Script started on Fri May 16 13:42:31 1997
% cl
Allegro CL 4.1 [DEC MIPS; R1] (9/21/92 16:49)
Copyright (C) 1985-1992, Franz Inc., Berkeley, CA, USA.  All Rights Reserved.
;; Optimization settings: safety 1, space 1, speed 1, debug 2
;; For a complete description of all compiler switches given the current
;; optimization settings evaluate (EXPLAIN-COMPILER-SETTINGS).

USER(1): (quote (a b c))
(A B C)

USER(2): (reverse '(a b c))
(C B A)

USER(3): (cond ((= 1 2) 3) (4 5) (t 6))
5

USER(4): (equal '(worm slug) '(worm slug))
T

USER(5): (eq '(worm slug) '(worm slug))
NIL

USER(6): (mapcar #'(lambda (x) x) '(worm slug snail))
(WORM SLUG SNAIL)

USER(7): (car (cdr (cdr '(1 2 3 4 5 6 7 8))))
3

USER(8): (cdr (car (car '(1 2 3 4 5 6 7 8))))
Error: Attempt to take the car of 1 which is not a cons.
  [condition type: SIMPLE-ERROR]

USER(9):  (+ (length '(1 2)) 7 'cow))

Error: COW is an illegal argument to +
  [condition type: TYPE-ERROR]

USER(10): (defun snork (L) 
              (cond ((null L) L)
                    ((null (cdr L)) L)
                    (t (cons (car (cdr L))
                             (cons (car L) (snork (cdr (cdr L))))))))
SNORK

USER(11):  (snork '(1 2 3 4 5 6 7 8 9 10))
(2 1 4 3 6 5 8 7 10 9)

USER(12): (snork '(1 2 3 4 5 6 7 8 9 ))
(2 1 4 3 6 5 8 7 9)

USER(13): (lambda (x) (+ x 1)) 1)

Error: attempt to call `LAMBDA' which is an undefined function.
  [condition type: UNDEFINED-FUNCTION]

;; Oops a paren go lost - this is what I meant:

USER(14):  ((lambda (x) (+ x 1)) 1)
2

USER(15): (and (= 1 2) (+ 'bob tom))
NIL

USER(16): (defun bob (bill jill) (list jill bill))
BOB

USER(17): (setq bob 'rob)      
ROB

USER(18): (bob 'bob bob)
(ROB BOB)

USER(21): (defun every-other (L) 
   (cond ((null L) nil)
	 ((null (cdr L)) nil)
         (t (cons (cadr L) (every-other (cddr L))))))
EVERY-OTHER

USER(22): (every-other '(1 2 3 4 5 6 7 8))
(2 4 6 8)

USER(23): (defun increasing (L)
   (cond ((null L) t)
         ((null (cdr L)) t)
         ((> (car L) (cadr L)) nil)
         (t (increasing (cdr L)))))
INCREASING

;; Of course you can use the following functions instead 
;; of the traditional ones:
;;    (car L)      =>     (first L)
;;    (cdr L)      =>     (rest L)
;;    (cadr L)     =>     (second L)
;;    (cddr L)     =>     (rest (rest L))

USER(24): (increasing '(1 1 2 3 4 5 5 6 7 8 ))
T

USER(25): (increasing '(1 2 3 4 3 4 5))
NIL

USER(26): ^D
Really exit lisp [n]? y
; Exiting Lisp
% 
script done on Fri May 16 13:55:22 1997


=========================================

Script started on Fri May 16 14:48:59 1997
lynx% prolog
Quintus Prolog Release 3.1 (DECstation, Ultrix 4.x)
Copyright (C) 1990, Quintus Corporation.  All rights reserved.
2100 Geng Road, Palo Alto, California U.S.A. (415) 813-3800

| ?- [q2].
% compiling file /Sucia/staff/anderson/plog/q2
% q2 compiled in module user, 0.183 sec 1,836 bytes

| ?- rank(A, assoc, _), dept(A, math).

A = erdos ;

A = euler ;

no
| ?- chair(_, A), rank(A, B, _).

A = newton,
B = full ;

A = euler,
B = assoc ;

A = vonNeumann,
B = full ;

no

// colleague(P1, P2) :- dept(P1, Dept), dept(P2, Dept).

| ?- colleague(einstein, A).

A = newton ;

A = einstein ;

A = feynman ;

no

// boss(Chair, Prof) :- chair(Dept, Chair), dept(Prof, Dept).

| ?- boss(A, turing).

A = euler ;

A = vonNeumann ;

no

// highPay(P1) :- boss(P2, P1), rank(P2, _, S2), rank(P1,_,S1), S1 > S2.

| ?- highPay(A).

A = einstein ;

no

| ?- highPay(A), boss(A,_).

no

// Now if we add the fact dept(physics, vonNeumann), we get 
// the result that vonNeumann is a highly paid chairman, since vonNeumann
// is paid more than the chairman of the physics department

| ?- highPay(A), boss(A,_).

A = vonNeumann ;

A = vonNeumann ;

A = vonNeumann ;

A = vonNeumann ;

no


//shift(L1,L2) :- append(A,B,L1), append(B,A,L2).


| ?- shift([1,2,3,4], [3,4,1,2]).

yes
| ?- shift([2,1,3,4], [1,2,3,4]).

no
| ?- shift([1,2,3,4], [1,2,3]).

no
| ?- shift(A, [1,2,3,4]).

A = [1,2,3,4] ;

A = [4,1,2,3] ;

A = [3,4,1,2] ;

A = [2,3,4,1] ;

A = [1,2,3,4] ;

// and now it goes into an infinite loop . . .


| ?- shift([1,2,3,4], A).

A = [1,2,3,4] ;

A = [2,3,4,1] ;

A = [3,4,1,2] ;

A = [4,1,2,3] ;

A = [1,2,3,4] ;

no

// eo([],[]).
// eo([_],[]).
// eo([_,Y|L1], [Y|L2]) :- eo(L1, L2).

| ?- eo([1,2,3,4,5], A).

A = [2,4] ;

no
| ?- eo([1,2,3,4,5], [1,3,5]).

no
| ?- eo([1,2,3,4,5], [2,4]).

yes
| ?- eo(A, [2,4]).

A = [_9289,2,_9293,4] ;

A = [_9289,2,_9293,4,_9297] ;

no