This is Definition 10.9 of the book "A Course in Universal Algebra" by Burris and Sankappanavar (page 73, Millennium Edition).
http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf
Definition 10.9 Let $K$ be a family of algebras of type $\mathfrak{F}$. Given a set $X$ of variables define the congruence $\theta_K(X)$ on $T(X)$ by
$\theta_K(X)=\bigcap\Phi_K(X)$,
where $\Phi_K(X)=\{\phi\in \textrm{Con } T(X):T(X)/\phi \in IS(K)\}$; and then define $F_K(\overline{X})$, the $K$-free algebra over $\overline{X}$, by $F_K(\overline{X})=T(X)/\theta_K(X)$, where $\overline{X}=X/\theta_K(X)$.
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Now, my question is
Let $K=\{L\}$ where $L$ is a two element lattice. In this case, what is $\theta_K(X)$? What are the elements of $F_K(\bar{x},\bar{y})$ and why are they elements of $F_K(\bar{x},\bar{y})$. I used the UACalculator tool, but I am still having difficulty in understanding the nature of $F_K(\overline{X})$.
Thank you.
ps) This is my another question in the reply thread of my previous question, but I think it deserves a new question.
Note that in case $K = \{L\}$, where $L$ is the two-element lattice, $IS(K)$ consists of one-element and two-element lattices and their isomorphic copies. Hence a congruence $\phi \in \Phi_K(X)$ iff $T(X)/\phi$ is one-element or two-element lattice.
If $T(X)/\phi$ is one-element set we have $\phi = \nabla$, it occurs precisely when $x\phi y, \forall x,y \in X$.
When $T(X)/\phi$ is two-element set precisely two distinct elements $x, y \in X$ must satisfy $\lnot x\phi y$, because in case $\lnot x\phi y, \lnot x\phi z, \lnot y\phi z$ we will have three different elements $\overline{x}, \overline{y}, \overline{z} \in T(X)/\phi$. So under all such congruences precisely two elements of $X$ are not identified. Hence w.l.o.g we can consider $X = \{x, y\}$. Note also that any term $t \in T(X)$ is congruent to one of terms $x, y, x\wedge x, x\vee y$, because of idempotence and absorption laws in lattice $T(X)/\phi$. So $T(X)/\theta_K(X)$ contains at most four element. Now we will show that there are exactly four elements in this lattice.
Consider the kernel congruences $\phi_1, \phi_2$ of the homomorphisms extending the mappings $x \mapsto 0, y \mapsto 1$ and $x \mapsto 1, y \mapsto 0$, where $L = \{0, 1\}$ is the two-element lattice. We have $x\phi_1 x\wedge y, y\phi_1 x\vee y$ and $y\phi_2 x\wedge y, x\phi_2 x\vee y$, and also $\lnot x\phi_1 y, \lnot x\phi_2 y$, hence $\lnot x \phi_1 x\vee y, \lnot y\phi_1 x\wedge y$ and $\lnot y \phi_2 x\vee y, \lnot x\phi_2 x\wedge y$. Consider the congruence $\phi = \phi_1 \cap \phi_2$, we have that all elements $x, y, x\wedge y, x\vee y$ are pairwise non-congruent with repsect to $\phi$. Assume e.g. that $x \phi x \wedge y$, hence $x \phi_1 x \wedge y, x \phi_2 x \wedge y$, but from the considerations above we have $\lnot x \phi_2 x \wedge y$ and hence $\lnot x \phi x \wedge y$. Hence $|T(X)/\phi| \geqslant 4.$
We have $T(X)/\phi_1 \cong T(X)/\phi_2 \cong L$ and $\phi_1, \phi_2 \in \Phi_K(X)$, also $\theta_K(X) \subseteq \phi_1 \cap \phi_2$ and $4 \leqslant |T(X)/(\phi_1 \cap \phi_2)| \leqslant |T(X)/\theta_K(X)| \leqslant 4.$ Hence $|T(X)/\theta_K(X)| = 4$. It is not the four-element chain because if e.g. $\overline{x} \leqslant \overline{y}$ that is $\overline{x} = \overline{x}\wedge \overline{y}$ then $x \phi x\wedge y$ for all congruences $\phi \in \Phi_K(X)$, which is not true: consider $\phi = \phi_2$ above. In this case $\theta_K(X)$ is the congruence which has precisely four classes with representatives $x, y, x\wedge y, x\vee y$. It can be seen as if you consider terms from $T(X)$ as term functions and say that two terms are congruent modulo $\theta_K(X)$ iff they induce the same term functions for all $L \in K$. In your case there is only one $L \in K$.
Finally it means that $T(X)/\theta_K(X) \cong L\times L$, where $L$ is the two-element lattice and $F_K(\overline{x}, \overline{y}) = T(X)/\theta_K(X) = \{\overline{x}, \overline{y}, \overline{x\wedge y}, \overline{x\vee y}\}$ and we have the following covering relations (which uniquely define the lattice structure): $\overline{x} \leqslant \overline{x\vee y}, \overline{y} \leqslant \overline{x\vee y}, \overline{x\wedge y} \leqslant \overline{x}, \overline{x\wedge y} \leqslant \overline{y}$. Moreover, this lattice is a Boolean algebra with atoms $\overline{x}, \overline{y}$ and hence is isomorphic to the set of all binary vectors of length 2 with bitwise AND and OR operations, which is a better known object. Alternatively, it is isomoprhic to the set of all term functions on $L$ of two arguments.
Hope this helps!