let fib = \n. if (output n) <= 2 
                then 1
                else this (n + -2) + this (n + -1)
in let double = \fib. eval(fib) + eval(fib)
   in double [| $(lift fib) 7 |] 
// dereferences outer `fib` once, as part of application of `double`
// applies the resulting lambda twice, thus computing the 7th fib number twice
// can be solved by replacing first inner $fib by !fib