TPTP Problem File: SEV094^5.p

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%------------------------------------------------------------------------------
% File     : SEV094^5 : TPTP v9.0.0. Released v4.0.0.
% Domain   : Set Theory (Relations)
% Problem  : TPS problem from RELN-THMS
% Version  : Especial.
% English  :

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_1063 [Bro09]

% Status   : Theorem
% Rating   : 0.25 v9.0.0, 0.40 v8.2.0, 0.38 v8.1.0, 0.36 v7.5.0, 0.29 v7.4.0, 0.56 v7.3.0, 0.67 v7.2.0, 0.62 v7.1.0, 0.75 v7.0.0, 0.71 v6.4.0, 0.83 v6.3.0, 0.80 v6.2.0, 1.00 v6.1.0, 0.86 v5.5.0, 0.83 v5.4.0, 0.60 v5.1.0, 0.80 v5.0.0, 0.60 v4.1.0, 0.33 v4.0.1, 0.67 v4.0.0
% Syntax   : Number of formulae    :    2 (   0 unt;   1 typ;   0 def)
%            Number of atoms       :    6 (   6 equ;   0 cnn)
%            Maximal formula atoms :    6 (   6 avg)
%            Number of connectives :   31 (   0   ~;   0   |;   8   &;  16   @)
%                                         (   0 <=>;   7  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   15 (  15 avg)
%            Number of types       :    2 (   1 usr)
%            Number of type conns  :    4 (   4   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    1 (   0 usr;   0 con; 2-2 aty)
%            Number of variables   :   12 (   0   ^;   8   !;   4   ?;  12   :)
% SPC      : TH0_THM_EQU_NAR

% Comments : This problem is from the TPS library. Copyright (c) 2009 The TPS
%            project in the Department of Mathematical Sciences at Carnegie
%            Mellon University. Distributed under the Creative Commons copyleft
%            license: http://creativecommons.org/licenses/by-sa/3.0/
%------------------------------------------------------------------------------
thf(a_type,type,
    a: $tType ).

thf(cEQP1_1B_pme,conjecture,
    ! [Xx: a > $o,Xy: a > $o] :
      ( ? [Xs: a > a] :
          ( ! [Xx0: a] :
              ( ( Xx @ Xx0 )
             => ( Xy @ ( Xs @ Xx0 ) ) )
          & ! [Xy0: a] :
              ( ( Xy @ Xy0 )
             => ? [Xx0: a] :
                  ( ( Xx @ Xx0 )
                  & ( Xy0
                    = ( Xs @ Xx0 ) )
                  & ! [Xz: a] :
                      ( ( ( Xx @ Xz )
                        & ( Xy0
                          = ( Xs @ Xz ) ) )
                     => ( Xz = Xx0 ) ) ) ) )
     => ? [Xs: a > a] :
          ( ! [Xx0: a] :
              ( ( Xy @ Xx0 )
             => ( Xx @ ( Xs @ Xx0 ) ) )
          & ! [Xy0: a] :
              ( ( Xx @ Xy0 )
             => ? [Xx0: a] :
                  ( ( Xy @ Xx0 )
                  & ( Xy0
                    = ( Xs @ Xx0 ) )
                  & ! [Xz: a] :
                      ( ( ( Xy @ Xz )
                        & ( Xy0
                          = ( Xs @ Xz ) ) )
                     => ( Xz = Xx0 ) ) ) ) ) ) ).

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