TSTP Solution File: SYN332-1 by SPASS---3.9
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- Process Solution
%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SYN332-1 : TPTP v8.1.0. Released v1.2.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n022.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 12:20:24 EDT 2022
% Result : Unsatisfiable 0.22s 0.42s
% Output : Refutation 0.22s
% Verified :
% SZS Type : Refutation
% Derivation depth : 8
% Number of leaves : 7
% Syntax : Number of clauses : 19 ( 4 unt; 7 nHn; 19 RR)
% Number of literals : 38 ( 0 equ; 15 neg)
% Maximal clause size : 3 ( 2 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( f(u,v)
| f(u,z__dfg(u,v)) ),
file('SYN332-1.p',unknown),
[] ).
cnf(2,axiom,
( f(u,v)
| f(v,z__dfg(v,u)) ),
file('SYN332-1.p',unknown),
[] ).
cnf(5,axiom,
( ~ f(u,z__dfg(u,v))
| f(z__dfg(u,v),u) ),
file('SYN332-1.p',unknown),
[] ).
cnf(7,axiom,
( ~ f(u,z__dfg(u,v))
| f(v,z__dfg(u,v)) ),
file('SYN332-1.p',unknown),
[] ).
cnf(8,axiom,
( ~ f(u,v)
| ~ f(z__dfg(v,u),z__dfg(v,u))
| f(v,u) ),
file('SYN332-1.p',unknown),
[] ).
cnf(13,axiom,
( ~ f(z__dfg(u,v),v)
| f(v,u)
| f(u,v) ),
file('SYN332-1.p',unknown),
[] ).
cnf(14,axiom,
( ~ f(u,v)
| ~ f(v,u)
| ~ f(z__dfg(v,u),u) ),
file('SYN332-1.p',unknown),
[] ).
cnf(16,plain,
( f(u,v)
| f(u,z__dfg(v,u)) ),
inference(res,[status(thm),theory(equality)],[2,7]),
[iquote('0:Res:2.1,7.0')] ).
cnf(17,plain,
( f(u,v)
| f(v,z__dfg(u,v)) ),
inference(res,[status(thm),theory(equality)],[1,7]),
[iquote('0:Res:1.1,7.0')] ).
cnf(27,plain,
( ~ f(u,z__dfg(u,u))
| f(u,u)
| f(u,u) ),
inference(res,[status(thm),theory(equality)],[5,13]),
[iquote('0:Res:5.1,13.0')] ).
cnf(29,plain,
( ~ f(u,z__dfg(u,u))
| f(u,u) ),
inference(obv,[status(thm),theory(equality)],[27]),
[iquote('0:Obv:27.1')] ).
cnf(30,plain,
f(u,u),
inference(mrr,[status(thm)],[29,17]),
[iquote('0:MRR:29.0,17.1')] ).
cnf(31,plain,
( ~ f(u,v)
| f(v,u) ),
inference(mrr,[status(thm)],[8,30]),
[iquote('0:MRR:8.1,30.0')] ).
cnf(33,plain,
( ~ f(z__dfg(u,v),v)
| f(u,v) ),
inference(mrr,[status(thm)],[13,31]),
[iquote('0:MRR:13.1,31.0')] ).
cnf(34,plain,
( ~ f(u,v)
| ~ f(z__dfg(u,v),v) ),
inference(mrr,[status(thm)],[14,31]),
[iquote('0:MRR:14.0,31.1')] ).
cnf(38,plain,
~ f(z__dfg(u,v),v),
inference(mrr,[status(thm)],[34,33]),
[iquote('0:MRR:34.0,33.1')] ).
cnf(41,plain,
( f(u,v)
| f(z__dfg(v,u),u) ),
inference(res,[status(thm),theory(equality)],[16,31]),
[iquote('0:Res:16.1,31.0')] ).
cnf(44,plain,
f(u,v),
inference(mrr,[status(thm)],[41,38]),
[iquote('0:MRR:41.1,38.0')] ).
cnf(45,plain,
$false,
inference(unc,[status(thm)],[44,38]),
[iquote('0:UnC:44.0,38.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.13 % Problem : SYN332-1 : TPTP v8.1.0. Released v1.2.0.
% 0.13/0.14 % Command : run_spass %d %s
% 0.14/0.35 % Computer : n022.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 12 01:21:10 EDT 2022
% 0.22/0.35 % CPUTime :
% 0.22/0.42
% 0.22/0.42 SPASS V 3.9
% 0.22/0.42 SPASS beiseite: Proof found.
% 0.22/0.42 % SZS status Theorem
% 0.22/0.42 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.22/0.42 SPASS derived 26 clauses, backtracked 0 clauses, performed 0 splits and kept 27 clauses.
% 0.22/0.42 SPASS allocated 63101 KBytes.
% 0.22/0.42 SPASS spent 0:00:00.06 on the problem.
% 0.22/0.42 0:00:00.03 for the input.
% 0.22/0.42 0:00:00.00 for the FLOTTER CNF translation.
% 0.22/0.42 0:00:00.00 for inferences.
% 0.22/0.42 0:00:00.00 for the backtracking.
% 0.22/0.42 0:00:00.00 for the reduction.
% 0.22/0.42
% 0.22/0.42
% 0.22/0.42 Here is a proof with depth 2, length 19 :
% 0.22/0.42 % SZS output start Refutation
% See solution above
% 0.22/0.42 Formulae used in the proof : clause1 clause2 clause5 clause7 clause8 clause13 clause14
% 0.22/0.42
%------------------------------------------------------------------------------