TSTP Solution File: SYN328-1 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SYN328-1 : TPTP v8.1.0. Released v1.2.0.
% 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 12:20:20 EDT 2022
% Result : Unsatisfiable 0.18s 0.40s
% Output : Refutation 0.18s
% Verified :
% SZS Type : Refutation
% Derivation depth : 9
% Number of leaves : 7
% Syntax : Number of clauses : 19 ( 4 unt; 6 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 : 4 ( 3 usr; 1 prp; 0-1 aty)
% Number of functors : 3 ( 3 usr; 1 con; 0-1 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( f(u)
| f(y__dfg(u)) ),
file('SYN328-1.p',unknown),
[] ).
cnf(3,axiom,
( ~ g(y__dfg(u))
| f(u) ),
file('SYN328-1.p',unknown),
[] ).
cnf(4,axiom,
( ~ g(u)
| ~ f(y__dfg(u))
| h(y__dfg(u)) ),
file('SYN328-1.p',unknown),
[] ).
cnf(6,axiom,
( ~ h(y__dfg(u))
| g(u) ),
file('SYN328-1.p',unknown),
[] ).
cnf(9,axiom,
( ~ f(y__dfg(u))
| h(u)
| g(y__dfg(u)) ),
file('SYN328-1.p',unknown),
[] ).
cnf(10,axiom,
( ~ h(y__dfg(u))
| h(u) ),
file('SYN328-1.p',unknown),
[] ).
cnf(11,axiom,
( ~ h(z__dfg(u))
| ~ g(z__dfg(u))
| ~ f(z__dfg(u)) ),
file('SYN328-1.p',unknown),
[] ).
cnf(28,plain,
( ~ f(y__dfg(u))
| h(u)
| f(u) ),
inference(res,[status(thm),theory(equality)],[9,3]),
[iquote('0:Res:9.2,3.0')] ).
cnf(29,plain,
( h(u)
| f(u) ),
inference(mrr,[status(thm)],[28,2]),
[iquote('0:MRR:28.0,2.1')] ).
cnf(30,plain,
( ~ g(u)
| h(y__dfg(u)) ),
inference(mrr,[status(thm)],[4,29]),
[iquote('0:MRR:4.1,29.1')] ).
cnf(31,plain,
( f(y__dfg(u))
| h(u) ),
inference(res,[status(thm),theory(equality)],[29,10]),
[iquote('0:Res:29.0,10.0')] ).
cnf(33,plain,
( h(u)
| g(y__dfg(u)) ),
inference(mrr,[status(thm)],[9,31]),
[iquote('0:MRR:9.0,31.0')] ).
cnf(35,plain,
( ~ g(u)
| h(u) ),
inference(res,[status(thm),theory(equality)],[30,10]),
[iquote('0:Res:30.1,10.0')] ).
cnf(37,plain,
( ~ g(y__dfg(u))
| h(u) ),
inference(res,[status(thm),theory(equality)],[35,10]),
[iquote('0:Res:35.1,10.0')] ).
cnf(39,plain,
h(u),
inference(mrr,[status(thm)],[37,33]),
[iquote('0:MRR:37.0,33.1')] ).
cnf(40,plain,
g(u),
inference(mrr,[status(thm)],[6,39]),
[iquote('0:MRR:6.0,39.0')] ).
cnf(41,plain,
( ~ g(z__dfg(u))
| ~ f(z__dfg(u)) ),
inference(mrr,[status(thm)],[11,39]),
[iquote('0:MRR:11.0,39.0')] ).
cnf(42,plain,
f(u),
inference(mrr,[status(thm)],[3,40]),
[iquote('0:MRR:3.0,40.0')] ).
cnf(43,plain,
$false,
inference(mrr,[status(thm)],[41,40,42]),
[iquote('0:MRR:41.0,41.1,40.0,42.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SYN328-1 : TPTP v8.1.0. Released v1.2.0.
% 0.03/0.12 % Command : run_spass %d %s
% 0.12/0.33 % Computer : n027.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 600
% 0.12/0.33 % DateTime : Mon Jul 11 23:55:04 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.18/0.40
% 0.18/0.40 SPASS V 3.9
% 0.18/0.40 SPASS beiseite: Proof found.
% 0.18/0.40 % SZS status Theorem
% 0.18/0.40 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.18/0.40 SPASS derived 15 clauses, backtracked 0 clauses, performed 0 splits and kept 20 clauses.
% 0.18/0.40 SPASS allocated 75590 KBytes.
% 0.18/0.40 SPASS spent 0:00:00.06 on the problem.
% 0.18/0.40 0:00:00.03 for the input.
% 0.18/0.40 0:00:00.00 for the FLOTTER CNF translation.
% 0.18/0.40 0:00:00.00 for inferences.
% 0.18/0.40 0:00:00.00 for the backtracking.
% 0.18/0.40 0:00:00.00 for the reduction.
% 0.18/0.40
% 0.18/0.40
% 0.18/0.40 Here is a proof with depth 2, length 19 :
% 0.18/0.40 % SZS output start Refutation
% See solution above
% 0.18/0.40 Formulae used in the proof : clause2 clause3 clause4 clause6 clause9 clause10 clause11
% 0.18/0.40
%------------------------------------------------------------------------------