R6RSのライブラリの「レベル」指定はなぜ必要か？ その2 - Scheme VM を書く

レベルが (meta 2) とか (meta 3) の例が理解できなくていろいろ探した結果、実際のコード例が見つかったのだけど全く分からないλ...。
レベルに関する記述なんて1ページなのになぜ理解できないのだろう。
examples.scm - Google Code Search

   ;;======================================================
   ;;
   ;; Further library tests:
   ;;
   ;;======================================================

   ;; Test meta-level resolution for chained imports:

   (library (foo)
     (export u)
     (import (rnrs))
     (define u 1))

   (library (bar)
     (export u v)
     (import (rnrs) (foo))
     (define-syntax v (lambda (e) (syntax u))))

   (library (baz)
     (export)
     (import (for (rnrs) (meta 2) expand run)
             (for (bar)  (meta 2)))
     (display
      (let-syntax ((m (lambda (e)
                        (let-syntax ((n (lambda (e) (+ u (v)))))
                          (n)))))
        (m))))

   (import (baz))    ;==> 2

   ;;======================================================
   ;;
   ;; Check that export levels compose correctly:
   ;;
   ;;======================================================

   (library (foo)
     (export x y)
     (import (rnrs))
     (define x 2)
     (define y 4))

   (library (baz)
     (export y)                      ;; exports y at level 1
     (import (rnrs) (for (foo) expand)))

   (library (bab)
     (export f)
     (import (for (rnrs) expand run)   ;; This also implicitly imports into (meta 2)
             (for (foo)  expand)       ;; imports x and y at level 1
             (for (baz)  expand))      ;; also imports y but at level expand + 1 = 2
     (define (f)
       (let-syntax ((foo (lambda (_)
                           (+ x                                   ;; level 1
                              y                                   ;; level 1
                              (let-syntax ((bar (lambda (_) y)))  ;; level 2
                                (bar))))))
         (foo))))

   (import (bab))
   (f)   ;==> 10

   ;;==========================================================
   ;;
   ;; Check that levels of reference are determined lexically:
   ;;
   ;;==========================================================

   (library (foo)
     (export f)
     (import (rnrs))
     (define (f) 1))

   (library (bar)
     (export g)
     (import (rnrs)
             (for (foo) expand))  ;; This is the wrong level !
     (define-syntax g
       (syntax-rules ()
         ((_) (f)))))

   ;; This *must* be an error:
   ;; The use of f in bar cannot be satisfied
   ;; by the import of foo into the client level 0 here.
   ;; That would violate lexical determination of
   ;; level of reference to f in bar.

   ;; (library (main)
   ;;   (export)
   ;;   (import (rnrs) (foo) (bar))
   ;;   (display (g)))

   ;; ==> Syntax violation: Attempt to use binding of f in library (bar) at invalid meta level 0.
   ;;     Binding is only valid at meta levels: 1

   ;; Example from http://www.r6rs.org/r6rs-editors/2006-August/001682.html

   (library (A)
     (export x)
     (import (rnrs))
     (define x 37))

   (library (B)
     (export)
     (import (A)))

   (library (C)
     (export foo)
     (import (rnrs) (for (A) expand))
     (define-syntax foo
       (syntax-rules ()
         ((_) x))))

   (library (D)
     (export foo)
     (import (rnrs) (C)))

   ;; This has to raise syntax error to avoid problem described in
   ;; above message.

   (library (E)
     (export)
     (import (rnrs) (B) (D))
     ;; (display (foo))  ; Attempt to use x in library (C) at invalid meta level 0.
     ;;                  ; Binding is only available at meta levels: 1
     )

   ;;==============================================================
   ;;
   ;; Importing into multiple and negative levels:
   ;;
   ;;==============================================================

   (library (foo)
     (export x)
     (import (rnrs))
     (define x 42))

   (library (bar)
     (export get-x)
     (import (rnrs)
             ;; Code in (syntax ...) expressions refer to bindings
             ;; at one lower level - for example, ordinary macros
             ;; are evaluated at level expand = 1, but manipulate
             ;; code that will run at level run = 0.
             ;; The occurrence of (syntax x) below is not in a macro
             ;; but rather at level 0.
             ;; The reference x in (syntax x) is therefore at level -1.
             ;; To make it refer to the x in foo, we need to import
             ;; the latter at level -1.
             (for (foo) (meta -1)))
     (define (get-x) (syntax x)))

   (library (baz)
     (export)
     (import (for (rnrs) (meta 3) (meta 2) expand run)
             (for (bar)  (meta 3) expand))

     (display
      (let-syntax ((m (lambda (ignore)
                        (get-x))))
        (m)))                                    ;==> 42

     (display
      (let-syntax ((m (lambda (ignore)
                        (let-syntax ((n (lambda (ignore)
                                          (let-syntax ((o (lambda (ignore)
                                                            (get-x))))
                                            (o)))))
                          (n)))))
        (m)))                                    ;==> 42

     ;; This should give a syntax error:

     ;; (display
     ;;  (let-syntax ((m (lambda (ignore)
     ;;                     (let-syntax ((n (lambda (ignore)
     ;;                                       (get-x))))
     ;;                       (n)))))
     ;;   (m)))              ;==> Syntax-violation: Attempt to use binding of get-x in library (baz) at invalid level 2
     ;;                      ;                      Binding is only valid at levels (1 3)

     ) ;; baz

   (import (baz))   ;==> 42 42