TSTP Solution File: NUN087+2 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : NUN087+2 : TPTP v8.1.0. Released v7.3.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n025.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 16:38:48 EDT 2022
% Result : Theorem 0.20s 0.44s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 7
% Syntax : Number of clauses : 18 ( 12 unt; 1 nHn; 18 RR)
% Number of literals : 25 ( 0 equ; 11 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 5 ( 4 usr; 1 prp; 0-3 aty)
% Number of functors : 6 ( 6 usr; 3 con; 0-1 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
r1(skf22(u)),
file('NUN087+2.p',unknown),
[] ).
cnf(2,axiom,
r1(skf24(u)),
file('NUN087+2.p',unknown),
[] ).
cnf(3,axiom,
r1(skf23(u)),
file('NUN087+2.p',unknown),
[] ).
cnf(8,axiom,
( equal(u,skc1)
| skP0(skc1,u) ),
file('NUN087+2.p',unknown),
[] ).
cnf(11,axiom,
r4(u,skf24(u),skf23(u)),
file('NUN087+2.p',unknown),
[] ).
cnf(13,axiom,
( ~ r1(u)
| ~ skP0(v,u) ),
file('NUN087+2.p',unknown),
[] ).
cnf(26,axiom,
( ~ r1(u)
| ~ r1(v)
| ~ r4(v,v,u) ),
file('NUN087+2.p',unknown),
[] ).
cnf(36,plain,
( ~ r1(u)
| ~ r4(skf22(v),skf22(v),u) ),
inference(res,[status(thm),theory(equality)],[1,26]),
[iquote('0:Res:1.0,26.0')] ).
cnf(41,plain,
( ~ r1(u)
| equal(u,skc1) ),
inference(res,[status(thm),theory(equality)],[8,13]),
[iquote('0:Res:8.1,13.1')] ).
cnf(43,plain,
equal(skf22(u),skc1),
inference(ems,[status(thm)],[41,1]),
[iquote('0:EmS:41.0,1.0')] ).
cnf(44,plain,
equal(skf24(u),skc1),
inference(ems,[status(thm)],[41,2]),
[iquote('0:EmS:41.0,2.0')] ).
cnf(45,plain,
equal(skf23(u),skc1),
inference(ems,[status(thm)],[41,3]),
[iquote('0:EmS:41.0,3.0')] ).
cnf(46,plain,
r1(skc1),
inference(rew,[status(thm),theory(equality)],[43,1]),
[iquote('0:Rew:43.0,1.0')] ).
cnf(49,plain,
( ~ r1(u)
| ~ r4(skc1,skc1,u) ),
inference(rew,[status(thm),theory(equality)],[43,36]),
[iquote('0:Rew:43.0,36.1')] ).
cnf(51,plain,
r4(u,skc1,skf23(u)),
inference(rew,[status(thm),theory(equality)],[44,11]),
[iquote('0:Rew:44.0,11.0')] ).
cnf(57,plain,
r4(u,skc1,skc1),
inference(rew,[status(thm),theory(equality)],[45,51]),
[iquote('0:Rew:45.0,51.0')] ).
cnf(73,plain,
~ r1(skc1),
inference(res,[status(thm),theory(equality)],[57,49]),
[iquote('0:Res:57.0,49.1')] ).
cnf(74,plain,
$false,
inference(ssi,[status(thm)],[73,46]),
[iquote('0:SSi:73.0,46.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : NUN087+2 : TPTP v8.1.0. Released v7.3.0.
% 0.11/0.13 % Command : run_spass %d %s
% 0.13/0.34 % Computer : n025.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 : Thu Jun 2 06:29:59 EDT 2022
% 0.13/0.34 % CPUTime :
% 0.20/0.44
% 0.20/0.44 SPASS V 3.9
% 0.20/0.44 SPASS beiseite: Proof found.
% 0.20/0.44 % SZS status Theorem
% 0.20/0.44 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.20/0.44 SPASS derived 39 clauses, backtracked 0 clauses, performed 0 splits and kept 58 clauses.
% 0.20/0.44 SPASS allocated 97687 KBytes.
% 0.20/0.44 SPASS spent 0:00:00.09 on the problem.
% 0.20/0.44 0:00:00.04 for the input.
% 0.20/0.44 0:00:00.03 for the FLOTTER CNF translation.
% 0.20/0.44 0:00:00.00 for inferences.
% 0.20/0.44 0:00:00.00 for the backtracking.
% 0.20/0.44 0:00:00.00 for the reduction.
% 0.20/0.44
% 0.20/0.44
% 0.20/0.44 Here is a proof with depth 2, length 18 :
% 0.20/0.44 % SZS output start Refutation
% See solution above
% 0.20/0.44 Formulae used in the proof : axiom_4a axiom_5a axiom_1 zerotimeszeroeqzero
% 0.20/0.44
%------------------------------------------------------------------------------