#lang sicp
#|
(define (add-interval x y)
  (make-interval (+ (lower-bound x) (lower-bound y))
                 (+ (upper-bound x) (upper-bound y))))

(define (sub-interval a b)
  (make-interval (- (lower-bound a) (upper-bound b))
                 (- (upper-bound a) (lower-bound b))))

Notice that in add-interval, the lower-bound is low-x + low-y.
Meanwhile the higher-bound is high-x + high-y = low-x + width-x
+ low-y + width-y = low-sum + (width-x + width-y).
So width-sum = width-x + width-y.

Similar logic applies to sub-interval.

For mul-interval it's enough to see that
(mul-interval (make-interval 1 2) (make-interval 3 4)) =
(3 . 8) -> width = 5
doesn't have the same width as:
(mul-interval (make-interval 3 4) (make-interval 3 4)) =
(9 . 16) -> width = 7
despite the fact that in both cases, the intervals have widths:
1 and 1.
|#

; This code is only here to pass hexlet's test, which I'm not sure is
; necessary.
(define (width x) (/ (- (upper-bound x) (lower-bound x))
                     2))