\chapter{Ordinary Congruence Identities}\label{rectangulation2}

In this chapter we will prove that 
a variety satisfies 
a nontrivial lattice identity 
as a congruence identity
if and only if it satisfies 
an idempotent 
Maltsev condition that fails in the variety
of semilattices.
In view of the results of 
Chapter~\ref{rectangulation1}, 
this shows that the class of varieties that
satisfy a congruence identity is definable
by a Maltsev condition (for example, the Maltsev 
condition of Theorem~\ref{equiv}).

\section{A Rank for Solvability Obstructions}

The 
%complexity of the type of 
congruence identity that we describe
depends on the number of variables 
in a Hobby--McKenzie term for the variety,
so let us now fix that arity.
We assume in this chapter that
$\vr V$ is a variety with a fixed
Hobby--McKenzie
term $F(x_1,\ldots,x_n)$, and
we let $N = \{1, 2, \ldots, n\}$.
By Corollary~\ref{justify_weak_cor},
$\vr V$ also has a weak difference term, so we
are free to use the results of Chapter~\ref{solvability}
in this chapter.

\begin{df}
If $I[\delta,\theta]$ is a 
%congruence interval
solvability obstruction
and $T$ is a tolerance, then 
$T$ {\bf separates} 
\index{tolerance!separates a solvability obstruction}
\index{separate!a tolerance ---s a solvability obstruction}
%the interval
$I := I[\delta,\theta]$ if 
$T\subseteq \theta$ and $T\not\subseteq \delta$.
We write 
$I[\delta,\theta]_T$ or $I_T$ to refer
to the 
%interval 
obstruction $I$ in a manner
that indicates that $T$ separates $I$.

If $f$ and $g$ are $m$-ary terms, then
{\bf $I_T$ supports the local 
equation
$f(x_1,\ldots,x_m)\approx_{I_T} g(x_1,\ldots,x_m)$}
\index{local equation}
\index{support!an interval ---s a local equation}
if 
$f(x_1,\ldots,x_m)\equiv_{\delta} g(x_1,\ldots,x_m)$
holds whenever all $x_i\in \{a,b\}$
for each $(a,b)\in T$.
\end{df}


\begin{df}
Assume that $\m a\in {\vr V}$, that
$I = I[\delta,\theta]$ is 
%an interval in $\Conb(\m a)$.
%If $I[\delta,\theta]$ is
a solvability
obstruction, and that $T$ is a tolerance
that separates $I$.
The {\bf rank of $I_T$}
\index{rank of a congruence interval} 
is the set
$$
{\rm Rank}(I_T):=
\{U\subseteq N\;|\;F_U(x,y)\approx_{I_T} x\}.
$$
\index{Rank@${\rm Rank}(I_T)$}
\end{df}

\begin{lm} \label{rank}
Assume that $\m a\in {\vr V}$ and that
$I_T:=I[\delta,\theta]_T$ is a separated solvability
obstruction in $\Conb(\m a)$. Then
\begin{enumerate}
\item[(1)] $N\in {\rm Rank}(I_T)$.
\item[(2)] $\emptyset\not\in {\rm Rank}(I_T)$.
\item[(3)] ${\rm Rank}(I_T)$ is a closed
subset of ${\mathcal B}(F)$ that is not a lattice
filter.
\end{enumerate}
\end{lm}

\begin{proof}
For (1), $N\in {\rm Rank}(I_{T})$ if and only
if 
$$
F(x,x,\ldots,x) = F_N(x,y)\approx_{I_T} x.
$$
That this is true follows from the fact 
that $F$ is idempotent.

For (2), if $\emptyset\in {\rm Rank}(I_{T})$,
then 
$$
y = F(y,y,\ldots,y) = F_{\emptyset}(x,y)\approx_{I_{T}}x.
$$
This statement means that $y\equiv_{\delta} x$ for
all $(x,y)\in T$, contradicting
the fact that $T$ separates $I[\delta,\theta]$.
Thus $\emptyset\not\in {\rm Rank}(I_{T})$,

To prove (3), notice that 
if ${\vr V}\models F_U(x,y)\approx F_V(x,y)$,
then $U\in {\rm Rank}(I_T)$ if and only if
$V\in {\rm Rank}(I_T)$. Therefore 
${\rm Rank}(I_T)$ is a closed
subset of ${\mathcal B}(F)$.
It is not a lattice filter since, by parts (2)
and (3), it is proper and nonempty,
and ${\mathcal B}(F)$ has no
closed, proper, nonempty lattice filter.
\end{proof}

A certain configuration involving
a pair of separated solvability obstructions
forces one obstruction to have rank 
that properly contains
the rank of the other obstruction.
We identify this configuration now,
and introduce a shorthand notation for it.

\begin{df} \label{leadsto}
{\bf $I[\mu,\nu]_S\leadsto I[\delta,\theta]_T$}
means that
\index{$\leadsto$}
\begin{enumerate}
\item[(a)] $S\cap(\delta:T)\subseteq \mu$, 
\item[(b)] $T\cap(S\circ(T\cap\delta)\circ(S\cap
\mu))\subseteq\delta$, and
\item[(c)] $S\cap((T\cap\delta)\circ(S\cap
\mu)\circ(T\cap\delta))\subseteq\mu$.
\end{enumerate}
\end{df}
The three conditions of this definition
describe a property
of the tolerance
$$\overline{T} :=T\cap(S\circ(T\cap\delta)\circ S).$$
This tolerance consists of all pairs
$(c,d)$ for which there exist elements
$a$ and $b$ related as in the 
quadrangle in Figure~\ref{fig26}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{.8mm}
\begin{picture}(20,33)
\put(0,6){\circle*{1.2}}
\put(20,6){\circle*{1.2}}
\put(0,26){\circle*{1.2}}
\put(20,26){\circle*{1.2}}
\put(0,6){\line(1,0){20}}
\put(0,6){\line(0,1){20}}
\put(20,26){\line(-1,0){20}}
\put(20,26){\line(0,-1){20}}
\put(-5,26){$a$}
\put(22,26){$b$}
\put(-5,3){$c$}
\put(22,3){$d$}
\put(9,0){$T$}
\put(5,29){$T\cap\delta$}
\put(-5,15){$S$}
\put(22,15){$S$}
\end{picture}
\end{center}
\caption{ \ }\label{fig26}
\end{figure}
Condition (a) implies that $\C C(S,T;\delta)$ fails
(since $S\not\subseteq \mu$). 
An $S,T$--matrix 
$
\left[
\begin{array}{cc}a&b\\c&d
\end{array}
\right]
$
witnessing this failure produces a pair
$(c,d)\in \overline{T}$ with $(c,d)\not\in\delta$.
Thus, if Condition (a) holds, 
the tolerance $\overline{T}\subseteq T$
also separates $I[\delta,\theta]$.
Condition (b) implies
that if $a, b, c$ and $d$ are any four elements
related as in the above quadrangle,
and $(b,d)\in\mu$, then $(c,d)\in \delta$.
%similarly, if $(a,c)\in\mu$, then $(c,d)\in \delta$.
If this happens, and Condition (c) holds,
then $(a,c)\in\mu$. 
In particular, if 
$I[\mu,\nu]_S\leadsto I[\delta,\theta]_T$, 
then there will exist quadrangles
related as above with $(c,d)\not\in\delta$,
%(and in particular $\overline{T}$ will be a tolerance
%that separates $I[\delta,\theta]$), 
but any such quadrangle must have
$(a,c), (b,d)\not\in\mu$.

\begin{lm} \label{lm0}
Assume that $\m a\in {\vr V}$, that
$I_S = I[\mu,\nu]_S$ and $J_T = I[\delta,\theta]_T$ 
are separated solvability obstructions in
$\Conb(\m a)$,
and that $I_S\leadsto J_T$.
If $\overline{S}\subseteq S$
also separates $I[\mu,\nu]$ and 
$\overline{T} = 
T\cap(\overline{S}\circ(T\cap\delta)\circ\overline{S})$, then 
\begin{enumerate}
\item[(1)] $I_{\overline{S}}\leadsto J_T$, and
\item[(2)] $\overline{T}$ separates $I[\delta,\theta]$.
\end{enumerate}
\end{lm}

\begin{proof}
For item (1), suppose that $\overline{S}\subseteq S$
and that $\overline{S}$ separates $I$.
Then the three conditions of the definition
of $\leadsto$ hold with $\overline{S}$ in place of $S$,
because the symbol $S$ appears only on the left
side of each inclusion and the left side of each
inclusion is an expression that is monotone 
in $S$ with respect to inclusion.

Item (2) follows from 
$I_{\overline{S}}\leadsto J_T$ as we explained 
in the remarks preceding the statement of this lemma
(using $S$ instead of $\overline{S}$).
\end{proof}


\begin{lm} \label{lm1}
Assume that $\m a\in {\vr V}$, that
$I_S = I[\mu,\nu]_S$ and $J_T = I[\delta,\theta]_T$ 
are separated solvability obstructions 
in $\Conb(\m a)$,
and that $I_S\leadsto J_T$.
Let $\overline{T} = 
T\cap(S\circ(T\cap\delta)\circ S)$.
If $f(x,y,z)$ is an idempotent
term of $\m a$, and $I_S$ supports
the local equation
$f(x,y,y)\approx_{I_S} x$, then
both of the local equations
$$
f(x,x,y)\approx_{J_{\overline{T}}} x
\hspace{.1in}\mbox{\rm and}\hspace{.1in}
f(x,y,x)\approx_{J_{\overline{T}}} x
$$
are supported by $J_{\overline{T}}$.
\end{lm}

\begin{proof}
The fact that the tolerance
$\overline{T}:= T\cap(S\circ(T\cap\delta)\circ S)$
separates $J$ follows from
Condition (a) defining the relation
$I_S\leadsto J_T$, as we remarked
prior to the statement of Lemma~\ref{lm0}
(or, it follows from Lemma~\ref{lm0}~(2)).
Now, to prove this lemma, it suffices 
to show that if $I_S$
supports the local equation 
$f(x,y,y)\approx_{I_S} x$, 
then $J_{\overline{T}}$
supports the local equation 
$f(x,x,y)\approx_{J_{\overline{T}}} x$.
The same argument applied
to $f'(x,y,z) = f(x,z,y)$ shows that 
if the local equation $f(x,y,y)\approx_{I_S} x$
is supported by $I_S$, then $f(x,y,x)\approx_{J_{\overline{T}}} x$ 
is supported by $J_{\overline{T}}$.

To show that 
$J_{\overline{T}}$
supports the local equation 
$f(x,x,y)\approx_{J_{\overline{T}}} x$, 
choose a pair
$(0,1)\in \overline{T} = T\cap(S\circ(T\cap\delta)\circ S)$. 
There exist elements $u$ and $v$ such that 
$0\equiv_S u\equiv_{T\cap \delta} v\equiv_S 1$.
The relationships between $0, 1, u$ and $v$
are described in the left hand square
in Figure~\ref{fig27}. 
The right hand square
of this figure describes the relationships
between four other elements. 

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(105,30)
\put(5,5){\circle*{1.1}}
\put(25,5){\circle*{1.1}}
\put(5,25){\circle*{1.1}}
\put(25,25){\circle*{1.1}}
\put(65,5){\circle*{1.1}}
\put(85,5){\circle*{1.1}}
\put(65,25){\circle*{1.1}}
\put(85,25){\circle*{1.1}}
\put(5,5){\line(1,0){20}}
\put(5,5){\line(0,1){20}}
\put(25,25){\line(-1,0){20}}
\put(25,25){\line(0,-1){20}}
\put(65,5){\line(1,0){20}}
\put(65,5){\line(0,1){20}}
\put(85,25){\line(-1,0){20}}
\put(85,25){\line(0,-1){20}}
\put(0,2){$0$}
\put(0,25){$1$}
\put(27,2){$u$}
\put(27,25){$v$}
\put(14,0){$S$}
\put(14,27){$S$}
\put(0,14){$T$}
\put(27,14){$T\cap\delta$}
\put(47,2){$f(0,0,0)$}
\put(47,25){$f(0,0,1)$}
\put(87,2){$f(0,u,u)$}
\put(87,25){$f(0,u,v)$}
\put(71,0){$S\cap\mu$}
\put(74,27){$S$}
\put(60,14){$T$}
\put(87,14){$T\cap\delta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig27}
\end{figure}

Let's convince ourselves that 
the information indicated
in the right hand square of Figure~\ref{fig27} is correct.
First, 
$$f(0,0,0)\equiv_{T} f(0,0,1)$$
follows from the fact that 
$(0,0), (0,1)\in T$. Second, 
$$f(0,u,u)\equiv_{T\cap \delta} f(0,u,v)$$
follows from the fact that 
$(0,0), (u,u), (u,v)\in T\cap \delta$.
Third, 
$$f(0,0,1)\equiv_{S} f(0,u,v)\;\;
\mbox{\rm and}\;\; f(0,0,0)\equiv_{S} f(0,u,u)$$
follow from $(0,0), (0,u), (1,v)\in S$,
Finally, 
$$f(0,0,0)\equiv_{\mu} f(0,u,u)$$
since both sides are $\mu$-related to $0$:
the left hand side equals $0$ since $f$
is idempotent and the right hand side is
$\mu$-related to $0$ since 
$(0,u)\in S$ and 
$I[\mu,\nu]_S$ supports 
$f(x,y,y)\approx_{I[\mu,\nu]_S} x$.

The relations that we have just verified
imply that 
$$
(f(0,0,1),f(0,0,0))\in 
T\cap(S\circ(T\cap\delta)\circ (S\cap\mu)).
$$
By Condition (b) of the definition of
$I_S\leadsto J_T$ (Definition~\ref{leadsto})
this implies that 
%$$(f(0,0,1),f(0,0,0))\in \delta,$$ 
%so
$$f(0,0,1)\equiv_{\delta} f(0,0,0) \;\;( = 0)$$
holds for any pair $(0,1)\in \overline{T}$.
Hence $J_{\overline{T}}$ supports the local equation
$f(x,x,y)\approx_{J_{\overline{T}}} x$,
and the proof is complete.
\end{proof}

\begin{lm} \label{lm2}
Assume that $\m a\in {\vr V}$, that
$I_S = I[\mu,\nu]_S$ and $J_T = I[\delta,\theta]_T$ 
are separated solvability obstructions in
$\Conb(\m a)$,
and that $I_S\leadsto J_T$.
Let $\overline{T} = 
T\cap(S\circ(T\cap\delta)\circ S)$.
If $f(x,y,z)$ is an idempotent
term of $\m a$, and $I_S$ supports
the local equations
$$
f(x,x,y)\approx_{I_S} x
\hspace{.1in}\mbox{\rm and}\hspace{.1in}
f(x,y,x)\approx_{I_S} x,
$$
then $f(x,y,y)\approx_{J_{\overline{T}}} x$
is supported by $J_{\overline{T}}$.
\end{lm}

\begin{proof}
As in the last proof, 
choose $(0,1)\in \overline{T} = 
T\cap(S\circ(T\cap\delta)\circ S)$, and 
let $u$ and $v$ denote
elements of $A$ for which 
$0\equiv_S u\equiv_{T\cap \delta} v\equiv_S 1$.
As before, relationships between $0, 1, u$ and $v$
are expressed in Figure~\ref{fig28}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(5,5){\circle*{1.1}}
\put(25,5){\circle*{1.1}}
\put(5,25){\circle*{1.1}}
\put(25,25){\circle*{1.1}}
\put(5,5){\line(1,0){20}}
\put(5,5){\line(0,1){20}}
\put(25,25){\line(-1,0){20}}
\put(25,25){\line(0,-1){20}}
\put(0,2){$0$}
\put(0,25){$1$}
\put(27,2){$u$}
\put(27,25){$v$}
\put(14,0){$S$}
\put(14,27){$S$}
\put(0,14){$T$}
\put(27,14){$T\cap \delta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig28}
\end{figure}
In this proof we will be interested in the relationships
between some of the elements depicted in Figure~\ref{fig29}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(100,60)
\put(5,5){\circle*{1.1}}
\put(35,5){\circle*{1.1}}
\put(65,5){\circle*{1.1}}
\put(95,5){\circle*{1.1}}
\put(5,20){\circle*{1.1}}
\put(35,20){\circle*{1.1}}
\put(65,20){\circle*{1.1}}
\put(95,20){\circle*{1.1}}
\put(5,35){\circle*{1.1}}
\put(35,35){\circle*{1.1}}
\put(65,35){\circle*{1.1}}
\put(95,35){\circle*{1.1}}
\put(5,50){\circle*{1.1}}
\put(35,50){\circle*{1.1}}
\put(65,50){\circle*{1.1}}
\put(95,50){\circle*{1.1}}
\put(5,5){\line(1,0){90}}
\put(5,5){\line(0,1){45}}
\put(5,20){\line(1,0){90}}
\put(35,5){\line(0,1){45}}
\put(5,35){\line(1,0){90}}
\put(65,5){\line(0,1){45}}
\put(5,50){\line(1,0){90}}
\put(95,5){\line(0,1){45}}
\bezier{500}(5,5),(50,-2.5),(95,5)
\bezier{500}(5,20),(50,12.5),(95,20)
\bezier{500}(5,35),(50,27.5),(95,35)
\bezier{500}(5,50),(50,42.5),(95,50)
\bezier{500}(5,5),(10,27.5),(5,50)
\bezier{500}(35,5),(40,27.5),(35,50)
\bezier{500}(65,5),(70,27.5),(65,50)
\bezier{500}(95,5),(100,27.5),(95,50)
\put(16,52){$S\cap\mu$}
\put(44,52){$T\cap\delta$}
\put(79,52){$S$}
\put(19,36){$S$}
\put(44,36){$T\cap\delta$}
\put(79,36){$S$}
\put(19,21){$S$}
\put(44,21){$T\cap\delta$}
\put(79,21){$S$}
\put(19,6){$S$}
\put(44,6){$T\cap\delta$}
\put(79,6){$S$}
\put(0.5,11){$S$}
\put(-5.5,26){$T\cap\delta$}
\put(-5.5,41){$S\cap\mu$}
\put(30.5,11){$S$}
\put(24.5,26){$T\cap\delta$}
\put(30.5,41){$S$}
\put(60.5,11){$S$}
\put(54.5,26){$T\cap\delta$}
\put(60.5,41){$S$}
\put(90.5,11){$S$}
\put(84.5,26){$T\cap\delta$}
\put(90.5,41){$S$}
\put(-13,53){$f(0,0,0)$}
\put(27,54){$f(0,0,u)$}
\put(57,54){$f(0,0,v)$}
\put(99,53){$f(0,0,1)$}
\put(-13,3){$f(0,1,0)$}
\put(-13,19){$f(0,v,0)$}
\put(-13,34){$f(0,u,0)$}
\put(99,3){$f(0,1,1)$}
\put(99,19){$f(0,v,1)$}
\put(99,34){$f(0,u,1)$}
\end{picture}
\end{center}
\caption{ \ }\label{fig29}
\end{figure}
In this figure each element has the form $f(0,a,b)$.
As one moves row-by-row from top to bottom the
value of $a$ changes from $0$ to $u$ to $v$ to $1$.
As one moves 
column-by-column from left to right the
value of $b$ changes from $0$ to $u$ to $v$ to $1$.
All but two of the indicated relations follow from
$(0,u), (1,v)\in S$, 
$(u,v)\in T\cap\delta$, and that 
$S$ and $T\cap\delta$ are compatible. 
The exceptions are the
two instances of
$S\cap\mu$ in the upper left hand corner
of Figure~\ref{fig29}. The relevant 
pairs are clearly $S$-related, since $(0,u)\in S$
and $S$ is compatible, so we must justify the claim that
$$
f(0,u,0) \equiv_{\mu} f(0,0,0) \equiv_{\mu} f(0,0,u).
$$
%for which we must justify the $\mu$-relation.
For this one can use the facts that 
$I[\mu,\nu]_S$ supports
$$f(x,y,x)\approx_{I[\mu,\nu]_S} x = f(x,x,x) 
\approx_{I[\mu,\nu]_S} f(x,x,y),$$ 
that $(0,u)\in S$, and that $S$ is 
a tolerance.
The curved lines in the figure represent
$T$-relations, and all follow from 
$(0,1)\in T$ and the fact that $T$
is a tolerance. We must prove
that $f(0,1,1)\equiv_{\delta} 0$.
We will accomplish this by showing that the
curved lines of Figure~xxx that appear
across the top and
along the right hand side represent
$T\cap \delta$-relations, and not merely
$T$-relations.

We begin our work by examining the top row of Figure~\ref{fig29}.
Following the straight line from right to left we see
that $(f(0,0,1),f(0,0,0))\in 
S\circ(T\cap\delta)\circ(S\cap\mu)$, while following the
curved path we see that these elements are related by $T$.
By Condition (b) of the definition of
$I_S\leadsto J_T$ we get that 
$(f(0,0,1),f(0,0,0))\in \delta$.
We can apply this information to the rectangle
formed from the leftmost and rightmost elements
of the first two rows, as shown in Figure~\ref{fig30}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(100,25)
\put(5,5){\circle*{1.1}}
\put(35,5){\circle*{1.1}}
\put(65,5){\circle*{1.1}}
\put(95,5){\circle*{1.1}}
\put(5,20){\circle*{1.1}}
\put(35,20){\circle*{1.1}}
\put(65,20){\circle*{1.1}}
\put(95,20){\circle*{1.1}}
\put(5,5){\line(1,0){90}}
\put(5,5){\line(0,1){15}}
\put(5,20){\line(1,0){90}}
\put(35,5){\line(0,1){15}}
\put(65,5){\line(0,1){15}}
\put(95,5){\line(0,1){15}}
\bezier{500}(5,5),(50,-2.5),(95,5)
\bezier{500}(5,20),(50,12.5),(95,20)
\put(-5.5,11){$S\cap\mu$}
\put(91,11){$S$}
\put(48,-3){$T$}
\put(45,12){$T\cap\delta$}
\put(-13,18){$f(0,0,0)$}
\put(98,18){$f(0,0,1)$}
\put(-13,3){$f(0,u,0)$}
\put(98,3){$f(0,u,1)$}
\end{picture}
\end{center}
\caption{ \ }\label{fig30}
\end{figure}
Condition (b) of the definition of
$I_S\leadsto J_T$ can be applied again to show that 
\[(f(0,u,1),f(0,u,0))\in T\cap(S\circ(T\cap\delta)\circ(S\cap\mu))
\subseteq\delta,\]
so the bottom curved line in this figure represents
a $\delta$-relation. Now we see that the two elements
on the right hand side are related by 
$S\cap((T\cap\delta)\circ(S\cap\mu)\circ(T\cap\delta))$.
Condition (c) of the definition of
$I_S\leadsto J_T$ asserts that 
$S\cap((T\cap\delta)\circ(S\cap\mu)\circ(T\cap\delta))\subseteq\mu$,
so $((f(0,0,1),f(0,u,1))\in\mu$.

Returning to Figure~\ref{fig29}, we examine the four elements 
of the right hand side:
\[f(0,0,1)\;\equiv_{S\cap\mu}\;
f(0,u,1)\;\equiv_{T\cap\delta}\;
f(0,v,1)\;\equiv_{S}\;
f(0,1,1),\]
and $f(0,0,1)\;\equiv_{T}\;f(0,1,1)$.
Hence
\[(f(0,1,1),f(0,0,1))\in T\cap (S\circ(T\cap\delta)\circ(S\cap\mu))
\subseteq \delta.\]
We have shown that the curved lines across the
top and down the right side are $\delta$-relations,
so we have
\[f(0,0,0)\;\equiv_{\delta}\;f(0,0,1)\;\equiv_{\delta}\;f(0,1,1).\]
Since $f$ is idempotent, this proves that 
$f(0,1,1)\;\equiv_{\delta}\;f(0,0,0) = 0$, as desired.
\end{proof}

Next is the promised result showing
that instances of $\leadsto$ are related
to proper increases in rank.

\begin{thm} \label{rank_increase}
Assume that $\m a\in {\vr V}$, that
$I_S = I[\mu,\nu]_S$ and $J_T = I[\delta,\theta]_T$ 
are separated solvability obstructions in
$\Conb(\m a)$,
and that $I_S\leadsto J_T$.
If $\overline{T} = 
T\cap({S}\circ(T\cap\delta)\circ{S})$,
then ${\rm Rank}(I_S)\subsetneqq
{\rm Rank}(J_{\overline{T}})$.
In fact,
\begin{enumerate}
\item[(1)] if $U\in {\rm Rank}(I_S)$ and
$U\subseteq V\subseteq N$, then 
$V\in {\rm Rank}(J_{\overline{T}})$, and
\item[(2)] if $U, V\in {\rm Rank}(I_S)$ and
$U\cup V = N$, then 
$U\cap V\in {\rm Rank}(J_{\overline{T}})$.
\end{enumerate}
\end{thm}

\begin{proof}
It follows from Lemma~\ref{lm0} that 
$\overline{T}$ separates $J$, so 
${\rm Rank}(J_{\overline{T}})$ is defined.
We will prove that (1) and (2) hold first,
and then we will deduce that 
$
{\rm Rank}(I_{S})\subsetneqq
{\rm Rank}(J_{\overline{T}}).
$
The following notation will be helpful:
Given a partition of 
$N = \{1,\ldots,n\}$ into three sets $P, Q$
and $R$, one or two of which may be empty, we write
$F_{\mbox{\tiny $P/Q/R$}}(x,y,z)$
to denote the term 
that results from substituting $x$ for $x_i$
in the Hobby--McKenzie term $F(x_1,\ldots,x_n)$
when $i\in P$, $y$ for $x_i$ when $i\in Q$, and $z$
for $x_i$ when $i\in R$. Any 
$F_{\mbox{\tiny $P/Q/R$}}(x,y,z)$
is idempotent, since
$F$ is.

We begin the proof of (1).
Assume that 
$U\in {\rm Rank}(I_{S})$ and
$U\subseteq V\subseteq N$.
Partition $N$ as $P/Q/R$ with 
$P = U, Q = V-U$ and $R = N-V$. The fact that 
$U\in {\rm Rank}(I_{S})$ means
precisely that the local equation
$$
F_{\mbox{\tiny $U/(V-U)/(N-V)$}}(x,y,y) = F_U(x,y) \approx_{I_{S}} x
$$
is supported by $I_{S}$. 
%From Lemma~\ref{lm0}~(1) we get that 
%$I_{S}\leadsto J_{T}$, and so 
From 
Lemma~\ref{lm1} we get that $J_{\overline{T}}$ supports
$$F_V(x,y) = F_{\mbox{\tiny $U/(V-U)/(N-V)$}}(x,x,y) 
\approx_{J_{\overline{T}}} x.
$$
Hence $V\in {\rm Rank}(J_{\overline{T}})$.

Now we prove (2).
Assume that $U, V\in {\rm Rank}(I_{S})$ and
$U\cup V=N$, Partition $N$ as $P/Q/R$ with $P = U\cap V,
Q = U-V$ and $R = V-U$. 
%This is a partition
%of $N$ since $U\cup V = N$. 
The fact that
$U, V\in {\rm Rank}(I_{S})$ translates into
$$
F_{\mbox{\tiny $(U\cap V)/(U-V)/(V-U)$}}(x,x,y) = F_U(x,y) 
\approx_{I_{S}} x
$$
and
$$
F_{\mbox{\tiny $(U\cap V)/(U-V)/(V-U)$}}(x,y,x) = F_V(x,y) 
\approx_{I_{S}} x,
$$
so Lemma~\ref{lm2} proves that 
\[
F_{U\cap V}(x,y) =
F_{\mbox{\tiny $(U\cap V)/(U-V)/(V-U)$}}(x,y,y)\approx_{J_{\overline{T}}} x
\]
is supported by $J_{\overline{T}}$.
Hence $U\cap V\in {\rm Rank}(J_{\overline{T}})$.

Now we use (1) and (2) to prove that 
$
{\rm Rank}(I_{S})\subsetneqq
{\rm Rank}(J_{\overline{T}}).
$
It follows from (1) that 
$U\in {\rm Rank}(I_{S})$ implies
$U\in {\rm Rank}(J_{\overline{T}})$, so we certainly
have 
$
{\rm Rank}(I_{S})\subseteq
{\rm Rank}(J_{\overline{T}}).
$
Suppose that 
$
{\rm Rank}(I_{S})={\rm Rank}(J_{\overline{T}}).
$
Then from (1) we get that 
${\rm Rank}(I_{S})$
is an order filter in 
${\mathcal B}(F)$.
If $U, V\in {\rm Rank}(I_{S})$
and $U\not\subseteq V$, then since $V$
is the intersection of the $(n-1)$--element
sets that contain it there must be an
$(n-1)$--element set $W\supseteq V$
such that $U\not\subseteq W$. For this set
$W$ we have $W\in {\rm Rank}(I_{S})$
and $U\cup W = N$, so by (2) we
get that $U\cap W\in {\rm Rank}(J_{\overline{T}})
={\rm Rank}(I_{S})$.
Since $U\cap W$ is a proper subset of $U$,
and is in ${\rm Rank}(I_{S})$, 
this shows that if $U$ is not the least element 
of ${\rm Rank}(I_{S})$, then $U$ is not minimal
under inclusion in ${\rm Rank}(I_{S})$.
It follows that ${\rm Rank}(I_{S})$
is a principal order filter, hence
a lattice filter in ${\mathcal B}(F)$.
But by Lemma~\ref{rank}~(3) the set
${\rm Rank}(I_{S})$ is not a lattice
filter, hence
$
{\rm Rank}(I_{S})={\rm Rank}(J_{\overline{T}})
$
is impossible. This completes the proof that
$
{\rm Rank}(I_{S})\subsetneqq{\rm Rank}(J_{\overline{T}}).
$
\end{proof}

Since an instance of the $\leadsto$--relation
leads to a proper increase in rank,
and there are only finitely many ranks,
one expects a bound on the length
of $\leadsto$--sequences.
That bound is provided by the next theorem.

\begin{thm} \label{leadsto_sequence}
Assume that $\m a\in {\vr V}$.
$\Conb(\m a)$ does not contain a 
$\leadsto$--sequence
of separated solvability obstructions 
$$
I^0_{S^0}\leadsto I^1_{S^1}\leadsto I^2_{S^2}\leadsto
\cdots \leadsto I^k_{S^k}
$$
of length $k\geq 2n-4$.
\end{thm}

\begin{proof}
Suppose otherwise, and that $I^j = I[\mu^j,\nu^j]$.
Define tolerances $\overline{S}^j\subseteq S^j$
according to the rules
\begin{enumerate}
\item[(i)] $\overline{S}^0:= S^0$, and
\item[(ii)] $\overline{S}^{j+1}:=
S^{j+1}\cap(\overline{S}^j\circ(S^{j+1}\cap\mu^{j+1})\circ
\overline{S}^j)$.
\end{enumerate}
Now consider two sequences of claims:
\begin{enumerate}
\item[$(a)^j$] $I^j_{\overline{S}^j}\leadsto
I^{j+1}_{S^{j+1}}$.
\item[$(b)^j$] $\overline{S}^j$ separates $I^j$.
\end{enumerate}
Claims $(a)^0$ and $(b)^0$ hold since
$I^0_{S^0}\leadsto I^1_{S^1}$ and 
$\overline{S}^0=S^0$ separates $I^0$.
Claim~$(a)^j$  implies Claim~$(b)^{j+1}$
by Lemma~\ref{lm0}~(2). Claim~$(b)^{j}$ implies
Claim~$(a)^{j}$ by Lemma~\ref{lm0}~(1)
and the facts that 
$\overline{S}^j\subseteq S^j$ and 
$I^j_{S^j}\leadsto I^{j+1}_{S^{j+1}}$.
By induction, Claims $(a)^j$ and $(b)^j$ hold
for all $j$. Applying Theorem~\ref{rank_increase}
to Claim $(a)^j$ yields that 
$
{\rm Rank}(I^{j}_{\overline{S}^{j}})\subsetneqq
{\rm Rank}(I^{j+1}_{\overline{S}^{j+1}})$
for each $j<k$.
This shows that 
$$
I^0_{S^0}\leadsto I^1_{S^1}\leadsto I^2_{S^2}\leadsto
\cdots \leadsto I^k_{S^k}
$$
implies that 
$$
{\rm Rank}(I^{0}_{\overline{S}^{0}})\subsetneqq
{\rm Rank}(I^{1}_{\overline{S}^{1}})\subsetneqq
{\rm Rank}(I^{2}_{\overline{S}^{2}})\subsetneqq
\cdots \subsetneqq
{\rm Rank}(I^{k}_{\overline{S}^{k}})
$$
where the tolerance $\overline{S}^j$ 
is contained in $S^j$ yet still separates $I^j$.

Let ${\mathcal U}^j:= {\rm Rank}(I^{j}_{\overline{S}^{j}})$.
In the last paragraph we showed that 
$$
{\mathcal U}^0\subsetneqq{\mathcal U}^1\subsetneqq
{\mathcal U}^2\subsetneqq\cdots\subsetneqq
{\mathcal U}^k.
$$
Other properties of the sets ${\mathcal U}^j$
include:
\begin{itemize}
\item Each ${\mathcal U}^j$ is a 
closed subset of ${\mathcal B}(F)$ 
that is not a lattice filter (by Lemma~\ref{rank}~(3)).
\item $N\in {\mathcal U}^j$ and 
$\emptyset\not\in {\mathcal U}^j$ for all $j$
(by Lemma~\ref{rank}~(1) and (2)).
\item ${\mathcal U}^{j+1}$ contains the 
order 
filter generated by ${\mathcal U}^j$
(by Lemma~\ref{rank_increase}~(1)).
\item ${\mathcal U}^{j+1}$ contains the intersections
of comaximal pairs from ${\mathcal U}^j$
(by Lemma~\ref{rank_increase}~(2)).
\end{itemize}
We now argue that such a sequence
of sets cannot exist if $k\geq 2n-4$.
First, ${\mathcal U}^0$ contains $N$, is closed,
and is not a lattice filter, so either
${\mathcal U}^0$ contains a set of size $\leq n-2$
or else at least $2$ sets of size $n-1$.
In either case, since
${\mathcal U}^1$ contains the order filter generated
by ${\mathcal U}^0$ and contains the intersections
of comaximal pairs in ${\mathcal U}^0$, the set
${\mathcal U}^1$ properly contains
an order filter generated by a set $U$
of size $n-2$. 
Now suppose that for some $r\geq 0$
the set ${\mathcal U}^{2r+1}$ contains
a principal 
order filter generated by a set $U$ of size 
$\leq n-2-r$.
If $U$ is not empty, then there is some 
$W\in {\mathcal U}^{2r+1}$ 
such that $W\not\supseteq U$ (since 
${\mathcal U}^{2r+1}$ is closed).
The set $V:=U\cup (N-W)$ is in
${\mathcal U}^{2r+1}$ 
since $V\supseteq U$, and $V\cup W = N$
since $V\supseteq N-W$.
Thus $V\cap W = U\cap W\in {\mathcal U}^{2r+2} = 
{\mathcal U}^{2(r+1)}$.
Since $W\not\supseteq U$, the set $U\cap W$
has size $\leq n-2-(r+1)$. The set 
${\mathcal U}^{2(r+1)+1}$ contains 
the order filter generated by $U\cap W$.
By induction we see that 
for all $r$ the set ${\mathcal U}^{2r+1}$ contains
a principal 
order filter generated by a set of size 
$\leq n-2-r$. For $r=n-3$ we get that 
${\mathcal U}^{2n-5}$ contains a principal
order filter generated by a set $U'$ of size $1$.
The containment is proper, since 
${\mathcal U}^{2n-5}$ is not a lattice filter.
If $W'\in {\mathcal U}^{2n-5}$ does not contain
$U'$, then as above we get that 
$V' := U'\cup (N-W')\in {\mathcal U}^{2n-5}$ 
and $V'\cap W' = \emptyset\in 
{\mathcal U}^{2n-4}\subseteq
{\mathcal U}^k$. This is a contradiction,
since no ${\mathcal U}^j$ contains
the empty set. This completes
the proof that there is no sequence
$$
I^0_{S^0}\leadsto I^1_{S^1}\leadsto I^2_{S^2}\leadsto
\cdots \leadsto I^k_{S^k}
$$
with $k\geq 2n-4$.
\end{proof}

Now that we have shown that there is 
a bound on the length
of $\leadsto$--sequences in congruence lattices
of algebras in $\vr V$, 
our next goal is to determine how to 
recognize $\leadsto$-related intervals 
lattice-theoretically.

\begin{lm} \label{leadsto_cond}
Assume that $\m a\in {\vr V}$, and
that $I_S = I[\mu,\nu]_S$ and $J_T = I[\delta,\theta]_T$ 
are separated solvability obstructions in $\Conb(\m a)$.
Choose and fix congruences $\delta'$ and $\mu'$
satisfying 
$T\cap \delta\subseteq \delta'\subseteq \Cg a (T)$
and
$S\cap \mu\subseteq \mu'\subseteq \Cg a (S)$.
%(For example, one may choose $\delta' = \Cg a(T\cap\delta)$
%and $\mu' = \Cg a(S\cap \mu)$.) 
If the conditions
\begin{enumerate}
\item[(d)] $\Cg a(T)\cap \Cg a(S)\subseteq \delta'\vee\mu'$,
\item[(e)] $T\cap (\delta'\vee\mu')\subseteq\delta$, and
\item[(f)] $S\cap (\delta'\vee\mu')\subseteq\mu$
\end{enumerate}
are satisfied, then the conditions
\begin{enumerate}
\item[(b)] $T\cap(S\circ(T\cap\delta)\circ(S\cap
\mu))\subseteq\delta$, and
\item[(c)] $S\cap((T\cap\delta)\circ(S\cap
\mu)\circ(T\cap\delta))\subseteq\mu$
\end{enumerate}
from Definition~\ref{leadsto} are also satisfied.
\end{lm}

\begin{proof}
Condition (c) follows immediately from Condition
(f), since $T\cap \delta\subseteq \delta'$
and $S\cap \mu\subseteq \mu'$.

We will prove that (b) follows from (d) and (e)
by showing that 
$T\cap(S\circ(T\cap\delta)\circ(S\cap\mu))\subseteq
T\cap (\delta'\vee\mu')$. Choose 
$(c,d)\in T\cap(S\circ(T\cap\delta)\circ(S\cap\mu))$.
There exist $a$ and $b$ such that 
$c\wrel{\equiv_S} a
\wrel{\equiv_{T\cap \delta}} b
\wrel{\equiv_{S\cap\mu}} d$,
as shown in Figure~\ref{fig31}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{.8mm}
\begin{picture}(20,30)
\put(0,5){\circle*{1.2}}
\put(20,5){\circle*{1.2}}
\put(0,25){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
\put(0,5){\line(1,0){20}}
\put(0,5){\line(0,1){20}}
\put(20,25){\line(-1,0){20}}
\put(20,25){\line(0,-1){20}}
\put(-5,25){$a$}
\put(22,25){$b$}
\put(-5,2){$c$}
\put(22,2){$d$}
\put(9,0){$T$}
\put(5,27){$T\cap\delta$}
\put(-5,14){$S$}
\put(22,14){$S\cap\mu$}
\end{picture}
\end{center}
\caption{ \ }\label{fig31}
\end{figure}
The pair $(c,b)$ belongs to 
$S\circ (T\cap\delta)\subseteq \Cg a(S)\circ\delta'$
and also to $T\circ (S\cap\mu)\subseteq \Cg a(T)\circ \mu'$.
Using Theorem~\ref{equiv}~(2) (to go
from the first line below to the second)
and Condition (d) of this theorem 
we get that
$$
\begin{array}{rl}
(c,b)&\in 
(\Cg a(T)\circ\mu')\cap(\Cg a(S)\circ \delta')\\
&\subseteq ((\Cg a(T)\vee\delta')\wedge(\Cg a(S)\vee\mu'))
\vee\delta'\vee\mu'\\
&\leq (\Cg a(T)\wedge\Cg a(S))
\vee\delta'\vee\mu'\\
&\leq \delta'\vee\mu'.
\end{array}
$$
(Here we used the facts that 
$\Cg a(T)\supseteq \delta'$ and
$\Cg a(S)\supseteq \mu'$
to replace 
$(\Cg a(T)\vee\delta')\wedge(\Cg a(S)\vee\mu')$
by $\Cg a(T)\wedge\Cg a(S)$.)
Hence
$c\wrel{\equiv_{\delta'\vee\mu'}}b
\wrel{\equiv_{\mu'}}d$, so 
$(c,d)\in T\cap (\delta'\vee\mu')$.
Since $(c,d)\in 
T\cap(S\circ(T\cap\delta)\circ(S\cap\mu))$
was chosen arbitrarily, we get that
$$
T\cap(S\circ(T\cap\delta)\circ(S\cap\mu))
\subseteq T\cap (\delta'\vee\mu').
$$
With this and Condition (e) we get 
Condition (b). This completes the proof.
\end{proof}

The following lemma concerns the behavior
of the $\obst$-relation in varieties
with a weak difference term, so this lemma could
have appeared in Chapter~\ref{solvability}.
It appears here instead because its 
primary function is to 
aid in the proof of the succeeding lemma,
which further simplifies the task
of recognizing some instances of the $\leadsto$
relation.

\begin{lm}\label{pseudocomplement}
Let $\vr U$ be a variety with a weak difference term,
$\m a\in {\vr U}$, and $I[\delta,\theta]_T$
a separated solvability obstruction in $\Conb(\m a)$.
\begin{enumerate}
\item[(1)] $(\delta : T)$ is the largest congruence
$\xi\in \Conb(\m a)$ such that $\xi\geq \delta$ and 
$\xi\cap T\subseteq \delta$. 
\item[(2)] 
%$(\delta : T)\wrel{\obst} 1$. (Equivalently,
$(\delta:T)$ is a radical congruence.
\item[(3)] If $\rho$ is a radical congruence and
$\nu\not\leq\rho$, then $\nu\wedge \rho \wrel{\obst}\nu$.
\end{enumerate}
\end{lm}

\begin{proof}
Recall that $(\delta:T)$ is defined to be the largest 
congruence $\xi$ such that $\C C(\xi,T;\delta)$ holds,
so in particular $\C C((\delta:T),T;\delta)$ holds.
If $T':= (\delta:T)\cap T$, then the monotonicity
of the centralizer relation guarantees that 
$\C C(T',T';\delta)$ holds. If $T'':= \delta\circ T'\circ \delta$
is the $\delta$-closure of $T'$, then 
Lemma~\ref{tol=cong} proves that $T''$ is
a congruence containing $\delta$ for which
$\C C(T'',T'';\delta)$ holds. Therefore 
$\delta\wrel{\lhd} T''\subseteq \Cg a(T\cup\delta)\subseteq \theta$,
which (since $I[\delta,\theta]$
is a solvability obstruction) forces $T''=\delta$.
Hence 
$(\delta:T)\cap T = T'\subseteq T'' = \delta$, proving
$(\delta:T)$ is one of the congruences 
above $\delta$ whose intersection with $T$
is contained in $\delta$.

Now suppose that $\xi$ is any congruence above $\delta$
whose intersection with $T$ is contained
in $\delta$. These properties of $\xi$
may be used to show that 
$$
T\cap (\xi \circ (T\cap\delta)\circ \xi)\subseteq T\cap\xi\subseteq 
\delta.
$$
By Theorem~\ref{basic_centrality}~(7) it follows
that $\C C(\xi,T;\delta)$, or $\xi\leq (\delta:T)$.
This proves item (1).

For item (2) assume instead that $(\delta:T) \wrel{\lhd} \alpha \leq 1$
for some congruence $\alpha$.
% > (\delta:T)$. 
%From item (1) it follows that the tolerance 
%$T':= \alpha\cap T$ separates $I[\delta,\theta]$.
Let $T' = \alpha\cap T$.
Since $\C C(\alpha,\alpha;(\delta:T))$ holds,
we also have $\C C(T',T';(\delta:T))$ by
monotonicity of the centralizer.
We will derive that $\C C(T',T';\delta)$
holds using Theorem~\ref{basic_centrality}~(4).
To do this we need to prove that 
$T'\cap (\delta:T) = T'\cap \delta$.
The inclusion $\supseteq$ follows from 
$(\delta:T)\supseteq \delta$, while the inclusion $\subseteq$
follows from $T'\cap (\delta:T)\subseteq 
T\cap (\delta:T)\subseteq \delta$. 
Hence $\C C(T',T';\delta)$ holds
indeed.  
Now, using an argument
like the one in the first paragraph of this proof,
this contradicts the fact that $I[\delta,\theta]$
is a solvability obstruction unless $T'\subseteq \delta$.
But then $\alpha$ is a congruence above $\delta$
for which $\alpha\cap T = T'\subseteq \delta$,
so by item (1) we get that $\alpha\leq (\delta:T)$.
This proves that $(\delta:T)\wrel{\obst} 1$,
and therefore that $(\delta:T)$ is a radical congruence.

For item (3), if $\nu\not\leq \rho$, then
we certainly have $(\nu\wedge \rho) < \nu$. 
Given any $\lambda$ satisfying
$(\nu\wedge \rho) \wrel{\lhd} \lambda\leq  \nu$
we may join with $\rho$ to obtain
$\rho\wrel{\lhd}(\rho\vee \lambda)\leq (\rho\vee\nu)$, 
according to Lemma~\ref{meet+join}~(1).
Since $\rho\wrel{\obst} 1$, it follows that 
$\lambda\leq \rho$. Since 
%$(\nu\wedge \rho)\leq
$\lambda\leq \nu$
as well, we must have $\lambda \leq (\nu\wedge\rho)$.
This proves that $(\nu\wedge \rho) \wrel{\obst} \nu$. 
\end{proof}

Because of Lemma~\ref{pseudocomplement}~(1), we
may refer to $(\delta : T)$ as the 
{\bf pseudocomplement of $T$ over $\delta$}
\index{pseudocomplement of $T$ over $\delta$}
when $I[\delta,\theta]_T$ is a separated 
solvability obstruction.

The following set of 
criteria for producing $\leadsto$-related
intervals is simpler than the definition of
$\leadsto$ or the criteria of Lemma~\ref{leadsto_cond},
but is still strong enough for all of our intended applications.

\begin{lm}\label{new_mu}
Suppose that $I[\mu,\nu]_S$ and $I[\delta,\theta]_T$
are separated solvability obstructions in $\Conb(\m a)$.
If
\begin{enumerate}
\item[(d)'] $\Cg a(S)\cap \Cg a(T) \subseteq 
(\Cg a(S)\wedge (\delta:T))\vee(\Cg a(T)\wedge(\delta:T))$
and
\item[(g)\hphantom{'}] 
$S\cap \mu = S\cap (\delta:T)$,
%\subseteq (\mu:S)$, 
\end{enumerate}
then $I[\mu,\nu]_S\leadsto I[\delta,\theta]_T$.

In particular, assume that $I[\delta,\theta]_T$ 
is any separated solvability obstruction, 
$\nu\not\leq (\delta:T)$, and $\mu:= \nu\wedge (\delta:T)$.
Then $I[\mu,\nu]_{\nu}$ is a separated solvability obstruction, 
and if $\nu\cap \Cg a(T)\subseteq \mu$ then
$I[\mu,\nu]_{\nu}\leadsto I[\delta,\theta]_T$.
\end{lm}

\begin{proof}
Assume that (d)' and (g) hold. 
We will argue that Condition (a)
of Definition~\ref{leadsto} holds, 
and Conditions (d), (e) and (f)
of Lemma~\ref{leadsto_cond} all hold in the case
where $\delta' = \Cg a(T)\cap(\delta:T)$ and
$\mu' = \Cg a(S)\cap (\delta:T)$. This choice
for $\delta'$ 
satisfies the required condition
$T\cap \delta \subseteq \delta'\subseteq \Cg a(T)$
because $T\subseteq \Cg a(T)$ and $\delta\subseteq (\delta:T)$.
The choice for $\mu'$ satisfies the
condition 
$S\cap \mu \subseteq \mu'\subseteq \Cg a(S)$
because $S\subseteq \Cg a(S)$, while 
$S\cap \mu\subseteq (\delta:T)$ is part of assumption (g).

From Condition~(g) we get 
$S\cap (\delta:T)=S\cap \mu \subseteq \mu$ as 
Condition~(a) requires.

Condition~(d)' is simply Condition~(d)
of Lemma~\ref{leadsto_cond} restricted to the case
where  $\delta' = \Cg a(T)\cap(\delta:T)$ and
$\mu' = \Cg a(S)\cap (\delta:T)$. 

Condition~(e) asserts that $T\cap (\delta'\vee \mu')\subseteq \delta$.
For this, observe that $\delta'\vee\mu'\leq (\delta:T)$
from our choice of primed congruences, so
from Lemma~\ref{pseudocomplement}~(1) we get that 
$T\cap (\delta'\vee\mu') \subseteq 
T\cap (\delta:T) \subseteq  \delta$.

Condition~(f) asserts that $S\cap (\delta'\vee \mu')\subseteq \mu$.
According to the previous paragraph we have 
$\delta'\vee\mu'\leq (\delta:T)$. Together with (g),
this yields 
$S\cap (\delta'\vee \mu')\subseteq S\cap (\delta:T)=
S\cap \mu \subseteq \mu$.

For the second paragraph of the lemma statement, assume
that $I[\delta,\theta]_T$ is any separated solvability 
obstruction. If 
$\nu\not\leq (\delta:T)$ and $\mu:= \nu\wedge (\delta:T)$,
then Lemma~\ref{pseudocomplement}~(2) and (3)
prove that $I[\mu,\nu]$ is a solvability obstruction.
It is certainly separated by $S:=\nu$. To see that (d)'
holds under the assumption 
$\nu\cap \Cg a(T)\subseteq \mu$, observe that
since $S=\nu$ we have
$$
\begin{array}{rl}
\Cg a(S)\cap \Cg a(T) &= \nu\cap \Cg a(T)\\
&\subseteq \mu \qquad\text{\rm (by the assumption)}\\
&:= \nu\cap (\delta:T)\\
&= \Cg a(S)\cap (\delta:T)\\
&\subseteq (\Cg a(S)\cap (\delta:T))\vee (\Cg a(T)\cap (\delta:T)).
\end{array}
$$
For (g), $S\cap \mu = \nu\cap \mu = \mu := \nu\cap (\delta:T)
= S\cap (\delta:T)$.
\end{proof}

\section{Congruence Identities}\label{sec_cong_id}

The first theorem of this section shows
that solvability obstructions of minimal rank
cannot be contained in failure of the distributive law.
It is pretty easy to combine this with our
bound on the length of $\leadsto$-sequences to derive 
a nontrivial congruence identity, which is
something that we will do in Theorem~\ref{main1}.

\begin{thm} \label{transfer1}
Assume that $\m a\in {\vr V}$
and that $J_T:=I[\delta,\theta]_T$
is a separated solvability obstruction
in $\Conb(\m a)$. 
If $\alpha, \beta$, and $\gamma$
are congruences on $\m a$ such that 
$$(\alpha\wedge\beta)\vee(\alpha\wedge\gamma)\leq
\delta < \theta\leq \alpha\wedge(\beta\vee\gamma),
$$
then there is a 
separated solvability obstruction 
$I_S$ that is a subinterval
of either $I[\alpha\wedge\beta,\beta]$
or $I[\alpha\wedge\gamma,\gamma]$
such that 
$I_S\leadsto J_T$.
\end{thm}

\begin{proof}
We will apply Lemma~\ref{new_mu}.
The pseudocomplement 
%$(\delta:T)$ 
of $T$ over $\delta$ cannot lie above
both $\beta$ and $\gamma$, for if it did
we would have 
$$
T\subseteq \theta\leq \beta\vee\gamma \leq
(\delta:T),
$$
which is in conflict with the properties
$T\cap (\delta:T)\subseteq \delta$
and $T\not\subseteq \delta$.
We may therefore assume that $\gamma\not\leq (\delta:T)$.
If $\mu:=\gamma\wedge (\delta:T)$, then
the second paragraph of the statement
of Lemma~\ref{new_mu} proves
that $I[\mu,\gamma]$ is a solvability obstruction.
Since $\alpha\wedge\gamma\leq \delta\leq (\delta:T)$,
we have $\alpha\wedge \gamma\leq \gamma\wedge (\delta:T) = \mu$,
so $I[\mu,\nu]$ is contained in $I[\alpha\wedge\gamma,\gamma]$.
Using 
the second paragraph of the statement
of Lemma~\ref{new_mu} again we obtain from
$$
\gamma \cap \Cg a(T) \leq \gamma \cap (\alpha\wedge (\beta\vee\gamma))
= \alpha\wedge\gamma\leq \mu
$$
%so by the second paragraph of Lemma~\ref{new_mu}
%we have 
that $I[\mu,\nu]_{\nu}\leadsto I[\delta,\theta]_T$.
\end{proof}


\begin{cor} \label{transfer1_cor}
Assume that $\m a\in {\vr V}$, and that 
$\Conb(\m a)$ has a sublattice of congruences 
labeled as in Figure~\ref{fig32}.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{.8mm}
\begin{picture}(20,35)
\put(0,20){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,35){\circle*{1.2}}
\put(20,15){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
\put(0,20){\line(2,3){10}}
\put(20,25){\line(-1,1){10}}
\put(10,5){\line(1,1){10}}
\bezier{200}(20,15)(16,20)(20,25)
\bezier{200}(20,15)(24,20)(20,25)
\bezier{200}(10,5)(0,9)(0,20)
\bezier{200}(10,5)(10,16)(0,20)
\put(22,14){$\beta$}
\put(22,24){$\alpha$}
\put(-5,19){$\gamma$}
\put(9,0){$\sigma$}
\end{picture}
\end{center}
\caption{ \ }\label{fig32}
\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 $I[\sigma,\gamma]$ such that
$I_S\leadsto J_T$.
\end{cor}

\begin{proof}
The hypotheses of Theorem~\ref{transfer1} are met, since
$(\alpha\wedge\beta)\vee(\alpha\wedge\gamma) = \beta\vee\sigma
=\beta\leq
\delta < \theta\leq \alpha=\alpha\wedge(\beta\vee\gamma).
$
Therefore there is a separated solvability obstruction
$I_S$ contained in either
$I[\alpha\wedge\beta,\beta] = I[\beta,\beta]$ or
$I[\alpha\wedge\gamma,\gamma] = I[\sigma,\gamma]$ 
such that $I_S\leadsto J_T$.
The first of these intervals is trivial, so 
$I_S$ must be a subinterval of the second one.
\end{proof}


\begin{thm} \label{main1}
Let ${\vr V}$ be a variety.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] ${\vr V}$ satisfies a nontrivial congruence
identity.
\item[(2)] ${\vr V}$ satisfies an idempotent Maltsev
condition that fails in the variety of semilattices.
\item[(3)] There exists $\ell < \omega$ such that the lattice 
$\N {\ell+5}$,\index{$\m n_{\ell+5}$}
depicted in Figure~\ref{fig33}, cannot be embedded into 
$\Conb(\m a)$ for any $\m a\in {\vr V}$.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{.6mm}
\begin{picture}(120,94)
\put(0,45){\circle*{1.6}}
\put(15,45){\circle*{1.6}}
\put(30,45){\circle*{1.6}}
\put(60,5){\circle*{1.6}}
\put(60,85){\circle*{1.6}}
\put(67.5,10){\circle*{1.6}}
\put(67.5,80){\circle*{1.6}}
\put(75,15){\circle*{1.6}}
\put(75,75){\circle*{1.6}}
\put(90,45){\circle*{1.6}}
\put(105,35){\circle*{1.6}}
\put(105,55){\circle*{1.6}}
\put(112.5,40){\circle*{1.6}}
\put(112.5,50){\circle*{1.6}}
\put(0,45){\line(3,2){60}}
\put(0,45){\line(3,-2){60}}
\put(15,45){\line(3,2){52.5}}
\put(15,45){\line(3,-2){52.5}}
\put(30,45){\line(3,2){45}}
\put(30,45){\line(3,-2){45}}
\put(90,45){\line(3,2){15}}
\put(90,45){\line(3,-2){15}}
\put(60,5){\line(3,2){22.5}}
\put(60,85){\line(3,-2){22.5}}
\put(112.5,40){\line(-3,-2){15}}
\put(112.5,50){\line(-3,2){15}}
\put(112.5,50){\line(0,-1){10}}
\put(-9,44){$\gamma_{0}$}
\put(7,44){$\gamma_{1}$}
\put(82,44){$\gamma_{\ell}$}
\put(57,-1){$\sigma_{0}$}
\put(57,89){$\tau_{0}$}
\put(67,5){$\sigma_{1}$}
\put(67,83){$\tau_{1}$}
\put(107,32){$\sigma_{\ell}$}
\put(107,55){$\tau_{\ell}$}
\put(115.5,38.5){$\beta$}
\put(115.5,48.5){$\alpha$}
\put(90,25){\circle*{.9}}
\put(90,65){\circle*{.9}}
\put(93,27){\circle*{.9}}
\put(93,63){\circle*{.9}}
\put(87,23){\circle*{.9}}
\put(87,67){\circle*{.9}}
\put(60,45){\circle*{.9}}
\put(67.5,45){\circle*{.9}}
\put(52.5,45){\circle*{.9}}
\end{picture}
\end{center}
\caption{The lattice $\N {\ell+5}$}\label{fig33}
\end{figure}
\end{enumerate}
\end{thm}


\begin{proof}
%Corollary~\ref{ident_implies_HM} proves
%that statement (1) implies that
%$\vr V$ has a Hobby--McKenzie term,
%and this property is equivalent to statement (2)
%by Theorem~\ref{rect_Malt_cond2}.    ***************************

If $\vr V$ satisfies a nontrivial congruence identity,
$\varepsilon$, then any variety satisfying the same idempotent
Maltsev conditions as $\vr V$ must also satisfy $\varepsilon$,
by the Pixley-Wille algorithm. But
Ralph Freese and J.~B.Nation proved in \cite{freese-nation1}
that the variety of semilattices satisfies no nontrivial congruence
identity. Therefore $\vr V$ satisfies an idempotent Maltsev
condition that fails in the variety of semilattices.

Assume that (2) holds.
Then $\vr V$ has a Hobby--McKenzie term $F$
of arity $n$ for some $n<\omega$.
We shall argue that for $\ell = 2n-4$ the lattice 
$\N {\ell+5}$ cannot be embedded into $\Conb(\m a)$
for any $\m a\in {\vr V}$. 
The interval $I[\sigma_{\ell},\tau_{\ell}]$
contains a sublattice 
$\{\sigma_{\ell},\alpha,\beta,\gamma_{\ell},\tau_{\ell}\}$ 
that is isomorphic
to $\m n_5$, so this interval is nonmodular.
By Theorem~\ref{sim_cong}, it follows that 
$I[\sigma_{\ell},\tau_{\ell}]$ is not $\XXX$--solvable.
Therefore, by Theorem~\ref{solv=omit_obst},
this interval contains a solvability obstruction
$I[\mu_{\ell},\nu_{\ell}]$. Choose a tolerance $S_{\ell}$
that separates this interval (e.g., $S_{\ell} = \nu_{\ell}$).
By Corollary~\ref{transfer1_cor}, applied to
the sublattice 
$\{\sigma_{\ell}, \sigma_{\ell-1},\gamma_{\ell-1},
\tau_{\ell-1},\tau_{\ell}\}$,
there is a separated
solvability obstruction 
$I[\mu_{\ell-1},\nu_{\ell-1}]_{S_{\ell-1}}$
contained in the interval $I[\sigma_{\ell-1},\gamma_{\ell-1}]$
such that 
$$
I[\mu_{\ell-1},\nu_{\ell-1}]_{S_{\ell-1}}\wrel{\leadsto}
I[\mu_{\ell},\nu_{\ell}]_{S_{\ell}}.
$$
Similarly, for each $i = 1, 2, \ldots, \ell$ successively
we can apply 
Corollary~\ref{transfer1_cor} to 
the sublattice 
$\{\sigma_{\ell-i+1}, \sigma_{\ell-i},\gamma_{\ell-i},
\tau_{\ell-i},\tau_{\ell-i+1}\}$ to obtain 
a separated
solvability obstruction 
$I[\mu_{\ell-i},\nu_{\ell-i}]_{S_{\ell-i}}$
contained in the interval $I[\sigma_{\ell-i},\gamma_{\ell-i}]$
such that 
$$
I[\mu_{\ell-i},\nu_{\ell-i}]_{S_{\ell-i}}\wrel{\leadsto}
I[\mu_{\ell-i+1},\nu_{\ell-i+1}]_{S_{\ell-i+1}}.
$$
When we reach $i=\ell$ we will have a sequence
$$
I[\mu_{0},\nu_{0}]_{S_{0}}\wrel{\leadsto}
I[\mu_{1},\nu_{1}]_{S_{1}}\wrel{\leadsto}
I[\mu_{2},\nu_{2}]_{S_{2}}\wrel{\leadsto}\cdots \wrel{\leadsto}
I[\mu_{\ell},\nu_{\ell}]_{S_{\ell}}.
$$
According to Theorem~\ref{leadsto_sequence},
there is no such sequence if $\ell\geq 2n-4$.
This completes the proof of $(2)\mimplies(3)$.


To prove that (3)$\mimplies$(1), we observe that:
\begin{enumerate}
\item[(i)] Each $\N {\ell+5}$ is subdirectly irreducible.
%(Well known, and easy to check.)
\item[(ii)] Each $\N {\ell+5}$ is projective in the variety
of lattices. (See \cite{mckenzie1}.)
\end{enumerate}
By Theorem~\ref{proj+si_implies_splitting},
there is an identity $\varepsilon_{\ell+5}$ that
holds in a lattice $\m l$ if and only if $\m l$ has no sublattice
isomorphic to $\N {\ell+5}$. 
Since $(3)$ asserts that $\N {\ell+5}$
is not embeddable in $\Conb(\m a)$ for 
any $\m a\in {\vr V}$, 
$\varepsilon_{\ell+5}$ is a congruence identity
of $\vr V$. Since $\varepsilon_{\ell+5}$
fails in $\N {\ell+5}$, it is a nontrivial congruence 
identity.
\end{proof}

\begin{cor} (Cf.\ \cite{lipparini1})
Any congruence $n$--permutable variety 
satisfies a nontrivial congruence identity.
\end{cor}

\begin{proof}
It suffices to observe that congruence $n$-permutability
can be characterized by an idempotent Maltsev condition,
for example by the Maltsev condition
of J.~Hagemann and A.~Mitschke from \cite{hagemann-mitschke}, 
and that the variety of semilattices is not congruence 
$n$-permutable for any $n$.
\end{proof}

\begin{cor} 
Any congruence join semidistributive variety 
satisfies a nontrivial congruence identity.
\end{cor}

\begin{proof}
Congruence join
semidistributivity
can be characterized by a set of 
idempotent Maltsev conditions, 
as is shown in \cite{czedli},
and the variety of semilattices is not congruence 
join semidistributive.
\end{proof}

See Theorem~\ref{sdj-thm}~(7) for explicit 
congruence identities for congruence
join semidistributive varieties.
