TSTP Solution File: RNG105+2 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : RNG105+2 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n032.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 20:41:29 EDT 2022
% Result : Theorem 0.15s 0.47s
% Output : Refutation 0.15s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 9
% Syntax : Number of clauses : 18 ( 10 unt; 0 nHn; 18 RR)
% Number of literals : 30 ( 0 equ; 19 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 4 ( 3 usr; 2 prp; 0-2 aty)
% Number of functors : 10 ( 10 usr; 8 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(6,axiom,
aElement0(xz),
file('RNG105+2.p',unknown),
[] ).
cnf(7,axiom,
aElement0(xu),
file('RNG105+2.p',unknown),
[] ).
cnf(8,axiom,
aElement0(xv),
file('RNG105+2.p',unknown),
[] ).
cnf(41,axiom,
equal(sdtasdt0(xc,sdtpldt0(xu,xv)),sdtpldt0(xx,xy)),
file('RNG105+2.p',unknown),
[] ).
cnf(42,axiom,
equal(sdtasdt0(xc,sdtasdt0(xu,xz)),sdtasdt0(xz,xx)),
file('RNG105+2.p',unknown),
[] ).
cnf(43,axiom,
( ~ aElement0(u)
| ~ aElement0(v)
| aElement0(sdtpldt0(v,u)) ),
file('RNG105+2.p',unknown),
[] ).
cnf(44,axiom,
( ~ aElement0(u)
| ~ aElement0(v)
| aElement0(sdtasdt0(v,u)) ),
file('RNG105+2.p',unknown),
[] ).
cnf(54,axiom,
( ~ aElement0(u)
| ~ equal(sdtasdt0(xc,u),sdtpldt0(xx,xy))
| skC0 ),
file('RNG105+2.p',unknown),
[] ).
cnf(56,axiom,
( ~ aElement0(u)
| ~ skC0
| ~ equal(sdtasdt0(xc,u),sdtasdt0(xz,xx)) ),
file('RNG105+2.p',unknown),
[] ).
cnf(135,plain,
( ~ aElement0(sdtasdt0(xu,xz))
| ~ skC0 ),
inference(res,[status(thm),theory(equality)],[42,56]),
[iquote('0:Res:42.0,56.1')] ).
cnf(144,plain,
( ~ aElement0(u)
| ~ equal(sdtasdt0(xc,u),sdtpldt0(xx,xy)) ),
inference(spt,[spt(split,[position(s1)])],[54]),
[iquote('1:Spt:54.0,54.1')] ).
cnf(181,plain,
( ~ aElement0(sdtpldt0(xu,xv))
| ~ equal(sdtpldt0(xx,xy),sdtpldt0(xx,xy)) ),
inference(spl,[status(thm),theory(equality)],[41,144]),
[iquote('1:SpL:41.0,144.1')] ).
cnf(182,plain,
~ aElement0(sdtpldt0(xu,xv)),
inference(obv,[status(thm),theory(equality)],[181]),
[iquote('1:Obv:181.1')] ).
cnf(249,plain,
( ~ aElement0(xu)
| ~ aElement0(xv) ),
inference(sor,[status(thm)],[182,43]),
[iquote('1:SoR:182.0,43.2')] ).
cnf(254,plain,
$false,
inference(ssi,[status(thm)],[249,8,7]),
[iquote('1:SSi:249.1,249.0,8.0,7.0')] ).
cnf(257,plain,
skC0,
inference(spt,[spt(split,[position(s2)])],[54]),
[iquote('1:Spt:254.0,54.2')] ).
cnf(258,plain,
~ skC0,
inference(ssi,[status(thm)],[135,44,7,6]),
[iquote('0:SSi:135.0,44.0,7.0,6.2')] ).
cnf(259,plain,
$false,
inference(mrr,[status(thm)],[258,257]),
[iquote('1:MRR:258.0,257.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.05/0.10 % Problem : RNG105+2 : TPTP v8.1.0. Released v4.0.0.
% 0.05/0.10 % Command : run_spass %d %s
% 0.10/0.30 % Computer : n032.cluster.edu
% 0.10/0.30 % Model : x86_64 x86_64
% 0.10/0.30 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.30 % Memory : 8042.1875MB
% 0.10/0.30 % OS : Linux 3.10.0-693.el7.x86_64
% 0.10/0.30 % CPULimit : 300
% 0.10/0.30 % WCLimit : 600
% 0.10/0.30 % DateTime : Mon May 30 18:10:57 EDT 2022
% 0.10/0.30 % CPUTime :
% 0.15/0.47
% 0.15/0.47 SPASS V 3.9
% 0.15/0.47 SPASS beiseite: Proof found.
% 0.15/0.47 % SZS status Theorem
% 0.15/0.47 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.15/0.47 SPASS derived 98 clauses, backtracked 53 clauses, performed 3 splits and kept 172 clauses.
% 0.15/0.47 SPASS allocated 99444 KBytes.
% 0.15/0.47 SPASS spent 0:00:00.16 on the problem.
% 0.15/0.47 0:00:00.03 for the input.
% 0.15/0.47 0:00:00.10 for the FLOTTER CNF translation.
% 0.15/0.47 0:00:00.00 for inferences.
% 0.15/0.47 0:00:00.00 for the backtracking.
% 0.15/0.47 0:00:00.01 for the reduction.
% 0.15/0.47
% 0.15/0.47
% 0.15/0.47 Here is a proof with depth 3, length 18 :
% 0.15/0.47 % SZS output start Refutation
% See solution above
% 0.15/0.47 Formulae used in the proof : m__1933 m__1956 m__1979 m__2010 m__2043 mSortsB mSortsB_02 m__
% 0.15/0.47
%------------------------------------------------------------------------------