This is an example from [A Course in Universal Algebra][1]. Sorry, I can't copy all necessary definitions here; there are a lot of them.
Let $\mathscr{F}$ be a type of algebras and $K$ be a class of algebras of type $\mathscr{F}$. $\mathbf{T}(X)$ denotes the term algebra of type $\mathscr{F}$ over $X$. $\mathbf{F}_K(\overline{X})$ denotes the $K$-free algebra of type $\mathscr{F}$ over $\overline{X}$.
In general $\mathbf{F}_K(\overline{X})$ is not isomorphic to a member of $K$ (for example, let $K=\{\mathbf{L}\}$ where $\mathbf{L}$ is a two-element lattice; then $\mathbf{F}_K(\overline{x}, \overline{y})\not\in I(K)$). However $\mathbf{F}_K(\overline{X})$ can be embedded in a product of members of $K$.
Page 68. I suppose that it means that $\mathbf{F}_K(\overline{X})$ is not isomorphic for any set $X$. $\mathbf{F}_K(\overline{X})$ was defined by $$\mathbf{F}_K(\overline{X}) = \mathbf{T}(X)/\theta_K(X),$$ $$\theta_K(X) = \bigcap\Phi_K(X),$$ $$\Phi_K(X) = \{\phi\in\operatorname{Con}\mathbf{T}(X) : \mathbf{T}(X)/\phi\in IS(K)\}.$$
Let $X=\{0, 1\}$. An obvious member of $\Phi_K(X)$ is $\ker\nu$ where $\nu$ is the homomorphism given by the universal mapping property of $\mathbf{T}(X)$ (Theorem 10.8). If it was the only member of $\Phi_K(X)$, then $\mathbf{F}_K(\overline{X})$ would be isomorphic to $\mathbf{L}$. I can't come up with another member that will make $\theta_K(X)$ smaller. That member $\phi$ must induce exactly 2 equivalence classes. If more, $\mathbf{T}(X)/\phi\in IS(K)$ would not hold.
Update. Can I choose $\phi$ in the following way? $\phi$ induces 2 equivalence classes, let's call them Class 0 and Class 1, such that Class 0 consists of variables, and Class 1 of all other terms. (In my case, there are exactly 2 variables: Variable 0 and Variable 1.) As the meet and join of $\mathbf{T}(X)$ produce a term that is not a variable, $\phi$ is a congruence.
Update 2 2018-03-17 11:05:23+00:00. With such $\phi$, $\mathbf{T}(X)/\phi$ would not be isomorphic to a subalgebra of $\mathbf{L}$, so $\phi\not\in\Phi_K(X)$. I still do not understand what is $\theta_K(X)$ and what can $\phi$ be.
[1]: Burris, Stanley, and H.P. Sankappanavar. A Course in Universal Algebra. 2012 ed. New York: Springer-Verlag, 1981. Web. Stan's Home Page. Graduate Texts in Mathematics 78. 10 Mar. 2018. <https://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra2012.pdf >
The free lattice on a two element set is the four element lattice that is not a chain, that is, $\{x\wedge y,x,y,x\vee y\}$, with $x$ and $y$ incomparable.
Let us call $\mathbf M_2$ to this lattice.
To see this, let $f:\{x,y\}\to L$, where $\mathbf L$ is the two element lattice (the only member of $K$).
Then $\varphi:\mathbf M_2 \to \mathbf L$, defined by $$x \wedge y \mapsto f(x) \wedge f(y), \quad x \mapsto f(x), \quad y \mapsto f(y), \quad x \vee y \mapsto f(x) \vee f(y),$$ is a homomorphism (trivially), and it extends $f$, so $\mathbf M_2$ has the Universal Mapping Property for $K$ over $\{x,y\}$, and thus it is the free lattice in $K$ over $\{x,y\}$.
Clearly, $\mathbf M_2 \notin K$.