TSTP Solution File: SET877+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET877+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n019.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 13:14:27 EDT 2022
% Result : Theorem 1.68s 1.89s
% Output : Refutation 1.68s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 6
% Syntax : Number of clauses : 10 ( 8 unt; 0 nHn; 8 RR)
% Number of literals : 13 ( 9 equ; 4 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 2 con; 0-2 aty)
% Number of variables : 8 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( A != singleton(B)
| ~ in(C,A)
| C = B ),
file('SET877+1.p',unknown),
[] ).
cnf(5,axiom,
( set_intersection2(A,singleton(B)) != singleton(B)
| in(B,A) ),
file('SET877+1.p',unknown),
[] ).
cnf(7,axiom,
dollar_c4 != dollar_c3,
file('SET877+1.p',unknown),
[] ).
cnf(9,axiom,
A = A,
file('SET877+1.p',unknown),
[] ).
cnf(10,axiom,
set_intersection2(A,B) = set_intersection2(B,A),
file('SET877+1.p',unknown),
[] ).
cnf(15,axiom,
set_intersection2(singleton(dollar_c4),singleton(dollar_c3)) = singleton(dollar_c4),
file('SET877+1.p',unknown),
[] ).
cnf(58,plain,
set_intersection2(singleton(dollar_c3),singleton(dollar_c4)) = singleton(dollar_c4),
inference(para_into,[status(thm),theory(equality)],[15,10]),
[iquote('para_into,15.1.1,10.1.1')] ).
cnf(70,plain,
in(dollar_c4,singleton(dollar_c3)),
inference(hyper,[status(thm)],[58,5]),
[iquote('hyper,58,5')] ).
cnf(83,plain,
dollar_c4 = dollar_c3,
inference(hyper,[status(thm)],[70,2,9]),
[iquote('hyper,70,2,9')] ).
cnf(85,plain,
$false,
inference(binary,[status(thm)],[83,7]),
[iquote('binary,83.1,7.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SET877+1 : TPTP v8.1.0. Released v3.2.0.
% 0.07/0.12 % Command : otter-tptp-script %s
% 0.13/0.33 % Computer : n019.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Wed Jul 27 10:47:37 EDT 2022
% 0.13/0.33 % CPUTime :
% 1.68/1.89 ----- Otter 3.3f, August 2004 -----
% 1.68/1.89 The process was started by sandbox2 on n019.cluster.edu,
% 1.68/1.89 Wed Jul 27 10:47:37 2022
% 1.68/1.89 The command was "./otter". The process ID is 20555.
% 1.68/1.89
% 1.68/1.89 set(prolog_style_variables).
% 1.68/1.89 set(auto).
% 1.68/1.89 dependent: set(auto1).
% 1.68/1.89 dependent: set(process_input).
% 1.68/1.89 dependent: clear(print_kept).
% 1.68/1.89 dependent: clear(print_new_demod).
% 1.68/1.89 dependent: clear(print_back_demod).
% 1.68/1.89 dependent: clear(print_back_sub).
% 1.68/1.89 dependent: set(control_memory).
% 1.68/1.89 dependent: assign(max_mem, 12000).
% 1.68/1.89 dependent: assign(pick_given_ratio, 4).
% 1.68/1.89 dependent: assign(stats_level, 1).
% 1.68/1.89 dependent: assign(max_seconds, 10800).
% 1.68/1.89 clear(print_given).
% 1.68/1.89
% 1.68/1.89 formula_list(usable).
% 1.68/1.89 all A (A=A).
% 1.68/1.89 all A B (in(A,B)-> -in(B,A)).
% 1.68/1.89 all A B (set_intersection2(A,B)=set_intersection2(B,A)).
% 1.68/1.89 all A B (B=singleton(A)<-> (all C (in(C,B)<->C=A))).
% 1.68/1.89 all A B (set_intersection2(A,A)=A).
% 1.68/1.89 all A B (set_intersection2(A,singleton(B))=singleton(B)->in(B,A)).
% 1.68/1.89 exists A empty(A).
% 1.68/1.89 exists A (-empty(A)).
% 1.68/1.89 -(all A B (set_intersection2(singleton(A),singleton(B))=singleton(A)->A=B)).
% 1.68/1.89 end_of_list.
% 1.68/1.89
% 1.68/1.89 -------> usable clausifies to:
% 1.68/1.89
% 1.68/1.89 list(usable).
% 1.68/1.89 0 [] A=A.
% 1.68/1.89 0 [] -in(A,B)| -in(B,A).
% 1.68/1.89 0 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.68/1.89 0 [] B!=singleton(A)| -in(C,B)|C=A.
% 1.68/1.89 0 [] B!=singleton(A)|in(C,B)|C!=A.
% 1.68/1.89 0 [] B=singleton(A)|in($f1(A,B),B)|$f1(A,B)=A.
% 1.68/1.89 0 [] B=singleton(A)| -in($f1(A,B),B)|$f1(A,B)!=A.
% 1.68/1.89 0 [] set_intersection2(A,A)=A.
% 1.68/1.89 0 [] set_intersection2(A,singleton(B))!=singleton(B)|in(B,A).
% 1.68/1.89 0 [] empty($c1).
% 1.68/1.89 0 [] -empty($c2).
% 1.68/1.89 0 [] set_intersection2(singleton($c4),singleton($c3))=singleton($c4).
% 1.68/1.89 0 [] $c4!=$c3.
% 1.68/1.89 end_of_list.
% 1.68/1.89
% 1.68/1.89 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.68/1.89
% 1.68/1.89 This ia a non-Horn set with equality. The strategy will be
% 1.68/1.89 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.68/1.89 deletion, with positive clauses in sos and nonpositive
% 1.68/1.89 clauses in usable.
% 1.68/1.89
% 1.68/1.89 dependent: set(knuth_bendix).
% 1.68/1.89 dependent: set(anl_eq).
% 1.68/1.89 dependent: set(para_from).
% 1.68/1.89 dependent: set(para_into).
% 1.68/1.89 dependent: clear(para_from_right).
% 1.68/1.89 dependent: clear(para_into_right).
% 1.68/1.89 dependent: set(para_from_vars).
% 1.68/1.89 dependent: set(eq_units_both_ways).
% 1.68/1.89 dependent: set(dynamic_demod_all).
% 1.68/1.89 dependent: set(dynamic_demod).
% 1.68/1.89 dependent: set(order_eq).
% 1.68/1.89 dependent: set(back_demod).
% 1.68/1.89 dependent: set(lrpo).
% 1.68/1.89 dependent: set(hyper_res).
% 1.68/1.89 dependent: set(unit_deletion).
% 1.68/1.89 dependent: set(factor).
% 1.68/1.89
% 1.68/1.89 ------------> process usable:
% 1.68/1.89 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.68/1.89 ** KEPT (pick-wt=10): 2 [] A!=singleton(B)| -in(C,A)|C=B.
% 1.68/1.89 ** KEPT (pick-wt=10): 3 [] A!=singleton(B)|in(C,A)|C!=B.
% 1.68/1.89 ** KEPT (pick-wt=14): 4 [] A=singleton(B)| -in($f1(B,A),A)|$f1(B,A)!=B.
% 1.68/1.89 ** KEPT (pick-wt=10): 5 [] set_intersection2(A,singleton(B))!=singleton(B)|in(B,A).
% 1.68/1.89 ** KEPT (pick-wt=2): 6 [] -empty($c2).
% 1.68/1.89 ** KEPT (pick-wt=3): 7 [] $c4!=$c3.
% 1.68/1.89
% 1.68/1.89 ------------> process sos:
% 1.68/1.89 ** KEPT (pick-wt=3): 9 [] A=A.
% 1.68/1.89 ** KEPT (pick-wt=7): 10 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.68/1.89 ** KEPT (pick-wt=14): 11 [] A=singleton(B)|in($f1(B,A),A)|$f1(B,A)=B.
% 1.68/1.89 ** KEPT (pick-wt=5): 12 [] set_intersection2(A,A)=A.
% 1.68/1.89 ---> New Demodulator: 13 [new_demod,12] set_intersection2(A,A)=A.
% 1.68/1.89 ** KEPT (pick-wt=2): 14 [] empty($c1).
% 1.68/1.89 ** KEPT (pick-wt=8): 15 [] set_intersection2(singleton($c4),singleton($c3))=singleton($c4).
% 1.68/1.89 ---> New Demodulator: 16 [new_demod,15] set_intersection2(singleton($c4),singleton($c3))=singleton($c4).
% 1.68/1.89 Following clause subsumed by 9 during input processing: 0 [copy,9,flip.1] A=A.
% 1.68/1.89 Following clause subsumed by 10 during input processing: 0 [copy,10,flip.1] set_intersection2(A,B)=set_intersection2(B,A).
% 1.68/1.89 >>>> Starting back demodulation with 13.
% 1.68/1.89 >>>> Starting back demodulation with 16.
% 1.68/1.89
% 1.68/1.89 ======= end of input processing =======
% 1.68/1.89
% 1.68/1.89 =========== start of search ===========
% 1.68/1.89
% 1.68/1.89 -------- PROOF --------
% 1.68/1.89
% 1.68/1.89 ----> UNIT CONFLICT at 0.00 sec ----> 85 [binary,83.1,7.1] $F.
% 1.68/1.89
% 1.68/1.89 Length of proof is 3. Level of proof is 3.
% 1.68/1.89
% 1.68/1.89 ---------------- PROOF ----------------
% 1.68/1.89 % SZS status Theorem
% 1.68/1.89 % SZS output start Refutation
% See solution above
% 1.68/1.89 ------------ end of proof -------------
% 1.68/1.89
% 1.68/1.89
% 1.68/1.89 Search stopped by max_proofs option.
% 1.68/1.89
% 1.68/1.89
% 1.68/1.89 Search stopped by max_proofs option.
% 1.68/1.89
% 1.68/1.89 ============ end of search ============
% 1.68/1.89
% 1.68/1.89 -------------- statistics -------------
% 1.68/1.89 clauses given 10
% 1.68/1.89 clauses generated 124
% 1.68/1.89 clauses kept 80
% 1.68/1.89 clauses forward subsumed 58
% 1.68/1.89 clauses back subsumed 0
% 1.68/1.89 Kbytes malloced 976
% 1.68/1.89
% 1.68/1.89 ----------- times (seconds) -----------
% 1.68/1.89 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.89 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.89 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.68/1.89
% 1.68/1.89 That finishes the proof of the theorem.
% 1.68/1.89
% 1.68/1.89 Process 20555 finished Wed Jul 27 10:47:38 2022
% 1.68/1.89 Otter interrupted
% 1.68/1.89 PROOF FOUND
%------------------------------------------------------------------------------