%------------------------------------------------------------------------------
% File     : SnakeForV-SAT---1.0
% Problem  : SYN968+1 : TPTP v8.1.0. Released v3.1.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_sat --cores 0 -t %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  : 300s
% DateTime : Wed Aug 31 19:46:13 EDT 2022

% Result   : Theorem 1.52s 0.55s
% Output   : Refutation 1.52s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    5
%            Number of leaves      :    2
% Syntax   : Number of formulae    :    8 (   3 unt;   0 def)
%            Number of atoms       :   15 (   0 equ)
%            Maximal formula atoms :    4 (   1 avg)
%            Number of connectives :   13 (   6   ~;   0   |;   4   &)
%                                         (   0 <=>;   3  =>;   0  <=;   0 <~>)
%            Maximal formula depth :    6 (   4 avg)
%            Maximal term depth    :    2 (   1 avg)
%            Number of predicates  :    2 (   1 usr;   1 prp; 0-1 aty)
%            Number of functors    :    1 (   1 usr;   0 con; 1-1 aty)
%            Number of variables   :   11 (   7   !;   4   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(f8,plain,
    $false,
    inference(resolution,[],[f6,f7]) ).

fof(f7,plain,
    ! [X0] : p(X0),
    inference(cnf_transformation,[],[f5]) ).

fof(f5,plain,
    ! [X0] :
      ( p(X0)
      & ~ p(sK0(X0)) ),
    inference(skolemisation,[status(esa),new_symbols(skolem,[sK0])],[f3,f4]) ).

fof(f4,plain,
    ! [X0] :
      ( ? [X1] :
          ( p(X0)
          & ~ p(X1) )
     => ( p(X0)
        & ~ p(sK0(X0)) ) ),
    introduced(choice_axiom,[]) ).

fof(f3,plain,
    ! [X0] :
    ? [X1] :
      ( p(X0)
      & ~ p(X1) ),
    inference(ennf_transformation,[],[f2]) ).

fof(f2,negated_conjecture,
    ~ ? [X0] :
      ! [X1] :
        ( p(X0)
       => p(X1) ),
    inference(negated_conjecture,[],[f1]) ).

fof(f1,conjecture,
    ? [X0] :
    ! [X1] :
      ( p(X0)
     => p(X1) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',prove_this) ).

fof(f6,plain,
    ! [X0] : ~ p(sK0(X0)),
    inference(cnf_transformation,[],[f5]) ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem    : SYN968+1 : TPTP v8.1.0. Released v3.1.0.
% 0.07/0.13  % Command    : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_sat --cores 0 -t %d %s
% 0.13/0.34  % Computer : n010.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit   : 300
% 0.13/0.34  % WCLimit    : 300
% 0.13/0.34  % DateTime   : Tue Aug 30 22:25:59 EDT 2022
% 0.13/0.34  % CPUTime    : 
% 0.19/0.53  % (30241)ott+2_1:1_fsr=off:gsp=on:i=50:si=on:rawr=on:rtra=on_0 on theBenchmark for (3000ds/50Mi)
% 0.19/0.54  % (30233)ott+4_1:1_av=off:bd=off:nwc=5.0:s2a=on:s2at=2.0:slsq=on:slsqc=2:slsql=off:slsqr=1,2:sp=frequency:i=37:si=on:rawr=on:rtra=on_0 on theBenchmark for (3000ds/37Mi)
% 1.52/0.54  % (30241)First to succeed.
% 1.52/0.55  % (30241)Refutation found. Thanks to Tanya!
% 1.52/0.55  % SZS status Theorem for theBenchmark
% 1.52/0.55  % SZS output start Proof for theBenchmark
% See solution above
% 1.52/0.55  % (30241)------------------------------
% 1.52/0.55  % (30241)Version: Vampire 4.7 (commit 807e37dd9 on 2022-08-23 09:55:27 +0200)
% 1.52/0.55  % (30241)Linked with Z3 4.8.13.0 f03d756e086f81f2596157241e0decfb1c982299 z3-4.8.4-5390-gf03d756e0
% 1.52/0.55  % (30241)Termination reason: Refutation
% 1.52/0.55  
% 1.52/0.55  % (30241)Memory used [KB]: 5373
% 1.52/0.55  % (30241)Time elapsed: 0.119 s
% 1.52/0.55  % (30241)Instructions burned: 1 (million)
% 1.52/0.55  % (30241)------------------------------
% 1.52/0.55  % (30241)------------------------------
% 1.52/0.55  % (30230)Success in time 0.196 s
%------------------------------------------------------------------------------