let fib = \n. if (output n) <= 2 
                then 1
                else this (n + -2) + this (n + -1)
in let double = \n. !n + (let fib = \x. 0 in eval(n))
   in double [| fib 7 |] // computes 7th fibno once, snd eval(n) gives 13, yields 26