TSTP Solution File: SWC036+1 by SnakeForV---1.0

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%------------------------------------------------------------------------------
% File     : SnakeForV---1.0
% Problem  : SWC036+1 : TPTP v8.1.0. Released v2.4.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_uns --cores 0 -t %d %s

% Computer : n023.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  : 300s
% DateTime : Wed Aug 31 18:38:21 EDT 2022

% Result   : Theorem 0.20s 0.50s
% Output   : Refutation 0.20s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :   10
%            Number of leaves      :    2
% Syntax   : Number of formulae    :   17 (   8 unt;   0 def)
%            Number of atoms       :   68 (  59 equ)
%            Maximal formula atoms :   12 (   4 avg)
%            Number of connectives :   79 (  28   ~;  19   |;  23   &)
%                                         (   0 <=>;   9  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   16 (   5 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    3 (   1 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   5 con; 0-0 aty)
%            Number of variables   :   24 (  12   !;  12   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(f113,plain,
    $false,
    inference(subsumption_resolution,[],[f111,f112]) ).

fof(f112,plain,
    sK1 = sK3,
    inference(trivial_inequality_removal,[],[f110]) ).

fof(f110,plain,
    ( sK1 = sK3
    | sK1 != sK1 ),
    inference(definition_unfolding,[],[f107,f109,f109,f104]) ).

fof(f104,plain,
    sK0 = sK3,
    inference(cnf_transformation,[],[f103]) ).

fof(f103,plain,
    ( sK2 = sK1
    & ( nil != sK1
      | nil = sK0 )
    & nil != sK3
    & nil = sK2
    & sK0 = sK3 ),
    inference(skolemisation,[status(esa),new_symbols(skolem,[sK0,sK1,sK2,sK3])],[f101,f102]) ).

fof(f102,plain,
    ( ? [X0,X1,X2,X3] :
        ( X1 = X2
        & ( nil != X1
          | nil = X0 )
        & nil != X3
        & nil = X2
        & X0 = X3 )
   => ( sK2 = sK1
      & ( nil != sK1
        | nil = sK0 )
      & nil != sK3
      & nil = sK2
      & sK0 = sK3 ) ),
    introduced(choice_axiom,[]) ).

fof(f101,plain,
    ? [X0,X1,X2,X3] :
      ( X1 = X2
      & ( nil != X1
        | nil = X0 )
      & nil != X3
      & nil = X2
      & X0 = X3 ),
    inference(rectify,[],[f100]) ).

fof(f100,plain,
    ? [X2,X3,X1,X0] :
      ( X1 = X3
      & ( nil != X3
        | nil = X2 )
      & nil != X0
      & nil = X1
      & X0 = X2 ),
    inference(ennf_transformation,[],[f98]) ).

fof(f98,plain,
    ~ ! [X0,X1,X2,X3] :
        ( X1 != X3
        | nil != X1
        | X0 != X2
        | ( nil != X2
          & nil = X3 )
        | nil = X0 ),
    inference(pure_predicate_removal,[],[f97]) ).

fof(f97,negated_conjecture,
    ~ ! [X0] :
        ( ssList(X0)
       => ! [X1] :
            ( ssList(X1)
           => ! [X2] :
                ( ssList(X2)
               => ! [X3] :
                    ( ssList(X3)
                   => ( X1 != X3
                      | nil != X1
                      | X0 != X2
                      | ( nil != X2
                        & nil = X3 )
                      | nil = X0 ) ) ) ) ),
    inference(negated_conjecture,[],[f96]) ).

fof(f96,conjecture,
    ! [X0] :
      ( ssList(X0)
     => ! [X1] :
          ( ssList(X1)
         => ! [X2] :
              ( ssList(X2)
             => ! [X3] :
                  ( ssList(X3)
                 => ( X1 != X3
                    | nil != X1
                    | X0 != X2
                    | ( nil != X2
                      & nil = X3 )
                    | nil = X0 ) ) ) ) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',co1) ).

fof(f109,plain,
    nil = sK1,
    inference(definition_unfolding,[],[f105,f108]) ).

fof(f108,plain,
    sK2 = sK1,
    inference(cnf_transformation,[],[f103]) ).

fof(f105,plain,
    nil = sK2,
    inference(cnf_transformation,[],[f103]) ).

fof(f107,plain,
    ( nil != sK1
    | nil = sK0 ),
    inference(cnf_transformation,[],[f103]) ).

fof(f111,plain,
    sK1 != sK3,
    inference(definition_unfolding,[],[f106,f109]) ).

fof(f106,plain,
    nil != sK3,
    inference(cnf_transformation,[],[f103]) ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem    : SWC036+1 : TPTP v8.1.0. Released v2.4.0.
% 0.03/0.12  % Command    : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_uns --cores 0 -t %d %s
% 0.12/0.33  % Computer : n023.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    : 300
% 0.12/0.33  % DateTime   : Tue Aug 30 18:26:05 EDT 2022
% 0.12/0.34  % CPUTime    : 
% 0.20/0.49  % (6675)lrs+2_1:1_lcm=reverse:lma=on:sos=all:spb=goal_then_units:ss=included:urr=on:i=39:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/39Mi)
% 0.20/0.49  % (6675)First to succeed.
% 0.20/0.49  % (6683)lrs+10_1:1_drc=off:sp=reverse_frequency:spb=goal:to=lpo:i=7:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/7Mi)
% 0.20/0.50  % (6675)Refutation found. Thanks to Tanya!
% 0.20/0.50  % SZS status Theorem for theBenchmark
% 0.20/0.50  % SZS output start Proof for theBenchmark
% See solution above
% 0.20/0.50  % (6675)------------------------------
% 0.20/0.50  % (6675)Version: Vampire 4.7 (commit 807e37dd9 on 2022-08-23 09:55:27 +0200)
% 0.20/0.50  % (6675)Linked with Z3 4.8.13.0 f03d756e086f81f2596157241e0decfb1c982299 z3-4.8.4-5390-gf03d756e0
% 0.20/0.50  % (6675)Termination reason: Refutation
% 0.20/0.50  
% 0.20/0.50  % (6675)Memory used [KB]: 6012
% 0.20/0.50  % (6675)Time elapsed: 0.005 s
% 0.20/0.50  % (6675)Instructions burned: 2 (million)
% 0.20/0.50  % (6675)------------------------------
% 0.20/0.50  % (6675)------------------------------
% 0.20/0.50  % (6667)Success in time 0.153 s
%------------------------------------------------------------------------------