TSTP Solution File: SWC103+1 by Etableau---0.67

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Etableau---0.67
% Problem  : SWC103+1 : TPTP v8.1.0. Released v2.4.0.
% 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 : n006.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 20:29:29 EDT 2022

% Result   : Theorem 0.18s 0.38s
% Output   : CNFRefutation 0.18s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    7
%            Number of leaves      :    1
% Syntax   : Number of formulae    :   14 (   7 unt;   0 def)
%            Number of atoms       :   58 (  14 equ)
%            Maximal formula atoms :   15 (   4 avg)
%            Number of connectives :   69 (  25   ~;  20   |;  16   &)
%                                         (   0 <=>;   8  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   18 (   5 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    5 (   3 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   5 con; 0-0 aty)
%            Number of variables   :    8 (   0 sgn   8   !;   0   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(co1,conjecture,
    ! [X1] :
      ( ssList(X1)
     => ! [X2] :
          ( ssList(X2)
         => ! [X3] :
              ( ssList(X3)
             => ! [X4] :
                  ( ssList(X4)
                 => ( X2 != X4
                    | X1 != X3
                    | ~ neq(X2,nil)
                    | ( nil != X3
                      & nil = X4 )
                    | ( neq(X1,nil)
                      & frontsegP(X2,X1) )
                    | ( neq(X4,nil)
                      & ( ~ neq(X3,nil)
                        | ~ frontsegP(X4,X3) ) ) ) ) ) ) ),
    file('/export/starexec/sandbox2/benchmark/theBenchmark.p',co1) ).

fof(c_0_1,negated_conjecture,
    ~ ! [X1] :
        ( ssList(X1)
       => ! [X2] :
            ( ssList(X2)
           => ! [X3] :
                ( ssList(X3)
               => ! [X4] :
                    ( ssList(X4)
                   => ( X2 != X4
                      | X1 != X3
                      | ~ neq(X2,nil)
                      | ( nil != X3
                        & nil = X4 )
                      | ( neq(X1,nil)
                        & frontsegP(X2,X1) )
                      | ( neq(X4,nil)
                        & ( ~ neq(X3,nil)
                          | ~ frontsegP(X4,X3) ) ) ) ) ) ) ),
    inference(assume_negation,[status(cth)],[co1]) ).

fof(c_0_2,negated_conjecture,
    ( ssList(esk48_0)
    & ssList(esk49_0)
    & ssList(esk50_0)
    & ssList(esk51_0)
    & esk49_0 = esk51_0
    & esk48_0 = esk50_0
    & neq(esk49_0,nil)
    & ( nil = esk50_0
      | nil != esk51_0 )
    & ( ~ neq(esk48_0,nil)
      | ~ frontsegP(esk49_0,esk48_0) )
    & ( neq(esk50_0,nil)
      | ~ neq(esk51_0,nil) )
    & ( frontsegP(esk51_0,esk50_0)
      | ~ neq(esk51_0,nil) ) ),
    inference(distribute,[status(thm)],[inference(skolemize,[status(esa)],[inference(variable_rename,[status(thm)],[inference(fof_nnf,[status(thm)],[inference(fof_simplification,[status(thm)],[c_0_1])])])])]) ).

cnf(c_0_3,negated_conjecture,
    neq(esk49_0,nil),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_4,negated_conjecture,
    esk49_0 = esk51_0,
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_5,negated_conjecture,
    ( ~ neq(esk48_0,nil)
    | ~ frontsegP(esk49_0,esk48_0) ),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_6,negated_conjecture,
    esk48_0 = esk50_0,
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_7,negated_conjecture,
    ( frontsegP(esk51_0,esk50_0)
    | ~ neq(esk51_0,nil) ),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_8,negated_conjecture,
    neq(esk51_0,nil),
    inference(rw,[status(thm)],[c_0_3,c_0_4]) ).

cnf(c_0_9,negated_conjecture,
    ( ~ frontsegP(esk51_0,esk50_0)
    | ~ neq(esk50_0,nil) ),
    inference(rw,[status(thm)],[inference(rw,[status(thm)],[inference(rw,[status(thm)],[c_0_5,c_0_6]),c_0_4]),c_0_6]) ).

cnf(c_0_10,negated_conjecture,
    frontsegP(esk51_0,esk50_0),
    inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_7,c_0_8])]) ).

cnf(c_0_11,negated_conjecture,
    ( neq(esk50_0,nil)
    | ~ neq(esk51_0,nil) ),
    inference(split_conjunct,[status(thm)],[c_0_2]) ).

cnf(c_0_12,negated_conjecture,
    ~ neq(esk50_0,nil),
    inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_9,c_0_10])]) ).

cnf(c_0_13,negated_conjecture,
    $false,
    inference(sr,[status(thm)],[inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_11,c_0_8])]),c_0_12]),
    [proof] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12  % Problem  : SWC103+1 : TPTP v8.1.0. Released v2.4.0.
% 0.04/0.12  % Command  : etableau --auto --tsmdo --quicksat=10000 --tableau=1 --tableau-saturation=1 -s -p --tableau-cores=8 --cpu-limit=%d %s
% 0.12/0.33  % Computer : n006.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 12 07:18:12 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 0.18/0.38  # No SInE strategy applied
% 0.18/0.38  # Auto-Mode selected heuristic G_E___208_C18_F1_SE_CS_SP_PS_S4c
% 0.18/0.38  # and selection function SelectCQPrecWNTNp.
% 0.18/0.38  #
% 0.18/0.38  # Presaturation interreduction done
% 0.18/0.38  
% 0.18/0.38  # Proof found!
% 0.18/0.38  # SZS status Theorem
% 0.18/0.38  # SZS output start CNFRefutation
% See solution above
% 0.18/0.38  # Training examples: 0 positive, 0 negative
%------------------------------------------------------------------------------