TSTP Solution File: SYN398+1 by SnakeForV---1.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : SnakeForV---1.0
% Problem  : SYN398+1 : TPTP v8.1.0. Released v2.0.0.
% 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 : n013.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 19:26:29 EDT 2022

% Result   : Theorem 0.20s 0.52s
% Output   : Refutation 0.20s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :   12
%            Number of leaves      :    3
% Syntax   : Number of formulae    :   18 (   3 unt;   0 def)
%            Number of atoms       :   70 (   0 equ)
%            Maximal formula atoms :    8 (   3 avg)
%            Number of connectives :   86 (  34   ~;  26   |;  20   &)
%                                         (   3 <=>;   2  =>;   0  <=;   1 <~>)
%            Maximal formula depth :    6 (   4 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    3 (   2 usr;   2 prp; 0-1 aty)
%            Number of functors    :    2 (   2 usr;   2 con; 0-0 aty)
%            Number of variables   :   26 (  18   !;   8   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(f21,plain,
    $false,
    inference(subsumption_resolution,[],[f20,f12]) ).

fof(f12,plain,
    ! [X2,X3] :
      ( f(X2)
      | f(X3) ),
    inference(cnf_transformation,[],[f10]) ).

fof(f10,plain,
    ( ( ~ p
      | ~ f(sK0)
      | ~ f(sK1)
      | ~ p )
    & ( ! [X2] :
          ( p
          & f(X2) )
      | ( ! [X3] : f(X3)
        & p ) ) ),
    inference(skolemisation,[status(esa),new_symbols(skolem,[sK0,sK1])],[f7,f9,f8]) ).

fof(f8,plain,
    ( ? [X0] :
        ( ~ p
        | ~ f(X0) )
   => ( ~ p
      | ~ f(sK0) ) ),
    introduced(choice_axiom,[]) ).

fof(f9,plain,
    ( ? [X1] : ~ f(X1)
   => ~ f(sK1) ),
    introduced(choice_axiom,[]) ).

fof(f7,plain,
    ( ( ? [X0] :
          ( ~ p
          | ~ f(X0) )
      | ? [X1] : ~ f(X1)
      | ~ p )
    & ( ! [X2] :
          ( p
          & f(X2) )
      | ( ! [X3] : f(X3)
        & p ) ) ),
    inference(rectify,[],[f6]) ).

fof(f6,plain,
    ( ( ? [X1] :
          ( ~ p
          | ~ f(X1) )
      | ? [X0] : ~ f(X0)
      | ~ p )
    & ( ! [X1] :
          ( p
          & f(X1) )
      | ( ! [X0] : f(X0)
        & p ) ) ),
    inference(flattening,[],[f5]) ).

fof(f5,plain,
    ( ( ? [X1] :
          ( ~ p
          | ~ f(X1) )
      | ? [X0] : ~ f(X0)
      | ~ p )
    & ( ! [X1] :
          ( p
          & f(X1) )
      | ( ! [X0] : f(X0)
        & p ) ) ),
    inference(nnf_transformation,[],[f4]) ).

fof(f4,plain,
    ( ( ! [X0] : f(X0)
      & p )
  <~> ! [X1] :
        ( p
        & f(X1) ) ),
    inference(ennf_transformation,[],[f3]) ).

fof(f3,plain,
    ~ ( ! [X1] :
          ( p
          & f(X1) )
    <=> ( ! [X0] : f(X0)
        & p ) ),
    inference(rectify,[],[f2]) ).

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

fof(f1,conjecture,
    ( ( p
      & ! [X1] : f(X1) )
  <=> ! [X0] :
        ( p
        & f(X0) ) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',kalish215) ).

fof(f20,plain,
    ~ f(sK1),
    inference(subsumption_resolution,[],[f19,f12]) ).

fof(f19,plain,
    ( ~ f(sK0)
    | ~ f(sK1) ),
    inference(subsumption_resolution,[],[f17,f16]) ).

fof(f16,plain,
    p,
    inference(duplicate_literal_removal,[],[f13]) ).

fof(f13,plain,
    ( p
    | p ),
    inference(cnf_transformation,[],[f10]) ).

fof(f17,plain,
    ( ~ f(sK0)
    | ~ p
    | ~ f(sK1) ),
    inference(duplicate_literal_removal,[],[f15]) ).

fof(f15,plain,
    ( ~ p
    | ~ f(sK0)
    | ~ f(sK1)
    | ~ p ),
    inference(cnf_transformation,[],[f10]) ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem    : SYN398+1 : TPTP v8.1.0. Released v2.0.0.
% 0.07/0.13  % Command    : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_uns --cores 0 -t %d %s
% 0.13/0.35  % Computer : n013.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit   : 300
% 0.13/0.35  % WCLimit    : 300
% 0.13/0.35  % DateTime   : Tue Aug 30 21:47:47 EDT 2022
% 0.13/0.35  % CPUTime    : 
% 0.20/0.51  % (21857)dis+21_1:1_av=off:sos=on:sp=frequency:ss=included:to=lpo:i=15:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/15Mi)
% 0.20/0.51  % (21876)dis+10_1:1_av=off:sos=on:sp=reverse_arity:ss=included:st=2.0:to=lpo:urr=ec_only:i=45:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/45Mi)
% 0.20/0.51  % (21876)First to succeed.
% 0.20/0.51  % (21868)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)
% 0.20/0.51  % (21868)Also succeeded, but the first one will report.
% 0.20/0.52  % (21876)Refutation found. Thanks to Tanya!
% 0.20/0.52  % SZS status Theorem for theBenchmark
% 0.20/0.52  % SZS output start Proof for theBenchmark
% See solution above
% 0.20/0.52  % (21876)------------------------------
% 0.20/0.52  % (21876)Version: Vampire 4.7 (commit 807e37dd9 on 2022-08-23 09:55:27 +0200)
% 0.20/0.52  % (21876)Linked with Z3 4.8.13.0 f03d756e086f81f2596157241e0decfb1c982299 z3-4.8.4-5390-gf03d756e0
% 0.20/0.52  % (21876)Termination reason: Refutation
% 0.20/0.52  
% 0.20/0.52  % (21876)Memory used [KB]: 1407
% 0.20/0.52  % (21876)Time elapsed: 0.058 s
% 0.20/0.52  % (21876)Instructions burned: 1 (million)
% 0.20/0.52  % (21876)------------------------------
% 0.20/0.52  % (21876)------------------------------
% 0.20/0.52  % (21847)Success in time 0.161 s
%------------------------------------------------------------------------------