To find a general formula for $x \diamond y$ given the properties of the binary function $\diamond$, we will use the provided properties and recursive definition.

1. Base Cases:
$$0 \diamond a = 1 \quad \text{for any non-negative integer } a$$
$$a \diamond a = 0 \quad \text{for any non-negative integer } a$$

2. Recursive Case:
For $a < b$:
$$a \diamond b = (b - a) \left[ (a - 1) \diamond (b - 1) \right]$$

3. General Case:
We need to find a general formula for $x \diamond y$ assuming $y \ge x > 0$.

4. Recursive Expansion:
Let's expand the recursive definition step-by-step:
$$x \diamond y = (y - x) \left[ (x - 1) \diamond (y - 1) \right]$$
Applying the recursive definition again:
$$(x - 1) \diamond (y - 1) = (y - 1 - (x - 1)) \left[ (x - 2) \diamond (y - 2) \right]$$
Simplifying:
$$(x - 1) \diamond (y - 1) = (y - x) \left[ (x - 2) \diamond (y - 2) \right]$$
Substituting back:
$$x \diamond y = (y - x) \left[ (y - x) \left[ (x - 2) \diamond (y - 2) \right] \right]$$
Continuing this process, we see a pattern emerging:
$$x \diamond y = (y - x)^{n} \left[ (x - n) \diamond (y - n) \right]$$
where $n$ is the number of recursive steps taken.

5. Termination of Recursion:
The recursion terminates when $x - n = 0$:
$$x - n = 0 \implies n = x$$
At this point:
$$0 \diamond (y - x) = 1$$
Therefore:
$$x \diamond y = (y - x)^{x} \cdot 1 = (y - x)^{x}$$

Conclusion:
The general formula for $x \diamond y$ when $y \ge x > 0$ is:
$$x \diamond y = (y - x)^x$$

The final answer is $\boxed{ (y - x)^x }$.