TSTP Solution File: SYN011-1 by SPASS---3.9
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- Process Solution
%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SYN011-1 : TPTP v8.1.0. Released v1.1.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n027.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 : Thu Jul 21 12:17:18 EDT 2022
% Result : Unsatisfiable 0.19s 0.40s
% Output : Refutation 0.19s
% Verified :
% SZS Type : Refutation
% Derivation depth : 8
% Number of leaves : 8
% Syntax : Number of clauses : 24 ( 15 unt; 4 nHn; 24 RR)
% Number of literals : 35 ( 0 equ; 16 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 0 ( 0 avg)
% Number of predicates : 8 ( 7 usr; 8 prp; 0-0 aty)
% Number of functors : 0 ( 0 usr; 0 con; --- aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ n
| ~ t ),
file('SYN011-1.p',unknown),
[] ).
cnf(2,axiom,
( m
| q
| n ),
file('SYN011-1.p',unknown),
[] ).
cnf(3,axiom,
( ~ m
| l ),
file('SYN011-1.p',unknown),
[] ).
cnf(4,axiom,
( ~ q
| l ),
file('SYN011-1.p',unknown),
[] ).
cnf(5,axiom,
( ~ p
| ~ l ),
file('SYN011-1.p',unknown),
[] ).
cnf(6,axiom,
( r
| p
| n ),
file('SYN011-1.p',unknown),
[] ).
cnf(7,axiom,
( ~ l
| ~ r ),
file('SYN011-1.p',unknown),
[] ).
cnf(8,axiom,
t,
file('SYN011-1.p',unknown),
[] ).
cnf(9,plain,
~ n,
inference(mrr,[status(thm)],[1,8]),
[iquote('0:MRR:1.1,8.0')] ).
cnf(10,plain,
( p
| r ),
inference(mrr,[status(thm)],[6,9]),
[iquote('0:MRR:6.2,9.0')] ).
cnf(11,plain,
( q
| m ),
inference(mrr,[status(thm)],[2,9]),
[iquote('0:MRR:2.2,9.0')] ).
cnf(12,plain,
p,
inference(spt,[spt(split,[position(s1)])],[10]),
[iquote('1:Spt:10.0')] ).
cnf(13,plain,
~ l,
inference(mrr,[status(thm)],[5,12]),
[iquote('1:MRR:5.0,12.0')] ).
cnf(14,plain,
~ q,
inference(mrr,[status(thm)],[4,13]),
[iquote('1:MRR:4.1,13.0')] ).
cnf(15,plain,
~ m,
inference(mrr,[status(thm)],[3,13]),
[iquote('1:MRR:3.1,13.0')] ).
cnf(16,plain,
m,
inference(mrr,[status(thm)],[11,14]),
[iquote('1:MRR:11.0,14.0')] ).
cnf(17,plain,
$false,
inference(mrr,[status(thm)],[15,16]),
[iquote('1:MRR:15.0,16.0')] ).
cnf(18,plain,
~ p,
inference(spt,[spt(split,[position(sa)])],[17,12]),
[iquote('1:Spt:17.0,10.0,12.0')] ).
cnf(19,plain,
r,
inference(spt,[spt(split,[position(s2)])],[10]),
[iquote('1:Spt:17.0,10.1')] ).
cnf(20,plain,
~ l,
inference(mrr,[status(thm)],[7,19]),
[iquote('1:MRR:7.1,19.0')] ).
cnf(21,plain,
~ m,
inference(mrr,[status(thm)],[3,20]),
[iquote('1:MRR:3.1,20.0')] ).
cnf(22,plain,
~ q,
inference(mrr,[status(thm)],[4,20]),
[iquote('1:MRR:4.1,20.0')] ).
cnf(23,plain,
q,
inference(mrr,[status(thm)],[11,21]),
[iquote('1:MRR:11.1,21.0')] ).
cnf(24,plain,
$false,
inference(mrr,[status(thm)],[22,23]),
[iquote('1:MRR:22.0,23.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SYN011-1 : TPTP v8.1.0. Released v1.1.0.
% 0.03/0.13 % Command : run_spass %d %s
% 0.12/0.34 % Computer : n027.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 600
% 0.12/0.34 % DateTime : Tue Jul 12 00:00:04 EDT 2022
% 0.12/0.34 % CPUTime :
% 0.19/0.40
% 0.19/0.40 SPASS V 3.9
% 0.19/0.40 SPASS beiseite: Proof found.
% 0.19/0.40 % SZS status Theorem
% 0.19/0.40 Problem: /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.19/0.40 SPASS derived 10 clauses, backtracked 7 clauses, performed 1 splits and kept 24 clauses.
% 0.19/0.40 SPASS allocated 63080 KBytes.
% 0.19/0.40 SPASS spent 0:00:00.05 on the problem.
% 0.19/0.40 0:00:00.03 for the input.
% 0.19/0.40 0:00:00.00 for the FLOTTER CNF translation.
% 0.19/0.40 0:00:00.00 for inferences.
% 0.19/0.40 0:00:00.00 for the backtracking.
% 0.19/0.40 0:00:00.00 for the reduction.
% 0.19/0.40
% 0.19/0.40
% 0.19/0.40 Here is a proof with depth 1, length 24 :
% 0.19/0.40 % SZS output start Refutation
% See solution above
% 0.19/0.40 Formulae used in the proof : clause_1 clause_2 clause_3 clause_4 clause_5 clause_6 clause_7 clause_8
% 0.19/0.40
%------------------------------------------------------------------------------