TSTP Solution File: NUM436+3 by SPASS---3.9
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- Process Solution
%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : NUM436+3 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n006.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 14:26:11 EDT 2022
% Result : Theorem 0.20s 0.46s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 8
% Syntax : Number of clauses : 17 ( 10 unt; 0 nHn; 17 RR)
% Number of literals : 27 ( 0 equ; 17 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; 7 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(5,axiom,
aInteger0(xp),
file('NUM436+3.p',unknown),
[] ).
cnf(6,axiom,
aInteger0(xq),
file('NUM436+3.p',unknown),
[] ).
cnf(8,axiom,
aInteger0(xm),
file('NUM436+3.p',unknown),
[] ).
cnf(30,axiom,
( ~ aInteger0(u)
| ~ aInteger0(v)
| aInteger0(sdtasdt0(v,u)) ),
file('NUM436+3.p',unknown),
[] ).
cnf(35,axiom,
equal(sdtasdt0(xp,sdtasdt0(xq,xm)),sdtpldt0(xa,smndt0(xb))),
file('NUM436+3.p',unknown),
[] ).
cnf(36,axiom,
equal(sdtasdt0(xq,sdtasdt0(xp,xm)),sdtpldt0(xa,smndt0(xb))),
file('NUM436+3.p',unknown),
[] ).
cnf(38,axiom,
( ~ aInteger0(u)
| ~ equal(sdtasdt0(xp,u),sdtpldt0(xa,smndt0(xb)))
| skC0 ),
file('NUM436+3.p',unknown),
[] ).
cnf(42,axiom,
( ~ aInteger0(u)
| ~ skC0
| ~ equal(sdtasdt0(xq,u),sdtpldt0(xa,smndt0(xb))) ),
file('NUM436+3.p',unknown),
[] ).
cnf(73,plain,
( ~ aInteger0(sdtasdt0(xp,xm))
| ~ skC0 ),
inference(res,[status(thm),theory(equality)],[36,42]),
[iquote('0:Res:36.0,42.1')] ).
cnf(74,plain,
( ~ aInteger0(u)
| ~ equal(sdtasdt0(xp,u),sdtpldt0(xa,smndt0(xb))) ),
inference(spt,[spt(split,[position(s1)])],[38]),
[iquote('1:Spt:38.0,38.1')] ).
cnf(125,plain,
( ~ aInteger0(sdtasdt0(xq,xm))
| ~ equal(sdtpldt0(xa,smndt0(xb)),sdtpldt0(xa,smndt0(xb))) ),
inference(spl,[status(thm),theory(equality)],[35,74]),
[iquote('1:SpL:35.0,74.1')] ).
cnf(126,plain,
~ aInteger0(sdtasdt0(xq,xm)),
inference(obv,[status(thm),theory(equality)],[125]),
[iquote('1:Obv:125.1')] ).
cnf(139,plain,
( ~ aInteger0(xq)
| ~ aInteger0(xm) ),
inference(sor,[status(thm)],[126,30]),
[iquote('1:SoR:126.0,30.2')] ).
cnf(151,plain,
$false,
inference(ssi,[status(thm)],[139,8,6]),
[iquote('1:SSi:139.1,139.0,8.0,6.0')] ).
cnf(156,plain,
skC0,
inference(spt,[spt(split,[position(s2)])],[38]),
[iquote('1:Spt:151.0,38.2')] ).
cnf(158,plain,
~ skC0,
inference(ssi,[status(thm)],[73,30,5,8]),
[iquote('0:SSi:73.0,30.0,5.0,8.2')] ).
cnf(159,plain,
$false,
inference(mrr,[status(thm)],[158,156]),
[iquote('1:MRR:158.0,156.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : NUM436+3 : TPTP v8.1.0. Released v4.0.0.
% 0.11/0.13 % Command : run_spass %d %s
% 0.13/0.33 % Computer : n006.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 : 600
% 0.13/0.33 % DateTime : Tue Jul 5 09:03:52 EDT 2022
% 0.13/0.34 % CPUTime :
% 0.20/0.46
% 0.20/0.46 SPASS V 3.9
% 0.20/0.46 SPASS beiseite: Proof found.
% 0.20/0.46 % SZS status Theorem
% 0.20/0.46 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.20/0.46 SPASS derived 89 clauses, backtracked 28 clauses, performed 3 splits and kept 114 clauses.
% 0.20/0.46 SPASS allocated 97782 KBytes.
% 0.20/0.46 SPASS spent 0:00:00.11 on the problem.
% 0.20/0.46 0:00:00.04 for the input.
% 0.20/0.46 0:00:00.03 for the FLOTTER CNF translation.
% 0.20/0.46 0:00:00.00 for inferences.
% 0.20/0.46 0:00:00.00 for the backtracking.
% 0.20/0.46 0:00:00.01 for the reduction.
% 0.20/0.46
% 0.20/0.46
% 0.20/0.46 Here is a proof with depth 3, length 17 :
% 0.20/0.46 % SZS output start Refutation
% See solution above
% 0.20/0.46 Formulae used in the proof : m__979 m__1032 mIntMult m__1071 m__
% 0.20/0.46
%------------------------------------------------------------------------------