Question about partial application in OCaml

I have a question about OCaml:Assume the following function declarations:

let secret x y =   let secret' x y z = x + y - z in       let z = if x mod 2 = 0 then y else x - y    in secret' x z ;;

print_int (secret 3 2 (-5));;

The output will be 9.

Why is the result 9?

Is this the way its calculations are performed?

let secret 3 2 =   let secret' 3 2 z = 3 + 2 - z in       let z = if 3 mod 2 = 0 then 2 else 3 - 2    in secret' 3 1 

Why doesn't it stop at secret' 3 1 but instead comes up with the result 9? Computes it secret' 3 1 (-5) = 3 + 1 -(-5) ? Why?