Let us analyze the properties:

i. $(a+b) \Delta b = a \Delta b + 1$

ii. $(a \Delta b) \cdot (b \Delta a) = 0$

From i., for fixed $b$, the function $f(a) = a \Delta b$ satisfies $f(a+1) = f(a) + 1$ for $a \geq 1$. This means $f(a)$ is an affine function in $a$:

Let $f(a) = a \Delta b$. Then $f(a+1) = f(a) + 1$ implies $f(a) = a + c$ for some constant $c$ depending on $b$.

But let's check the initial value. For $a = 1$, $f(1) = 1 \Delta b = d$ for some $d \geq 0$.

Then $f(a) = d + (a-1)$, so $a \Delta b = a - 1 + d$.

But now consider property ii: $(a \Delta b) \cdot (b \Delta a) = 0$ for all $a, b$.

This means for any $a, b$, at least one of $a \Delta b$ or $b \Delta a$ is zero.

Suppose $a \Delta b = 0$. Then by i., $(a+1) \Delta b = a \Delta b + 1 = 1$, $(a+2) \Delta b = 2$, etc. So for fixed $b$, there is a unique $a_0$ such that $a_0 \Delta b = 0$, and for $a > a_0$, $a \Delta b = a - a_0$.

Similarly, for fixed $a$, there is a unique $b_0$ such that $b_0 \Delta a = 0$.

But for all $a, b$, at least one of $a \Delta b$ or $b \Delta a$ is zero. This is only possible if for all $a \neq b$, either $a \Delta b = 0$ or $b \Delta a = 0$.

Let us try to construct such a function. Suppose $x \Delta y = 0$ if $x \leq y$, and $x \Delta y = x - y$ if $x > y$.

Check property i:

If $a > b$, then $(a+b) > b$, so $(a+b) \Delta b = (a+b) - b = a$.

$a \Delta b = a - b$, so $a \Delta b + 1 = a - b + 1$.

But $(a+b) \Delta b = a$, $a \Delta b + 1 = a - b + 1$. These are equal only if $b = 1$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x = y$, $x \Delta y = x - y + 1$ if $x > y$.

Check property i:

If $a > 0$, $b > 0$.

Case 1: $a + b > b$ so $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a > b$, $a \Delta b = a - b + 1$, so $a \Delta b + 1 = a - b + 2$.
- If $a = b$, $a \Delta b = 1$, $a \Delta b + 1 = 2$.
- If $a < b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.

But $(a+b) \Delta b = a + 1$ always.

So for $a < b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = a + 1$.

So only possible if $a = 0$.

Alternatively, try $x \Delta y = 0$ if $x \leq y$, $x \Delta y = 1$ if $x > y$.

Check property i:

If $a > b$, $(a+b) > b$, so $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x \leq y$, $x \Delta y = 1$ if $x > y$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, so $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, so $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x \leq y$, $x \Delta y = 1$ if $x > y$.

Check property i:

If $a > b$, $(a+b) > b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.

Check property i:

If $a \geq b$, $(a+b) \geq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = 1$ if $x \neq y$.

Check property i:

If $a \neq b$, $(a+b) \neq b$, $(a+b) \Delta b = 1$.
$a \Delta b = 1$, $a \Delta b + 1 = 2$.

But $(a+b) \Delta b = 1$, $a \Delta b + 1 = 2$.

Alternatively, try $x \Delta y = 0$ if $x = y$, $x \Delta y = x - y + 1$ if $x \neq y$.

If $x = y$, $x \Delta y = 0$.
If $x \neq y$, $x \Delta y = x - y + 1$.

Check property i:

$(a+b) \Delta b$:
- If $a + b = b$, $a = 0$, not possible for positive integers.
- If $a + b \neq b$, $(a+b) \Delta b = a + b - b + 1 = a + 1$.

$a \Delta b$:
- If $a = b$, $a \Delta b = 0$, $a \Delta b + 1 = 1$.
- If $a \neq b$, $a \Delta b = a - b + 1$, $a \Delta b + 1 = a - b + 2$.

But $(a+b) \Delta b = a + 1$.

If $a = b$, $a \Delta b + 1 = 1$, $(a+b) \Delta b = b + 1$.

So only possible if $b = 0$.

Alternatively, try $x \Delta y = 0$ if $x < y$, $x \Delta y = 1$ if $x \geq y$.