TSTP Solution File: NUM482+1 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : NUM482+1 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n009.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 14:26:39 EDT 2022
% Result : Theorem 0.18s 0.47s
% Output : Refutation 0.18s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 7
% Syntax : Number of clauses : 16 ( 6 unt; 0 nHn; 16 RR)
% Number of literals : 37 ( 0 equ; 25 neg)
% Maximal clause size : 5 ( 2 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 5 ( 4 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 5 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
isPrime0(xk),
file('NUM482+1.p',unknown),
[] ).
cnf(3,axiom,
aNaturalNumber0(sz10),
file('NUM482+1.p',unknown),
[] ).
cnf(4,axiom,
aNaturalNumber0(xk),
file('NUM482+1.p',unknown),
[] ).
cnf(15,axiom,
( ~ aNaturalNumber0(u)
| equal(sdtasdt0(u,sz10),u) ),
file('NUM482+1.p',unknown),
[] ).
cnf(19,axiom,
( ~ isPrime0(u)
| ~ aNaturalNumber0(u)
| ~ doDivides0(u,xk) ),
file('NUM482+1.p',unknown),
[] ).
cnf(21,axiom,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| aNaturalNumber0(sdtasdt0(v,u)) ),
file('NUM482+1.p',unknown),
[] ).
cnf(46,axiom,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| ~ aNaturalNumber0(w)
| ~ equal(u,sdtasdt0(v,w))
| doDivides0(v,u) ),
file('NUM482+1.p',unknown),
[] ).
cnf(76,plain,
( ~ aNaturalNumber0(xk)
| ~ doDivides0(xk,xk) ),
inference(res,[status(thm),theory(equality)],[1,19]),
[iquote('0:Res:1.0,19.1')] ).
cnf(88,plain,
~ doDivides0(xk,xk),
inference(mrr,[status(thm)],[76,4]),
[iquote('0:MRR:76.0,4.0')] ).
cnf(397,plain,
( ~ aNaturalNumber0(sdtasdt0(u,v))
| ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| doDivides0(u,sdtasdt0(u,v)) ),
inference(eqr,[status(thm),theory(equality)],[46]),
[iquote('0:EqR:46.3')] ).
cnf(405,plain,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| doDivides0(u,sdtasdt0(u,v)) ),
inference(ssi,[status(thm)],[397,21]),
[iquote('0:SSi:397.0,21.2')] ).
cnf(444,plain,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(sz10)
| doDivides0(u,u) ),
inference(spr,[status(thm),theory(equality)],[15,405]),
[iquote('0:SpR:15.1,405.2')] ).
cnf(457,plain,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(sz10)
| doDivides0(u,u) ),
inference(obv,[status(thm),theory(equality)],[444]),
[iquote('0:Obv:444.0')] ).
cnf(458,plain,
( ~ aNaturalNumber0(u)
| doDivides0(u,u) ),
inference(ssi,[status(thm)],[457,3]),
[iquote('0:SSi:457.1,3.0')] ).
cnf(482,plain,
~ aNaturalNumber0(xk),
inference(res,[status(thm),theory(equality)],[458,88]),
[iquote('0:Res:458.1,88.0')] ).
cnf(487,plain,
$false,
inference(ssi,[status(thm)],[482,1,4]),
[iquote('0:SSi:482.0,1.0,4.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.12 % Problem : NUM482+1 : TPTP v8.1.0. Released v4.0.0.
% 0.10/0.12 % Command : run_spass %d %s
% 0.13/0.32 % Computer : n009.cluster.edu
% 0.13/0.32 % Model : x86_64 x86_64
% 0.13/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.32 % Memory : 8042.1875MB
% 0.13/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.32 % CPULimit : 300
% 0.13/0.32 % WCLimit : 600
% 0.13/0.32 % DateTime : Thu Jul 7 05:43:37 EDT 2022
% 0.13/0.32 % CPUTime :
% 0.18/0.47
% 0.18/0.47 SPASS V 3.9
% 0.18/0.47 SPASS beiseite: Proof found.
% 0.18/0.47 % SZS status Theorem
% 0.18/0.47 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.18/0.47 SPASS derived 268 clauses, backtracked 0 clauses, performed 0 splits and kept 146 clauses.
% 0.18/0.47 SPASS allocated 98013 KBytes.
% 0.18/0.47 SPASS spent 0:00:00.14 on the problem.
% 0.18/0.47 0:00:00.04 for the input.
% 0.18/0.47 0:00:00.04 for the FLOTTER CNF translation.
% 0.18/0.47 0:00:00.00 for inferences.
% 0.18/0.47 0:00:00.00 for the backtracking.
% 0.18/0.47 0:00:00.03 for the reduction.
% 0.18/0.47
% 0.18/0.47
% 0.18/0.47 Here is a proof with depth 3, length 16 :
% 0.18/0.47 % SZS output start Refutation
% See solution above
% 0.18/0.47 Formulae used in the proof : m__ mSortsC_01 m__1716 m_MulUnit mSortsB_02 mDefDiv
% 0.18/0.47
%------------------------------------------------------------------------------