Условный выход из функции: PROG RETURN

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (return
      (cond ((and (> x e) (< x r)) 'yes)
            ((and (> x r) (< x e)) 'no))))) 
 
==> XI
(xi 2)
 
==> NIL

(defun xip (x)
  (< 3 x 10))

(if (eql (xi x) 'yes) ...)

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (return
      (when (and (> x e) (< x r)) "yes")
      (when (or (> x r) (< x e)) "No"))))

 (load "c:/ox.lisp")
 
; file: c:/ox.lisp
; in: DEFUN XI
;     (RETURN (WHEN (AND (> X E) (< X R)) "yes") (WHEN (OR (> X R) (< X E)) "No"
))
;
; caught ERROR:
;   during macroexpansion of (RETURN (WHEN # "yes") (WHEN # "No")). Use
;   *BREAK-ON-SIGNALS* to intercept.
;
;    error while parsing arguments to DEFMACRO RETURN:
;      too many elements in
;        ((WHEN (AND (> X E) (< X R)) "yes") (WHEN (OR (> X R) (< X E)) "No"))
;      to satisfy lambda list
;        (&OPTIONAL (SB-IMPL::VALUE)):
;      between 0 and 1 expected, but got 2
 
;     (DEFUN XI (X)
;       (PROG (E R)
;         (SETF E 3)
;         (SETF R 10)
;         (RETURN (WHEN (AND # #) "yes") (WHEN (OR # #) "No"))))
;
; caught STYLE-WARNING:
;   The variable X is defined but never used.
;
; compilation unit finished
;   caught 1 ERROR condition
;   caught 1 STYLE-WARNING condition
T

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (return
      (cond ((and (> x e) (< x r)) 'yes)
            ((or (>= x r) (<= x e)) 'no)))))

(load "c:/ox.lisp")
T
* (xi 4)
YES
* (xi 1)
NO
* (xi 15)
NO
*

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (return
      (if (and (> x e) (< x r)) "yes" "No"))))

(print "Vi vveli 7")

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (return
      (cond 
          (if (= x 7) (go L))
          ((and (> x e) (< x r)) 'yes)
          ((or (>= x r) (<= x e)) 'no)
 L        (print "Vi vveli 7")))))

(defun xi (x)
  (prog (e r)
    (setf e 3)
    (setf r 10)
    (when (= x 7) (print "Вы ввели 7"))  
    (return (cond ((and (> x e) (< x r)) 'yes)
                  ((or (>= x r) (<= x e)) 'no)))))
 
==> XI
 
(xi 5)
==> YES
 
(xi -5)
==> NO
 
(xi 7)
"Вы ввели 7"
==> YES