let
  b = 1;
  fu0 = (x: x);
  fu1 = (x: y: x + y) 4;
  fu2 = (x: y: (2 * x) + y);
in
rec {
  ex00 = fu0 4;     # must return 4
  ex01 = (fu1) 1;   # must return 5
  ex02 = (fu2 3 ) 1; # must return 7
  ex03 = (fu2 3 );   # must return <LAMBDA>
  ex04 = ex03 1;    # must return 7
  ex05 = (n: x: (fu2 x n)) 1 3; # must return 7
}