TSTP Solution File: SYN733+1 by Etableau---0.67

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%------------------------------------------------------------------------------
% File     : Etableau---0.67
% Problem  : SYN733+1 : TPTP v8.1.0. Released v2.5.0.
% 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 : n015.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 06:13:41 EDT 2022

% Result   : Theorem 0.14s 0.32s
% Output   : CNFRefutation 0.14s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    5
%            Number of leaves      :    1
% Syntax   : Number of formulae    :    8 (   3 unt;   0 def)
%            Number of atoms       :   22 (   0 equ)
%            Maximal formula atoms :    5 (   2 avg)
%            Number of connectives :   20 (   6   ~;   6   |;   6   &)
%                                         (   0 <=>;   2  =>;   0  <=;   0 <~>)
%            Maximal formula depth :    8 (   4 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    3 (   2 usr;   1 prp; 0-1 aty)
%            Number of functors    :    1 (   1 usr;   1 con; 0-0 aty)
%            Number of variables   :   13 (   3 sgn   5   !;   4   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(y2141,conjecture,
    ( ! [X1] :
      ? [X2] :
        ( p(X1)
        & ( q(X2)
          | q(X1) ) )
   => ? [X3] :
        ( p(X3)
        & q(X3) ) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',y2141) ).

fof(c_0_1,negated_conjecture,
    ~ ( ! [X1] :
        ? [X2] :
          ( p(X1)
          & ( q(X2)
            | q(X1) ) )
     => ? [X3] :
          ( p(X3)
          & q(X3) ) ),
    inference(assume_negation,[status(cth)],[y2141]) ).

fof(c_0_2,negated_conjecture,
    ! [X4,X6,X7] :
      ( p(X4)
      & ( q(esk1_0)
        | q(X6) )
      & ( ~ p(X7)
        | ~ q(X7) ) ),
    inference(shift_quantors,[status(thm)],[inference(skolemize,[status(esa)],[inference(variable_rename,[status(thm)],[inference(shift_quantors,[status(thm)],[inference(fof_nnf,[status(thm)],[c_0_1])])])])]) ).

cnf(c_0_3,negated_conjecture,
    ( ~ p(X1)
    | ~ q(X1) ),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_4,negated_conjecture,
    p(X1),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_5,negated_conjecture,
    ( q(esk1_0)
    | q(X1) ),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_6,negated_conjecture,
    ~ q(X1),
    inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_3,c_0_4])]) ).

cnf(c_0_7,negated_conjecture,
    $false,
    inference(sr,[status(thm)],[inference(sr,[status(thm)],[c_0_5,c_0_6]),c_0_6]),
    [proof] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.09  % Problem  : SYN733+1 : TPTP v8.1.0. Released v2.5.0.
% 0.00/0.09  % Command  : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s
% 0.08/0.29  % Computer : n015.cluster.edu
% 0.08/0.29  % Model    : x86_64 x86_64
% 0.08/0.29  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.08/0.29  % Memory   : 8042.1875MB
% 0.08/0.29  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.08/0.29  % CPULimit : 300
% 0.08/0.29  % WCLimit  : 600
% 0.08/0.29  % DateTime : Mon Jul 11 19:21:24 EDT 2022
% 0.08/0.29  % CPUTime  : 
% 0.14/0.32  # No SInE strategy applied
% 0.14/0.32  # Auto-Mode selected heuristic G_E___208_C18_F1_SE_CS_SP_PS_S5PRR_RG_S04AN
% 0.14/0.32  # and selection function SelectComplexExceptUniqMaxHorn.
% 0.14/0.32  #
% 0.14/0.32  # Presaturation interreduction done
% 0.14/0.32  
% 0.14/0.32  # Proof found!
% 0.14/0.32  # SZS status Theorem
% 0.14/0.32  # SZS output start CNFRefutation
% See solution above
% 0.14/0.32  # Training examples: 0 positive, 0 negative
%------------------------------------------------------------------------------