Proving the LOOP function

We must demonstrate that our guess is good, i.e.

   f = [WHILE]
     = ( R < Y --> ident ) | ( R >= Y and Y > 0 --> Q,R := Q+R/Y, R mod Y )

1. show f behaves like identity when the loop boolean is false
   
   In this case, we constructed f exactly so this will happen
   ( ~( R >= Y ) --> ident ) is identity function restricted to the loop boolean false
   rewrite as ( R < Y --> ident ), this is exactly one clause of f