TSTP Solution File: PUZ131+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : PUZ131+1 : TPTP v8.1.0. Released v4.1.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n016.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Wed Jul 27 13:11:40 EDT 2022
% Result : Theorem 1.75s 1.95s
% Output : Refutation 1.75s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 8
% Syntax : Number of clauses : 11 ( 9 unt; 0 nHn; 11 RR)
% Number of literals : 17 ( 1 equ; 7 neg)
% Maximal clause size : 6 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 8 ( 6 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 3 con; 0-1 aty)
% Number of variables : 4 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(10,axiom,
( ~ course(A)
| teaches(coordinatorof(A),A) ),
file('PUZ131+1.p',unknown),
[] ).
cnf(11,axiom,
( ~ student(A)
| ~ course(B)
| ~ enrolled(A,B)
| ~ professor(C)
| ~ teaches(C,B)
| taughtby(A,C) ),
file('PUZ131+1.p',unknown),
[] ).
cnf(12,axiom,
~ taughtby(michael,victor),
file('PUZ131+1.p',unknown),
[] ).
cnf(17,axiom,
student(michael),
file('PUZ131+1.p',unknown),
[] ).
cnf(18,axiom,
professor(victor),
file('PUZ131+1.p',unknown),
[] ).
cnf(19,axiom,
course(csc410),
file('PUZ131+1.p',unknown),
[] ).
cnf(20,axiom,
enrolled(michael,csc410),
file('PUZ131+1.p',unknown),
[] ).
cnf(22,axiom,
coordinatorof(csc410) = victor,
file('PUZ131+1.p',unknown),
[] ).
cnf(37,plain,
teaches(victor,csc410),
inference(demod,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[19,10]),22]),
[iquote('hyper,19,10,demod,22')] ).
cnf(72,plain,
taughtby(michael,victor),
inference(hyper,[status(thm)],[37,11,17,19,20,18]),
[iquote('hyper,37,11,17,19,20,18')] ).
cnf(73,plain,
$false,
inference(binary,[status(thm)],[72,12]),
[iquote('binary,72.1,12.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : PUZ131+1 : TPTP v8.1.0. Released v4.1.0.
% 0.11/0.13 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n016.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Wed Jul 27 02:09:22 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.75/1.95 ----- Otter 3.3f, August 2004 -----
% 1.75/1.95 The process was started by sandbox on n016.cluster.edu,
% 1.75/1.95 Wed Jul 27 02:09:22 2022
% 1.75/1.95 The command was "./otter". The process ID is 19821.
% 1.75/1.95
% 1.75/1.95 set(prolog_style_variables).
% 1.75/1.95 set(auto).
% 1.75/1.95 dependent: set(auto1).
% 1.75/1.95 dependent: set(process_input).
% 1.75/1.95 dependent: clear(print_kept).
% 1.75/1.95 dependent: clear(print_new_demod).
% 1.75/1.95 dependent: clear(print_back_demod).
% 1.75/1.95 dependent: clear(print_back_sub).
% 1.75/1.95 dependent: set(control_memory).
% 1.75/1.95 dependent: assign(max_mem, 12000).
% 1.75/1.95 dependent: assign(pick_given_ratio, 4).
% 1.75/1.95 dependent: assign(stats_level, 1).
% 1.75/1.95 dependent: assign(max_seconds, 10800).
% 1.75/1.95 clear(print_given).
% 1.75/1.95
% 1.75/1.95 formula_list(usable).
% 1.75/1.95 all A (A=A).
% 1.75/1.95 exists A student(A).
% 1.75/1.95 exists A professor(A).
% 1.75/1.95 exists A course(A).
% 1.75/1.95 student(michael).
% 1.75/1.95 professor(victor).
% 1.75/1.95 course(csc410).
% 1.75/1.95 all A (course(A)->professor(coordinatorof(A))).
% 1.75/1.95 all X (student(X)-> (exists Y (course(Y)&enrolled(X,Y)))).
% 1.75/1.95 all X (professor(X)-> (exists Y (course(Y)&teaches(X,Y)))).
% 1.75/1.95 all X (course(X)-> (exists Y (student(Y)&enrolled(Y,X)))).
% 1.75/1.95 all X (course(X)-> (exists Y (professor(Y)&teaches(Y,X)))).
% 1.75/1.95 all X (course(X)->teaches(coordinatorof(X),X)).
% 1.75/1.95 all X Y (student(X)&course(Y)-> (enrolled(X,Y)-> (all Z (professor(Z)-> (teaches(Z,Y)->taughtby(X,Z)))))).
% 1.75/1.95 enrolled(michael,csc410).
% 1.75/1.95 coordinatorof(csc410)=victor.
% 1.75/1.95 -taughtby(michael,victor).
% 1.75/1.95 end_of_list.
% 1.75/1.95
% 1.75/1.95 -------> usable clausifies to:
% 1.75/1.95
% 1.75/1.95 list(usable).
% 1.75/1.95 0 [] A=A.
% 1.75/1.95 0 [] student($c1).
% 1.75/1.95 0 [] professor($c2).
% 1.75/1.95 0 [] course($c3).
% 1.75/1.95 0 [] student(michael).
% 1.75/1.95 0 [] professor(victor).
% 1.75/1.95 0 [] course(csc410).
% 1.75/1.95 0 [] -course(A)|professor(coordinatorof(A)).
% 1.75/1.95 0 [] -student(X)|course($f1(X)).
% 1.75/1.95 0 [] -student(X)|enrolled(X,$f1(X)).
% 1.75/1.95 0 [] -professor(X)|course($f2(X)).
% 1.75/1.95 0 [] -professor(X)|teaches(X,$f2(X)).
% 1.75/1.95 0 [] -course(X)|student($f3(X)).
% 1.75/1.95 0 [] -course(X)|enrolled($f3(X),X).
% 1.75/1.95 0 [] -course(X)|professor($f4(X)).
% 1.75/1.95 0 [] -course(X)|teaches($f4(X),X).
% 1.75/1.95 0 [] -course(X)|teaches(coordinatorof(X),X).
% 1.75/1.95 0 [] -student(X)| -course(Y)| -enrolled(X,Y)| -professor(Z)| -teaches(Z,Y)|taughtby(X,Z).
% 1.75/1.95 0 [] enrolled(michael,csc410).
% 1.75/1.95 0 [] coordinatorof(csc410)=victor.
% 1.75/1.95 0 [] -taughtby(michael,victor).
% 1.75/1.95 end_of_list.
% 1.75/1.95
% 1.75/1.95 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=6.
% 1.75/1.95
% 1.75/1.95 This is a Horn set with equality. The strategy will be
% 1.75/1.95 Knuth-Bendix and hyper_res, with positive clauses in
% 1.75/1.95 sos and nonpositive clauses in usable.
% 1.75/1.95
% 1.75/1.95 dependent: set(knuth_bendix).
% 1.75/1.95 dependent: set(anl_eq).
% 1.75/1.95 dependent: set(para_from).
% 1.75/1.95 dependent: set(para_into).
% 1.75/1.95 dependent: clear(para_from_right).
% 1.75/1.95 dependent: clear(para_into_right).
% 1.75/1.95 dependent: set(para_from_vars).
% 1.75/1.95 dependent: set(eq_units_both_ways).
% 1.75/1.95 dependent: set(dynamic_demod_all).
% 1.75/1.95 dependent: set(dynamic_demod).
% 1.75/1.95 dependent: set(order_eq).
% 1.75/1.95 dependent: set(back_demod).
% 1.75/1.95 dependent: set(lrpo).
% 1.75/1.95 dependent: set(hyper_res).
% 1.75/1.95 dependent: clear(order_hyper).
% 1.75/1.95
% 1.75/1.95 ------------> process usable:
% 1.75/1.95 ** KEPT (pick-wt=5): 1 [] -course(A)|professor(coordinatorof(A)).
% 1.75/1.95 ** KEPT (pick-wt=5): 2 [] -student(A)|course($f1(A)).
% 1.75/1.95 ** KEPT (pick-wt=6): 3 [] -student(A)|enrolled(A,$f1(A)).
% 1.75/1.95 ** KEPT (pick-wt=5): 4 [] -professor(A)|course($f2(A)).
% 1.75/1.95 ** KEPT (pick-wt=6): 5 [] -professor(A)|teaches(A,$f2(A)).
% 1.75/1.95 ** KEPT (pick-wt=5): 6 [] -course(A)|student($f3(A)).
% 1.75/1.95 ** KEPT (pick-wt=6): 7 [] -course(A)|enrolled($f3(A),A).
% 1.75/1.95 ** KEPT (pick-wt=5): 8 [] -course(A)|professor($f4(A)).
% 1.75/1.95 ** KEPT (pick-wt=6): 9 [] -course(A)|teaches($f4(A),A).
% 1.75/1.95 ** KEPT (pick-wt=6): 10 [] -course(A)|teaches(coordinatorof(A),A).
% 1.75/1.95 ** KEPT (pick-wt=15): 11 [] -student(A)| -course(B)| -enrolled(A,B)| -professor(C)| -teaches(C,B)|taughtby(A,C).
% 1.75/1.95 ** KEPT (pick-wt=3): 12 [] -taughtby(michael,victor).
% 1.75/1.95
% 1.75/1.95 ------------> process sos:
% 1.75/1.95 ** KEPT (pick-wt=3): 13 [] A=A.
% 1.75/1.95 ** KEPT (pick-wt=2): 14 [] student($c1).
% 1.75/1.95 ** KEPT (pick-wt=2): 15 [] professor($c2).
% 1.75/1.95 ** KEPT (pick-wt=2): 16 [] course($c3).
% 1.75/1.95 ** KEPT (pick-wt=2): 17 [] student(michael).
% 1.75/1.95 ** KEPT (pick-wt=2): 18 [] professor(victor).
% 1.75/1.95 ** KEPT (pick-wt=2): 19 [] course(csc410).
% 1.75/1.95 ** KEPT (pick-wt=3): 20 [] enrolled(michael,csc410).
% 1.75/1.95 ** KEPT (pick-wt=4): 21 [] coordinatorof(csc410)=victor.
% 1.75/1.95 ---> New Demodulator: 22 [new_demod,21] coordinatorof(csc410)=victor.
% 1.75/1.95 Following clause subsumed by 13 during input processing: 0 [copy,13,flip.1] A=A.
% 1.75/1.95 >>>> Starting back demodulation with 22.
% 1.75/1.95
% 1.75/1.95 ======= end of input processing =======
% 1.75/1.95
% 1.75/1.95 =========== start of search ===========
% 1.75/1.95
% 1.75/1.95 -------- PROOF --------
% 1.75/1.95
% 1.75/1.95 ----> UNIT CONFLICT at 0.00 sec ----> 73 [binary,72.1,12.1] $F.
% 1.75/1.95
% 1.75/1.95 Length of proof is 2. Level of proof is 2.
% 1.75/1.95
% 1.75/1.95 ---------------- PROOF ----------------
% 1.75/1.95 % SZS status Theorem
% 1.75/1.95 % SZS output start Refutation
% See solution above
% 1.75/1.95 ------------ end of proof -------------
% 1.75/1.95
% 1.75/1.95
% 1.75/1.95 Search stopped by max_proofs option.
% 1.75/1.95
% 1.75/1.95
% 1.75/1.95 Search stopped by max_proofs option.
% 1.75/1.95
% 1.75/1.95 ============ end of search ============
% 1.75/1.95
% 1.75/1.95 -------------- statistics -------------
% 1.75/1.95 clauses given 18
% 1.75/1.95 clauses generated 55
% 1.75/1.95 clauses kept 71
% 1.75/1.95 clauses forward subsumed 6
% 1.75/1.95 clauses back subsumed 0
% 1.75/1.95 Kbytes malloced 976
% 1.75/1.95
% 1.75/1.95 ----------- times (seconds) -----------
% 1.75/1.95 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.95 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.95 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.75/1.95
% 1.75/1.95 That finishes the proof of the theorem.
% 1.75/1.95
% 1.75/1.95 Process 19821 finished Wed Jul 27 02:09:23 2022
% 1.75/1.95 Otter interrupted
% 1.75/1.95 PROOF FOUND
%------------------------------------------------------------------------------