Axiomatic set theory: definitions. In his book axiomatic set theory, Supped writes:

An equivalence P introducing a new $n$-place operation symbol O is a proper definition if and only if P is of the form $O(v_1, v_2, v_3, \dots, v_n) = w\iff Q$

Where $v_1,v_2,v_3,\dots,v_n$ are distinct variables, $Q$ has no free variables other than $v_1,\dots,v_n, w, Q$ is a formula in which the only non-logical constants are primitive or previously defined, and the formula "there exists exactly one $w$ such that $Q$" is derivable from the axioms.
The questions are why must $v_1,\dots,v_n$ be distinct? Surely that stops us from defining a symbol such as the multiplication sign which still works for say $3*3$? And why can't there be more free variables in $Q$? Any help would be appreciated.
 A: Questioning definitions is usually a fruitless activity, but I think this question is nevertheless worth thinking about. I'll treat them in order.
$$\fbox{1}$$
The requirement that $v_1,...,v_n$ be distinct doesn't mean that they cannot take on identical values. The multiplication operator can be defined, for example, as follows:
$$\times(a,b) = \begin{cases}0 & b = 0, \\ a+\times(a, b-1) & \text{otherwise}. \end{cases}$$
We're assuming, of course, that the addition operator is given. Our definition meets the criterion of having distinct formal parameters (viz. $a$ and $b$), but it can still successfully be applied to identical pairs of arguments. For example:
$$\begin{align}\times(3,3) & = 3 +\times(3,2) \\ & = 3 + 3 + \times(3,1) \\ & = 3 + 3 + 3 +\times(3, 0) \\ & = 3 + 3 + 3 + 0 \\ & = 9.\end{align}$$
The distinctness condition is imposed because in classical logic it's assumed that identical variables have identical values, so if we allowed definitions to contain non-distinct formal parameters, we could define a binary operator like the following:
$$\sf{weird}_1(a,a) ~=~ a.$$
To see what's weird about this operator, consider the following application:
$$\sf{weird}_1(3, 4) ~=~?$$
What's the value of $\sf{weird}_1(3, 4)$? Is it $3$ or $4$? You get the idea..
$$\fbox{2}$$
Suppose we allowed other free variables to occur in "proper definitions". Then the following would constitute a proper definition of operator $\sf{weird}_2$:
$$\sf{weird}_2(a) ~=~ z$$
Variable $z$ is free in that formula. To see what's wrong with that definition, try to evaluate:
$$\sf{weird}_2(3) ~=~?$$
The answer, of course, is whatever the value of $z$ is. But the problem is that the value of $z$ varies depending on the context of evaluation (or environment), so the definition above fails to define a particular operator. It defines, if you will, a class of operators, one for each member of the domain $z$ is allowed to range over.
