The easiest way to think about this, as Tim mentions in his article, is that in Common Lisp, you can write nil as '(), so if foo is defined as the list (d e f),

The easiest way to think about this, as Tim mentions in his article, is that in Common Lisp, you can write nil as '(), so if foo is defined as the list (d e f), (append nil foo) is equivalent to (append '() '(d e f)) which more obviously evaluates to (d e f), just as (append '(a b c) '(d e f)) evaluates to (a b c d e f). It's better to think of Lisp append's first argument being appended onto its second: (append '(a b) '(c d)) => (cons 'a (append '(b) '(c d))) => (cons 'a (cons 'b (append '() '(c d)))) => (cons 'a (cons 'b '(c d))) => (cons 'a '(b c d)), which evaluates to (a b c d). The lists '(a b) and '(c d) are still unaltered at the end of the computation, and a new list '(a b c d) is returned by the call to append.