I wonder if there is any way to bind a free symbol in a lambda in order to
pass the lambda to a defined function (for example)

What I want to do is something like this:

(de myf (f l) (f l))

(let l 99 (myf '((x) (+ (car x) l)) (1 2)))

I want it to return 100 but it fails with an error (1 2) number expected

this is because free symbol l in lambda ((x) (+ (car x) l)) is not bind by
let as pretended because of quoting of lambda

In other words, I think the problem is quoting avoids let binding as in:

(setq f (let n 10 '((x) (+ x n)))) -> ((x) (+ x n))

but I want it to return -> ((x) (+ x 10))

or using the typicall example:

(de adder (n) '((x) (+ x n)))  ->  ((n) '((x) (+ x n)))

so (adder 1) should return ((x) (+ x 1)) but it returns ((x) (+ x n))

Is there any way to manage this?  Something similar to "expand" in newlisp
will do the job:

   (define (badadder n) (lambda (x) (+ x n)))
   (badadder 3) -> (lambda (x) (+ x n))
   (define (adder n) (expand (lambda (x) (+ x n)) 'n))
   (adder 3) -> (lambda (x) (+ x 3))

The fist example in newlisp will be:

(define (myf f l) (f l))

(let ((l 99)) (myf (lambda (x) (+ (first x) l)) '(1 2)))

which fails for same reason picolisp fails but in newlisp the solution is:

(define (myf f l) (f l))

(let ((l 99)) (myf (expand (lambda (x) (+ (first x) l)) 'l) '(1 2)))  -> 100


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