System of equations for operations Given a system with multiple equations, where we know the values and the result, but not the operations between the values:
\begin{cases} 3 ⊕ 5 ⊙ 2 = 13 \\ 7 ⊕ 2 ⊙ 4 = 10 \\ 4 ⊕ 3 ⊙ 3 = 9 \end{cases}
Is there an algorithmic way to deduce $⊕$ and $⊙$ (which in this case would be multiplication and subtraction, respectively)? Like a system of equations where the unknowns are the operations themselves? Does this have a name? Are there ways of calculating it?
 A: There is, but it is complicated, and you would need to know more. You would have to know what type of operations $⊕$ and $⊙$ are. Let's start simple. If we were told that $x⊙y$ was of form $ax+by$ (which is is; ⊙ represents subtraction and we would just plug in 1 for a and -1 for b), we could solve a system of equations for $a$ and $b$. Let's say we were given that
$$7⊙2=5$$
$$6⊙4=2$$
$$9⊙1=8$$
Since we are only solving for two variables, we only need two equations. Let's pick $7⊙2=5$ and $6⊙4=2$.
We have that
$$7a+2b=5$$
$$6a+4b=2$$
I am going to solve this system without substitution, because there is an easier method. We start by multiplying the second equation by $-\frac{1}{2}$, and we get $-3a-2b=-1$.
Now let's add the equations:
$$7a+2b=5$$
$$\underline{-3a-2b=-1}$$
$$4a=4$$
From here, we can see that $a=1$. Plugging this in to any equation, let's take the first, gives us $7+2b=5$, and from here we can solve for $b$ to get that $b=-1$.
Thus $a⊙b=(1)a+(-1)b=a+(-b)=a-b$, so we have that $a⊙b=a-b$. Now we know that $⊙$ represents subtraction.  
Now let's say we wanted to solve for $⊕$. Since this is multiplication, we would be told that $x⊕y=pxy$, from which we could solve for $p$ (which would be $1$). Or, we could have a definition like $x⊕y=pxy+qx+ry$, where $p$ would be $1$, and $q$ and $r$ would each be $0$. We could even have $x⊕y=pxy+qx+ry+s$, and we would have to solve for all of these (we would need more equations).  
Now, bringing in the two operations, we have
$$3⊕5⊙2=13$$
$$7⊕2⊙4=10$$
$$4⊕3⊙3=9$$
And thus
$$(p(3*5))⊙2=13$$
$$(p(2*7))⊙4=10$$
$$(p(4*3))⊙3=9$$
Going further,
$$15pa+2b=13$$
$$14pa+4b=10$$
$$12pa+3b=9$$
And we would have to solve for $a$, $b$, and $p$. We would get that they are $1, -1,$ and $1$, respectively, and then we could determine the operations.  
Adding in exponents, and things would get crazy. You'd be probably better off with trial and error.
