TSTP Solution File: SEU148+3 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU148+3 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n022.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 13:14:55 EDT 2022
% Result : Theorem 1.70s 1.91s
% Output : Refutation 1.70s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 7
% Syntax : Number of clauses : 11 ( 8 unt; 1 nHn; 9 RR)
% Number of literals : 17 ( 11 equ; 7 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 4 ( 2 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 3 con; 0-1 aty)
% Number of variables : 11 ( 1 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( A != singleton(B)
| ~ in(C,A)
| C = B ),
file('SEU148+3.p',unknown),
[] ).
cnf(3,axiom,
( A != singleton(B)
| in(C,A)
| C != B ),
file('SEU148+3.p',unknown),
[] ).
cnf(5,axiom,
singleton(A) != empty_set,
file('SEU148+3.p',unknown),
[] ).
cnf(6,axiom,
( ~ subset(A,singleton(B))
| A = empty_set
| A = singleton(B) ),
file('SEU148+3.p',unknown),
[] ).
cnf(10,axiom,
dollar_c4 != dollar_c3,
file('SEU148+3.p',unknown),
[] ).
cnf(12,axiom,
A = A,
file('SEU148+3.p',unknown),
[] ).
cnf(17,axiom,
subset(singleton(dollar_c4),singleton(dollar_c3)),
file('SEU148+3.p',unknown),
[] ).
cnf(19,plain,
in(A,singleton(A)),
inference(hyper,[status(thm)],[12,3,12]),
[iquote('hyper,12,3,12')] ).
cnf(58,plain,
singleton(dollar_c4) = singleton(dollar_c3),
inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[17,6]),5]),
[iquote('hyper,17,6,unit_del,5')] ).
cnf(60,plain,
dollar_c4 = dollar_c3,
inference(hyper,[status(thm)],[58,2,19]),
[iquote('hyper,58,2,19')] ).
cnf(62,plain,
$false,
inference(binary,[status(thm)],[60,10]),
[iquote('binary,60.1,10.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SEU148+3 : TPTP v8.1.0. Released v3.2.0.
% 0.07/0.13 % Command : otter-tptp-script %s
% 0.12/0.34 % Computer : n022.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Wed Jul 27 07:45:50 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.70/1.91 ----- Otter 3.3f, August 2004 -----
% 1.70/1.91 The process was started by sandbox2 on n022.cluster.edu,
% 1.70/1.91 Wed Jul 27 07:45:50 2022
% 1.70/1.91 The command was "./otter". The process ID is 1457.
% 1.70/1.91
% 1.70/1.91 set(prolog_style_variables).
% 1.70/1.91 set(auto).
% 1.70/1.91 dependent: set(auto1).
% 1.70/1.91 dependent: set(process_input).
% 1.70/1.91 dependent: clear(print_kept).
% 1.70/1.91 dependent: clear(print_new_demod).
% 1.70/1.91 dependent: clear(print_back_demod).
% 1.70/1.91 dependent: clear(print_back_sub).
% 1.70/1.91 dependent: set(control_memory).
% 1.70/1.91 dependent: assign(max_mem, 12000).
% 1.70/1.91 dependent: assign(pick_given_ratio, 4).
% 1.70/1.91 dependent: assign(stats_level, 1).
% 1.70/1.91 dependent: assign(max_seconds, 10800).
% 1.70/1.91 clear(print_given).
% 1.70/1.91
% 1.70/1.91 formula_list(usable).
% 1.70/1.91 all A (A=A).
% 1.70/1.91 all A B (in(A,B)-> -in(B,A)).
% 1.70/1.91 all A B (B=singleton(A)<-> (all C (in(C,B)<->C=A))).
% 1.70/1.91 empty(empty_set).
% 1.70/1.91 all A (singleton(A)!=empty_set).
% 1.70/1.91 all A B (subset(A,singleton(B))<->A=empty_set|A=singleton(B)).
% 1.70/1.91 exists A empty(A).
% 1.70/1.91 exists A (-empty(A)).
% 1.70/1.91 all A B subset(A,A).
% 1.70/1.91 -(all A B (subset(singleton(A),singleton(B))->A=B)).
% 1.70/1.91 end_of_list.
% 1.70/1.91
% 1.70/1.91 -------> usable clausifies to:
% 1.70/1.91
% 1.70/1.91 list(usable).
% 1.70/1.91 0 [] A=A.
% 1.70/1.91 0 [] -in(A,B)| -in(B,A).
% 1.70/1.91 0 [] B!=singleton(A)| -in(C,B)|C=A.
% 1.70/1.91 0 [] B!=singleton(A)|in(C,B)|C!=A.
% 1.70/1.91 0 [] B=singleton(A)|in($f1(A,B),B)|$f1(A,B)=A.
% 1.70/1.91 0 [] B=singleton(A)| -in($f1(A,B),B)|$f1(A,B)!=A.
% 1.70/1.91 0 [] empty(empty_set).
% 1.70/1.91 0 [] singleton(A)!=empty_set.
% 1.70/1.91 0 [] -subset(A,singleton(B))|A=empty_set|A=singleton(B).
% 1.70/1.91 0 [] subset(A,singleton(B))|A!=empty_set.
% 1.70/1.91 0 [] subset(A,singleton(B))|A!=singleton(B).
% 1.70/1.91 0 [] empty($c1).
% 1.70/1.91 0 [] -empty($c2).
% 1.70/1.91 0 [] subset(A,A).
% 1.70/1.91 0 [] subset(singleton($c4),singleton($c3)).
% 1.70/1.91 0 [] $c4!=$c3.
% 1.70/1.91 end_of_list.
% 1.70/1.91
% 1.70/1.91 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.70/1.91
% 1.70/1.91 This ia a non-Horn set with equality. The strategy will be
% 1.70/1.91 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.70/1.91 deletion, with positive clauses in sos and nonpositive
% 1.70/1.91 clauses in usable.
% 1.70/1.91
% 1.70/1.91 dependent: set(knuth_bendix).
% 1.70/1.91 dependent: set(anl_eq).
% 1.70/1.91 dependent: set(para_from).
% 1.70/1.91 dependent: set(para_into).
% 1.70/1.91 dependent: clear(para_from_right).
% 1.70/1.91 dependent: clear(para_into_right).
% 1.70/1.91 dependent: set(para_from_vars).
% 1.70/1.91 dependent: set(eq_units_both_ways).
% 1.70/1.91 dependent: set(dynamic_demod_all).
% 1.70/1.91 dependent: set(dynamic_demod).
% 1.70/1.91 dependent: set(order_eq).
% 1.70/1.91 dependent: set(back_demod).
% 1.70/1.91 dependent: set(lrpo).
% 1.70/1.91 dependent: set(hyper_res).
% 1.70/1.91 dependent: set(unit_deletion).
% 1.70/1.91 dependent: set(factor).
% 1.70/1.91
% 1.70/1.91 ------------> process usable:
% 1.70/1.91 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.70/1.91 ** KEPT (pick-wt=10): 2 [] A!=singleton(B)| -in(C,A)|C=B.
% 1.70/1.91 ** KEPT (pick-wt=10): 3 [] A!=singleton(B)|in(C,A)|C!=B.
% 1.70/1.91 ** KEPT (pick-wt=14): 4 [] A=singleton(B)| -in($f1(B,A),A)|$f1(B,A)!=B.
% 1.70/1.91 ** KEPT (pick-wt=4): 5 [] singleton(A)!=empty_set.
% 1.70/1.91 ** KEPT (pick-wt=11): 6 [] -subset(A,singleton(B))|A=empty_set|A=singleton(B).
% 1.70/1.91 ** KEPT (pick-wt=7): 7 [] subset(A,singleton(B))|A!=empty_set.
% 1.70/1.91 ** KEPT (pick-wt=8): 8 [] subset(A,singleton(B))|A!=singleton(B).
% 1.70/1.91 ** KEPT (pick-wt=2): 9 [] -empty($c2).
% 1.70/1.91 ** KEPT (pick-wt=3): 10 [] $c4!=$c3.
% 1.70/1.91
% 1.70/1.91 ------------> process sos:
% 1.70/1.91 ** KEPT (pick-wt=3): 12 [] A=A.
% 1.70/1.91 ** KEPT (pick-wt=14): 13 [] A=singleton(B)|in($f1(B,A),A)|$f1(B,A)=B.
% 1.70/1.91 ** KEPT (pick-wt=2): 14 [] empty(empty_set).
% 1.70/1.91 ** KEPT (pick-wt=2): 15 [] empty($c1).
% 1.70/1.91 ** KEPT (pick-wt=3): 16 [] subset(A,A).
% 1.70/1.91 ** KEPT (pick-wt=5): 17 [] subset(singleton($c4),singleton($c3)).
% 1.70/1.91 Following clause subsumed by 12 during input processing: 0 [copy,12,flip.1] A=A.
% 1.70/1.91
% 1.70/1.91 ======= end of input processing =======
% 1.70/1.91
% 1.70/1.91 =========== start of search ===========
% 1.70/1.91
% 1.70/1.91 -------- PROOF --------
% 1.70/1.91
% 1.70/1.91 ----> UNIT CONFLICT at 0.00 sec ----> 62 [binary,60.1,10.1] $F.
% 1.70/1.91
% 1.70/1.91 Length of proof is 3. Level of proof is 2.
% 1.70/1.91
% 1.70/1.91 ---------------- PROOF ----------------
% 1.70/1.91 % SZS status Theorem
% 1.70/1.91 % SZS output start Refutation
% See solution above
% 1.70/1.91 ------------ end of proof -------------
% 1.70/1.91
% 1.70/1.91
% 1.70/1.91 Search stopped by max_proofs option.
% 1.70/1.91
% 1.70/1.91
% 1.70/1.91 Search stopped by max_proofs option.
% 1.70/1.91
% 1.70/1.91 ============ end of search ============
% 1.70/1.91
% 1.70/1.91 -------------- statistics -------------
% 1.70/1.91 clauses given 9
% 1.70/1.91 clauses generated 70
% 1.70/1.91 clauses kept 59
% 1.70/1.91 clauses forward subsumed 28
% 1.70/1.91 clauses back subsumed 0
% 1.70/1.91 Kbytes malloced 976
% 1.70/1.91
% 1.70/1.91 ----------- times (seconds) -----------
% 1.70/1.91 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.70/1.91 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.70/1.91 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.70/1.91
% 1.70/1.91 That finishes the proof of the theorem.
% 1.70/1.91
% 1.70/1.91 Process 1457 finished Wed Jul 27 07:45:52 2022
% 1.70/1.91 Otter interrupted
% 1.70/1.91 PROOF FOUND
%------------------------------------------------------------------------------