From Osher Doctorow

Remember y/x having a difficulty when x = 0? Let's consider a new
type of equation which "erases" terms with 0 denominators. To avoid
proliferating notation, I won't use special notation for such
equations, as long as they're clearly identified as "erasure
convention".

So let's consider the equation:

1) P* = a(1 + y - x) + b(1 + z - x) + c(y/x) + d(1 + x - y) + f(1 + x
- z) + g(x/y) + h(z/x) + k(x/z) + ly + mz

P* is taken to be a probability here in a "Multiverse", with its
arguments explicitly written as:

2) P*(x, y, z) (for example, x = P(A), y = P(AB), z = P(B))

Let's calculate P*(x, x, x):

3) P*(x, x, x) = a + b + c + d + f + g + h + k + l + m

where these coefficients are constant. We might as well set their
sum equal to 1 by analogy with the fact that P' (A-->B) = 1 + z - x =
1 when x = z.

Notice what happens in P*(0, 1, 1):

4) P*(0, 1, 1) = 2a + 2b + l + m

so that we meet the expression 2 again from earlier. For P*(0, 0, 1)
we get:

5) P*(0, 0, 1) = a + 2b + d + m

Readers can try various other combinations.

Osher Doctorow