%------------------------------------------------------------------------------
% File     : KRS162+1 : TPTP v9.0.0. Released v3.1.0.
% Domain   : Knowledge Representation (Semantic Web)
% Problem  : Entailment for three natural numbers
% Version  : Especial.
% English  : This entailment can be replicated for any three natural numbers
%            i, j, k such that i+j >= k. In this example, they are chosen as
%            2, 3 and 5.

% Refs     : [Bec03] Bechhofer (2003), Email to G. Sutcliffe
%          : [TR+04] Tsarkov et al. (2004), Using Vampire to Reason with OW
% Source   : [Bec03]
% Names    : positive_description-logic-Manifest901 [Bec03]

% Status   : Theorem
% Rating   : 0.00 v5.3.0, 0.27 v5.2.0, 0.00 v4.1.0, 0.30 v4.0.0, 0.33 v3.7.0, 0.14 v3.5.0, 0.11 v3.4.0, 0.08 v3.3.0, 0.22 v3.2.0, 0.33 v3.1.0
% Syntax   : Number of formulae    :   20 (   0 unt;   0 def)
%            Number of atoms       :   78 (  26 equ)
%            Maximal formula atoms :   28 (   3 avg)
%            Number of connectives :   77 (  19   ~;   0   |;  39   &)
%                                         (   2 <=>;  17  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   25 (   6 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :   10 (   9 usr;   0 prp; 1-2 aty)
%            Number of functors    :    0 (   0 usr;   0 con; --- aty)
%            Number of variables   :   54 (  44   !;  10   ?)
% SPC      : FOF_THM_EPR_SEQ

% Comments : Sean Bechhofer says there are some errors in the encoding of
%            datatypes, so this problem may not be perfect. At least it's
%            still representative of the type of reasoning required for OWL.
%------------------------------------------------------------------------------
fof(cA_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & cA(A) )
     => cA(B) ) ).

fof(cB_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & cB(A) )
     => cB(B) ) ).

fof(cowlNothing_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & cowlNothing(A) )
     => cowlNothing(B) ) ).

fof(cowlThing_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & cowlThing(A) )
     => cowlThing(B) ) ).

fof(rp_substitution_1,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rp(A,C) )
     => rp(B,C) ) ).

fof(rp_substitution_2,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rp(C,A) )
     => rp(C,B) ) ).

fof(rq_substitution_1,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rq(A,C) )
     => rq(B,C) ) ).

fof(rq_substitution_2,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rq(C,A) )
     => rq(C,B) ) ).

fof(rr_substitution_1,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rr(A,C) )
     => rr(B,C) ) ).

fof(rr_substitution_2,axiom,
    ! [A,B,C] :
      ( ( A = B
        & rr(C,A) )
     => rr(C,B) ) ).

fof(xsd_integer_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & xsd_integer(A) )
     => xsd_integer(B) ) ).

fof(xsd_string_substitution_1,axiom,
    ! [A,B] :
      ( ( A = B
        & xsd_string(A) )
     => xsd_string(B) ) ).

%----Thing and Nothing
fof(axiom_0,axiom,
    ! [X] :
      ( cowlThing(X)
      & ~ cowlNothing(X) ) ).

%----String and Integer disjoint
fof(axiom_1,axiom,
    ! [X] :
      ( xsd_string(X)
    <=> ~ xsd_integer(X) ) ).

%----Range: rp
fof(axiom_2,axiom,
    ! [X,Y] :
      ( rp(X,Y)
     => cA(Y) ) ).

%----Range: rq
fof(axiom_3,axiom,
    ! [X,Y] :
      ( rq(X,Y)
     => cB(Y) ) ).

fof(axiom_4,axiom,
    ! [X] :
      ~ ( cB(X)
        & cA(X) ) ).

fof(axiom_5,axiom,
    ! [X,Y] :
      ( rq(X,Y)
     => rr(X,Y) ) ).

fof(axiom_6,axiom,
    ! [X,Y] :
      ( rp(X,Y)
     => rr(X,Y) ) ).

%----Thing and Nothing
%----String and Integer disjoint
fof(the_axiom,conjecture,
    ( ! [X] :
        ( cowlThing(X)
        & ~ cowlNothing(X) )
    & ! [X] :
        ( xsd_string(X)
      <=> ~ xsd_integer(X) )
    & ! [X] :
        ( ( ? [Y0,Y1] :
              ( rp(X,Y0)
              & rp(X,Y1)
              & Y0 != Y1 )
          & ? [Y0,Y1,Y2] :
              ( rq(X,Y0)
              & rq(X,Y1)
              & rq(X,Y2)
              & Y0 != Y1
              & Y0 != Y2
              & Y1 != Y2 ) )
       => ? [Y0,Y1,Y2,Y3,Y4] :
            ( rr(X,Y0)
            & rr(X,Y1)
            & rr(X,Y2)
            & rr(X,Y3)
            & rr(X,Y4)
            & Y0 != Y1
            & Y0 != Y2
            & Y0 != Y3
            & Y0 != Y4
            & Y1 != Y2
            & Y1 != Y3
            & Y1 != Y4
            & Y2 != Y3
            & Y2 != Y4
            & Y3 != Y4 ) ) ) ).

%------------------------------------------------------------------------------