TSTP Solution File: KLE086+1 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : KLE086+1 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n015.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 : Sun Jul 17 02:28:26 EDT 2022
% Result : Theorem 0.20s 0.45s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 6
% Syntax : Number of clauses : 15 ( 15 unt; 0 nHn; 15 RR)
% Number of literals : 15 ( 0 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 4 ( 2 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 7 ( 7 usr; 3 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
~ equal(domain__dfg(zero),zero),
file('KLE086+1.p',unknown),
[] ).
cnf(2,axiom,
equal(addition(u,zero),u),
file('KLE086+1.p',unknown),
[] ).
cnf(4,axiom,
equal(multiplication(u,one),u),
file('KLE086+1.p',unknown),
[] ).
cnf(8,axiom,
equal(multiplication(antidomain(u),u),zero),
file('KLE086+1.p',unknown),
[] ).
cnf(9,axiom,
equal(antidomain(antidomain(u)),domain__dfg(u)),
file('KLE086+1.p',unknown),
[] ).
cnf(13,axiom,
equal(addition(antidomain(antidomain(u)),antidomain(u)),one),
file('KLE086+1.p',unknown),
[] ).
cnf(24,plain,
equal(addition(domain__dfg(u),antidomain(u)),one),
inference(rew,[status(thm),theory(equality)],[9,13]),
[iquote('0:Rew:9.0,13.0')] ).
cnf(39,plain,
equal(antidomain(one),zero),
inference(spr,[status(thm),theory(equality)],[8,4]),
[iquote('0:SpR:8.0,4.0')] ).
cnf(50,plain,
equal(antidomain(zero),domain__dfg(one)),
inference(spr,[status(thm),theory(equality)],[39,9]),
[iquote('0:SpR:39.0,9.0')] ).
cnf(58,plain,
equal(addition(domain__dfg(one),zero),one),
inference(spr,[status(thm),theory(equality)],[39,24]),
[iquote('0:SpR:39.0,24.0')] ).
cnf(61,plain,
equal(domain__dfg(one),one),
inference(rew,[status(thm),theory(equality)],[2,58]),
[iquote('0:Rew:2.0,58.0')] ).
cnf(62,plain,
equal(antidomain(zero),one),
inference(rew,[status(thm),theory(equality)],[61,50]),
[iquote('0:Rew:61.0,50.0')] ).
cnf(69,plain,
equal(antidomain(one),domain__dfg(zero)),
inference(spr,[status(thm),theory(equality)],[62,9]),
[iquote('0:SpR:62.0,9.0')] ).
cnf(72,plain,
equal(domain__dfg(zero),zero),
inference(rew,[status(thm),theory(equality)],[39,69]),
[iquote('0:Rew:39.0,69.0')] ).
cnf(73,plain,
$false,
inference(mrr,[status(thm)],[72,1]),
[iquote('0:MRR:72.0,1.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.13 % Problem : KLE086+1 : TPTP v8.1.0. Released v4.0.0.
% 0.03/0.13 % Command : run_spass %d %s
% 0.13/0.35 % Computer : n015.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 600
% 0.13/0.35 % DateTime : Thu Jun 16 07:56:11 EDT 2022
% 0.13/0.35 % CPUTime :
% 0.20/0.45
% 0.20/0.45 SPASS V 3.9
% 0.20/0.45 SPASS beiseite: Proof found.
% 0.20/0.45 % SZS status Theorem
% 0.20/0.45 Problem: /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.20/0.45 SPASS derived 40 clauses, backtracked 0 clauses, performed 0 splits and kept 32 clauses.
% 0.20/0.45 SPASS allocated 85142 KBytes.
% 0.20/0.45 SPASS spent 0:00:00.08 on the problem.
% 0.20/0.45 0:00:00.03 for the input.
% 0.20/0.45 0:00:00.03 for the FLOTTER CNF translation.
% 0.20/0.45 0:00:00.00 for inferences.
% 0.20/0.45 0:00:00.00 for the backtracking.
% 0.20/0.45 0:00:00.00 for the reduction.
% 0.20/0.45
% 0.20/0.45
% 0.20/0.45 Here is a proof with depth 3, length 15 :
% 0.20/0.45 % SZS output start Refutation
% See solution above
% 0.20/0.45 Formulae used in the proof : goals additive_identity multiplicative_right_identity domain1 domain4 domain3
% 0.20/0.45
%------------------------------------------------------------------------------