TSTP Solution File: SEU169+2 by SPASS---3.9
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%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SEU169+2 : TPTP v8.1.0. Released v3.3.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n012.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 : Tue Jul 19 14:34:40 EDT 2022
% Result : Theorem 0.45s 0.68s
% Output : Refutation 0.45s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 8
% Syntax : Number of clauses : 18 ( 10 unt; 2 nHn; 18 RR)
% Number of literals : 31 ( 0 equ; 19 neg)
% Maximal clause size : 4 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 5 ( 4 usr; 1 prp; 0-2 aty)
% Number of functors : 9 ( 9 usr; 7 con; 0-1 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(3,axiom,
in(skc7,skc6),
file('SEU169+2.p',unknown),
[] ).
cnf(7,axiom,
element(skc6,powerset(skc5)),
file('SEU169+2.p',unknown),
[] ).
cnf(8,axiom,
~ in(skc7,skc5),
file('SEU169+2.p',unknown),
[] ).
cnf(28,axiom,
equal(union(powerset(u)),u),
file('SEU169+2.p',unknown),
[] ).
cnf(45,axiom,
( ~ empty(u)
| ~ in(v,u) ),
file('SEU169+2.p',unknown),
[] ).
cnf(62,axiom,
( ~ element(u,v)
| empty(v)
| in(u,v) ),
file('SEU169+2.p',unknown),
[] ).
cnf(64,axiom,
( ~ empty(u)
| ~ element(v,u)
| empty(v) ),
file('SEU169+2.p',unknown),
[] ).
cnf(157,axiom,
( ~ in(u,v)
| ~ in(w,u)
| ~ equal(x,union(v))
| in(w,x) ),
file('SEU169+2.p',unknown),
[] ).
cnf(211,plain,
~ empty(skc6),
inference(res,[status(thm),theory(equality)],[3,45]),
[iquote('0:Res:3.0,45.1')] ).
cnf(236,plain,
( ~ in(u,v)
| ~ in(skc7,u)
| ~ equal(union(v),skc5) ),
inference(res,[status(thm),theory(equality)],[157,8]),
[iquote('0:Res:157.3,8.0')] ).
cnf(244,plain,
( ~ empty(powerset(skc5))
| empty(skc6) ),
inference(res,[status(thm),theory(equality)],[7,64]),
[iquote('0:Res:7.0,64.1')] ).
cnf(245,plain,
( in(skc6,powerset(skc5))
| empty(powerset(skc5)) ),
inference(res,[status(thm),theory(equality)],[7,62]),
[iquote('0:Res:7.0,62.0')] ).
cnf(247,plain,
~ empty(powerset(skc5)),
inference(mrr,[status(thm)],[244,211]),
[iquote('0:MRR:244.1,211.0')] ).
cnf(248,plain,
in(skc6,powerset(skc5)),
inference(mrr,[status(thm)],[245,247]),
[iquote('0:MRR:245.1,247.0')] ).
cnf(258,plain,
( ~ in(skc6,u)
| ~ equal(union(u),skc5) ),
inference(res,[status(thm),theory(equality)],[3,236]),
[iquote('0:Res:3.0,236.0')] ).
cnf(824,plain,
~ equal(union(powerset(skc5)),skc5),
inference(res,[status(thm),theory(equality)],[248,258]),
[iquote('0:Res:248.0,258.0')] ).
cnf(830,plain,
~ equal(skc5,skc5),
inference(rew,[status(thm),theory(equality)],[28,824]),
[iquote('0:Rew:28.0,824.0')] ).
cnf(831,plain,
$false,
inference(obv,[status(thm),theory(equality)],[830]),
[iquote('0:Obv:830.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SEU169+2 : TPTP v8.1.0. Released v3.3.0.
% 0.07/0.12 % Command : run_spass %d %s
% 0.12/0.33 % Computer : n012.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 : Sun Jun 19 04:21:38 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.45/0.68
% 0.45/0.68 SPASS V 3.9
% 0.45/0.68 SPASS beiseite: Proof found.
% 0.45/0.68 % SZS status Theorem
% 0.45/0.68 Problem: /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.45/0.68 SPASS derived 600 clauses, backtracked 0 clauses, performed 0 splits and kept 401 clauses.
% 0.45/0.68 SPASS allocated 100315 KBytes.
% 0.45/0.68 SPASS spent 0:00:00.33 on the problem.
% 0.45/0.68 0:00:00.04 for the input.
% 0.45/0.68 0:00:00.22 for the FLOTTER CNF translation.
% 0.45/0.68 0:00:00.00 for inferences.
% 0.45/0.68 0:00:00.00 for the backtracking.
% 0.45/0.68 0:00:00.03 for the reduction.
% 0.45/0.68
% 0.45/0.68
% 0.45/0.68 Here is a proof with depth 3, length 18 :
% 0.45/0.68 % SZS output start Refutation
% See solution above
% 0.45/0.68 Formulae used in the proof : l3_subset_1 t99_zfmisc_1 t7_boole d2_subset_1 d4_tarski t9_tarski
% 0.45/0.68
%------------------------------------------------------------------------------