How to show that an algebra is contained in a clone My question:
If we have a be the binary operation on 2 = {0, 1}, denoted $\overline{\land}$, defined by
\begin{array}{|c|c|c|}
\hline
\overline{\land} &0 & 1  \\ \hline
 0&1 & 0\\ 
\hline
1 &  0 &0 \\ 
\hline
\end{array}
How do I prove that $\mathcal{A}$ = {¬, ∧, ∨} ⊆ $\mathcal{C}$ = Clo($(2, \overline{\land})$)?
My thoughts:
The $\mathcal{A}$ is probably some lattice, however, I am confused about what should be the universe here. I understand the operations {¬, ∧, ∨} as inverse, meet and join respectively.
The $\mathcal{C}$ is the clone of term operations. The clone $\mathcal{C}$ on $(2, \overline{\land})$ should be defined a set of operations on $(2, \overline{\land})$ which contains all projections and is closed under
composition of functions.
So I think maybe I should just show that any two elements that give 0 or 1 after applying the operations from $\mathcal{A}$ have to automatically give 0 or 1 after applying the operation $\overline{\land}$? But I am not very sure how would I show that if this was the correct process.
Why I am interested in this:
We were assigned this as an exercise in our university course and I am following mostly the book Universal Algebra: Fundamentals and Selected Topics by Clifford Bergman. In the book, there is only similar exercise (4.10.4 in particular), but no solutions.
I would love to learn to work with clones, not only study theorems and proofs, so any advice on this or any learning sources with more exercises are very appreciated.
 A: 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 $\mathcal{C} = Clo((2, \overline{\land}))$.
