\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 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 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} 
  $I := I[\delta,\theta]$ if
  $T\subseteq \theta$ and $T\not\subseteq \delta$.  We write
  $I[\delta,\theta]_T$ or $I_T$ to refer to the 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 a solvability obstruction, and that $T$ is a tolerance
  that separates $I$.  The 
  {\bf rank of~$I_T$}\index{rank of a congruence interval}\index{Rank@${\rm Rank}(I_T)$} 
  is the set
\[
{\rm Rank}(I_T):=
\{U\subseteq N\mid F_U(x,y)\approx_{I_T} x\}\,.
\]
\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)] $\emptyset\notin {\rm Rank}(I_T)$.
  \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 $\emptyset\notin {\rm
    Rank}(I_{T})$.

  To prove item (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)] $T\cap\big(S\circ(T\cap\delta)\circ(S\cap
    \mu)\big)\subseteq\delta$, and
  \item[(c)] $S\cap\big((T\cap\delta)\circ(S\cap
    \mu)\circ(T\cap\delta)\big)\subseteq\mu$.
  \end{enumerate}
\end{df}

The three conditions of this definition describe a property
of the tolerance
\[
\overline{T} :=T\cap\big(S\circ(T\cap\delta)\circ S\big)\,.
\]
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 $(c,d)\notin\delta$.  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
$(c,d)\in \delta$.  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 $(c,d)\notin\delta$,
but any such quadrangle must have $(a,c), (b,d)\notin\mu$.

\begin{lm}\label{lm0}
  Assume that\/ $I_S = I[\mu,\nu]_S$ and\/ $J_T =
  I[\delta,\theta]_T$ are separated solvability obstructions
  in\/~$\Conb(\m a)$, $\m a\in {\vr V}$, and that\/
  $I_S\leadsto J_T$.  If\/ $\overline{S}\subseteq S$ also
  separates $I[\mu,\nu]$ and\/ $\overline{T} =
  T\cap\big(\overline{S}\circ(T\cap\delta)\circ
  \overline{S}\big)$, 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\/ $I_S = I[\mu,\nu]_S$ and\/ $J_T =
  I[\delta,\theta]_T$ are separated solvability obstructions
  in\/~$\Conb(\m a)$, $\m a\in {\vr V}$, and that\/
  $I_S\leadsto J_T$.  Let\/ $\overline{T} =
  T\cap\big(S\circ(T\cap\delta)\circ S\big)$.  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
  \[
  f(x,x,y)\approx_{J_{\overline{T}}} x
  \quad\mbox{\rm and}\quad
  f(x,y,x)\approx_{J_{\overline{T}}} x
  \]
  are supported by $J_{\overline{T}}$\,.
\end{lm}

\begin{proof}
  The fact that the tolerance $\overline{T}:=
  T\cap\big(S\circ(T\cap\delta)\circ S\big)$ 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 $(0,1)\in \overline{T} =
  T\cap\big(S\circ(T\cap\delta)\circ S\big)$.  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 us 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,
\[
f(0,0,1)\equiv_{S} f(0,u,v)
\quad\mbox{\rm and}\quad
f(0,0,0)\equiv_{S} f(0,u,u)
\]
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
\[
\big(f(0,0,1),f(0,0,0)\big)\in 
T\cap\big(S\circ(T\cap\delta)\circ (S\cap\mu)\big)\,.
\]
By Condition~(b) of the definition of~$I_S\leadsto J_T$
(Definition~\ref{leadsto}) this implies that
\[
f(0,0,1)\equiv_{\delta} f(0,0,0) \;\;( = 0)
\]
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\/ $I_S = I[\mu,\nu]_S$ and\/ $J_T =
  I[\delta,\theta]_T$ are separated solvability obstructions
  in\/~$\Conb(\m a)$, $\m a\in {\vr V}$, and that\/
  $I_S\leadsto J_T$.  Let\/ $\overline{T} =
  T\cap\big(S\circ(T\cap\delta)\circ S\big)$.  If\/
  $f(x,y,z)$ is an idempotent term of\/~$\m a$, and\/ $I_S$
  supports the local equations
  \[
  f(x,x,y)\approx_{I_S} x 
  \quad\mbox{\rm and}\quad
  f(x,y,x)\approx_{I_S} x\,,
  \]
  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 $(0,1)\in \overline{T} =
  T\cap\big(S\circ(T\cap\delta)\circ S\big)$, 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(34,5){\circle*{1.1}}
\put(63,5){\circle*{1.1}}
\put(92,5){\circle*{1.1}}
\put(5,20){\circle*{1.1}}
\put(34,20){\circle*{1.1}}
\put(63,20){\circle*{1.1}}
\put(92,20){\circle*{1.1}}
\put(5,35){\circle*{1.1}}
\put(34,35){\circle*{1.1}}
\put(63,35){\circle*{1.1}}
\put(92,35){\circle*{1.1}}
\put(5,50){\circle*{1.1}}
\put(34,50){\circle*{1.1}}
\put(63,50){\circle*{1.1}}
\put(92,50){\circle*{1.1}}
\put(5,5){\line(1,0){87}}
\put(5,5){\line(0,1){45}}
\put(5,20){\line(1,0){87}}
\put(34,5){\line(0,1){45}}
\put(5,35){\line(1,0){87}}
\put(63,5){\line(0,1){45}}
\put(5,50){\line(1,0){87}}
\put(92,5){\line(0,1){45}}
\bezier{500}(5,5),(48,-2.5),(92,5)
\bezier{500}(5,20),(48,12.5),(92,20)
\bezier{500}(5,35),(48,27.5),(92,35)
\bezier{500}(5,50),(48,42.5),(92,50)
\bezier{500}(5,5),(10,27.5),(5,50)
\bezier{500}(34,5),(39,27.5),(34,50)
\bezier{500}(63,5),(68,27.5),(63,50)
\bezier{500}(92,5),(97,27.5),(92,50)
\put(12,52){$S\cap\mu$}
\put(44,52){$T\cap\delta$}
\put(76,52){$S$}
\put(16,36){$S$}
\put(44,36){$T\cap\delta$}
\put(76,36){$S$}
\put(16,21){$S$}
\put(44,21){$T\cap\delta$}
\put(76,21){$S$}
\put(16,6){$S$}
\put(44,6){$T\cap\delta$}
\put(76,6){$S$}
\put(0,11){$S$}
\put(-6.5,26){$T\cap\delta$}
\put(-6.5,41){$S\cap\mu$}
\put(29.5,11){$S$}
\put(22.5,26){$T\cap\delta$}
\put(29.5,41){$S$}
\put(58.5,11){$S$}
\put(51.5,26){$T\cap\delta$}
\put(58.5,41){$S$}
\put(87.5,11){$S$}
\put(80.5,26){$T\cap\delta$}
\put(87.5,41){$S$}
\put(-12,53){$f(0,0,0)$}
\put(26,54){$f(0,0,u)$}
\put(57,54){$f(0,0,v)$}
\put(95,53){$f(0,0,1)$}
\put(-12,3){$f(0,1,0)$}
\put(-12,19){$f(0,v,0)$}
\put(-12,34){$f(0,u,0)$}
\put(95,3){$f(0,1,1)$}
\put(95,19){$f(0,v,1)$}
\put(95,34){$f(0,u,1)$}
\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 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
\[
\big(f(0,0,1),f(0,0,0)\big)\in
S\circ(T\cap\delta)\circ(S\cap\mu)\,,
\]
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
$\big(f(0,0,1),f(0,0,0)\big)\in \delta$.  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(-6.5,11){$S\cap\mu$}
\put(91,11){$S$}
\put(48,-3){$T$}
\put(45,12){$T\cap\delta$}
\put(-12,18){$f(0,0,0)$}
\put(96,18){$f(0,0,1)$}
\put(-12,3){$f(0,u,0)$}
\put(96,3){$f(0,u,1)$}
\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
\[
\big(f(0,u,1),f(0,u,0)\big)\in
T\cap\big(S\circ(T\cap\delta)\circ(S\cap\mu)\big)
\subseteq\delta\,,
\]
so the bottom curved line in this figure represents a
$\delta$-relation. Now we see that the two elements on the
right hand side of this figure are related by
$S\cap\big((T\cap\delta)\circ(S\cap\mu)\circ
(T\cap\delta)\big)$.  Condition~(c) of the definition
of~$I_S\leadsto J_T$ asserts that
$S\cap\big((T\cap\delta)\circ
(S\cap\mu)\circ(T\cap\delta)\big)\subseteq\mu$, therefore
$\big((f(0,0,1),f(0,u,1)\big)\in\mu$.

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
\[
\big(f(0,1,1),f(0,0,1)\big)\in T\cap
\big(S\circ(T\cap\delta)\circ(S\cap\mu)\big)
\subseteq \delta\,.
\]
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\/ $I_S = I[\mu,\nu]_S$ and\/ $J_T =
  I[\delta,\theta]_T$ are separated solvability obstructions
  in\/~$\Conb(\m a)$, $\m a\in {\vr V}$, and\/ ${I_S\leadsto
    J_T}$.  If\/~$\overline{T} =
  T\cap\big({S}\circ(T\cap\delta)\circ{S}\big)$, 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 first prove that (1) and (2) hold, 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.

  To prove (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{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 $R = V-U$.  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\/ $\m a\in {\vr V}$. Then\/ $\Conb(\m a)$ 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)] $\overline{S}^{j+1}:=
    S^{j+1}\cap\big(\overline{S}^j\circ
    (S^{j+1}\cap\mu^{j+1})\circ
    \overline{S}^j\big)$.
  \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 $\emptyset\notin {\mathcal
      U}^j$ for all $j$ (by Lemma~\ref{rank}~(1) and (2)).
  \item ${\mathcal U}^{j+1}$ contains the order filter
    generated by ${\mathcal U}^j$ (according to
    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 $V\cup W = N$
  because $V\supseteq N-W$.  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}

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
  $S\cap \mu\subseteq \mu'\subseteq \Cg a (S)$.  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)] $T\cap\big(S\circ(T\cap\delta)\circ (S\cap
    \mu)\big)\subseteq\delta$, and
  \item[(c)] $S\cap\big((T\cap\delta)\circ(S\cap
    \mu)\circ(T\cap\delta)\big)\subseteq\mu$
  \end{enumerate}
  from Definition~\ref{leadsto} are also satisfied.
\end{lm}

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

  We will prove that (b) follows from (d) and (e) by showing
  that $T\cap\big(S\circ(T\cap\delta)\circ
  (S\cap\mu)\big)\subseteq T\cap (\delta'\vee\mu')$. Choose
  \[
  (c,d)\in T\cap\big(S\circ(T\cap\delta)\circ
  (S\cap\mu)\big)\,.  
  \]
  There exist $a$ and $b$ such that $c\equiv_S a
  \equiv_{T\cap\,\delta} b \equiv_{S\cap\,\mu} d$, 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
\[
\begin{array}{r@{}l}
(c,b)&{} \in 
\big(\Cg a(T)\circ\mu'\big)\cap
\big(\Cg a(S)\circ \delta'\big)\\
&{} \subseteq \big((\Cg a(T)\vee\delta'\big)\wedge
\big(\Cg a(S)\vee\mu')\big)
\vee\delta'\vee\mu'\\
&{} \leq \big(\Cg a(T)\wedge\Cg a(S)\big)
\vee\delta'\vee\mu'\\
&{} \leq \delta'\vee\mu'\,.
\end{array}
\]
(In this argument we used the facts that $\Cg a(T)\supseteq
\delta'$ and ${\Cg a(S)\supseteq \mu'}$ to replace $\big(\Cg
a(T)\vee\delta'\big)\wedge\big(\Cg a(S)\vee\mu'\big)$ by
$\Cg a(T)\wedge\Cg a(S)$.)  Hence
${c\equiv_{\delta'\vee\mu'} b \equiv_{\mu'} d}$, so
$(c,d)\in T\cap (\delta'\vee\mu')$.  Since $(c,d)$ was
chosen arbitrarily in
$T\cap\big(S\circ(T\cap\delta)\circ(S\cap\mu)\big)$, we get
that
\[
T\cap\big(S\circ(T\cap\delta)\circ(S\cap\mu)\big) \subseteq
T\cap (\delta'\vee\mu')\,.
\]
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 and
  $I[\delta,\theta]_T$ a separated solvability obstruction
  in~$\Conb(\m a)$,\, ${\m a\in {\vr U}}$.
  \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)] $(\delta:T)$ is a radical congruence.
  \item[(3)] If $\rho$ is a radical congruence and $\nu$ is
    a congruence such that $\nu\not\leq\rho$, then
    $\nu\wedge \rho \obst \nu$.
  \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 $\C C\big((\delta:T),T;\delta\big)$ 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 $\delta\lhd
  T''\subseteq \Cg a(T\cup\delta)\subseteq \theta$, 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
  \[
  T\cap \big(\xi \circ (T\cap\delta)\circ \xi\big)
  \subseteq T\cap\xi\subseteq \delta\,.
  \]
  By Theorem~\ref{basic_centrality}~(7) it follows that $\C
  C(\xi,T;\delta)$, that is, $\xi\leq (\delta:T)$.  This
  proves item (1).

  For item (2) assume instead that $(\delta:T) \lhd \alpha
  \leq 1$ for some congruence~$\alpha$.  Let $T' = \alpha
  \cap T$.  Since $\C C\big(\alpha,\alpha;(\delta:T)\big)$
  holds, we also have~$\C C\big(T',T';(\delta:T)\big)$ by
  the 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 $\subseteq$ 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
  $(\delta:T)\obst 1$, and therefore that $(\delta:T)$ is a
  radical congruence.

  To prove item (3), if $\nu\not\leq \rho$, then we
  certainly have $(\nu\wedge \rho) < \nu$.  Given
  any~$\lambda$ satisfying $(\nu\wedge \rho) \lhd
  \lambda\leq \nu$ we may join with $\rho$ to obtain
  $\rho\lhd (\rho\vee \lambda)\leq (\rho\vee\nu)$, according
  to Lemma~\ref{meet+join}~(1).  Since $\rho\obst 1$, it
  follows that $\lambda\leq \rho$. Since $\lambda\leq \nu$
  as well, we must have $\lambda \leq (\nu\wedge\rho)$.
  This proves that $(\nu\wedge \rho) \obst \nu$.
\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[$\textrm{(d)}'$] $\Cg a(S)\cap \Cg a(T) \subseteq
    \big(\Cg a(S)\wedge (\delta:T)\big)\vee
    \big(\Cg a(T)\wedge(\delta:T)\big)$ and
  \item[$\textrm{(g)}\hphantom{'}$] $S\cap \mu = S\cap
    (\delta:T)$,
  \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
  \[
  \begin{array}{r@{}l}
    \Cg a(S)\cap \Cg a(T) &{} = \nu\cap \Cg a(T)\\
    &{} \subseteq \mu \qquad\text{\rm (by the assumption)}\\
    &{} := \nu\cap (\delta:T)\\
    &{} = \Cg a(S)\cap (\delta:T)\\
    &{} \subseteq \big(\Cg a(S)\cap (\delta:T)\big)\vee
    \big(\Cg a(T)\cap (\delta:T)\big)\,.
  \end{array}
  \]
  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 easy to combine this
with our bound on the length of $\leadsto$-sequences to
derive a nontrivial congruence identity, which we 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
  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
  \[
  \gamma \cap \Cg a(T) \leq \gamma \cap
  \big(\alpha\wedge (\beta\vee\gamma)\big) =
  \alpha\wedge\gamma\leq \mu
  \]
  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 satisfied,
  because $(\alpha\wedge\beta)\vee (\alpha\wedge\gamma) =
  \beta\vee\sigma =\beta\leq \delta < \theta\leq
  \alpha=\alpha\wedge(\beta\vee\gamma)$.
  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}
  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.
  \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 $\m a\in {\vr
    V}$, this identity~$\varepsilon_{\ell+5}$ 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}{\upshape (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} 
  Every 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.
