TSTP Solution File: GRP665-10 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : GRP665-10 : TPTP v8.1.0. Released v8.1.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n003.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 12:57:36 EDT 2022
% Result : Unsatisfiable 1.59s 1.79s
% Output : Refutation 1.59s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 4
% Syntax : Number of clauses : 10 ( 10 unt; 0 nHn; 4 RR)
% Number of literals : 10 ( 9 equ; 3 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 4 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 3 con; 0-2 aty)
% Number of variables : 10 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
mult(op_c,mult(x0,x1)) != mult(mult(op_c,x0),x1),
file('GRP665-10.p',unknown),
[] ).
cnf(2,plain,
mult(mult(op_c,x0),x1) != mult(op_c,mult(x0,x1)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
[iquote('copy,1,flip.1')] ).
cnf(6,axiom,
ld(A,mult(A,B)) = B,
file('GRP665-10.p',unknown),
[] ).
cnf(19,axiom,
mult(mult(A,B),C) = mult(mult(A,C),ld(C,mult(B,C))),
file('GRP665-10.p',unknown),
[] ).
cnf(20,axiom,
mult(op_c,A) = mult(A,op_c),
file('GRP665-10.p',unknown),
[] ).
cnf(22,plain,
mult(A,op_c) = mult(op_c,A),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[20])]),
[iquote('copy,20,flip.1')] ).
cnf(56,plain,
ld(op_c,mult(A,op_c)) = A,
inference(para_from,[status(thm),theory(equality)],[20,6]),
[iquote('para_from,20.1.1,6.1.1.2')] ).
cnf(57,plain,
mult(mult(x0,op_c),x1) != mult(op_c,mult(x0,x1)),
inference(para_from,[status(thm),theory(equality)],[20,2]),
[iquote('para_from,20.1.1,2.1.1.1')] ).
cnf(71,plain,
mult(mult(A,op_c),B) = mult(op_c,mult(A,B)),
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[19,22]),56])]),
[iquote('para_into,19.1.1,22.1.1,demod,56,flip.1')] ).
cnf(73,plain,
$false,
inference(binary,[status(thm)],[71,57]),
[iquote('binary,71.1,57.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : GRP665-10 : TPTP v8.1.0. Released v8.1.0.
% 0.11/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n003.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 05:08:11 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.59/1.79 ----- Otter 3.3f, August 2004 -----
% 1.59/1.79 The process was started by sandbox2 on n003.cluster.edu,
% 1.59/1.79 Wed Jul 27 05:08:11 2022
% 1.59/1.79 The command was "./otter". The process ID is 25778.
% 1.59/1.79
% 1.59/1.79 set(prolog_style_variables).
% 1.59/1.79 set(auto).
% 1.59/1.79 dependent: set(auto1).
% 1.59/1.79 dependent: set(process_input).
% 1.59/1.79 dependent: clear(print_kept).
% 1.59/1.79 dependent: clear(print_new_demod).
% 1.59/1.79 dependent: clear(print_back_demod).
% 1.59/1.79 dependent: clear(print_back_sub).
% 1.59/1.79 dependent: set(control_memory).
% 1.59/1.79 dependent: assign(max_mem, 12000).
% 1.59/1.79 dependent: assign(pick_given_ratio, 4).
% 1.59/1.79 dependent: assign(stats_level, 1).
% 1.59/1.79 dependent: assign(max_seconds, 10800).
% 1.59/1.79 clear(print_given).
% 1.59/1.79
% 1.59/1.79 list(usable).
% 1.59/1.79 0 [] A=A.
% 1.59/1.79 0 [] mult(A,ld(A,B))=B.
% 1.59/1.79 0 [] ld(A,mult(A,B))=B.
% 1.59/1.79 0 [] mult(rd(A,B),B)=A.
% 1.59/1.79 0 [] rd(mult(A,B),B)=A.
% 1.59/1.79 0 [] mult(A,unit)=A.
% 1.59/1.79 0 [] mult(unit,A)=A.
% 1.59/1.79 0 [] mult(A,mult(B,C))=mult(rd(mult(A,B),A),mult(A,C)).
% 1.59/1.79 0 [] mult(mult(A,B),C)=mult(mult(A,C),ld(C,mult(B,C))).
% 1.59/1.79 0 [] mult(op_c,A)=mult(A,op_c).
% 1.59/1.79 0 [] mult(op_c,mult(x0,x1))!=mult(mult(op_c,x0),x1).
% 1.59/1.79 end_of_list.
% 1.59/1.79
% 1.59/1.79 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.59/1.79
% 1.59/1.79 All clauses are units, and equality is present; the
% 1.59/1.79 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.59/1.79
% 1.59/1.79 dependent: set(knuth_bendix).
% 1.59/1.79 dependent: set(anl_eq).
% 1.59/1.79 dependent: set(para_from).
% 1.59/1.79 dependent: set(para_into).
% 1.59/1.79 dependent: clear(para_from_right).
% 1.59/1.79 dependent: clear(para_into_right).
% 1.59/1.79 dependent: set(para_from_vars).
% 1.59/1.79 dependent: set(eq_units_both_ways).
% 1.59/1.79 dependent: set(dynamic_demod_all).
% 1.59/1.79 dependent: set(dynamic_demod).
% 1.59/1.79 dependent: set(order_eq).
% 1.59/1.79 dependent: set(back_demod).
% 1.59/1.79 dependent: set(lrpo).
% 1.59/1.79
% 1.59/1.79 ------------> process usable:
% 1.59/1.79 ** KEPT (pick-wt=11): 2 [copy,1,flip.1] mult(mult(op_c,x0),x1)!=mult(op_c,mult(x0,x1)).
% 1.59/1.79
% 1.59/1.79 ------------> process sos:
% 1.59/1.79 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.59/1.79 ** KEPT (pick-wt=7): 4 [] mult(A,ld(A,B))=B.
% 1.59/1.79 ---> New Demodulator: 5 [new_demod,4] mult(A,ld(A,B))=B.
% 1.59/1.79 ** KEPT (pick-wt=7): 6 [] ld(A,mult(A,B))=B.
% 1.59/1.79 ---> New Demodulator: 7 [new_demod,6] ld(A,mult(A,B))=B.
% 1.59/1.79 ** KEPT (pick-wt=7): 8 [] mult(rd(A,B),B)=A.
% 1.59/1.79 ---> New Demodulator: 9 [new_demod,8] mult(rd(A,B),B)=A.
% 1.59/1.79 ** KEPT (pick-wt=7): 10 [] rd(mult(A,B),B)=A.
% 1.59/1.79 ---> New Demodulator: 11 [new_demod,10] rd(mult(A,B),B)=A.
% 1.59/1.79 ** KEPT (pick-wt=5): 12 [] mult(A,unit)=A.
% 1.59/1.79 ---> New Demodulator: 13 [new_demod,12] mult(A,unit)=A.
% 1.59/1.79 ** KEPT (pick-wt=5): 14 [] mult(unit,A)=A.
% 1.59/1.79 ---> New Demodulator: 15 [new_demod,14] mult(unit,A)=A.
% 1.59/1.79 ** KEPT (pick-wt=15): 17 [copy,16,flip.1] mult(rd(mult(A,B),A),mult(A,C))=mult(A,mult(B,C)).
% 1.59/1.79 ---> New Demodulator: 18 [new_demod,17] mult(rd(mult(A,B),A),mult(A,C))=mult(A,mult(B,C)).
% 1.59/1.79 ** KEPT (pick-wt=15): 19 [] mult(mult(A,B),C)=mult(mult(A,C),ld(C,mult(B,C))).
% 1.59/1.79 ** KEPT (pick-wt=7): 20 [] mult(op_c,A)=mult(A,op_c).
% 1.59/1.79 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.59/1.79 >>>> Starting back demodulation with 5.
% 1.59/1.79 >>>> Starting back demodulation with 7.
% 1.59/1.79 >>>> Starting back demodulation with 9.
% 1.59/1.79 >>>> Starting back demodulation with 11.
% 1.59/1.79 >>>> Starting back demodulation with 13.
% 1.59/1.79 >>>> Starting back demodulation with 15.
% 1.59/1.79 >>>> Starting back demodulation with 18.
% 1.59/1.79 ** KEPT (pick-wt=15): 21 [copy,19,flip.1] mult(mult(A,B),ld(B,mult(C,B)))=mult(mult(A,C),B).
% 1.59/1.79 ** KEPT (pick-wt=7): 22 [copy,20,flip.1] mult(A,op_c)=mult(op_c,A).
% 1.59/1.79 Following clause subsumed by 19 during input processing: 0 [copy,21,flip.1] mult(mult(A,B),C)=mult(mult(A,C),ld(C,mult(B,C))).
% 1.59/1.79 Following clause subsumed by 20 during input processing: 0 [copy,22,flip.1] mult(op_c,A)=mult(A,op_c).
% 1.59/1.79
% 1.59/1.79 ======= end of input processing =======
% 1.59/1.79
% 1.59/1.79 =========== start of search ===========
% 1.59/1.79
% 1.59/1.79 -------- PROOF --------
% 1.59/1.79
% 1.59/1.79 ----> UNIT CONFLICT at 0.00 sec ----> 73 [binary,71.1,57.1] $F.
% 1.59/1.79
% 1.59/1.79 Length of proof is 5. Level of proof is 2.
% 1.59/1.79
% 1.59/1.79 ---------------- PROOF ----------------
% 1.59/1.79 % SZS status Unsatisfiable
% 1.59/1.79 % SZS output start Refutation
% See solution above
% 1.59/1.79 ------------ end of proof -------------
% 1.59/1.79
% 1.59/1.79
% 1.59/1.79 Search stopped by max_proofs option.
% 1.59/1.79
% 1.59/1.79
% 1.59/1.79 Search stopped by max_proofs option.
% 1.59/1.79
% 1.59/1.79 ============ end of search ============
% 1.59/1.79
% 1.59/1.79 -------------- statistics -------------
% 1.59/1.79 clauses given 16
% 1.59/1.79 clauses generated 94
% 1.59/1.79 clauses kept 40
% 1.59/1.79 clauses forward subsumed 75
% 1.59/1.79 clauses back subsumed 0
% 1.59/1.79 Kbytes malloced 976
% 1.59/1.79
% 1.59/1.79 ----------- times (seconds) -----------
% 1.59/1.79 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.59/1.79 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.59/1.79 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.59/1.79
% 1.59/1.79 That finishes the proof of the theorem.
% 1.59/1.79
% 1.59/1.79 Process 25778 finished Wed Jul 27 05:08:13 2022
% 1.59/1.79 Otter interrupted
% 1.59/1.79 PROOF FOUND
%------------------------------------------------------------------------------