reducable.ml

let reducable a b =
 {code for aA reducing to nothing etc}

let rec solve a x =
if a.length = 0 
  then match x with 
  # a :: b indicates that x is being split into a - the head and b - the tail of the list.
  # [] indicates an empty element 
    elem :: list_of_elems -> (solve elem list_of_elems)
  # If there's nothing left in the list, just return the element. 
  | elem :: [] -> elem
else 
  match x with 
    elem :: list_of_elems -> (
      match reducable a elem with 
        # There's no reduction, leave a in the list and work on the rest
        (a, b) -> (a :: (solve b list_of_elems))
        # There's a reduction to 0, so just let the upper function deal with it
        | (a when a.length = 0 ) -> (solve '' x)  
  | elem :: None -> elem
  # This is the end of the list, so only return the element we were using for comparison
  | [] -> a
    
solve '' x