TSTP Solution File: SYN056+1 by SPASS---3.9
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- Process Solution
%------------------------------------------------------------------------------
% File : SPASS---3.9
% Problem : SYN056+1 : TPTP v8.1.0. Released v2.0.0.
% Transfm : none
% Format : tptp
% Command : run_spass %d %s
% Computer : n016.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 12:17:43 EDT 2022
% Result : Theorem 0.22s 0.45s
% Output : Refutation 0.22s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 8
% Syntax : Number of clauses : 18 ( 7 unt; 1 nHn; 18 RR)
% Number of literals : 37 ( 0 equ; 22 neg)
% Maximal clause size : 4 ( 2 avg)
% Maximal term depth : 1 ( 1 avg)
% Number of predicates : 5 ( 4 usr; 1 prp; 0-1 aty)
% Number of functors : 6 ( 6 usr; 6 con; 0-0 aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
big_q(skc5),
file('SYN056+1.p',unknown),
[] ).
cnf(2,axiom,
big_p(skc4),
file('SYN056+1.p',unknown),
[] ).
cnf(3,axiom,
( ~ big_p(u)
| big_q(skc6) ),
file('SYN056+1.p',unknown),
[] ).
cnf(4,axiom,
( ~ big_q(u)
| big_p(skc7) ),
file('SYN056+1.p',unknown),
[] ).
cnf(5,axiom,
( ~ big_s(skc5)
| ~ big_r(skc4) ),
file('SYN056+1.p',unknown),
[] ).
cnf(6,axiom,
( ~ big_p(u)
| ~ big_q(v)
| big_r(u)
| big_s(v) ),
file('SYN056+1.p',unknown),
[] ).
cnf(7,axiom,
( ~ big_r(u)
| ~ big_p(u)
| ~ big_q(v)
| big_s(v) ),
file('SYN056+1.p',unknown),
[] ).
cnf(8,axiom,
( ~ big_s(u)
| ~ big_p(v)
| ~ big_q(u)
| big_r(v) ),
file('SYN056+1.p',unknown),
[] ).
cnf(9,plain,
( ~ big_q(u)
| ~ big_p(v)
| big_r(v) ),
inference(mrr,[status(thm)],[8,6]),
[iquote('0:MRR:8.0,6.3')] ).
cnf(10,plain,
( ~ big_q(u)
| ~ big_p(v)
| big_s(u) ),
inference(mrr,[status(thm)],[7,9]),
[iquote('0:MRR:7.0,9.2')] ).
cnf(17,plain,
big_p(skc7),
inference(ems,[status(thm)],[4,1]),
[iquote('0:EmS:4.0,1.0')] ).
cnf(18,plain,
big_q(skc6),
inference(ems,[status(thm)],[3,17]),
[iquote('0:EmS:3.0,17.0')] ).
cnf(20,plain,
( ~ big_p(u)
| big_r(u) ),
inference(ems,[status(thm)],[9,18]),
[iquote('0:EmS:9.0,18.0')] ).
cnf(22,plain,
( ~ big_q(u)
| big_s(u) ),
inference(ems,[status(thm)],[10,17]),
[iquote('0:EmS:10.1,17.0')] ).
cnf(24,plain,
( ~ big_q(skc5)
| ~ big_r(skc4) ),
inference(res,[status(thm),theory(equality)],[22,5]),
[iquote('0:Res:22.1,5.0')] ).
cnf(25,plain,
~ big_r(skc4),
inference(ssi,[status(thm)],[24,1]),
[iquote('0:SSi:24.0,1.0')] ).
cnf(26,plain,
~ big_p(skc4),
inference(res,[status(thm),theory(equality)],[20,25]),
[iquote('0:Res:20.1,25.0')] ).
cnf(27,plain,
$false,
inference(ssi,[status(thm)],[26,2]),
[iquote('0:SSi:26.0,2.0')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.14 % Problem : SYN056+1 : TPTP v8.1.0. Released v2.0.0.
% 0.14/0.14 % Command : run_spass %d %s
% 0.16/0.36 % Computer : n016.cluster.edu
% 0.16/0.36 % Model : x86_64 x86_64
% 0.16/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.36 % Memory : 8042.1875MB
% 0.16/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.16/0.36 % CPULimit : 300
% 0.16/0.36 % WCLimit : 600
% 0.16/0.36 % DateTime : Mon Jul 11 22:24:43 EDT 2022
% 0.22/0.36 % CPUTime :
% 0.22/0.45
% 0.22/0.45 SPASS V 3.9
% 0.22/0.45 SPASS beiseite: Proof found.
% 0.22/0.45 % SZS status Theorem
% 0.22/0.45 Problem: /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.22/0.45 SPASS derived 9 clauses, backtracked 0 clauses, performed 0 splits and kept 14 clauses.
% 0.22/0.45 SPASS allocated 97571 KBytes.
% 0.22/0.45 SPASS spent 0:00:00.08 on the problem.
% 0.22/0.45 0:00:00.03 for the input.
% 0.22/0.45 0:00:00.02 for the FLOTTER CNF translation.
% 0.22/0.45 0:00:00.00 for inferences.
% 0.22/0.45 0:00:00.00 for the backtracking.
% 0.22/0.45 0:00:00.00 for the reduction.
% 0.22/0.45
% 0.22/0.45
% 0.22/0.45 Here is a proof with depth 4, length 18 :
% 0.22/0.45 % SZS output start Refutation
% See solution above
% 0.22/0.45 Formulae used in the proof : pel26 pel26_1 pel26_2
% 0.22/0.45
%------------------------------------------------------------------------------