[r6rs-discuss] Comparison procedures' number of arguments

[Alan Bawden]

   Date: Mon, 20 Oct 2008 18:30:09 -0700
   From: Thomas Lord <lord at emf.net>
   Cc: r6rs-discuss at lists.r6rs.org

   Alan Bawden wrote:
   > I've deleted the message, but somebody wrote:
   >
   >      Is this a valid program transformation?
   >
   >      (apply = (append x y z))
   >
   >      <==>
   >
   >      (let ((x0 x)
   > 	   (y0 y)
   > 	   (z0 z))
   >        (let* ((xy (append x0 y0))
   > 	      (yz (append y0 z0)))
   > 	 (and (apply = xy)
   > 	      (apply = yz))))
   >
   > This is -not- a valid program transformation.  Consider the case where
   > y is the empty list.

   It wasn't meant as a question of "what does R6 do".

Stop being so argumentative and think it through.

I presume we all agree that:

   (= 1 1 2 2) ==> #F		[1]
   (= 1 1)     ==> #T		[2]
   (= 2 2)     ==> #T		[3]

Now play along:

    (apply = (append '(1 1) '() '(2 2)))

should be #F because of [1].

Following your transformation this becomes:

     (let ((x0 '(1 1))
	   (y0 '())
	   (z0 '(2 2)))
       (let* ((xy (append x0 y0))
	      (yz (append y0 z0)))
	 (and (apply = xy)
	      (apply = yz))))

which reduces to:

	 (and (apply = '(1 1))
	      (apply = '(2 2)))

which must be #T according to [2] and [3].

Note that `=' was not applied to less than two arguments anywhere in this
example.

I repeat: This is -not- a valid program transformation.