The question is asking for the free algebra over $X=\{x,y\}$ in the variety $\sf V = {\sf H}{\sf S}{\sf P}(\mathcal A)$. This free algebra has 130 elements, so I doubt you are expected to actually construct it. Rather, it seems likely that they only want you to explain the procedure for constructing it.
One standard construction for ${\mathcal F}_{\sf V}(\{x,y\})$ is as a subalgebra of the algebra ${\mathcal A}^{{\mathcal A}^2}$. The $\underline{\textrm{set}}$ $A^{A^2}$ consists of all functions $f\colon A^2\to A$. We may regard $x$ and $y$ as elements of this set of functions by identifying $x$ with binary first-coordinate projection ($x(u,v) = u$) and $y$ with binary second-coordinate projection ($y(u,v)=v$). The $\underline{\textrm{algebra}}$ ${\mathcal A}^{{\mathcal A}^2}$ is this set of functions under pointwise operations from $\mathcal A$. One construction of ${\mathcal F}_{\sf V}(\{x,y\})$ is as the subalgebra of ${\mathcal A}^{{\mathcal A}^2}$ generated by the subset $\{x,y\}$.
More concretely, order the elements of $A^2$ lexicographically as
$$\langle (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2)\rangle.$$
These 9 elements can be taken as indices of coordinates of tuples for elements of $A^{A^2}$. With this representation, the function table for first projection $x$ may be represented as $[x]:=\langle 0,0,0,1,1,1,2,2,2\rangle\in A^9$, and the function table for the second projection $y$ may be represented as $[y]:=\langle 0,1,2,0,1,2,0,1,2\rangle\in A^9$. To get ${\mathcal F}_{\sf V}(\{x,y\})$, generate a subalgebra of ${\mathcal A}^{9}$ with $[x], [y]$.
This means, start with $[x]=\langle 0,0,0,1,1,1,2,2,2\rangle, [y]=\langle 0,1,2,0,1,2,0,1,2\rangle\in A^9$ and close under coordinatewise multiplication.
EDIT 1/8/22.
In response to a comment, I will explain why the $2$-generated free algebra in the variety generated by ${\mathcal A}$
has size $130$. Moreover I will explain how to
$*$-multiply any two elements of this $130$-element set.
In order to avoid writing too many bracketed variables, I will use the letter $a$ to denote $[x]$ from above, and the letter $b$ to denote $[y]$ from above.
Observe that
- any product of elements of $A$ lies in the subset $B = \{0,1\}$, and
- the product operation of $A$ restricted to $B=\{0,1\}$
is addition modulo $2$.
In particular, $B=\{0,1\}$
is the underlying set of a subalgebra of $\mathcal A$
and the algebra structure on $B$ is
equivalent to that of a $2$-element group,
or equivalent to that of a $1$-dimensional vector space
over $\mathbb F_2$.
These facts mean that the subalgebra $\mathcal F$
of ${\mathcal A}^9$
that is generated by
$a = \langle 0,0,0,1,1,1,2,2,2\rangle$
and $b=\langle 0,1,2,0,1,2,0,1,2\rangle$ has
the property that the product of any two
elements lies in $B^9 = \{0,1\}^9$, which is a
$9$-dimensional vector space over $\mathbb F_2$
under the $*$-operation. Hence the underlying set
of $\mathcal F$ must have the form $\{a,b\}\cup V$
for some $\mathbb F_2$-subspace $V\leq B^9$.
I claim that this $V$ is a $7$-dimensional subspace,
and I will tell you which one it is.
Write a typical element of $A^9$ as
$z = \langle z_1,z_2,z_3,z_4,z_5,z_6,z_7,z_8,z_9\rangle$.
Observe that if $z$ is one of the generators of $\mathcal F$,
$a = \langle 0,0,0,1,1,1,2,2,2\rangle$
or $b=\langle 0,1,2,0,1,2,0,1,2\rangle$, the
following hold:
- $z_1, z_2, z_4, z_5\in B$, and the following
linear relations hold among these coordinates:
- $z_1 = 0$, and
- $z_2+z_4+z_5 = 0$.
The relations are preserved under the $*$-operation
acting coordinatewise. (Reason:
the generators have coordinates $z_1, z_2, z_4, z_5\in B$,
so the projection of $\mathcal F$ onto these coordinates is a
vector subspace of $B^4$ with respect to $*$ and this vector space
is generated by vectors $\langle a_1,a_2,a_4,a_5\rangle = \langle 0, 0, 1, 1\rangle$ and $\langle b_1,b_2,b_4,b_5\rangle = \langle 0, 1, 0, 1\rangle$, which satisfy these linear relations.)
This implies that the underlying set
of $\mathcal F$ is $\{a, b\}\cup V$ where in every $z\in V$
we have $z_1=0$ and $z_2+z_4+z_5=0$. The two linear relations
guarantee that $V$ has codimension at least $2$
in the $9$-dimensional space $B^9$, so $V$ is at most
$7$-dimensional. But $V$ is exactly $7$-dimensional,
since I can exhibit $7$ independent vectors in $V$, namely:
- $a((ba)b) = \langle 0,0,1,0,0,0,0,0,0\rangle$
- $a(b(ab)) = \langle 0,0,0,0,0,1,0,0,0\rangle$
- $b((ab)a) = \langle 0,0,0,0,0,0,1,0,0\rangle$
- $b(a(ab)) = \langle 0,0,0,0,0,0,0,1,0\rangle$
- $(aa)(b(b(aa))) = \langle 0,0,0,0,0,0,0,0,1\rangle$
- $a(a(ba)) = \langle 0,1,0,1,0,0,0,0,0\rangle$
- $a((ab)(bb)) = \langle 0,1,0,0,1,0,0,0,0\rangle$
(In the products on the left of the equal signs I am abbreviating
$*$-products $p*q$ by $pq$.)
Altogether, this shows that the underlying set of $\mathcal F$
is $\{a, b\}\cup V$ where $V\leq B^9$ is the $7$-dimensional subspace
of all $z = \langle z_1,z_2,z_3,z_4,z_5,z_6,z_7,z_8,z_9\rangle$
satisfying $z_1=0$ and $z_2+z_4+z_5=0$.
This allows us to determine the size of $\mathcal F$:
it is $2+|V|=2+2^7 = 130$.
We can do more than determine the size of this algebra, we can even determine
how the $*$-operation interprets on the set $\{a,b\}\cup V$.
One may verify that:
- $aa = \langle 0,0,0,0,0,0,1,1,1\rangle\in V$
- $ab = \langle 0,1,1,1,0,0,0,0,1\rangle\in V$
- $ba = \langle 0,1,0,1,0,0,1,0,1\rangle\in V$
- $bb = \langle 0,0,1,0,0,1,0,0,1\rangle\in V$
This explains how to $*$-multiply one element of $\{a,b\}$
times another. One may also verify that if $z\in B^9$ then
- $az = \langle z_1,z_2,z_3,z_4',z_5',z_6',0,0,0\rangle\in V$
- $za = \langle z_1,z_2,z_3,z_4',z_5',z_6',z_7'z_8'z_9'\rangle\in V$
- $bz = \langle z_1,z_2',0,z_4,z_5',0,z_7,z_8',0\rangle\in V$
- $zb = \langle z_1,z_2',z_3'z_4,z_5',z_6',z_7,z_8',z_9'\rangle\in V$
Here I am writing primes on variables to mean: swap
$0$ and $1$ (that is, $0'=1, 1'=0$).
These four rules explain how to $*$-multiply an element of $\{a,b\}$
times an element $z\in V$. Finally, we already know how to $*$-multiply
two elements $z, w\in V$ since $*$ is addition modulo $2$ coordinatewise on $V$.
Altogether this identifies the $130$ tuples comprised by the subalgebra of ${\mathcal A}^9$ generated by $\{a,b\}$, and how the $*$-algebra acts on these $130$ tuples. Usually it is hard to give this much detail without the use of a computer, but we were lucky here that ${\mathcal A}$ is close
to being a $1$-dimensional $\mathbb F_2$-vector space.
Best Answer
Since I got some nice advice in the comments (thank you!), I will try to post an answer myself. Feel free to add anything.
A clone $\mathcal{C}$ is a set of operations on $(2, \overline{\land})$ which contains all projections and is closed under composition of functions.
Denote $\mathcal{A} = (\{0, 1\}, \overline{\land})$.
The unary operation $\neg$ ("not"), and the binary operations $\land$ ("and"), $\lor$ ("or") are operations defined on $\{0,1 \}$. We need to know how the operations $\neg, \land$ and $\lor$ behave to solve the problem. We create Cayley tables to observe how the operations can be generated from one another.
We will show that that all $\neg, \land$ and $\lor$ can be created by projections and compositions of the $\overline{\land}$ operator.
Proof that $\neg \in \mathcal{C}$
The $\neg$ operator is unary operator and therefore the Cayley table has outputs only for $(x,x)$.
$\begin{array}{|c|c|c|} \hline \neg &0 & 1 \\ \hline 0& 1 & -\\ \hline 1 & - & 0 \\ \hline \end{array}$
Observe that $\neg(x,x)$ is the same $x \overline{\land} x$. The $\neg(x,x)$ operator is like the $\overline{\land}$ operator restricted only to one variable instead of two. Hence, $\neg(x,x)$ is a part of the $\mathcal{C} = Clo((2, \overline{\land}))$
Proof that $\land \in \mathcal{C}$
$\begin{array}{|c|c|c|} \hline \land &0 & 1 \\ \hline 0& 0 & 0\\ \hline 1 & 0 &1 \\ \hline \end{array}$
First, I will generate $\land$ from $\lor$ and $\neg$. Then I will express $\lor$ and $\neg$ using $\overline{\land}$ and we are done.
Observe $\neg x \lor \neg y$.
$\begin{array}{|c|c|c|} \hline \neg x \lor \neg y &0 & 1 \\ \hline 0& 1 & 1\\ \hline 1 & 1 &0 \\ \hline \end{array}$
Then, observe that the table above is just negation of the $\land$ operator. Therefore $\land = \neg (\neg x \lor \neg y)$.
$\begin{array}{|c|c|c|} \hline \neg (\neg x \lor \neg y) &0 & 1 \\ \hline 0& 0 & 0\\ \hline 1 & 0 &1 \\ \hline \end{array}$
Since we have already shown that the $\lor$ and $\neg$ operators are in $Clo((2, \overline{\land}))$, it means $\land$ is also in the clone, because it can be created from these operators and therefore from the $\overline{\land}$ operator.
Proof that $\lor \in \mathcal{C}$
$\begin{array}{|c|c|c|} \hline \lor &0 & 1 \\ \hline 0& 0 & 1\\ \hline 1 & 1 &1 \\ \hline \end{array}$
Observe that $x \lor y = \neg(x \overline{\land} y)$ and hence equal to $(x \overline{\land} y) \overline{\land} (x \overline{\land} y)$. It means that it can be made as a composition of $ \overline{\land}$ operators, so it belongs to $\mathcal{C} = Clo((2, \overline{\land}))$.