You know that the binary function $\diamond$ takes in two non-negative integers and has the following properties:

\begin{align*}0\diamond a&=1\\ a\diamond a&=0\end{align*} 
$\text{If } a<b, \text{ then } a\diamond b\&=(b-a)[(a-1)\diamond (b-1)].$

Find a general formula for $x\diamond y$, assuming that $y\gex>0$.
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 \( \boxed{ (y - x)^x } \).