TSTP Solution File: NUM623+3 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : NUM623+3 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n013.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 : 600s
% DateTime : Mon Jul 18 14:28:14 EDT 2022
% Result : Theorem 1.18s 1.36s
% Output : Refutation 1.18s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 9
% Syntax : Number of clauses : 17 ( 9 unt; 0 nHn; 17 RR)
% Number of literals : 31 ( 0 equ; 20 neg)
% Maximal clause size : 5 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 4 ( 3 usr; 1 prp; 0-2 aty)
% Number of functors : 9 ( 9 usr; 8 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(30,axiom,
aElementOf0(xp,xQ),
file('NUM623+3.p',unknown),
[] ).
cnf(36,axiom,
aElementOf0(xx,szNzAzT0),
file('NUM623+3.p',unknown),
[] ).
cnf(38,axiom,
aElementOf0(xx,xQ),
file('NUM623+3.p',unknown),
[] ).
cnf(39,axiom,
~ equal(xp,xx),
file('NUM623+3.p',unknown),
[] ).
cnf(78,axiom,
aElementOf0(xp,sdtlpdtrp0(xN,xm)),
file('NUM623+3.p',unknown),
[] ).
cnf(108,axiom,
( ~ aElementOf0(u,xQ)
| aElementOf0(u,szNzAzT0) ),
file('NUM623+3.p',unknown),
[] ).
cnf(110,axiom,
( ~ aElementOf0(u,xQ)
| sdtlseqdt0(xp,u) ),
file('NUM623+3.p',unknown),
[] ).
cnf(141,axiom,
( ~ aElementOf0(u,sdtlpdtrp0(xN,xm))
| sdtlseqdt0(xx,u) ),
file('NUM623+3.p',unknown),
[] ).
cnf(272,axiom,
( ~ sdtlseqdt0(u,v)
| ~ sdtlseqdt0(v,u)
| ~ aElementOf0(u,szNzAzT0)
| ~ aElementOf0(v,szNzAzT0)
| equal(v,u) ),
file('NUM623+3.p',unknown),
[] ).
cnf(517,plain,
( ~ aElementOf0(xp,szNzAzT0)
| ~ sdtlseqdt0(xp,xx)
| ~ sdtlseqdt0(xx,xp)
| ~ aElementOf0(xx,szNzAzT0) ),
inference(res,[status(thm),theory(equality)],[272,39]),
[iquote('0:Res:272.4,39.0')] ).
cnf(536,plain,
( ~ sdtlseqdt0(xx,xp)
| ~ sdtlseqdt0(xp,xx)
| ~ aElementOf0(xp,szNzAzT0) ),
inference(mrr,[status(thm)],[517,36]),
[iquote('0:MRR:517.3,36.0')] ).
cnf(576,plain,
aElementOf0(xp,szNzAzT0),
inference(res,[status(thm),theory(equality)],[30,108]),
[iquote('0:Res:30.0,108.0')] ).
cnf(579,plain,
( ~ sdtlseqdt0(xx,xp)
| ~ sdtlseqdt0(xp,xx) ),
inference(mrr,[status(thm)],[536,576]),
[iquote('0:MRR:536.2,576.0')] ).
cnf(602,plain,
( ~ aElementOf0(xx,xQ)
| ~ sdtlseqdt0(xx,xp) ),
inference(res,[status(thm),theory(equality)],[110,579]),
[iquote('0:Res:110.1,579.1')] ).
cnf(603,plain,
~ sdtlseqdt0(xx,xp),
inference(mrr,[status(thm)],[602,38]),
[iquote('0:MRR:602.0,38.0')] ).
cnf(666,plain,
sdtlseqdt0(xx,xp),
inference(res,[status(thm),theory(equality)],[78,141]),
[iquote('0:Res:78.0,141.0')] ).
cnf(668,plain,
$false,
inference(mrr,[status(thm)],[666,603]),
[iquote('0:MRR:666.0,603.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : NUM623+3 : TPTP v8.1.0. Released v4.0.0.
% 0.07/0.13 % Command : run_spass %d %s
% 0.13/0.34 % Computer : n013.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 600
% 0.13/0.34 % DateTime : Tue Jul 5 16:28:44 EDT 2022
% 0.13/0.35 % CPUTime :
% 1.18/1.36
% 1.18/1.36 SPASS V 3.9
% 1.18/1.36 SPASS beiseite: Proof found.
% 1.18/1.36 % SZS status Theorem
% 1.18/1.36 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 1.18/1.36 SPASS derived 138 clauses, backtracked 0 clauses, performed 0 splits and kept 443 clauses.
% 1.18/1.36 SPASS allocated 105043 KBytes.
% 1.18/1.36 SPASS spent 0:00:01.00 on the problem.
% 1.18/1.36 0:00:00.04 for the input.
% 1.18/1.36 0:00:00.87 for the FLOTTER CNF translation.
% 1.18/1.36 0:00:00.00 for inferences.
% 1.18/1.36 0:00:00.00 for the backtracking.
% 1.18/1.36 0:00:00.04 for the reduction.
% 1.18/1.36
% 1.18/1.36
% 1.18/1.36 Here is a proof with depth 2, length 17 :
% 1.18/1.36 % SZS output start Refutation
% See solution above
% 1.18/1.36 Formulae used in the proof : m__5173 m__5365 m__5481 m__ m__5106 m__5147 m__5401 mLessASymm
% 1.18/1.36
%------------------------------------------------------------------------------