## TPTP Problem File: SEV195^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV195^5 : TPTP v7.5.0. Released v4.0.0.
% Domain   : Set Theory (Sets of sets)
% Problem  : TPS problem from S-THMS
% Version  : Especial.
% English  :

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_1034 [Bro09]

% Status   : Theorem
% Rating   : 0.45 v7.5.0, 0.14 v7.4.0, 0.33 v7.2.0, 0.25 v7.0.0, 0.43 v6.4.0, 0.50 v6.3.0, 0.40 v6.2.0, 0.71 v6.1.0, 0.57 v5.5.0, 1.00 v5.3.0, 0.80 v5.2.0, 1.00 v5.1.0, 0.80 v4.1.0, 0.67 v4.0.0
% Syntax   : Number of formulae    :    4 (   0 unit;   3 type;   0 defn)
%            Number of atoms       :   38 (   6 equality;  24 variable)
%            Maximal formula depth :   13 (   5 average)
%            Number of connectives :   26 (   1   ~;   1   |;   5   &;  15   @)
%                                         (   0 <=>;   4  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    3 (   3   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    5 (   3   :;   0   =)
%            Number of variables   :   13 (   0 sgn;  11   !;   2   ?;   0   ^)
%                                         (  13   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% 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
%------------------------------------------------------------------------------
thf(a_type,type,(
a: \$tType )).

thf(cP,type,(
cP: a > a > a )).

thf(cZ,type,(
cZ: a )).

thf(cS_LEM1D_pme,conjecture,
( ( ! [Xx: a,Xy: a] :
( ( cP @ Xx @ Xy )
!= cZ )
& ! [Xx: a,Xy: a,Xu: a,Xv: a] :
( ( ( cP @ Xx @ Xu )
= ( cP @ Xy @ Xv ) )
=> ( ( Xx = Xy )
& ( Xu = Xv ) ) )
& ! [X: a > \$o] :
( ( ( X @ cZ )
& ! [Xx: a,Xy: a] :
( ( ( X @ Xx )
& ( X @ Xy ) )
=> ( X @ ( cP @ Xx @ Xy ) ) ) )
=> ! [Xx: a] :
( X @ Xx ) ) )
=> ! [Xx: a] :
( ( Xx = cZ )
| ? [Xy: a,Xz: a] :
( Xx
= ( cP @ Xy @ Xz ) ) ) )).

%------------------------------------------------------------------------------
```