Why does "$[\beta,\beta]\leq\alpha$" imply that "$\alpha$ has $\beta,\beta$-term condition" in general algebras? Let $\mathbf A$ be an (universal) algebra and let us denote $\mathrm{Clo}(\mathbf A)$ the set of term operations of $\mathbf A$. Let $\alpha, \beta$ be congruences of $\mathbf A$. We say that $\alpha$ has $\beta,\beta$-term condition iff
$$
\forall t\in \mathrm{Clo}(\mathbf A)\ \ \forall (a,b)\in \beta\ \ \forall (\mathbf c,\mathbf d)\in \beta:\ \ (t(a,\mathbf c),t(a,\mathbf d))\in \alpha \Longrightarrow (t(b,\mathbf c),t(b,\mathbf d))\in\alpha
$$
We denote by $[\beta,\beta]$ the least congruence $\gamma$ such that $\gamma$ has $\beta,\beta$-term condition. It is easy to see that $[\beta,\beta]\leq \beta$.
Laet $\alpha\leq\beta$ be congruences on $\mathbf A$. We say that $\beta$ is Abelian over $\alpha$ iff $\alpha$ has $\beta,\beta$-term condition. This should be equivalent to $[\beta,\beta]\leq\alpha$. Indeed, left to right implication holds by definition of $[\beta,\beta]$. Why does the implication right to left hold? I seem to be overlooking something simple.
I use the terminology used in the Bergman's book "Universal Algebra". The fact in my question is noted on page 263 there.
 A: You are not overlooking anything.
There exist situations
where $[\beta,\beta]\leq\alpha$, but
$\beta$ is NOT abelian over $\alpha$.
Let me discuss one special case, where $\beta = 1_{\bf A}$
is the universal congruence on ${\bf A}$ and $[\beta,\beta]=0$.
This means: ${\bf A}$ is an abelian algebra.
If you could deduce that $\beta$ is abelian over
$\alpha$ for an arbitrary congruence
$\alpha$ on ${\bf A}$, you could go one more step
and deduce that $\beta/\alpha = 1_{{\bf A}/\alpha}$
is an abelian congruence on ${\bf A}/\alpha$;
i.e., that ${\bf A}/\alpha$ is an abelian algebra.
To summarize, if the statement you are trying to prove
was true, then it would imply that quotients of abelian
algebras are abelian.
Here is a counterexample.
Let ${\bf A} = \langle \{0,1,2,3\}; \ast\rangle$
be defined so that


* $3\ast 3 = 2$, 


* $3\ast x = x\ast 3 = 1$ if $x\neq 3$, and 


* $x\ast y = 0$ otherwise. 

You can check that ${\bf A}$ is an abelian algebra that has a nonabelian
quotient ${\bf A}/\alpha$, where $\alpha$ is the congruence
on ${\bf A}$ generated by $(0,1)$.
In fact, it is easy to see 
that nonabelian quotients of abelian
algebras must exist, even without
constructing a specific one:
Every algebra is a quotient of an
absolutely free algebra, and
every absolutely free algebra is abelian.
Thus nonabelian quotients 
of abelian algebras (like nonabelian groups) are easy to produce.
Having said this, what we are discussing belongs to the
pathology of commutator theory.
If


* ${\bf A}$ belongs to a congruence modular variety, or 


* ${\bf A}$ is finite and $\alpha$ is covered by $\beta$, 

then $[\beta,\beta]\leq\alpha$ DOES imply that 
$\beta$ is abelian over $\alpha$. (The proofs
of these statements are not trivial.)
