TSTP Solution File: NUM477+2 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : NUM477+2 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n029.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 14:26:36 EDT 2022
% Result : Theorem 0.20s 0.48s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 8
% Number of leaves : 9
% Syntax : Number of clauses : 17 ( 10 unt; 3 nHn; 17 RR)
% Number of literals : 31 ( 0 equ; 13 neg)
% Maximal clause size : 4 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 4 ( 3 usr; 1 prp; 0-2 aty)
% Number of functors : 7 ( 7 usr; 6 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
aNaturalNumber0(sz00),
file('NUM477+2.p',unknown),
[] ).
cnf(3,axiom,
aNaturalNumber0(xm),
file('NUM477+2.p',unknown),
[] ).
cnf(5,axiom,
aNaturalNumber0(skc1),
file('NUM477+2.p',unknown),
[] ).
cnf(7,axiom,
~ sdtlseqdt0(xm,xn),
file('NUM477+2.p',unknown),
[] ).
cnf(11,axiom,
~ equal(xn,sz00),
file('NUM477+2.p',unknown),
[] ).
cnf(12,axiom,
equal(sdtasdt0(xm,skc1),xn),
file('NUM477+2.p',unknown),
[] ).
cnf(19,axiom,
( ~ aNaturalNumber0(u)
| equal(sdtasdt0(sz00,u),sz00) ),
file('NUM477+2.p',unknown),
[] ).
cnf(26,axiom,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| equal(sdtasdt0(v,u),sdtasdt0(u,v)) ),
file('NUM477+2.p',unknown),
[] ).
cnf(29,axiom,
( ~ aNaturalNumber0(u)
| ~ aNaturalNumber0(v)
| equal(v,sz00)
| sdtlseqdt0(u,sdtasdt0(u,v)) ),
file('NUM477+2.p',unknown),
[] ).
cnf(281,plain,
( ~ aNaturalNumber0(xm)
| ~ aNaturalNumber0(skc1)
| equal(skc1,sz00)
| sdtlseqdt0(xm,xn) ),
inference(spr,[status(thm),theory(equality)],[12,29]),
[iquote('0:SpR:12.0,29.3')] ).
cnf(283,plain,
( equal(skc1,sz00)
| sdtlseqdt0(xm,xn) ),
inference(ssi,[status(thm)],[281,5,3]),
[iquote('0:SSi:281.1,281.0,5.0,3.0')] ).
cnf(284,plain,
equal(skc1,sz00),
inference(mrr,[status(thm)],[283,7]),
[iquote('0:MRR:283.1,7.0')] ).
cnf(286,plain,
equal(sdtasdt0(xm,sz00),xn),
inference(rew,[status(thm),theory(equality)],[284,12]),
[iquote('0:Rew:284.0,12.0')] ).
cnf(292,plain,
( ~ aNaturalNumber0(xm)
| ~ aNaturalNumber0(sz00)
| equal(sdtasdt0(sz00,xm),xn) ),
inference(spr,[status(thm),theory(equality)],[286,26]),
[iquote('0:SpR:286.0,26.2')] ).
cnf(300,plain,
( ~ aNaturalNumber0(xm)
| ~ aNaturalNumber0(sz00)
| equal(xn,sz00) ),
inference(rew,[status(thm),theory(equality)],[19,292]),
[iquote('0:Rew:19.1,292.2')] ).
cnf(301,plain,
equal(xn,sz00),
inference(ssi,[status(thm)],[300,1,3]),
[iquote('0:SSi:300.1,300.0,1.0,3.0')] ).
cnf(302,plain,
$false,
inference(mrr,[status(thm)],[301,11]),
[iquote('0:MRR:301.0,11.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.13 % Problem : NUM477+2 : TPTP v8.1.0. Released v4.0.0.
% 0.04/0.13 % Command : run_spass %d %s
% 0.14/0.35 % Computer : n029.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 600
% 0.14/0.35 % DateTime : Tue Jul 5 17:03:27 EDT 2022
% 0.14/0.35 % CPUTime :
% 0.20/0.48
% 0.20/0.48 SPASS V 3.9
% 0.20/0.48 SPASS beiseite: Proof found.
% 0.20/0.48 % SZS status Theorem
% 0.20/0.48 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.20/0.48 SPASS derived 153 clauses, backtracked 19 clauses, performed 1 splits and kept 122 clauses.
% 0.20/0.48 SPASS allocated 97840 KBytes.
% 0.20/0.48 SPASS spent 0:00:00.12 on the problem.
% 0.20/0.48 0:00:00.04 for the input.
% 0.20/0.48 0:00:00.04 for the FLOTTER CNF translation.
% 0.20/0.48 0:00:00.00 for inferences.
% 0.20/0.48 0:00:00.00 for the backtracking.
% 0.20/0.48 0:00:00.01 for the reduction.
% 0.20/0.48
% 0.20/0.48
% 0.20/0.48 Here is a proof with depth 1, length 17 :
% 0.20/0.48 % SZS output start Refutation
% See solution above
% 0.20/0.49 Formulae used in the proof : mSortsC m__1494 m__1494_04 m__ m_MulZero mMulComm mMonMul2
% 0.20/0.49
%------------------------------------------------------------------------------