\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\footnote{%
``Closed'' was defined after Example~\ref{maltsev_term}.}
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 
$I_S=I[\mu,\nu]_S$ supports 
$f(x,y,y)\approx_{I_S} x$.

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 
$I_S=I[\mu,\nu]_S$ supports
$$f(x,y,x)\approx_{I_S} x = f(x,x,x) 
\approx_{I_S} f(x,x,y),$$ 
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~\ref{fig29} 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
$\nu$ is a congruence such that 
$\nu\not\leq\rho$, 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[$\textrm{(d)}'$] $\Cg a(S)\cap \Cg a(T) \subseteq 
(\Cg a(S)\wedge (\delta:T))\vee(\Cg a(T)\wedge(\delta:T))$
and
\item[$\textrm{(g)}\hphantom{'}$] 
$S\cap \mu = S\cap (\delta:T)$,
%\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 $\textrm{(d)}'$ 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~$\textrm{(d)}'$ 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 $\textrm{(d)}'$
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 a 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.
