TPTP Problem File: SYO812+1.p

View Solutions - Solve Problem

%------------------------------------------------------------------------------
% File     : SYO812+1 : TPTP v9.0.0. Released v7.4.0.
% Domain   : Syntactic
% Problem  : Theory of immediate successor reduced problem spSUC30
% Version  : Especial.
% English  :

% Refs     : [Lam20] Lampert (2020), Email to Geoff Sutcliffe
%          : [LN20]  Lampert & Nakano (2020), Deciding Simple Infinity Axio
% Source   : [Lam20]
% Names    : spSUC30.p [Lam20]

% Status   : Satisfiable
% Rating   : 1.00 v7.4.0
% Syntax   : Number of formulae    :    8 (   1 unt;   0 def)
%            Number of atoms       :   42 (   0 equ)
%            Maximal formula atoms :    8 (   5 avg)
%            Number of connectives :   64 (  30   ~;  21   |;  13   &)
%                                         (   0 <=>;   0  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   11 (   8 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    1 (   1 usr;   0 prp; 2-2 aty)
%            Number of functors    :    0 (   0 usr;   0 con; --- aty)
%            Number of variables   :   41 (  30   !;  11   ?)
% SPC      : FOF_SAT_RFO_NEQ

% Comments : Has only infinite models.
%------------------------------------------------------------------------------
fof(axiom_1,axiom,
    ? [Y1] :
      ( ? [Y2] :
          ( ? [Y3] :
              ( f(Y1,Y3)
              & ~ f(Y2,Y3) )
          & ! [X1] : ~ f(X1,Y2) )
      & ! [X2] : ~ f(X2,Y1)
      & ! [X3] :
          ( ! [X4] :
              ( ! [X5] :
                  ( ~ f(X3,X5)
                  | f(X4,X5) )
              | ~ f(Y1,X4) )
          | ~ f(Y1,X3) ) ) ).

fof(axiom_2,axiom,
    ? [Y1] :
      ( ? [Y2] :
          ( f(Y2,Y1)
          & ! [X1] : ~ f(X1,Y2) )
      & ? [Y3] :
          ( ~ f(Y3,Y1)
          & ! [X2] : ~ f(X2,Y3) )
      & ! [X3] :
          ( ! [X4] :
              ( ~ f(X4,X3)
              | ~ f(X4,Y1) )
          | ! [X5] :
              ( f(X3,X5)
              | ~ f(Y1,X5) ) ) ) ).

fof(axiom_3,axiom,
    ? [Y1] :
      ( ? [Y2] :
          ( f(Y2,Y1)
          & ! [X1] : ~ f(X1,Y2) )
      & ? [Y3] :
          ( ~ f(Y3,Y1)
          & ! [X2] : ~ f(X2,Y3) )
      & ! [X3] :
          ( ! [X4] :
              ( ~ f(X4,X3)
              | ~ f(X4,Y1) )
          | ! [X5] :
              ( ~ f(X3,X5)
              | f(Y1,X5) ) ) ) ).

fof(axiom_4,axiom,
    ? [Y1] : f(Y1,Y1) ).

fof(axiom_5,axiom,
    ! [X1] :
    ? [Y1] :
      ( f(X1,Y1)
      & ! [X2] :
          ( ! [X3] :
              ( ! [X4] :
                  ( ~ f(X2,X4)
                  | f(X3,X4) )
              | ~ f(Y1,X3) )
          | ~ f(Y1,X2) ) ) ).

fof(axiom_6,axiom,
    ! [X1,X2] :
      ( ! [X3] :
          ( ~ f(X1,X3)
          | ~ f(X2,X3) )
      | ! [X4] :
          ( ~ f(X4,X1)
          | f(X4,X2) ) ) ).

fof(axiom_7,axiom,
    ! [X1,X2] :
      ( ! [X3] :
          ( ~ f(X3,X1)
          | ~ f(X3,X2) )
      | ! [X4] :
          ( ~ f(X1,X4)
          | f(X2,X4) ) ) ).

fof(axiom_8,axiom,
    ! [X1] :
      ( ! [X2] :
          ( ~ f(X1,X2)
          | ~ f(X2,X1) )
      | ! [X3] :
          ( ~ f(X1,X3)
          | f(X3,X1) ) ) ).

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