in {\vr V}$ 
has an abelian congruence $\gamma$, 
then $0 = [\gamma,\gamma] = \gamma\wedge\gamma = \gamma$.
Hence $\gamma$ is trivial, and (4) holds.
Conversely, assume that (4) holds.
It follows from Theorem~\ref{basic_centrality}~(8)
that $\C C(\alpha,\beta;\alpha\wedge\beta)$ holds,
so $[\alpha,\beta]\leq \alpha\wedge\beta$ holds
for any pair of congruences on any algebra $\m a$.
Moreover $\C C(\alpha,\beta;[\alpha,\beta])$
holds, according to the definition of the commutator, and
\[
\C C(\alpha,\beta;[\alpha,\beta])\mimplies
\C C(\alpha\wedge\beta,\alpha\wedge\beta;[\alpha,\beta])
\]
by Theorem~\ref{basic_centrality}~(1),
so $(\alpha\wedge\beta)/[\alpha,\beta]$ is an
abelian congruence on $\m a/[\alpha,\beta]$
by Theorem~\ref{basic_centrality}~(10).
Therefore, from (4) we derive (3).

Assume that (4) holds. Since 
$\vr V$ omits nontrivial abelian congruences, it
omits nontrivial strongly abelian congruences.
Therefore,  by Theorem~\ref{str_ab_thm1},
$\vr V$ satisfies a nontrivial idempotent Maltsev condition.
In this situation,
Theorem~\ref{abel_tol=abel_cong} guarantees
that an abelian tolerance generates
an abelian congruence. Since $\vr V$ omits
nontrivial abelian congruences, 
$\vr V$ omits nontrivial abelian tolerances.
This shows that (4)$\mimplies$(5).
The reverse implication is trivial.

To connect item (6) with the preceding conditions,
assume that (3) holds and that 
\begin{enumerate}
\item[(a)] $\C C(\alpha,\beta;\delta)$ holds.
\end{enumerate}
By Theorem~\ref{basic_centrality}~(8), 
\begin{enumerate}
\item[(b)] $\C C(\alpha,\beta;\beta)$ holds, 
\end{enumerate}
so by 
Theorem~\ref{basic_centrality}~(6) we derive 
from (a) and (b) that
\begin{enumerate}
\item[(c)] $\C C(\alpha,\beta;\beta\wedge\delta)$ holds.
\end{enumerate}
Again by 
Theorem~\ref{basic_centrality}~(8)
we get that 
\begin{enumerate}
\item[(d)] $\C C(\beta\wedge\delta,\beta;\beta\wedge\delta)$ holds,
\end{enumerate}
so by 
Theorem~\ref{basic_centrality}~(5)
we get from (c) and (d) that 
\begin{enumerate}
\item[(e)] $\C C(\alpha\vee(\beta\wedge\delta),\beta;\beta\wedge\delta)$ holds.
\end{enumerate}
From this and the definition of the commutator we get that 
\begin{enumerate}
\item[(f)] $[(\alpha\vee(\beta\wedge\delta)),\beta]\leq \beta\wedge\delta$ holds.
\end{enumerate}
Now from (f) and item (3) we derive that
\[
\beta\wedge(\alpha\vee(\beta\wedge\delta)) = 
[(\alpha\vee(\beta\wedge\delta)),\beta]\leq\beta\wedge\delta\leq\delta\,.
\]
Conversely, 
$\beta\wedge(\alpha\vee(\beta\wedge\delta)) \leq \delta$
implies $\C C(\alpha,\beta;\delta)$ for any
three congruences on any algebra, 
as Theorem~\ref{basic_centrality}~(8) proves.
This shows that (3) implies (6). 

On the other hand, to see that 
(6)$\mimplies$(4), take $\alpha = \beta = \gamma$
and $\delta = 0$ in (6). The result
is the assertion that $\gamma$ is abelian
if and only if $\gamma = 0$.

The equivalence of (1), (7) and (8) follows 
from Theorem~\ref{strange-thm}
and the fact that congruence lattices are
meet continuous.

Item (8) implies item (9), since $\m m_3$ is not 
meet semidistributive.

Assume that (9) holds. Then, since $\m m_3$
satisfies Whitman's condition $\W$, it follows from
Theorem~\ref{czedli-day_thm} that $\vr V$
satisfies a nontrivial idempotent Maltsev condition.
In this situation, a result from \cite{k-szendrei2}
explains how to construct from a nontrivial
abelian congruence on an algebra in $\vr V$
a copy of $\m m_3$ in
the congruence lattice of some other algebra in $\vr V$. 
Namely, suppose that
$\alpha$ is a nonzero abelian congruence on 
some $\m a\in {\vr V}$, that 
$\m a\times_{\alpha} \m a$ is the
subalgebra
of $\m a^2$ whose universe is $\alpha$,
that $\eta_1, \eta_2\in\Con(\m a\times_{\alpha}\m a)$ 
are the coordinate projection kernels, 
and that $\Delta$ is the congruence on
$\m a\times_{\alpha} \m a$ generated by
the pairs $\lb (a,a), (b,b)\rb$ with $(a,b)\in \alpha$.
It is a consequence of Theorem~3.5 of 
\cite{k-szendrei2} that 
$\eta_1, \eta_2$, and $\Delta$ generate
a sublattice of $\Conb(\m a\times_{\alpha}\m a)$
isomorphic to $\m m_3$.
Thus, if (9) holds, then (4) must hold.

The Pixley--Wille algorithm shows that 
the class of varieties
satisfying a congruence condition like the one in
item (2) can be defined by a single idempotent
Maltsev condition. 
Since (2) is equivalent to (4), this Maltsev condition
cannot be satisfied by any variety with nontrivial
abelian algebras, hence by any nontrivial variety of modules.
Thus (7) implies the stronger of the two statements in (10).

Finally, we show that the weaker of
the two statements in (10) implies (4).
Assume instead that $\vr V$ satisfies a family
$\vr F$
of idempotent Maltsev conditions that fail
in every nontrivial variety of modules,
but that some $\m a\in {\vr V}$ has a nontrivial
abelian congruence $\alpha$. 
We may assume, without loss of generality, that
$\vr V$ is an idempotent variety. For if we replace
$\vr V$ and $\m a$ by their idempotent reducts,
then we do not affect the idempotent Maltsev conditions
satisfied by $\vr V$, nor that $\alpha$ is an
abelian congruence on $\m a$. Thus, we make this assumption.
We may (and do) further assume that $\m a$ itself is abelian,
since if it is not we may simply replace it by one of its 
subalgebras that is supported by nontrivial $\alpha$-block.

At least one of the Maltsev conditions in $\vr F$ 
is nontrivial, since these conditions taken
together fail
in every nontrivial variety of modules.
Corollary~4.5
of \cite{k-szendrei2} proves that an
abelian algebra $\m a$ in a variety satisfying some
nontrivial idempotent Maltsev condition
is {\bf quasi-affine}, 
\index{quasi-affine}
which by definition means
that $\m a$ is a subalgebra of a reduct 
of an algebra that has the same universe and
polynomial operations as some module.
Since $\m a$ is both 
idempotent and quasi-affine, 
it is a subalgebra of an algebra
$\m m^*$ that is a reduct of some module $\m m$
over some ring $\m r$.
There is no harm in renaming the zero element
of $\m m$ so that it lies in $A$.
If $p$ is the module term $x-y+z$, 
then $p\colon \m m^3\to \m m$
is a homomorphism, and therefore
$p\colon \m a^3\to \m m^*$ is also a homomorphism.
The image $\m a':=p(\m a^3)$ contains
$A$, since $p(a,a,a)=a$. Moreover, 
$\m a'$ is a quotient of 
$\m a^3\in \vr V$, hence $\m a'\in \vr V$.
This produces a possibly larger subalgebra
$\m a\leq \m a'\leq \m m^*$. Iterating this,
$\m a\leq \m a'\leq \m a''\leq \cdots$,
and taking a union yields an algebra
$\m b\in \vr V$ that is a subalgebra
of $\m m^*$ and is closed under $p$.
Since $0\in A\subseteq B$, the set $B$
is closed under the abelian group operations
$x+y = p(x,0,y), -x = p(0,x,0)$, and $0$.
Thus $\widehat{\m b}:= \lb \m b; p, 0\rb$ 
is an additive subgroup of $\m m$ that is also 
a subalgebra of a 
reduct of $\m m$.
This implies that $\widehat{\m b}$ is term 
equivalent to a module over the subring
of $\m r$ consisting of all elements
$r\in R$ such that $rB\subseteq B$.
Clearly $\widehat{\m b}$ is nontrivial,
since $\m b$ contains $\m a$ as a subalgebra.
Moreover, $\widehat{\m b}$ satisfies the 
conditions in $\vr F$, since its reduct $\m b$ does.
Therefore the variety generated by $\widehat{\m B}$
is (up to term equivalence)
a nontrivial variety of modules that satisfies
the conditions in $\vr F$, contrary to
(10). This contradicts our assumption that
(4) fails to hold, so we are done.
\end{proof}

%Very little of Theorem~\ref{SD-meet-thm}
%is new. The equivalence of (1)---(5), (9) and (10)
%was proved for locally finite varieties in
%Theorem~9.10 of \cite{hobby-mckenzie}.
%The proofs of many of those implications 
%for locally finite varieties work
%for arbitrary varieties, at least in spirit.
%The one implication among these that is difficult
%to extend to nonlocally finite varieties
%is the implication (1)$\mimplies$(2),
%which was originally established in \cite{k-szendrei2}.
%
%Identity (\ref{SD-meet-identity}) of Lemma~\ref{SD-meet-lm}
%was linked to the
%meet semidistributive law in the proof of Theorem 6.1
%of \cite{jonsson-rival}. This 
%
%% in a way that 
%supplies the argument for the
%implication (1)$\mimplies$(7). 
%
%The proof that the reverse implication 
%holds is embedded 
%That the reverse implication holds can 
%be derived from the proof 
%%But the deeper 
%%fact that items (1) and (7) of Theorem~\ref{SD-meet-thm}
%%are equivalent for arbitrary varieties seems to have
%%first appeared implicitly in the proof 
%of the main theorem of 
%\cite{czedli1}.
%%and later explicitly in
%%Theorem~4.1 of \cite{lipparini4}. 
%
%The paper \cite{lipparini4}
%was also the first to explicitly state that item (9) is
%equivalent to (1) for arbitrary varieties.
%The fact that item (10) is equivalent to (1)
%for arbitrary varieties 
%is implicit in \cite{k-szendrei2} and \cite{lipparini4},
%%and perhaps \cite{finbas}, 
%although it seems to be 
%explicitly stated for the first time here.

\section{More on Congruence Identities}\label{8.2}

\begin{thm} \label{transfer2}
Let $\vr V$ be a variety with a Hobby-McKenzie term.
Assume that $\m a\in {\vr V}$, and that 
$\Conb(\m a)$ has a sublattice of congruences 
labeled as in Figure~\ref{fig34}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,35)
\put(0,15){\circle*{1.2}}
\put(10,0){\circle*{1.2}}
\put(10,30){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(20,20){\circle*{1.2}}
\put(0,15){\line(2,3){10}}
\put(0,15){\line(2,-3){10}}
\put(10,0){\line(1,1){10}}
\bezier{200}(20,10)(16,15)(20,20)
\bezier{200}(20,10)(24,15)(20,20)
\bezier{200}(20,20)(12,22)(10,30)
\bezier{200}(20,20)(18,28)(10,30)
\put(22,9){$\beta$}
\put(22,19){$\alpha$}
\put(-5,14){$\gamma$}
\put(9,32){$\tau$}
\end{picture}
\end{center}
\caption{ \ }\label{fig34}
\end{figure}
If $J_T$ is a separated solvability
obstruction in $I[\beta,\alpha]$, then
there is a separated solvability obstruction 
$I_S$ contained in $I[\alpha,\tau]$ 
and a $\leadsto$--sequence from $I_S$ to $J_T$.
\end{thm}

\begin{proof}
We will show first 
that if $J_T$ is a separated solvability
obstruction in $I[\beta,\alpha]$, then
there is a separated solvability obstruction 
$I'_{S'}$ contained in either $I[\beta,\alpha]$ or
$I[\alpha,\tau]$ 
and a $\leadsto$--sequence of length 
$1$ or $2$ from ${I'}_{S'}$ to $J_T$.
If it happens that $I'_{S'}$ is contained in 
$I[\alpha,\tau]$, then we are done.
If instead $I'_{S'}$ is contained in 
$I[\beta,\alpha]$, then we can repeat the argument
to find $I''_{S''}$ contained in either $I[\beta,\alpha]$ or
$I[\alpha,\tau]$ 
and a $\leadsto$--sequence from $I''_{S''}$ to $I'_{S'}$.
The argument cannot be repeated indefinitely
since concatenating $\leadsto$--sequences
forms longer $\leadsto$--sequences and
there is a bound on the length of 
$\leadsto$--sequences. Therefore the argument will produce
a separated solvability obstruction in 
$I[\alpha,\tau]$ that initiates a $\leadsto$--sequence
terminating at $J_T$.


Recall from the proof of Theorem~\ref{transfer1},
applied to the setting of Corollary~\ref{transfer1_cor},
that if $\mu:=\gamma \cap (\delta:T)$, then 
$I[\mu,\gamma]_{\gamma}\leadsto I[\delta,\theta]_T$.
Since $\mu\leq (\delta:T)$ and $\delta\leq (\delta:T)$,
we have 
$$
\gamma\cap (\delta\vee\mu)\leq \gamma\cap (\delta:T) = \mu\,,
$$
and therefore $\gamma\wedge (\delta\vee(\gamma\wedge \mu))\leq \mu$.
By Theorem~\ref{basic_centrality}~(8)
we get that
$\C C(\delta,\gamma;\mu)$ holds, or equivalently 
$\delta\leq (\mu:\gamma)$. 
If $\alpha\not\leq (\mu:\gamma)$,
then for $\lambda:=\alpha\wedge (\mu:\gamma)$
( $\geq \delta\geq \beta$)
we have that $I[\lambda,\alpha]$ is a nontrivial 
subinterval of $I[\beta,\alpha]$. 
According to 
the second paragraph of
Lemma~\ref{new_mu}, the
interval $I[\lambda,\alpha]$ is a solvability obstruction.
That paragraph further guarantees that, since 
$\alpha\cap \Cg a(\gamma)\leq \lambda$, we have
$I[\lambda,\alpha]_{\alpha}\leadsto I[\mu,\gamma]_{\gamma}$.
Therefore, in the case where $\alpha\not\leq (\mu:\gamma)$
we have located a separated solvability obstruction 
$I[\lambda,\alpha]_{\alpha}$
in $I[\beta,\alpha]$ and a $\leadsto$-sequence
$$
I[\lambda,\alpha]_{\alpha}\leadsto I[\mu,\gamma]_{\gamma}
\leadsto I[\delta,\theta]_T = J_T
$$
of length two.

Now consider the case where $\alpha\leq (\mu:\gamma)$.
Since $\gamma\not\leq (\mu:\gamma)$ and $\gamma\leq \tau$
we have $\tau\not\leq (\mu:\gamma)$, so $\kappa:=\tau\cap (\mu:\gamma)$
satisfies $\alpha \leq \kappa < \tau$ in this case. 
According to the second paragraph of
Lemma~\ref{new_mu}, the interval $I[\kappa,\tau]$
is a solvability obstruction. Since $\gamma\not\leq \kappa$,
$I[\kappa,\tau]_{\gamma}$ is a separated solvability
obstruction. We prove that 
$I[\kappa,\tau]_{\gamma}\leadsto I[\delta,\tau]_T$
by verifying criteria (d)' and (g)
from Lemma~\ref{new_mu}.
In our setting these statements are
\begin{enumerate}
\item[(d)'] $\gamma\cap \Cg a(T) \subseteq 
(\gamma\wedge (\delta:T))\vee(\Cg a(T)\wedge(\delta:T))$
and
\item[(g)\hphantom{'}] 
$\gamma\cap \kappa = \gamma\cap (\delta:T)$.
\end{enumerate}
Item (d)' follows from the observation that
$$
\gamma\cap \Cg a(T)\subseteq \gamma\cap \alpha
\subseteq \mu := \gamma\cap (\delta:T)\,,
$$
since $\gamma\cap (\delta:T)$ is the first
joinand in the right hand side of (d)'.
For item (g), observe that 
$$
\gamma\cap \kappa = \gamma\cap \tau \cap (\mu:\gamma)
= \gamma\cap (\mu:\gamma)
= \mu =\gamma\cap (\delta:T)\,.
$$
This proves that 
$I[\kappa,\tau]_{\gamma}\leadsto I[\delta,\tau]_T$.
\end{proof}

Let $\m l$ be a lattice. 
A {\bf herringbone} 
\index{herringbone}
in $\m l$ is
the union of three descending 
chains $\{\alpha^i\}\cup \{\beta^{2i}\}\cup 
\{\gamma^{2i+1}\}$ in 
$\m l$ where $\{\alpha^i\}\cup \{\beta^{2i}\}$ 
and $\{\alpha^i\}\cup \{\gamma^{2i+1}\}$ are
sublattices of $\m l$ ordered as in Figure~\ref{fig35}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,37)
\put(0,0){\circle*{1.2}}
\put(0,10){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(0,30){\circle*{1.2}}
\put(15,5){\circle*{1.2}}
\put(15,10){\circle*{1.2}}
\put(15,15){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(15,25){\circle*{1.2}}
\put(15,30){\circle*{1.2}}
\put(15,35){\circle*{1.2}}
\put(30,5){\circle*{1.2}}
\put(30,15){\circle*{1.2}}
\put(30,25){\circle*{1.2}}
\put(0,-2){\line(0,1){32}}
\put(15,0){\line(0,1){35}}
\put(30,-2){\line(0,1){27}}
\put(0,0){\line(3,1){15}}
\put(0,10){\line(3,1){15}}
\put(0,20){\line(3,1){15}}
\put(0,30){\line(3,1){15}}
\put(30,5){\line(-3,1){15}}
\put(30,15){\line(-3,1){15}}
\put(30,25){\line(-3,1){15}}
\put(16,35){$\alpha^{0}$}
\put(10,29){$\alpha^{1}$}
\put(16,24){$\alpha^{2}$}
\put(10,19){$\alpha^{3}$}
\put(16,14){$\alpha^{4}$}
\put(10,9){$\alpha^{5}$}
\put(16,4){$\alpha^{6}$}
\put(-5,29){$\beta^{0}$}
\put(-5,19){$\beta^{2}$}
\put(-5,9){$\beta^{4}$}
\put(-5,-1){$\beta^{6}$}
\put(31,24){$\gamma^{1}$}
\put(31,14){$\gamma^{3}$}
\put(31,4){$\gamma^{5}$}
\end{picture}
\end{center}
\caption{A herringbone}\label{fig35}
\end{figure}
In other words, a herringbone is a partial 
sublattice of $\m l$, ordered
as in Figure~\ref{fig35}, with 
\begin{enumerate}
\item[$(\beta)$] $\alpha^{2i+1}\wedge \beta^{2i} = \beta^{2i+2}$, 
$\alpha^{2i+2}\vee \beta^{2i} = \alpha^{2i}$, and
\item[$(\gamma)$] $\alpha^{2i+2}\wedge \gamma^{2i+1} = \gamma^{2i+3}$, 
$\alpha^{2i+3}\vee \gamma^{2i+1} = \alpha^{2i+1}$.
\end{enumerate}
The {\bf length}
\index{herringbone!length}
\index{length!of a herringbone}
of the herringbone 
is ${\infty}$ if there are infinitely many distinct $\alpha$'s.
Otherwise, the {\bf length} of 
the herringbone is the 
supremum of the superscripts $k$ such that 
$\alpha^0 > \alpha^1 > \cdots > \alpha^{k}$.
(It is easy to see from $(\beta)$ and $(\gamma)$
that if some $\alpha^k = \alpha^{k+1}$ then
$\alpha^k = \alpha^{k+1} = \alpha^{k+2} = \cdots$.)
Our first goal is to prove that if a 
variety $\vr V$ satisfies a
nontrivial congruence identity, 
%n idempotent Maltsev
%condition that fails in the variety of semilattices,
then there
is a fixed
%positive integer $N$ which 
bound on the length
of any herringbone that appears 
in the congruence lattice
of a member of $\vr V$. 

Let $x, y$ and $z$ be lattice variables.
Define lattice words by 
$y^0 = y$, $z^0 = z$,
$y^{n+1} = y\wedge(x\vee z^n)$, and
$z^{n+1} = z\wedge(x\vee y^n)$. 
The construction of these words is 
dual to the words $y_n$ and $z_n$,
which preceded the statement of 
Theorem~\ref{SD-meet-thm}.
As there, we will write 
$y^n(x,y,z)$ and $z^n(x,y,z)$ when it is necessary 
to indicate the order of the variables.

\begin{thm} \label{main2}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ satisfies a nontrivial congruence identity.
\item[(2)] There is a positive integer $N$ such that 
no algebra in $\vr V$ has a herringbone of length
greater than $N$ in its congruence lattice.
\item[(3)] $\vr V$ satisfies the congruence
identity 
\[
z^M\approx z^{M+1}
\]
for some $M$.
\end{enumerate}
\end{thm}

\begin{proof}
If $\vr V$ satisfies a nontrivial congruence identity,
then by Theorems~\ref{main1} and
\ref{rect_Malt_cond2} the variety
$\vr V$ has a Hobby--McKenzie term of
arity $n$ for some $n < \omega$.
We argue that no algebra in $\vr V$ has a herringbone of length
greater than $2n-3$ in its congruence lattice.
If this is not the case, then some
$\m a\in {\vr V}$ has a herringbone
$\{\alpha^i\}\cup\{\beta^{2i}\}\cup \{\gamma^{2i+1}\}$ with 
$\alpha^0 > \alpha^1 > \cdots > \alpha^{2n-3} > \alpha^{2n-2}$.
Figure~\ref{fig36} isolates five congruences 
near the bottom of the herringbone that form
a sublattice of $\Conb(\m a)$ isomorphic to $\m n_5$.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,32)
\put(0,0){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(20,20){\circle*{1.2}}
\put(20,30){\circle*{1.2}}
\put(0,0){\line(2,1){20}}
\put(0,20){\line(2,1){20}}
\put(0,0){\line(0,1){20}}
\put(20,10){\line(0,1){20}}
\put(-12.5,-1){$\beta^{2n-2}$}
\put(-12.5,19){$\beta^{2n-4}$}
\put(22,29){$\alpha^{2n-4}$}
\put(22,19){$\alpha^{2n-3}$}
\put(22,9){$\alpha^{2n-2}$}
\end{picture}
\end{center}
\caption{A piece near the bottom of the herringbone}\label{fig36}
\end{figure}
For the congruences in this figure, 
if $\alpha^{2n-3}\wrel{\lhdlhd}\alpha^{2n-4}$, then by 
Lemma~\ref{meet+join}~(2) we derive that 
$$
\beta^{2n-2} = \beta^{2n-4}\wedge\alpha^{2n-3}
\wrel{\lhdlhd} \beta^{2n-4}\wedge\alpha^{2n-4} = 
\beta^{2n-4}.
$$
But this was proved to be impossible in Theorem~\ref{weak_omit}~(1).
Therefore we cannot have
$\alpha^{2n-3}\wrel{\lhdlhd}\alpha^{2n-4}$.
By Theorem~\ref{solv=omit_obst} there is
a solvability obstruction $I[\mu_{2n-4},\nu_{2n-4}]$
contained in the interval
$I[\alpha^{2n-3},\alpha^{2n-4}]$.
This obstruction is separated by
$S_{2n-4}:=\nu_{2n-4}$. We are now in a position to
apply Theorem~\ref{transfer2} to a 
copy of $\m n_5$ that is
slightly higher up in this herringbone
(Figure~\ref{fig37}).
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,32)
\put(20,0){\circle*{1.2}}
\put(20,20){\circle*{1.2}}
\put(0,10){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(0,30){\circle*{1.2}}
\put(20,0){\line(-2,1){20}}
\put(20,20){\line(-2,1){20}}
\put(20,0){\line(0,1){20}}
\put(0,10){\line(0,1){20}}
\put(22,-1){$\gamma^{2n-3}$}
\put(22,19){$\gamma^{2n-5}$}
\put(-12.5,29){$\alpha^{2n-5}$}
\put(-12.5,19){$\alpha^{2n-4}$}
\put(-12.5,9){$\alpha^{2n-3}$}
\end{picture}
\end{center}
\caption{A piece higher up}\label{fig37}
\end{figure}
The separated solvability obstruction
$I[\mu_{2n-4},\nu_{2n-4}]_{S_{2n-4}}$
that is contained 
in $I[\alpha^{2n-3},\alpha^{2n-4}]$
plays the role of $J_T$ of that theorem.
The theorem guarantees the existence 
of a separated solvability obstruction
$I[\mu_{2n-5},\nu_{2n-5}]_{S_{2n-5}}$
in $I[\alpha^{2n-4},\alpha^{2n-5}]$
and a $\leadsto$--sequence
$$
I[\mu_{2n-5},\nu_{2n-5}]_{S_{2n-5}}
\leadsto\cdots\leadsto
I[\mu_{2n-4},\nu_{2n-4}]_{S_{2n-4}}.
$$
We can work our way up the herringbone,
alternately using
copies of $\m n_5$ on the right
and left and using
Theorem~\ref{transfer2} at each step,
to derive that for each $i = 4, 5, \ldots, 2n-1$,
if $I[\alpha^{2n-i+1},\alpha^{2n-i}]$ contains
a separated solvability obstruction
$I[\mu_{2n-i},\nu_{2n-i}]_{S_{2n-i}}$, 
then $I[\alpha^{2n-i},\alpha^{2n-i-1}]$
contains a separated solvability obstruction
$I[\mu_{2n-i-1},\nu_{2n-i-1}]_{S_{2n-i-1}}$
for which there exists a $\leadsto$--sequence
$$
I[\mu_{2n-i-1},\nu_{2n-i-1}]_{S_{2n-i-1}}
\leadsto\cdots\leadsto
I[\mu_{2n-i},\nu_{2n-i}]_{S_{2n-i}}.
$$
When we reach the top of the herringbone 
we will have located a $\leadsto$--sequence
$$
I[\mu_{0},\nu_{0}]_{S_{0}}
\leadsto\cdots\leadsto
I[\mu_{1},\nu_{1}]_{S_{1}}
\leadsto\cdots\cdots\leadsto
I[\mu_{2n-4},\nu_{2n-4}]_{S_{2n-4}}
$$
of length at least $2n-4$.
But we proved in Theorem~\ref{leadsto_sequence}
that there is no $\leadsto$--sequence
of this length.
This contradiction proves that there
is no herringbone of length greater
than $2n-3$ in $\Conb(\m a)$.


Now we prove that $(2)\Longrightarrow(3)$.
Start with three congruences $\alpha, \beta$
and $\gamma$, and 
define $\beta^n := y^n(\alpha,\beta,\gamma)$
and $\gamma^n := z^n(\alpha,\beta,\gamma)$ where
$y^n(x,y,z)$ and $z^n(x,y,z)$ are the terms
defined before the statement of this theorem.
Let $\alpha^n := \alpha\vee \beta^n$ if
$n$ is even, and $\alpha^n := \alpha\vee \gamma^n$
if $n$ is odd.
Since $\beta^0 = \beta\geq \beta\wedge(\alpha\vee\gamma^0) = 
\beta^1$, and $\gamma^0\geq \gamma^1$, it is easy to see 
inductively that
\[\beta^{n+1} = \beta\wedge (\alpha\vee\gamma^{n})\geq
\beta\wedge (\alpha\vee\gamma^{n+1}) = \beta^{n+2},\]
and $\gamma^{n+1}\geq \gamma^{n+2}$.
Thus the $\beta$ and $\gamma$-sequences
are descending chains, which forces the
$\alpha$-sequence to be a descending chain.
We claim that 
\[\{\alpha^n\;|\;\mbox{\rm all $n$}\}
\cup\{\beta^n\;|\;\mbox{\rm even $n$}\}\cup
\{\gamma^n\;|\;\mbox{\rm odd $n$}\}\]
is a herringbone.
To see this, we must verify 
conditions $(\beta)$ and $(\gamma)$ from the 
definition of a herringbone:
\begin{enumerate}
\item[$(\beta)$] $\alpha^{2i+1}\wedge \beta^{2i} = \beta^{2i+2}$, 
$\alpha^{2i+2}\vee \beta^{2i} = \alpha^{2i}$
\item[$(\gamma)$] $\alpha^{2i+2}\wedge \gamma^{2i+1} = \gamma^{2i+3}$, 
$\alpha^{2i+3}\vee \gamma^{2i+1} = \alpha^{2i+1}$.
\end{enumerate}
To show that $\alpha^{2i+1}\wedge \beta^{2i} = \beta^{2i+2}$,
observe that
\[\beta^{2i+2} = \beta\wedge (\alpha\vee\gamma^{2i+1})
 = \beta\wedge \alpha^{2i+1}\leq \alpha^{2i+1},\]
and (as observed earlier) $\beta^{2i+2}\leq \beta^{2i}$.
Thus $\beta^{2i+2}\leq \alpha^{2i+1}\wedge \beta^{2i}$.
Conversely, 
\[\alpha^{2i+1}\wedge \beta^{2i} =
(\alpha\vee\gamma^{2i+1})\wedge\beta^{2i} \leq
(\alpha\vee\gamma^{2i+1})\wedge\beta
= \beta^{2i+2}.\]
To show that $\alpha^{2i+2}\vee \beta^{2i} = \alpha^{2i}$,
observe that 
\[\alpha^{2i+2}\vee \beta^{2i} \geq \alpha\vee\beta^{2i}
= \alpha^{2i},\]
while $\alpha^{2i+2}\leq \alpha^{2i}$ and 
$\beta^{2i}\leq \alpha\vee\beta^{2i} = \alpha^{2i}$,
so $\alpha^{2i+2}\vee\beta^{2i}\leq \alpha^{2i}$.
This establishes $(\beta)$, and $(\gamma)$ 
can be established the same way.

It follows that if $N$ bounds the length of any 
herringbone in any congruence lattice of
a member of $\vr V$, then however we choose
our original three congruences $\alpha, \beta$
and $\gamma$, the sequences defined above
must satisfy $\alpha^N = \alpha^{N+1} = \alpha^{N+2} = \cdots$,
and therefore $\gamma^M = \gamma^{M+2}$ holds
for each $M\geq N$. Since the $\gamma$'s form a descending
chain, we conclude that $\gamma^M = \gamma^{M+1}$ holds
for each $M\geq N$. Therefore $z^M\approx z^{M+1}$
holds in all congruence lattices of members of $\vr V$.

Item (1) follows from (3) and the fact
that the identity $z^M\approx z^{M+1}$
is nontrivial. To see that it fails in some lattice,
take any lattice containing a herringbone
of length at least $M+1$. By setting
$\alpha:=\alpha^{M+1},
\beta:=\beta^0$ and $\gamma:=\gamma^1$, it is
easy to see that 
$z^M(\alpha,\beta,\gamma)\neq z^{M+1}(\alpha,\beta,\gamma)$.
\end{proof}

This argument shows that 
if $\vr V$ has a Hobby--McKenzie term
of arity $n$, then $\vr V$ satisfies
the congruence identity $z^{2n-3}\approx z^{2n-2}$.
These exponents are not optimal, since if 
$n=3$ then our proof shows that 
${\vr V}$ satisfies the congruence
identity $z^3\approx z^4$.
But it can be argued (by considering
all Hobby--Mckenzie terms of arity $3$) 
that when $n=3$
the variety $\vr V$ is congruence modular,
and consequently 
satisfies $z^1\approx z^2$
as a congruence identity.

Recall from Section~\ref{lattice_theory}
that the $\sdj$-configuration is the pair
$$(\m p(\sdj),p\approx j)$$ 
where $\m p(\sdj)$
is the partial lattice presented by $\lb G \mid R\rb$
where $G = \{p,q,r,j,m\}$ and $R$ consists
of the relations $p\vee q = j = p\vee r$ and 
$q\wedge r=m\leq p$.

\begin{lm}\label{sdj_projective}
%The $\sdj$-configuration 
$\m p(\sdj)$ is projective relative
to any variety of lattices satisfying 
an identity
of the form $z^M\approx z^{M+1}$.

In fact, if $\sigma\colon \m l\to \m k$
is a surjective
homomorphism in this variety
and $\varphi\colon \m p(\sdj)\to \m k$
is a partial lattice 
homomorphism,
%from the $\sdj$-configuration to $\m k$, 
then for any function
$f\colon \{p,q,r,s\}\to \m l$ satisfying 
$\sigma\circ f = \varphi$ on the domain of $f$
there is a homomorphism of partial lattices
$\overline{\varphi}\colon \m p(\sdj)\to \m l$
such that $\sigma\circ\overline{\varphi} =\varphi$
and $\overline{\varphi}(p)\geq f(p)$, 
$\overline{\varphi}(q)\leq f(q)$, and
$\overline{\varphi}(r)\leq f(r)$.
\end{lm}

\begin{proof}
Define $\m p:=\m p(\sdj)$, and 
suppose that $\varphi\colon \m p\to \m k$
is a homomorphism of partial lattices
and that 
$f\colon \{p,q,r,s\}\to \m l$ is a function 
satisfying $\sigma\circ f = \varphi$ on the
domain of $f$. Let 
$a = f(p), b = f(q)$ and $c = f(r)$.
Recursively define $b^0 := b,
c^0 := c, 
b^{n+1} := b\wedge (a\vee c^{n})$,
and 
$c^{n+1} := c\wedge (a\vee b^{n})$,
so that $b^n = y^n(a,b,c)$
and $c^n := z^n(a,b,c)$ where
$y^n(x,y,z)$ and $z^n(x,y,z)$ are the terms
defined for the proof of Theorem~\ref{main2}.
As we showed in the proof of Theorem~\ref{main2} (1)$\mimplies$(2),
$b=b^0\geq b^1\geq\cdots$
and 
$c =c^0\geq c^1\geq\cdots$.
From the identity 
$z^M\approx z^{M+1}$ we get that these chains terminate, 
in which case
$b^{M}= b^{M+1}= b\wedge(a\vee c^M)\leq
a\vee c^M$ and similarly
$c^M\leq a\vee b^M$.
Thus $b^M\leq b$, $c^M\leq c$ and 
$a\vee b^M = a\vee c^M$.
Define $\overline{\varphi}\colon \m p\to \m l$ by 
$\overline{\varphi}(p) = a\vee (b^M\wedge c^M)$
($\geq f(p)$), 
$\overline{\varphi}(q) = b^M$ ($\leq f(q)$), 
$\overline{\varphi}(r) = c^M$ ($\leq f(r)$), 
$\overline{\varphi}(j) = a\vee b^M$, and 
$\overline{\varphi}(m) = b^M\wedge c^M$.
It is clear that 
$$
\overline{\varphi}(j) = 
a\vee b^M = 
\overline{\varphi}(p)\vee \overline{\varphi}(q) = 
\overline{\varphi}(p)\vee \overline{\varphi}(r)
$$
and that 
$$
\overline{\varphi}(m) = b^M\wedge c^M = 
\overline{\varphi}(q)\wedge \overline{\varphi}(r)
\leq \overline{\varphi}(p).
$$
Therefore $\overline{\varphi}\colon \m p\to \m l$
is a homomorphism of partial lattices. To finish
the proof we must show that $\sigma\circ\overline{\varphi} = \varphi$.
It suffices to prove that $\sigma\circ \overline{\varphi}(x)
=\sigma\circ f(x)$ ($ = \varphi(x)$)
when $x\in \{p, q, r\}$ is
one of the generators of $\m p$.

By simultaneous 
induction, $\sigma(b^n) = \sigma(b)$
and $\sigma(c^n) = \sigma(c)$ for all $n$, 
where one of the induction steps is proved by
$$
\begin{array}{rl}
\sigma(b^{n+1}) 
&= \sigma(b)\wedge 
(\sigma(a)\vee\sigma(c^n))\\
&= 
\sigma(b)\wedge (\sigma(a)\vee \sigma(c))\\
&=
\sigma(b)\wedge (\sigma(a)\vee \sigma(b)) = \sigma(b).
\end{array}
$$
Thus $\sigma\circ f(q) = \sigma(b) = 
\sigma(b^M) = \sigma\circ \overline{\varphi}(q)$,
and similarly 
$\sigma\circ f(r) = \sigma(c) = 
\sigma(c^M) = \sigma\circ \overline{\varphi}(r)$.
Since $q\wedge r\leq p$ and $\varphi$ is a homomorphism
we have $\varphi(q)\wedge \varphi(r)\leq \varphi(p)$.
This yields the second equality in
$$
\begin{array}{rl}
\sigma\circ f(p) &= \sigma(a) \\
&= \sigma(a)\vee (\sigma(b)\wedge\sigma(c))\\
&= \sigma(a)\vee (\sigma(b^M)\wedge\sigma(c^M))\\
&=\sigma\circ \overline{\varphi}(p).
\end{array}
$$
The lemma is proved.
\end{proof}

Our first use of Lemma~\ref{sdj_projective}
will require the following consequence
of Theorem~\ref{main2}, which 
seems to be new even for locally finite varieties.

\begin{thm} \label{sdj_failure=solv}
Assume that $\vr V$ satisfies a nontrivial
congruence identity.
If $I$ is an $\sdj$-failure in 
$\Conb(\m a)$ for some $\m a\in {\vr V}$,
then $I$ is abelian.
\end{thm}

\begin{proof}
We proved in 
Corollary~\ref{sdj_failure=abel} that if
$\vr V$ has a weak difference term, then
$\XXX$-solvable $\sdj$-failures are abelian.
Since the assumption that $\vr V$ satisfies a 
nontrivial congruence identity implies that
$\vr V$ has a Hobby--McKenzie term,
%(Corollary~\ref{ident_implies_HM}),  *******************
and therefore
a weak difference term 
(Corollary~\ref{justify_weak_cor}),
to prove the present theorem it will suffice to show 
that all $\sdj$-failures are $\XXX$-solvable.
To set up notation for this, assume that 
$\m a\in {\vr V}$ has congruences
$\alpha, \beta$ and $\gamma$ satisfying
$\alpha\vee\beta = \alpha\vee\gamma$
and $\beta\wedge\gamma\leq\alpha$.
Our goal is to prove that 
$\alpha\wrel{\lhdlhd}\alpha\vee\beta$.
Factoring by $\beta\wedge\gamma$ if necessary,
we also assume that $\beta\wedge\gamma = 0$.

Our strategy will be to show that if
it is not true that 
$\alpha\wrel{\lhdlhd}\alpha\vee\beta$,
then $\Conb(\m a)$ contains an infinite
$\leadsto$--sequence. This will suffice
to prove the theorem, since Theorem~\ref{leadsto_sequence}
provides a finite bound on the length of 
$\leadsto$--sequences. 
Therefore, assume that $\alpha$ is 
{\em not} $\XXX$--solvably related to 
$\alpha\vee\beta=\alpha\vee\gamma$.
Define new congruences as follows:
\begin{itemize}
\item $\beta^0 := \beta\wedge\rad(\alpha)$, 
and $\gamma^0 := \gamma\wedge\rad(\alpha)$.
\item $\beta^{k+1}:= \beta\wedge(\gamma^k:\gamma)$ and
$\gamma^{k+1}:= \gamma\wedge(\beta^k:\beta)$.
%\item $\alpha^{k+1} = \alpha\vee\beta^k$ if $k$
%is even, and $\alpha^{k+1} = \alpha\vee\gamma^k$ if $k$
%is odd.
\end{itemize}
Our first goal will be to prove that 
$\beta^k\wrel{\obst}\beta$ and
$\gamma^k\wrel{\obst}\gamma$ for all $k$.
%We start with $k=0$.

\begin{clm} 
$\beta^0\wrel{\obst}\beta$ and
$\gamma^0\wrel{\obst}\gamma$.
\end{clm}

We have assumed that $\alpha$ is not
$\XXX$--solvably related to $\alpha\vee\beta$,
so $\alpha\vee\beta\not\leq \rad(\alpha)$.
Since $\alpha\leq \rad(\alpha)$, we conclude that 
$\beta\not\leq \rad(\alpha)$. Since $\rad(\alpha)$
is a radical congruence, Lemma~\ref{pseudocomplement}~(3)
guarantees that 
$\beta^0:= \beta\wedge \rad(\alpha)\wrel{\obst}\beta$
and similarly $\gamma^0\wrel{\obst}\gamma$.

\medskip

\begin{clm} \label{alpha_leq}
$\alpha\leq (\beta^k:\beta)$ and 
$\alpha\leq (\gamma^k:\gamma)$ for all $k$.
\end{clm}

From $\beta^0 := \beta\wedge\rad(\alpha)$ we derive
$\C C(\rad(\alpha),\beta;\beta^0)$ using
Theorem~\ref{basic_centrality}~(8).
Hence $\alpha\leq \rad(\alpha)\leq (\beta^0:\beta)$.
An identical argument proves that 
$\alpha\leq (\gamma^0:\gamma)$.

Now assume that $\alpha\leq (\beta^k:\beta)$
for some $k$. Since $\gamma^{k+1} =
\gamma\wedge(\beta^k:\beta)$ we compute that 
$$
\gamma^{k+1}\leq \gamma\wedge(\alpha\vee\gamma^{k+1})
\leq \gamma\wedge((\beta^k:\beta)\vee\gamma^{k+1}) =
\gamma\wedge(\beta^k:\beta) = \gamma^{k+1},
$$
so $\gamma^{k+1} = \gamma\wedge(\alpha\vee\gamma^{k+1})$.
By Theorem~\ref{basic_centrality}~(8)
we conclude that $\C c(\alpha,\gamma;\gamma^{k+1})$
holds, or $\alpha\leq (\gamma^{k+1}:\gamma)$.
This proof that 
$\alpha\leq (\beta^k:\beta)$ implies
$\alpha\leq (\gamma^{k+1}:\gamma)$ works for every $k$,
and for $\beta$ and $\gamma$ interchanged,
so the claim follows by induction.
%We proved that 
%$\alpha\leq (\beta^0:\beta)$ and 
%$\alpha\leq (\gamma^0:\gamma)$ in the last paragraph,
%so by induction $\alpha\leq (\beta^k:\beta)$ and 
%$\alpha\leq (\gamma^k:\gamma)$ for all $k$.

\medskip

\begin{clm}\label{obst_clm}
If $\beta^k\wrel{\obst}\beta$, then 
$\gamma^{k+1} \wrel{\obst} \gamma$. Similarly, if 
$\gamma^k\wrel{\obst}\gamma$, then 
$\beta^{k+1} \wrel{\obst} \beta$.
\end{clm}

If $\beta^k\wrel{\obst}\beta$, then
$I[\beta^k,\beta]_{\beta}$ is a separated
solvability obstruction, so according to
Lemma~\ref{pseudocomplement}~(2) the pseudocomplement
$(\beta^k:\beta)$ is a radical congruence.
We proved in Claim~\ref{alpha_leq} that 
$\alpha\leq (\beta^k:\beta)$. 
If it is also the case that
$\gamma\leq (\beta^k:\beta)$, then 
$$\beta\leq \alpha\vee\beta = \alpha\vee\gamma\leq (\beta^k:\beta),$$
forcing $\C C(\beta,\beta;\beta^k)$,
or equivalently $\beta^k\lhd \beta$. 
This is contrary to $\beta^k\wrel{\obst}\beta$,
so it must be that $\gamma\not\leq (\beta^k:\beta)$.
Now Lemma~\ref{pseudocomplement}~(3) guarantees that
$$
\gamma^{k+1}:= \gamma\wedge (\beta^k:\beta) \wrel{\obst}\gamma.
$$
The proof that 
$\gamma^k\wrel{\obst}\gamma$ implies 
$\beta^{k+1} \wrel{\obst} \beta$ is the same.

\medskip

\begin{clm}
$I[\gamma^{k+1},\gamma]_{\gamma}
\leadsto I[\beta^{k},\beta]_{\beta}$
and
$I[\beta^{k+1},\beta]_{\beta}\leadsto
I[\gamma^{k},\gamma]_{\gamma}$ 
for all $k$.
\end{clm}

The second paragraph of Lemma~\ref{new_mu}
describes a method to produce $\leadsto$-related
intervals, which we employ now.
Since $I[\beta^k,\beta]_{\beta}$ is a 
separated solvability obstruction
and $\gamma \not\leq (\beta^k:\beta)$, then
for $\gamma^{k+1}:=\gamma\wedge (\beta^k:\beta)$
we have that 
$I[\gamma^{k+1},\gamma]_{\gamma}$ is also a separated
solvability obstruction. 
Moreover, as is shown in Lemma~\ref{new_mu},
to establish that $I[\gamma^{k+1},\gamma]_{\gamma}
\leadsto I[\beta^{k},\beta]_{\beta}$ it is sufficient
to verify that $\gamma\cap \Cg a(\beta)\subseteq \gamma^{k+1}$.
But $\gamma\cap \Cg a(\beta) = \gamma\cap \beta = 0$,
so this certainly holds.

\medskip

Under the assumption that $\alpha$ is not 
$\XXX$--solvably related
to $\alpha\vee\beta$ we have produced an infinite
$\leadsto$--sequence
$$
\cdots\cdots
\leadsto I[\beta^{4},\beta]_{\beta}
\leadsto I[\gamma^{3},\gamma]_{\gamma}
\leadsto I[\beta^{2},\beta]_{\beta}
\leadsto I[\gamma^{1},\gamma]_{\gamma}
\leadsto I[\beta^{0},\beta]_{\beta}.
$$
This completes the proof of the theorem,
as we explained in the second paragraph of this proof.
\end{proof}

Let $\vr V$ be a variety satisfying a nontrivial
congruence identity.
Theorem~\ref{sdj_failure=solv} shows that if $\m a\in {\vr V}$,
then the $\sdj$-failures
in $\Conb(\m a)$ are abelian. 
Since $\vr V$ has a weak difference term, these 
abelian intervals consist of
permuting equivalence relations,
hence these intervals are modular.
It turns out that this property
characterizes the class of varieties
satisfying nontrivial congruence identities.

\begin{thm} \label{SD-join/Mod}
The following are equivalent
for a variety $\vr V$.
\begin{enumerate}
\item[(1)] ${\vr V}$ satisfies a nontrivial congruence identity.
\item[(2)] For every $\m a\in {\vr V}$, $\Conb(\m a)$
has an $\sdj/\Mod$ factorization 
in the category of meet continuous lattices.
\item[(3)] For every $\m a\in {\vr V}$, $\Conb(\m a)$
has an $\sdj/\Mod$ factorization 
in the category of ordinary lattices.
\item[(4)] For every $\m a\in {\vr V}$,
each $\sdj$-failure in $\Conb(\m a)$
is modular.
\item[(5)] If $\m l$ belongs to $\CON({\vr V})$, 
%congruence variety of $\vr V$, 
then each $\sdj$-failure in $\m l$ is modular.
\end{enumerate}
\end{thm}

\begin{proof}
If (1) holds, then from 
Theorem~\ref{main1} (1)$\mimplies$(2), 
Theorem~\ref{rect_Malt_cond2} (2)$\mimplies$(3), 
and Corollary~\ref{justify_weak_cor}
it can be deduced that $\vr V$ has a weak difference term.
Therefore the natural homomorphism 
$$
\nu\colon \Conb(\m a)\to \Conb(\m a)/\solv
$$
provides a factorization of $\Conb(\m a)$
of the form $\sdm/\Mod$ in the category
of meet continuous lattices. We need to
argue that the quotient
$\Conb(\m a)/\solv$ is not only 
meet semidistributive, but is also join semidistributive.

%By the remarks preceding Lemma~\ref{sdj_projective},
$\sdj$-failures in $\Conb(\m a)/\solv$ 
are each of the form $I = I[\varphi(p),\varphi(j)]$
for some homomorphism of partial lattices 
$\varphi\colon \m p\to \Conb(\m a)/\solv$ 
from the partial lattice $\m p := \m p(\sdj)$ of the
$\sdj$-configuration. By the projectivity of $\m p$,
%the $\sdj$-configuration, 
proved
in Lemma~\ref{sdj_projective},
to each such homomorphism there is 
a homomorphism $\overline{\varphi}\colon \m p\to \Conb(\m a)$
such that $\nu\circ \overline{\varphi}=\varphi$
where $\nu\colon \Conb(\m a)\to \Conb(\m a)/\solv$
is the natural map.
The homomorphism $\overline{\varphi}$ 
identifies an $\sdj$-failure 
$I[\overline{\varphi}(p), \overline{\varphi}(j)]$
in $\Conb(\m a)$, which by Theorem~\ref{sdj_failure=solv}
must be abelian. But then
$\overline{\varphi}(p)\solv \overline{\varphi}(j)$,
so since $\solv$ is contained in $\ker(\nu)$ we derive that
$$
\varphi(p) = \nu\circ \overline{\varphi}(p) = 
\nu\circ \overline{\varphi}(j) = \varphi(j).
$$
This shows that our original $\sdj$-failure
$I[\varphi(p),\varphi(j)]$ in $\Conb(\m a)/\solv$ 
is trivial. Since it was chosen arbitrarily, 
$\Conb(\m a)/\solv$ is join semidistributive.

As noted when proving the dual result
(Theorem~\ref{HMa}) the implications 
(2)$\mimplies$(3)$\mimplies$(4) are straightforward.

We prove the contrapositive of (4)$\mimplies$(5).
If there is a $\m k\in \CON({\vr V})$
that has a nonmodular $\sdj$-failure,
then since $\CON({\vr V})$ consists of the homomorphic
images of lattices in the congruence
prevariety of $\vr V$ there is an algebra
$\m a\in {\vr V}$, a sublattice
$\m l\leq \Conb(\m a)$, and surjective
homomorphism $\sigma\colon \m l\to \m k$.
The nonmodular $\sdj$-failure $I$
in $\m k$ can be lifted to an $\sdj$-failure
$\overline{I}$ in $\m l$ using the projectivity of the
$\sdj$-configuration. The fact that 
%the $\sdj$-failure 
$I$ 
%in $\m k$ 
is nonmodular
means that it contains a sublattice isomorphic
to the pentagon, $\m n_5$. The pentagon 
is a projective lattice, so it can be lifted 
to a pentagon in $\overline{I}$, establishing that 
$\overline{I}$ is a nonmodular $\sdj$-failure
in $\m l$, hence the interval in $\Conb(\m a)$
generated by $\overline{I}$ is also 
a nonmodular $\sdj$-failure.
This proves that (4)$\mimplies$(5).

The implication (5)$\mimplies$(4) is trivial, 
so we complete the proof by showing that 
(4)$\mimplies$(1).
Assume that (4) holds.
This fact can be expressed by a quasi-identity
valid in congruence lattices of $\vr V$.
We have already discussed how to write
down a quasi-identity for the dual property
that $\sdm$-failures are modular in Remark~\ref{sdm_rem},
so it is easy to write down the one we need
now. Namely, set $t:=p\vee(q\wedge r)$,
and for each $i$ set $x_i^*:=(x_i\wedge s)\vee t$.
A quasi-identity expressing the fact
that $\sdj$-failures are modular is 
\begin{equation}\label{sdm_mod_q}
((p\vee q) \approx s) \wrel{\&} ((p\vee r)\approx s)\to
u(x_1^*,x_2^*,x_3^*)\approx v(x_1^*,x_2^*,x_3^*)
\end{equation}
where $u\approx v$ is the modular law.
As explained in the paragraph
after Definition~\ref{quasi_sat_W},
the fact that the premises of (\ref{sdm_mod_q})
are meet-free implies that 
(\ref{sdm_mod_q}) satisfies $\W$.
By Theorem~\ref{czedli-day-W}
this means that the class of varieties
for which item (4) is true is definable
by a set of idempotent Maltsev conditions.
At least one of these must fail in the variety
of semilattices, since, as we now prove,
the variety of semilattices does not satisfy item (4).
Let $\m 2 = \lb \{0,1\}; +\rb$ be the $2$-element join
semilattice, and let $\m b = \m 2^2$.
We claim that $\Conb(\m b^2)$ 
%( = $\Conb(\m 2^4)$)
has a nonmodular
$\sdj$-failure. To see this, let $\theta$ be the
kernel of the homomorphism $+\colon \m b^2\to \m b$.
That is, $\theta  = \{\lb (a,b), (c,d)\rb\;|\;a+b=c+d\}$.
Let $\eta_1 = \{\lb (a,b), (c,d)\rb\;|\;a=c\}$
and $\eta_2 = \{\lb (a,b), (c,d)\rb\;|\;b=d\}$
be the coordinate projection kernels.
For any two pairs $(p,q), (r,s)\in B^2$
it is the case that 
$$
(p,q)\;\eta_1\;
(p,p+q+r+s) \;\theta\;
(p+q+r+s,p+q+r+s), 
$$
and similarly $(p+q+r+s,p+q+r+s)\;\theta\circ\eta_1\;(r,s)$,
so $(p,q)$ is related to $(r,s)$ by $\theta\vee \eta_1$.
The pairs were arbitrary, so $\theta\vee\eta_1 = 1$;
similarly $\theta\vee\eta_2 = 1$. 
This proves that $I[\theta\vee(\eta_1\wedge\eta_2),1] =
I[\theta,1]$ is an $\sdj$-failure. But $\m b^2/\theta$
is isomorphic to $\m b$, since $+$ is a surjective homomorphism,
so $I[\theta,1]\cong \Conb(\m b)=\Conb(\m 2^2)\cong \m d_2$,
which is a nonmodular lattice.
This proves that any set of idempotent Maltsev conditions
defining the class of varieties satisfying item (4)
includes one that fails in the variety of semilattices, 
hence from Theorem~\ref{main1} any variety satisfying (4)
must satisfy (1).
\end{proof}

The next result is an analogue of 
%should be compared with 
the $\m d_1$-version of Theorem~\ref{omit_d1_sub}.

\begin{thm} \label{omit_d2_sub}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ satisfies an idempotent Maltsev
condition that fails in the variety of semilattices.
\item[(2)] $\vr V$ satisfies a nontrivial congruence identity.
\item[(3)] $\m d_2$ does not appear as
a sublattice of $\Conb(\m a)$ for any 
$\m a\in {\vr V}$.
\item[(4)] $\m d_2\not\in \CON(\vr V)$.
\end{enumerate}
\end{thm}

\begin{proof}
We proved the equivalence of (1) and (2)
in Theorem~\ref{main1}.

If item (3) holds,
%$\m d_2$ does not appear as
%a sublattice of $\Conb(\m a)$ for any 
%$\m a\in {\vr V}$, 
then $\vr V$ omits
special $\m d_2$'s. According to 
Theorems~\ref{equiv},
this implies that $\vr V$ satisfies an idempotent Maltsev
condition that fails in the variety of semilattices, 
so (3)$\mimplies$(1).

Now suppose that item (2) holds and that item (3) fails
to hold. The failure of (3) implies 
that some $\m a\in {\vr V}$ had a copy of $\m d_2$
in its congruence lattice.
Choose a copy and label
it as in Figure~\ref{fig38}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(0,10){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(15,0){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(15,30){\circle*{1.2}}
\put(30,10){\circle*{1.2}}
\put(30,20){\circle*{1.2}}

\put(0,10){\line(3,-2){15}}
\put(0,10){\line(3,2){15}}
\put(0,10){\line(0,1){10}}
\put(15,30){\line(-3,-2){15}}
\put(15,30){\line(0,-1){10}}
\put(15,30){\line(3,-2){15}}
\put(30,10){\line(-3,-2){15}}
\put(30,10){\line(-3,2){15}}
\put(30,10){\line(0,1){10}}

\put(14,15){$\alpha$}
%\put(14,32){$\tau$}
%\put(-4.5,9){$\mu$}
\put(-4.5,19){$\beta$}
%\put(32,9){$\delta$}
\put(32,19){$\gamma$}
\end{picture}
\end{center}
\caption{ \ }\label{fig38}
\end{figure}
Then $\alpha\vee\beta = \alpha\vee\gamma$,
and $\alpha\vee(\beta\wedge\gamma) = \alpha$.
Therefore, since item (2) holds, 
Theorem~\ref{sdj_failure=solv} guarantees that
$\alpha\wrel{\lhd}\alpha\vee\beta$.
But this is forbidden by Theorem~\ref{taylor_forbidden}~(2).
Thus (1), (2) and (3) are equivalent.

Both implications (4)$\mimplies$(2) and
(4)$\mimplies$(3) are trivial, so to complete
the proof we will show that items (2)
and (3) jointly imply (4).
Assume that items (2) and (3) are true for $\vr V$.
%but that $\m d_2\in \CON({\vr V})$.
From (2) and Theorem~\ref{main2} we obtain that 
there is some positive integer $M$ such that 
the identity $z^M\approx z^{M+1}$ holds in
$\CON({\vr V})$. 
Let ${\vr Z}^M$ denote the variety of lattices axiomatized 
by $z^M\approx z^{M+1}$. 
It is proved in Theorem~2.4 (5)$\mimplies$(6) 
of \cite{sdj} that $\m d_2$ is projective
in any ${\vr Z}^M$ that contains $\m d_2$.
Since $\m d_2$ is projective and subdirectly
irreducible in ${\vr Z}^M$, 
the subclass ${\vr U}\subseteq {\vr Z}^M$
of lattices that have no sublattice isomorphic to $\m d_2$
is a variety. By item (3), $\vr U$ contains the congruence
lattices of members of $\vr V$,
hence ${\vr U}$ contains the variety generated by
these lattices. Since $\CON(\vr V)\subseteq {\vr U}$
and $\m d_2\notin {\vr U}$, (4) holds.
\end{proof}

The statement of the next theorem,
which characterizes
the class of varieties satisfying congruence identities
in yet another way,
should be compared to the statement 
of Theorem~\ref{typ1condcor},
which characterizes 
the class of varieties satisfying 
meet continuous congruence identities.

\begin{thm}\label{typ1condcor_dual}
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 semilattices.
\item[(2)] The quasi-identity 
\begin{equation}\label{weak_W_dual}
((p\vee q)\approx s) 
\wrel{\&}
((p\vee r)\approx s) 
\wrel{\&}
(p^{[2]}\approx s) 
\to
((p\vee (q\wedge r))\approx s)\,.
\end{equation}
holds in the congruence lattices of
algebras in $\vr v$.
\item[(3)] Quasi-identity (\ref{weak_W_dual})
holds in $\CON({\vr V})$.
\item[(4)] $\vr V$ satisfies a 
nontrivial congruence identity.
\end{enumerate}
\end{thm}

Quasi-identity (\ref{weak_W_dual}) in item (2) is the
dual of quasi-identity (\ref{weak_W}) of Theorem~\ref{typ1condcor}~(2).

\begin{proof}
We proved the equivalence of (1) and (4)
in Theorem~\ref{main1}. 
The implication (3)$\mimplies$(2) is obvious.
We have previously noted that $\m d_1$
fails quasi-identity (\ref{weak_W}) of Theorem~\ref{typ1condcor}~(2),
so of course $\m d_2$ fails the dual quasi-identity of
the present theorem. So, if the congruence lattices
of algebras in $\vr V$ satisfy quasi-identity (\ref{weak_W_dual}), 
then $\m d_2$ is not embeddable in the congruence
lattice of any algebra in $\vr V$.
From Theorem~\ref{omit_d2_sub} 
%(3)$\mimplies$(2)
it follows that $\vr V$ satisfies a nontrivial 
congruence identity. This shows that (2)$\mimplies$(4).
In summary, these observations
show that (3)$\mimplies$(2)$\mimplies$(4)$\miff$(1).
The rest of the proof will be devoted to
showing that (4)$\mimplies$(3).
We argue by contradiction, so
assume that $\vr V$ satisfies a nontrivial
congruence identity, but $\CON({\vr V})$
fails to satisfy quasi-identity 
(\ref{weak_W_dual}). 

Quasi-identity (\ref{weak_W_dual}) is equivalent
modulo the axioms of lattice theory to the following quasi-identity:
\begin{equation}\label{weak_W_dual_equiv}
((p\vee q)\approx s) 
\&
((p\vee r)\approx s) 
\&
(p^{[2]}\approx s) 
\&
((q\wedge r)\leq p)
\to
(p\approx s)\,.
\end{equation}
This is because any assignment of variables 
in some lattice that fails (\ref{weak_W_dual_equiv})
also fails (\ref{weak_W_dual}), while, if 
$p\mapsto a, q\mapsto b, r\mapsto c, s\mapsto d$
is an assignment that fails 
(\ref{weak_W_dual}), then the reassignment
$p\mapsto a':=a\vee(b\wedge c), q\mapsto b, r\mapsto c, s\mapsto d$
clearly satisfies the first, second and fourth premise
of (\ref{weak_W_dual_equiv}) and fails the conclusion.
To see that this reassignment 
also satisfies the third
premise of (\ref{weak_W_dual_equiv}) 
we must verify that $(a')^{[2]}(a',b,c) = d$.
The fact that the first two premises are satisfied 
implies that 
%$a'\vee b = a'\vee c = d$, so 
$d$ is the largest element 
of the sublattice generated by $\{a, b, c\}$.
This shows that $(a')^{[2]}(a',b,c)\leq d$. But since the lattice
operations are monotone, and $a\leq a'$, 
we get $d = a^{[2]}(a,b,c)\leq (a')^{[2]}(a',b,c)$.
Thus $(a')^{[2]}(a',b,c) = d$, and the third premise is satisfied.

Let $Q$ denote quasi-identity (\ref{weak_W_dual_equiv}),
and let $(\m p(Q),p\approx s)$ be the $Q$-configuration.
Since we have assumed that some $\m k\in \CON({\vr V})$
fails $Q$, there is a lattice $\m k\in \CON({\vr V})$
and a homomorphism of partial lattices 
$\varphi\colon \m p(Q)\to \m k$ such that 
$\varphi(p) < \varphi(s)$. 
For concreteness, let $a:=\varphi(p),
b:=\varphi(q), c:=\varphi(r)$ and $d:=\varphi(s)$.
Since $\m k\in \CON({\vr V})$
there is an algebra $\m a\in {\vr V}$, a sublattice
$\m l\leq \Conb(\m a)$ and a surjective homomorphism
$\sigma\colon \m l\to\m k$. Choose a function
$f\colon \{p,q,r,s\}\to \m l$ such that 
$\sigma\circ f = \varphi$ on the domain of $f$.
Since $\varphi$ is a homomorphism of partial lattices
and $q\wedge r\leq p, q, r, s$ in $\m p(Q)$,
the element 
%$\varphi(s) = d$ is the largest element of the
%sublattice generated by $\{\varphi(p),\varphi(q),\varphi(r)\}
%=\{a, b, c\}$ and 
$\varphi(q)\wedge \varphi(r)=b\wedge c$
is the smallest element of the sublattice generated
by $\{a, b, c, d\}$. Thus, we may assume that 
$f$ is chosen so that 
$f(q)\wedge f(r)\leq f(p), f(q), f(r), f(s)$.

The function $f$ is an assignment of the variables 
of $Q$ in $\m l$. We wish to modify this assignment
to a better one. 
%Although we do not claim that $\m p(Q)$ is projective
%relative to $\CON({\vr V})$, we may nevertheless
%apply Lemma~\ref{sdj_projective} to obtain 
%the desired modification of $f$.
%That is, 
Since $\m p(Q)$ may be obtained 
(up to isomorphism) from $\m p(\sdj)$
by adding more relations, there is a
(surjective) homomorphism
$\psi\colon \m p(\sdj)\to \m p(Q)$ that is the
identity on the generators. Applying 
%the second statement of 
Lemma~\ref{sdj_projective} to
the assignment $f\circ \psi\colon \{p,q,r,s\}\to \m l$
we obtain a homomorphism of partial lattices
$\overline{\psi}\colon \m p(\sdj)\to \m l$
such that $\sigma\circ \overline{\psi} = \varphi\circ \psi$
(Figure~\ref{fig39}).
\begin{figure}[ht]
$$
\xymatrix{
&\ar@{-->}@/_/[ddl]_*+{\overline{\psi}}\ar[d]^*+{\psi} {\m p}(\sdj)\\
& {\m p}(Q)\ar[d]^*+{\varphi}\ar[ld]|*+{f}\\
  {\m L} \ar@{->>}[r]_*+{\sigma}   & {\m K}
}
$$
\caption{ \ }\label{fig39}
\end{figure}
The second clause of 
Lemma~\ref{sdj_projective} guarantees that 
\begin{equation}\label{weird_stuff}
\begin{array}{rl}
\overline{\psi}(q)\wedge 
\overline{\psi}(r)&\leq 
(f\circ \psi(q))\wedge (f\circ \psi(r))\\
&= f(q)\wedge f(r)\leq f(p)\\
&= f\circ \psi(p)\\
&\leq \overline{\psi}(p),
\end{array}
\end{equation}
so if $\overline{\psi}$ is taken to
define an assignment of the variables
of $Q$ in $\m l$,
$p\mapsto \alpha, q\mapsto \beta, r\mapsto \gamma, s\mapsto \delta$,
then 
\begin{enumerate}
\item[(i)] $\alpha\vee \beta = \delta$ and $\alpha\vee \gamma = \delta$, 
(by the projectivity of $\m p(\sdj)$)
\item[(ii)] $\beta\wedge \gamma\leq \alpha$ (by (\ref{weird_stuff})), and
\item[(iii)] $\sigma(\alpha) = a < d = \sigma(\delta)$
(by the projectivity of $\m p(\sdj)$).
\end{enumerate}
This assignment is not necessarily a failure of $Q$
in $\m l$,
since the third premise
$\alpha^{[2]}=\delta$ of $Q$ may not be satisfied. But since
the third premise was satisfied by $\varphi\colon \m p(Q)\to \m k$
we at least have 
\begin{enumerate}
\item[(iv)] $\sigma(\alpha^{[2]}) = a^{[2]}=d=\sigma(\delta)$.
\end{enumerate}

The rest of the argument 
produces a contradiction from properties (i)---(iv).
From items (i) and (ii) 
we derive that $I[\alpha,\alpha\vee\beta]$
is an $\sdj$-failure, hence is an abelian interval
according to Theorem~\ref{sdj_failure=solv}.
By this and item (i) we get 
$\alpha\vee \beta\solv \alpha\solv \alpha\vee\gamma$,
which is hypothesis~(1)
of Lemma~\ref{sdj_failure_cent_lm}.
Item (ii) above is the same as hypothesis (2)
of Lemma~\ref{sdj_failure_cent_lm}, and
our current assumption that $\vr V$ satisfies a congruence
identity is strong enough to guarantee that $\vr V$
has a weak difference term. Since all the
hypotheses of Lemma~\ref{sdj_failure_cent_lm}
currently hold, its conclusion also holds, which is the equality
$$
(\beta\vee(\alpha\wedge \gamma))\wedge(\gamma\vee(\alpha\wedge \beta))=
(\alpha\wedge\beta)\vee(\alpha\wedge\gamma).
$$
Using this to move from the first line
to the second in
$$
\begin{array}{rl}
\alpha^{[2]}(\alpha,\beta,\gamma) &:= \alpha\vee ((\beta\vee(\alpha\wedge \gamma))\wedge(\gamma\vee(\alpha\wedge \beta)))\\
&= \alpha\vee ((\alpha\wedge \beta)\vee(\alpha\wedge \gamma))\\
&= \alpha,
\end{array}
$$
we derive that $\alpha^{[2]} = \alpha$.
But from item (iii) we have $\sigma(\alpha) < \sigma(\delta)$,
while from item (iv) we have $\sigma(\alpha^{[2]}) = \sigma(\delta)$,
which together conflict with $\alpha^{[2]} = \alpha$.
This contradiction completes the proof.
\end{proof}

The following result defines the class
of varieties satisfying 
a nontrivial congruence identity by
a 3-variable Maltsev condition.

\begin{thm} \label{cong_id2}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ satisfies a nontrivial congruence identity.
\item[(2)] ${\vr V}\models_{con} \alpha\cap(\beta\circ\gamma)
\subseteq 
(\gamma\vee(\alpha\wedge\beta))\circ(\beta\vee(\alpha\wedge\gamma))$.
\item[(3)] 
There is a positive integer $k$ and ternary terms 
$d_0$,\ldots, $d_{2k+1}$, $e_0$,\ldots, $e_{2k+1}$, $p$
such that $\mathcal V$ satisfies the following equations:
\begin{enumerate}
\item[(i)] $d_0(x,y,z)\approx p(x,y,z) \approx e_0(x,y,z)$;
\item[(ii)] $d_{i}(x,y,y)\approx d_{i+1}(x,y,y)$ 
and $e_{i}(x,x,y)\approx e_{i+1}(x,x,y)$ for even $i$;
\item[(iii)] $d_{i}(x,x,y)\approx d_{i+1}(x,x,y)$,
$d_{i}(x,y,x)\approx d_{i+1}(x,y,x)$,\hfil\hphantom{a}\linebreak
$e_{i}(x,y,y)\approx e_{i+1}(x,y,y)$ 
and $e_{i}(x,y,x)\approx e_{i+1}(x,y,x)$ for odd~$i$;
\item[(iv)] $d_{2k+1}(x,y,z)\approx x$ and
$e_{2k+1}(x,y,z)\approx z$.
\end{enumerate}
\end{enumerate}
\end{thm}

\begin{proof}
For locally finite varieties, this result is 
part of Theorem~9.8 of \cite{hobby-mckenzie}.
The proof given there shows that items (2) and
(3) are equivalent for any variety, 
and that the idempotent Maltsev condition in 
item (3)
fails in the variety of semilattices.
Therefore (3)$\mimplies$(1), according to
Theorem~\ref{main1}.

What remains to show is that (1)$\mimplies$(2).
Let $\m f = \m f_{\vr V}(x,y,z)$ and
set $\alpha := \Cg f(x,z), \beta:=\Cg f(x,y)$, 
and $\gamma:=\Cg f (y,z)$.
%and $\delta := \alpha\vee \beta = 
%\alpha\vee\gamma = \beta\vee\gamma$. 
Since $(x,z)$ is a generic element of 
$\alpha\cap(\beta\circ\gamma)$ it will suffice to prove that 
$(x,z)\in (\gamma\vee(\alpha\wedge\beta))\circ(\beta\vee(\alpha\wedge\gamma))$.
Since $\alpha\vee \beta = 
\alpha\vee\gamma = \beta\vee\gamma =: \delta$, 
%It follows from the definitions of $\alpha, \beta$ and $\gamma$ that 
the intervals $I[\beta\vee(\alpha\wedge\gamma),\delta]$ and
$I[\gamma\vee(\alpha\wedge\beta),\delta]$ 
are $\sdj$-failures, and therefore 
$\beta\vee(\alpha\wedge \gamma)\solv \delta\solv 
\gamma\vee(\alpha\wedge\beta)$ (Theorem~\ref{sdj_failure=solv}).
Since these congruences are solvably related,
they permute (Theorem~\ref{sim_cong}).
Since 
$(x,y)\in \beta\vee(\alpha\wedge \gamma)$
and 
$(y,z)\in \gamma\vee(\alpha\wedge\beta)$,
we get that 
$$
(x,z)\in 
(\beta\vee(\alpha\wedge \gamma))\circ (\gamma\vee(\alpha\wedge\beta))=
(\gamma\vee(\alpha\wedge\beta))\circ(\beta\vee(\alpha\wedge\gamma)).
$$
\end{proof}


\section{Congruence Join Semidistributivity}\label{8.3}

The main purpose of this section is to prove
that a variety is congruence join semidistributive
if and only if it is both 
congruence meet semidistributive 
and satisfies a nontrivial congruence identity.
Thus, congruence join semidistributive
varieties are exactly those varieties whose
members have no congruences that are `abelian' in 
any of the senses that have been introduced.

\begin{thm} \label{sdj-thm}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ is congruence join semidistributive.
\item[(2)] 
If $\m a\in {\vr V}$ and 
$\alpha, \beta, \gamma\in \Con(\m a)$, then
%it is the case that 
\[
\alpha\cap(\beta\circ\gamma)
\subseteq \beta\vee(\alpha\wedge\gamma)\,.
%\;\;( = \beta_1)
\]
%\beta\vee(\alpha\wedge\gamma)$.
\item[(3)] 
There is a positive integer $k$ and ternary terms 
$d_0,\ldots, d_{k}$
such that $\mathcal V$ satisfies the following equations:
\begin{enumerate}
\item[(i)] $d_0(x,y,z)\approx x$;
\item[(ii)] $d_{i}(x,y,y)\approx d_{i+1}(x,y,y)$ 
and $d_{i}(x,y,x)\approx d_{i+1}(x,y,x)$
for even $i<k$;
\item[(iii)] $d_{i}(x,x,y)\approx d_{i+1}(x,x,y)$,
for odd~$i<k$;
\item[(iv)] $d_{k}(x,y,z)\approx z$.
\end{enumerate}
\item[(4)] ${\vr V}$ satisfies an idempotent Maltsev
condition that fails in any nontrivial variety of modules
and in the variety of semilattices.
\item[(5)]
No member of $\vr V$ has a nontrivial abelian or
rectangular congruence.
\item[(6)] ${\vr V}$ is congruence meet semidistributive
and satisfies a nontrivial congruence identity.
\item[(7)] There is a positive integer
$N$ such that if $\m a\in {\vr V}$ and 
$\alpha, \beta, \gamma\in \Con(\m a)$, then 
%it is the case that 
\[
\alpha\vee(\beta\wedge\gamma)\approx
(\alpha\vee\beta^N)\wedge
(\alpha\vee\gamma^N)\,.
\]
\item[(8)] ${\rm CON}({\vr V})$ is a join semidistributive
variety of lattices.
\item[(9)] $\m m_3, \m d_2\not\in {\rm CON}({\vr V})$.
\item[(10)] $\m m_3$ and $\m d_2$ are not embeddable 
in $\Conb(\m a)$ for any $\m a\in {\vr V}$.
\end{enumerate}
\end{thm}

\begin{proof}
We prove (1)--(6) cyclically, and
then complete the proof by showing that
$((1)\wrel{\&}(6))\mimplies(7)
\mimplies(8)\mimplies(9)\mimplies(10)\mimplies(6)$.

Assume that $(1)$ holds and that $\m a\in {\vr V}$
has congruences $\alpha,\beta$ and $\gamma$.
Choose any $(a,c)\in \alpha\cap(\beta\circ\gamma)$.
There exists $b\in A$ such that 
$a\equiv_{\beta} b\equiv_{\gamma} c$.
Let $\alpha' = \Cg a(a,c), \beta' = \Cg a(a,b)$
and $\gamma' = \Cg a(b,c)$. Since the generating
pairs are related by $\alpha, \beta$ and $\gamma$
respectively, we get that $\alpha'\leq \alpha$,
$\beta'\leq \beta$ and $\gamma'\leq \gamma$.
The congruence $\beta'\vee\alpha'$ is generated
by $(a,c)$ and $(a,b)$, so it is the congruence
on $\m a$ generated by $X\times X$ where $X = \{a, b, c\}$.
Similarly $\beta'\vee\gamma'$ is the congruence
generated by $X\times X$, 
so $\beta'\vee\alpha' = \beta'\vee\gamma'$.
By $(1)$, we get that this congruence is also
$\beta'\vee (\alpha'\wedge\gamma')$. Since
this congruence contains $(a,c)$, we get that 
$$
(a,c)\in \beta'\vee(\alpha'\wedge\gamma')\leq 
\beta\vee(\alpha\wedge\gamma),
$$
where the inequality follows from
the monotonicity of the lattice operations.
This argument shows that 
$
(a,c)\in \beta\vee(\alpha\wedge\gamma)
$
for every $(a,c)\in \alpha\cap(\beta\circ\gamma)$,
so $\alpha\cap(\beta\circ\gamma)
\subseteq \beta\vee(\alpha\wedge\gamma)$ and
$(2)$ holds.

For $(2)\mimplies(3)$, let $p = \alpha\cap(\beta\circ\gamma)$,
$r = \beta\vee(\alpha\wedge\gamma)$, and 
let $r_i$ be the $\{\circ,\wedge\}$-word
obtained from $r$ by replacing $\vee$ with
$i$-fold relational product. Then the congruence
inclusion of (2) has the form
$p\subseteq \bigcup_{i\in \omega} r_i$
with the $r_i$'s increasing. By the remark after 
Theorem~\ref{pixley-wille_extension}, such a 
congruence inclusion is associated to 
a Maltsev condition. The method described
in the proof of Theorem~\ref{pixley-wille_extension}
for producing the Maltsev condition yields the one
in item (3).

To prove that $(3)\mimplies (4)$, we will show
that the idempotent Maltsev condition in $(3)$
fails in every nontrivial variety of modules
and in the variety of semilattices.
Assume that $\vr M$ is the variety
of all modules over the ring $\m r$,
and that $\vr M$ satisfies the 
Maltsev condition
in item (3). Since each $d_i$ is a
module term, it must be that 
$d_i(x,y,z) = a_ix+b_iy+c_iz$ for 
some $a_i, b_i, c_i\in R$. Since 
$d_i(x,x,x)\approx x$, we get that 
$a_ix+b_ix+c_ix \approx (a_i+b_i+c_i)x\approx x$
holds in any $\m m\in \vr M$, hence 
$a_i+b_i+c_i=1$ in $\m r$.
The first equation in (3)(ii) implies
that $a_i=a_{i+1}$ for even $i$, the
second equation in (3)(ii) implies that
$b_i=b_{i+1}$ for even $i$, and with the equations
of the type 
$a_i+b_i+c_i=1$ we derive that $c_i=c_{i+1}$
holds for all even $i$. The equation from
(3)(iii) implies that $c_i=c_{i+1}$ holds
for odd $i$, so $c_i=c_{i+1}$ for all $i$.
But the equation from (3)(i) implies that $c_0=0$,
while the equation from (3)(iv) implies that $c_k=1$.
Altogether we derive that $0=1$ in $\m r$,
so $\vr M$ is a trivial variety of modules.

Now suppose that $\vr S$ is the variety of semilattices,
and that $\vr S$ satisfies the Maltsev condition in (3).
Choose $k$ minimal for the condition in (3) to be satisfied
in $\vr S$. By equations (3)(i) and (3)(iv),
it must be that $k>0$. If $k>0$ is even, then by
equation (3)(iii) we find that 
$d_{k-1}(x,x,y)\approx d_k(x,x,y)\approx y$,
which for semilattices implies that $d_{k-1}(x,y,z)\approx z$.
Thus we can delete the term $d_k$ and have
a shorter sequence of terms satisfying 
the conditions in (3). If $k>0$ is odd, then
by the equations in (3)(ii) that 
$d_{k-1}$ does not depend on its
first variable (since 
$d_{k-1}(x,y,y)\approx d_{k}(x,y,y)\approx y$),
and does not depend on its
second variable (since 
$d_{k-1}(x,y,x)\approx d_{k}(x,y,x)\approx x$).
Therefore $d_{k-1}(x,y,z)\approx d_{k-1}(z,z,z)\approx z$,
and we can again shorten the sequence. 
It follows that the equations in (3) cannot be satisfied
by semilattice terms. Thus (3)$\mimplies$(4).

It follows
from Theorem~\ref{SD-meet-thm}
(10)$\mimplies$(4) that if $\vr V$ 
satisfies an idempotent Maltsev condition that 
fails in every nontrivial variety of modules,
then $\vr V$ omits abelian congruences.
It follows
from Theorem~\ref{rect_Malt_cond2}
%(2)$\mimplies$(1) 
that if $\vr V$ 
satisfies an idempotent Maltsev condition that 
fails in the variety of semilattices,
then $\vr V$ omits rectangular tolerances
(hence $\vr V$ omits rectangular congruences
by Theorem~\ref{rect_cong}).
Thus (4)$\mimplies$(5).

If (5) holds, then $\vr V$ has no abelian
congruence, so from (4)$\mimplies$(1)
of Theorem~\ref{SD-meet-thm} we derive
that $\vr V$ is congruence meet semidistributive.
$\vr V$ also has no rectangular congruence, or tolerance,
so by Theorems~\ref{rect_Malt_cond2} and \ref{main1}
we have that $\vr V$ satisfies a nontrivial congruence
identity. Hence (6) holds.

Now suppose that $\vr V$ is congruence meet semidistributive
and satisfies a nontrivial congruence identity.
The latter supposition implies that the
$\sdj$-failures in congruence lattices 
of members of $\vr V$ are abelian, according to
Theorem~\ref{sdj_failure=solv}.
But the former supposition implies that 
there are no abelian intervals
in congruence lattices of members of $\vr V$,
according to (1)$\mimplies$(4) of 
Theorem~\ref{SD-meet-thm}.
This shows that there are no $\sdj$-failures
at all, which means of course that
$\vr V$ is congruence join semidistributive.

Now that the first six items of the theorem
have been shown to be equivalent, we turn
to the seventh. Assume that (1) and
(the equivalent property) (6) hold.
From (6) and 
Theorem~\ref{main2} we know that there is a 
positive integer $M$ such that the identity
\[
z^M(x,y,z)\approx z^{M+1}(x,y,z)
\]
holds in all congruence lattices of members of $\vr V$.
Thus, if $\alpha, \beta$ and $\gamma$ are as
in the statement of item (7), 
and $\beta^k:= y^k(\alpha,\beta,\gamma)$ and
$\gamma^k:= z^k(\alpha,\beta,\gamma)$, then 
$\beta^{M} = \beta^{M+1}$ and 
$\gamma^{M} = \gamma^{M+1}$. 
This implies that 
$\alpha\vee \beta^M \geq \alpha\vee\gamma^{M+1} = \alpha\vee\gamma^M$,
and similarly that $\alpha\vee\gamma^M\geq \alpha\vee \beta^M$,
so $\alpha\vee \beta^M  = \alpha\vee\gamma^M$.
Since (1) holds, we have 
\[
\alpha\vee(\beta^M\wedge \gamma^M) = \alpha\vee \beta^M\,.
\]
But $\beta\wedge\gamma \leq \beta^M\wedge \gamma^M \leq 
\beta^M\leq \beta$
and $\beta\wedge\gamma \leq \beta^M\wedge \gamma^M \leq 
\gamma^M\leq \gamma$, so 
$\beta^M\wedge \gamma^M = \beta\wedge \gamma$.
Therefore 
\[
\alpha\vee(\beta\wedge \gamma) = \alpha\vee \beta^M\,,
\]
and similarly
\[
\alpha\vee(\beta\wedge \gamma) = \alpha\vee \gamma^M\,.
\]
This proves that $\alpha\vee(\beta\wedge \gamma) = 
(\alpha\vee \beta^M) \wedge (\alpha\vee \gamma^M)$
for the fixed value of $M$ provided by 
Theorem~\ref{main2}. 
%This completes the proof
%that (1) and (6) imply (7).

If item (7) holds, then ${\rm CON}({\vr V})$ satisfies the
weakened distributive law
$x\vee(y\wedge z)\approx (x\vee y^N)\wedge 
(x\vee z^N)$ for some $N$. We claim that any lattice
satisfying this law is join semidistributive.
Indeed, if $\m l$ is a lattice, 
$a, b, c\in L$, and 
$b^k:= y^k(a,b,c)$ and 
$c^k:= z^k(a,b,c)$, then it may be
shown by induction that 
if $a\vee b = a\vee c$, then 
$b^k = b$ and $c^k = c$ for all $k$.
Thus, if $d:= a\vee b = a\vee c$, the 
identity in (7) implies that
\[
a\vee (b\wedge c) = (a\vee b^N)\wedge (a\vee c^N) = 
(a\vee b)\wedge (a\vee c) = d\,.
\]
Since $a, b, c\in L$ were chosen arbitrarily,
$\m l$ is join semidistributive.

Item (8) implies item (9) because
$\m m_3$ and $\m d_2$ are not join semidistributive.

Item (9) implies item (10) because
sublattices of $\Conb(\m a)$
belong to ${\rm CON}({\vr V})$.

Assume that item (10) holds. Since $\m m_3$
is not embeddable in the congruence lattice 
of any member of $\vr V$, it follows from 
Theorem~\ref{SD-meet-thm} that
$\vr V$ is congruence meet semidistributive.
Since $\m d_2$ 
is not embeddable in the congruence lattice 
of any member of $\vr V$, it follows from 
Theorem~\ref{omit_d2_sub} that
$\vr V$ satisfies a nontrivial congruence identity.
Thus item (6) holds.
The proof is complete.
\end{proof}

This theorem provides a positive solution to 
Problem~2.18 of \cite{jonsson3}
from this theorem, which we extract as the
following corollary.

\begin{cor}
The class of congruence join semidistributive
varieties is definable by a(n idempotent) Maltsev condition.
\end{cor}

\begin{proof}
By Theorem~\ref{sdj-thm}~(1)$\miff$(3).
\end{proof}

%\begin{cor}
%Assume that $\vr V$ has a Hobby--McKenzie term
%of arity $n$, and that $\m a\in {\vr V}$.
%If $\m l = \Conb(\m a)/\solv$, then 
%$\m l$ is both join and meet semidistributive.
%Moreover, $\m l$ has no $D$--sequences of length 
%$\geq 2n-4$.
%\end{cor}
