I'm guessing two variables here; I didn't look too closely.
x=3, y=8
x*(y*(x/x)) = f(x,y)
x+((x*y)-x) = g(x,y)
These are pretty obviously both xy, but rather than solve it, find the maximum degree, which is 2, and make a few points:
(1,1) => 1(1(1/1)) = 1 = 1+((1*1)-1)
(1,0) => 0 for both
(1,2) => 2 etc.
It's an old technique; Alan Perlis taught it to me. If two polynomials of degree d agree on more than d points, then their difference (of degree <= d) has more than d roots, so their difference must be 0.
But your last sentence is only true among single-variable polynomials, isn't it? For example, f(x,y)=xy and g(x,y)=x^2 agree on infinitely many points.
Pick one value for x, now you have a single variable polynomial. Verify that it is equal with above mentioned technique.
Pick another value for x, do the same.
Pick a third...
I didn't prove this works, but I think it should. It is however exponential in the number of variables which may be seen as a downside.
Edit:
This can be proven by induction. Let V = (y, z, ...). Then the polynomial can be expressed as
p = (x-x2)(x-x3)f(V) + (x-x1)(x-3)g(V) + (x-x1)(x-x2)h(V), where x1, x2, x3 are the values we test for x. By inductive hypothesis we can prove each of f, g and h to be zero by testing. But if those are all zero then p is zero.
It seems to me that expanding everything to sum-of-products form and checking for equality directly would be simpler than your algorithm and would run about as fast.
Nice! Maybe rational functions rather than polynomials though, because division is allowed. Anyway, equality of rational functions is reducible to equality of polynomials.
