TSTP Solution File: SYN350-1 by SPASS---3.9
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- Process Solution
%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SYN350-1 : TPTP v8.1.0. Released v1.2.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n020.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 12:20:38 EDT 2022
% Result : Unsatisfiable 0.20s 0.99s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 8
% Number of leaves : 5
% Syntax : Number of clauses : 15 ( 4 unt; 4 nHn; 15 RR)
% Number of literals : 31 ( 0 equ; 17 neg)
% Maximal clause size : 3 ( 2 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 3 con; 0-2 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ f(a,z__dfg(u,v))
| f(z__dfg(u,v),a) ),
file('SYN350-1.p',unknown),
[] ).
cnf(2,axiom,
( ~ f(z__dfg(u,v),a)
| f(a,z__dfg(u,v)) ),
file('SYN350-1.p',unknown),
[] ).
cnf(3,axiom,
( f(u,z__dfg(u,v))
| f(v,z__dfg(u,v))
| f(a,z__dfg(u,v)) ),
file('SYN350-1.p',unknown),
[] ).
cnf(5,axiom,
( ~ f(u,v)
| f(v,z__dfg(u,v))
| f(a,z__dfg(u,v)) ),
file('SYN350-1.p',unknown),
[] ).
cnf(6,axiom,
( ~ f(u,v)
| ~ f(v,z__dfg(u,v))
| ~ f(a,z__dfg(u,v)) ),
file('SYN350-1.p',unknown),
[] ).
cnf(8,plain,
( ~ f(a,z__dfg(u,v))
| f(a,z__dfg(z__dfg(u,v),a))
| f(a,z__dfg(z__dfg(u,v),a)) ),
inference(res,[status(thm),theory(equality)],[1,5]),
[iquote('0:Res:1.1,5.0')] ).
cnf(9,plain,
( ~ f(a,z__dfg(u,v))
| f(a,z__dfg(z__dfg(u,v),a)) ),
inference(obv,[status(thm),theory(equality)],[8]),
[iquote('0:Obv:8.1')] ).
cnf(12,plain,
( f(u,z__dfg(a,u))
| f(a,z__dfg(a,u)) ),
inference(fac,[status(thm)],[3]),
[iquote('0:Fac:3.0,3.2')] ).
cnf(17,plain,
f(a,z__dfg(a,a)),
inference(fac,[status(thm)],[12]),
[iquote('0:Fac:12.0,12.1')] ).
cnf(29,plain,
( ~ f(a,z__dfg(u,v))
| ~ f(z__dfg(u,v),a)
| ~ f(a,z__dfg(z__dfg(u,v),a)) ),
inference(res,[status(thm),theory(equality)],[9,6]),
[iquote('0:Res:9.1,6.1')] ).
cnf(35,plain,
( ~ f(z__dfg(u,v),a)
| ~ f(a,z__dfg(z__dfg(u,v),a)) ),
inference(mrr,[status(thm)],[29,2]),
[iquote('0:MRR:29.0,2.1')] ).
cnf(39,plain,
( ~ f(a,z__dfg(u,v))
| ~ f(z__dfg(u,v),a) ),
inference(res,[status(thm),theory(equality)],[9,35]),
[iquote('0:Res:9.1,35.1')] ).
cnf(40,plain,
~ f(z__dfg(u,v),a),
inference(mrr,[status(thm)],[39,2]),
[iquote('0:MRR:39.0,2.1')] ).
cnf(41,plain,
~ f(a,z__dfg(u,v)),
inference(mrr,[status(thm)],[1,40]),
[iquote('0:MRR:1.1,40.0')] ).
cnf(42,plain,
$false,
inference(unc,[status(thm)],[41,17]),
[iquote('0:UnC:41.0,17.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.34 % Problem : SYN350-1 : TPTP v8.1.0. Released v1.2.0.
% 0.11/0.34 % Command : run_spass %d %s
% 0.16/0.91 % Computer : n020.cluster.edu
% 0.16/0.91 % Model : x86_64 x86_64
% 0.16/0.91 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.91 % Memory : 8042.1875MB
% 0.16/0.91 % OS : Linux 3.10.0-693.el7.x86_64
% 0.16/0.92 % CPULimit : 300
% 0.16/0.92 % WCLimit : 600
% 0.16/0.92 % DateTime : Tue Jul 12 01:48:34 EDT 2022
% 0.16/0.92 % CPUTime :
% 0.20/0.99
% 0.20/0.99 SPASS V 3.9
% 0.20/0.99 SPASS beiseite: Proof found.
% 0.20/0.99 % SZS status Theorem
% 0.20/0.99 Problem: /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.20/0.99 SPASS derived 29 clauses, backtracked 0 clauses, performed 0 splits and kept 23 clauses.
% 0.20/0.99 SPASS allocated 63122 KBytes.
% 0.20/0.99 SPASS spent 0:00:00.06 on the problem.
% 0.20/0.99 0:00:00.03 for the input.
% 0.20/0.99 0:00:00.00 for the FLOTTER CNF translation.
% 0.20/0.99 0:00:00.00 for inferences.
% 0.20/0.99 0:00:00.00 for the backtracking.
% 0.20/0.99 0:00:00.00 for the reduction.
% 0.20/0.99
% 0.20/0.99
% 0.20/0.99 Here is a proof with depth 3, length 15 :
% 0.20/0.99 % SZS output start Refutation
% See solution above
% 0.20/0.99 Formulae used in the proof : clause1 clause2 clause3 clause5 clause6
% 0.20/0.99
%------------------------------------------------------------------------------