$n$-ary operations written in terms of binary operations

Question

I found the following result and decided to try proving it.

Let $A$ be a set and $n \geq 2$ a natural number. Show that all $n$-ary operations (i.e. functions defined on $A^n$ with values in $A$, or elements of $A^{A^n}$) can be expressed by finite compositions of binary operations.

Let's first analyze a simpler case where $A$ has two elements and $n=2$. WLOG, assume $A=2=\{0, 1\}$. This can be interpreted as the fact that all logical operators can be expressed in terms of negation and conjunction. I will show my thought process in this context.

Assume there is some logical operator f, for which we write the truth table

\begin{array} {|r|r|}\hline x_1 & x_2 & f \\ \hline 0 & 0 & a \\ \hline 0 & 1 & b \\ \hline 1 & 0 & c \\ \hline 1 & 1 & d \\ \hline \end{array}

where $a, b, c, d$ are elements of $A$.

My first approach was to think of this formula $$( (\neg x_1 \land \neg x_2) \to a) \land ( (\neg x_1 \land x_2) \to b) \land ( (x_1 \land \neg x_2) \to c) \land ( (x_1 \land x_2) \to d)$$

where $\land$ is the logical and, $\neg$ is the logical negation and $\to$ is the logical implication. If either $a, b, c$ or $d$ is $0$ substitute it for $x_1 \land \neg x_2$ and $x_1 \lor \neg x_2$ if they are $1$.

Basically, I form $4$ units that have $1$ in the output except one place where there is $a, b, c$ or $d$. Then we unify them via conjunction, because $1 \land a \iff a$.

We could also define units that have $0$ in the output except one place which is some value defined by $f$ and then disjunct them.

We can interpret the logical operators as functions, like $a \land b$ is just $\land(a, b)$.

Therefore $f(x_1, x_2)$ is just $$ (\neg x_1 \land \neg x_2 \land a) \lor (\neg x_1 \land x_2 \land b) \lor (x_1 \land \neg x_2 \land c) \lor (x_1 \land x_2 \land d)$$ or $$\lor(\lor(\lor( \land( \land(\neg(x_1), \neg(x_2) ), a) , \land( \land(\neg(x_1), x_2), ) b), \land( \land(x_1, \neg(x_2)), c) ), \land( \land(x_1, x_2), d) )$$

For the sake of simplicity, I will write $a \lor b$ instead of $\lor(a, b)$ and $\bigvee_{b \in B}(b)$ mean the disjunction of all the elements in $B$. Analogously, $\bigwedge_{b \in B}(b)$.

I will also introduce a new function, namely $\leftrightarrow$, where $\leftrightarrow(x, y)$ or $x \leftrightarrow y$ is $1$ if and only if $x=y$, otherwise $0$.

We can rewrite $f(x_1, x_2)$ as $$( (0 \leftrightarrow x_1) \land (0 \leftrightarrow x_2) \land a) \lor ((0 \leftrightarrow x_1) \land (1 \leftrightarrow x_2) \land b) \lor ((1 \leftrightarrow x_1) \land (0 \leftrightarrow x_2) \land c) \lor ((1 \leftrightarrow x_1) \land (1 \leftrightarrow x_2) \land d)$$

or condensed

$$ \bigvee_{(a_1, a_2) \in A^2}( \bigwedge_{i=1}^n(a_i \leftrightarrow x_i) \land f(a_1, a_2) ) $$.

However, this generalizes nicely even when $n$ is bigger than $2$. We can check by induction that this formula works for any natural $n$.

Now we want to write all of $x \land y$ and $\neg x$ and $x \lor y$ and $x \leftrightarrow y$ in terms of one function. Let's call assume that such function exists and call it $g$.

Now $g(0, 0)$ has to be $1$. If it were $0$, we couldn't have composed $g$ with itself to get some function such that it maps $(0, 0)$ to $1$. Analogously, $g(1, 1)=0$. Now $g(0, 1)$ has to be equal to $g(1, 0)$, otherwise $g(x, y)$ would be equivalent to $\neg x$ or $\neg y$, a one variable function, which can't get us anywhere.
And indeed, the two functions,
$g_1(x, y)=0$ when $(x, y)=1$ and $1$ otherwise
and
$g_2(x, y)=1$ when $(x, y)=0$ and $0$ otherwise
can be used to create $\lor, \land$ and $\leftrightarrow$.

Assume $g(x, y) = \neg (x \land y)$. Then $x \land y = g(g(x, x), g(y, y))$, $\neg(x) = g(x, x)$, $\lor(x, y) = \neg (x \land y)$ and $x \leftrightarrow y = (x \land y) \lor (\neg x \land \neg y)$, so everything works, for $A=\{0, 1 \}$

Now, assume $A$ is some finite set. Again, WLOG, assume $A = \{ 0, 1, 2, ..., m-1\}$. Now let's define the following functions: $\bullet$ $x \leftrightarrow y = \leftrightarrow(x, y)$ is $1$ only when $x=y$ and $0$ othwerwise.
$\bullet$ $x \land y = \land(x, y)$ has to respect the following rules $0 \land a = 0$ and $1 \land a = a$ and values such as $2 \land 2$ can be fixed arbitrarily.
$\bullet$ $x \lor y = \lor(x, y)$ has to respect the following rules $0 \lor a = a$ and $1 \lor a = 1$ and values such as $2 \lor 2$ can be fixed arbitrarily.

By the same argument, an arbitrary function $f$ defined on $A^n$ taking values in $A$ can be expressed as

$$f(x_1, x_2, ..., x_n) = \bigvee_{(a_1, a_2, ..., a_n) \in A^n}(\bigwedge_{i=1}^n(a_i \leftrightarrow x_i) \land f(a_1, a_2, ..., a_n))$$

Now my intuition says that there is some binary function $g$ such that $\land, \lor$ and $\leftrightarrow$ can all be expressed in terms of $g$, however I did not seem to find an algorithm to restrict those functions, like I could when $A$ had $2$ elements.

Regarding the work I have done above I have a few questions:

$\bullet$ The first question:
How can I phrase my original question "Let $A$ be a set and $n \geq 2$ a natural number. Show that all $n$-ary operations can be expressed by finite compositions of binary operations." in mathematical terms. Like I just proved for the case that $A$ is finite that any $n$-ary function can be written in terms of finite compositions, but how do I express that in rigurous mathematical terms (I think the words I am looking for are first order logic in set theory).
Firstly, let's define $\text{function}(f, A, B) \iff ( f \subseteq A \times B \land \forall x \in A \colon \exists ! y \in B \colon (x, y) \in f )$.
Now, how do i say "any $n$-ary function can be written in terms of finite compositions"?I think i should write something like $\forall f (\text{function}(f, A^n, A) \implies \exists ...)$ I don't know what to write after the existential quantifier.

$\bullet$ The second question:
Consider $A=\{ 0, 1, ..., m-1 \}$ As I have stated in my proof above, how can I find all the functions $g$ such that $\lor, \land$ and $\leftrightarrow$ can be written only in terms of $g$. I haven't found any good algorithm so the only option I think it is left is that I have to try out almost all functions from $A^2$ to $A$.

$\bullet$ The third question:
The formula $$f(x_1, x_2, ..., x_n) = \bigvee_{(a_1, a_2, ..., a_n) \in A^n}(\bigwedge_{i=1}^n(a_i \leftrightarrow x_i) \land f(a_1, a_2, ..., a_n)) $$ that I found looks extremly closely related to Lagrange interpolation polynomials. Is there any connection/intuition between them? (I think the answer is yes, I can answer myself like 70%, but I just wanna share the beauty of math and the connection of multiple areas of mathematics).

$\bullet$ The fourth question:
How can I go about proving this in the case that $A$ is not finite. The same method does not work since I get an inifnite "composition" of functions. I am also wondering if $1$ binary function would be sufficient to express all the $n$-ary functions or we would need more binary functions. I assume I need to use the fact that if $A$ is an infinite cardinal then $A^n=A$. How can I prove my problem when $A$ is infinite.

$\bullet$ The fifth question:
Is this the reason why in abstract algebra we don't talk about ternary or $n$-ary operations on sets? Because we can reduce them to binary operations?

$\bullet$ The last question:
Can I draw some conclusion from the problem if instead of letting $n$ be a natural number we let $n$ be any ordinal? I do not have any ideea on how to even approach such a problem, since I didn't study $\infty$-ary operations before.

I am not necessarily looking for a proof for what I have written, but for the answers with some resources to find the proofs to my questions, although proofs written here would be gladly appreciated and would save me a lot of time.

Thank you!

Edit: I have looked this concept up and found out it is called currying, but I still didn't find any answer to my questions.

Answer 1

Q1: Your phrasing of the question is fine. Writing down something more detailed is pretty annoying because it's pretty annoying to write down explicitly the general form of composition of $n$-ary operations, and it's not necessary. "Show that every $n$-ary operation is a composition of a finite number of binary operations" is already perfectly clear.

Q2: I think your idea is a little messy although maybe it can be made to work. It's cleaner to argue by induction on $n$. The base case is $n = 2$ which is true by definition. In general if $n \ge 3$ and $f : A^n \to A$ is an $n$-ary operation then, writing $A = \{ 0, 1, \dots m-1 \}$ WLOG as you've done, $f$ is determined by the behavior of the $m$ functions $f_i(a_2, \dots a_n) = f(i, a_2, \dots a_n)$; it can just be written as

$$f(a_1, a_2, \dots a_n) = f_{a_1}(a_2, \dots a_n).$$

The point of doing this is that by the inductive hypothesis each of these $m$ functions are $(n-1)$-ary operations so they are each expressible as a composition of a finite number of binary operations. So now it suffices to show that we can assemble these functions to get $f$ using some more binary operations. There are probably lots of ways to do this but the first one that comes to mind is to write

$$f(a_1, a_2, \dots a_n) = \sum_{a \in A} [a_1 = a] f_a(a_2, \dots a_n)$$

where $[a_1 = a]$ is the Iverson bracket, which is equal to $1$ if $a_1 = a$ and $0$ otherwise. Technically this is a unary operation $A \to A$ (here we are thinking of $0$ and $1$ as being elements of $A$) but we can arbitrarily think of these as binary operations by picking any other $a_j$ for the operation to depend on and then ignoring it. So in other words it's harmless to toss in unary operators. The sum above is addition $\bmod m$ regarded as a repeated application of a binary operation $A^2 \to A$.

This is similar to what you've done but we only need to use induction and then a single sum over $A$.

Q3: Yes, it's related to the idea behind Lagrange interpolation. In Lagrange interpolation we're looking for polynomials which play the role of the Iverson brackets, so they're equal to $1$ at a specific point and $0$ otherwise.

Q4: Your idea in the infinite case is correct. Starting from a bijection $p : A^2 \to A$, repeated application of $p$ allows us to write down a bijection between $n$-ary operations $f : A^n \to A$ and binary operations $A^2 \to A$, by repeatedly pairing off copies of $A$ in $A^n$ until only two are left. Then $f$ can be expressed as the composite of the binary operation $A^2 \to A$ you get when you do this with a bunch of copies of $p$.

Q5: No. In abstract algebra we typically don't talk about $n$-ary operations because all the familiar examples are associative, so reduce to binary operations in a particularly simple way. In more complex parts of mathematics there are interesting $n$-ary operations that aren't associative, for example the $n$-ary operations used to define an $A_{\infty}$-algebra. These don't reduce to binary operations in any meaningful way I'm aware of.

Q6: $n$ can be replaced with an infinite set, there's no need for it to be well-ordered. But if you write down a finite composition of binary operations on $A^n$ for $n$ infinite then you can only use finitely many of the copies of $A$ in the domain, so you're unable to express any functions that depend on infinitely many of the copies.