Why isn't addition defined by $a+0=a$ and $a+succ(b)=succ(a)+b$? The rule for addition is stated in the Peano axioms as:
$a + 0 = a$
$a + succ(b) = succ(a + b)$
But why couldn't we define it as
$a + 0 = a$
$a + succ(b) = succ(a) + b$
for example $3 + 4 = 4 + 3 = 5 + 2 = 6 + 1 = 7 + 0 = 7$
What makes the usual definition the preferred one?
 A: Both definitions work fine. The usual definition has the advantage of fitting the "general form" of recursive definitions: when recursively defining a function $f$, we give a method for defining a "new value" $f(x_1,...,x_n)$ in terms of the "already defined" values $$\{f(y_1,...,y_n): y_1\le x_1,..., y_n\le x_n\mbox{ and some $1\le i\le n$ has }y_i<x_i\}.$$
The usual definition does this: it rewrites $Plus(a, Succ(b))$ in terms of $$Plus(a,b).$$ Your definition, however, rewrites $Plus(a, Succ(b))$ in terms of $$Plus(Succ(a),b).$$ It still works, of course, but it doesn't "strictly simplify" things: the ordered pair $(Succ(a), b)$ isn't really "smaller" than the ordered pair $(a, Succ(b))$.

Of course, this just pushes the question back one step: why is this approach preferred for general recursive definitions?
Well, the point is that reducing one input to zero doesn't generally trivialize things. In the specific case of addition we're comfortable asserting $Plus(a,0)=a$ for all values of $a$ right at the outset. But a general binary function $F$ might be quite complicated even when one of the coordinates is zero. In such a case, the approach you describe won't obviously help - by contrast, the approach described above always works since the relevant partial order on $n$-tuples is well-founded.
