TSTP Solution File: SYN387+1 by Etableau---0.67

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Etableau---0.67
% Problem  : SYN387+1 : TPTP v8.1.0. Released v2.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s

% Computer : n010.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 06:10:33 EDT 2022

% Result   : Theorem 0.15s 0.38s
% Output   : CNFRefutation 0.15s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    4
%            Number of leaves      :    1
% Syntax   : Number of formulae    :    6 (   3 unt;   0 def)
%            Number of atoms       :    9 (   0 equ)
%            Maximal formula atoms :    2 (   1 avg)
%            Number of connectives :    8 (   5   ~;   2   |;   1   &)
%                                         (   0 <=>;   0  =>;   0  <=;   0 <~>)
%            Maximal formula depth :    4 (   2 avg)
%            Maximal term depth    :    0 (   0 avg)
%            Number of predicates  :    2 (   1 usr;   2 prp; 0-0 aty)
%            Number of functors    :    0 (   0 usr;   0 con; --- aty)
%            Number of variables   :    0 (   0 sgn   0   !;   0   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(pel6,conjecture,
    ( p
    | ~ p ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',pel6) ).

fof(c_0_1,negated_conjecture,
    ~ ( p
      | ~ p ),
    inference(assume_negation,[status(cth)],[pel6]) ).

fof(c_0_2,negated_conjecture,
    ( ~ p
    & p ),
    inference(fof_nnf,[status(thm)],[inference(fof_simplification,[status(thm)],[c_0_1])]) ).

cnf(c_0_3,negated_conjecture,
    ~ p,
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_4,negated_conjecture,
    p,
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_5,negated_conjecture,
    $false,
    inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_3,c_0_4])]),
    [proof] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.13  % Problem  : SYN387+1 : TPTP v8.1.0. Released v2.0.0.
% 0.12/0.14  % Command  : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s
% 0.15/0.36  % Computer : n010.cluster.edu
% 0.15/0.36  % Model    : x86_64 x86_64
% 0.15/0.36  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.36  % Memory   : 8042.1875MB
% 0.15/0.36  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.15/0.36  % CPULimit : 300
% 0.15/0.36  % WCLimit  : 600
% 0.15/0.36  % DateTime : Tue Jul 12 02:26:30 EDT 2022
% 0.15/0.36  % CPUTime  : 
% 0.15/0.38  # No SInE strategy applied
% 0.15/0.38  # Auto-Mode selected heuristic G_E___107_C48_F1_PI_AE_Q4_CS_SP_PS_S0Y
% 0.15/0.38  # and selection function SelectMaxLComplexAvoidPosPred.
% 0.15/0.38  #
% 0.15/0.38  # Presaturation interreduction done
% 0.15/0.38  
% 0.15/0.38  # Proof found!
% 0.15/0.38  # SZS status Theorem
% 0.15/0.38  # SZS output start CNFRefutation
% See solution above
% 0.15/0.38  # Training examples: 0 positive, 0 negative
%------------------------------------------------------------------------------