## TPTP Problem File: SEV019^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV019^5 : TPTP v7.5.0. Released v4.0.0.
% Domain   : Set Theory (Relations)
% Problem  : TPS problem from EQUIVALENCE-RELATIONS-THMS
% Version  : Especial.
% English  :

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_0993 [Bro09]

% Status   : Theorem
% Rating   : 0.25 v7.4.0, 0.33 v7.3.0, 0.30 v7.2.0, 0.38 v7.1.0, 0.43 v7.0.0, 0.38 v6.4.0, 0.43 v6.3.0, 0.50 v6.0.0, 0.33 v5.5.0, 0.20 v5.4.0, 0.25 v5.1.0, 0.50 v5.0.0, 0.25 v4.1.0, 0.33 v4.0.0
% Syntax   : Number of formulae    :    3 (   0 unit;   2 type;   0 defn)
%            Number of atoms       :   29 (   0 equality;  22 variable)
%            Maximal formula depth :   12 (   6 average)
%            Number of connectives :   28 (   0   ~;   0   |;   5   &;  18   @)
%                                         (   1 <=>;   4  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    3 (   3   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    4 (   2   :;   0   =)
%            Number of variables   :   11 (   0 sgn;   9   !;   2   ?;   0   ^)
%                                         (  11   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% SPC      : TH0_THM_NEQ_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(cQ,type,(
cQ: a > a > \$o )).

thf(cTHM559_pme,conjecture,
( ! [Xx: a] :
? [Xp: a > \$o] :
( ? [Xz: a] :
( Xp @ Xz )
& ! [Xx0: a] :
( ( Xp @ Xx0 )
=> ! [Xy: a] :
( ( Xp @ Xy )
<=> ( cQ @ Xx0 @ Xy ) ) )
& ( Xp @ Xx ) )
=> ( ! [Xx: a] :
( cQ @ Xx @ Xx )
& ! [Xx: a,Xy: a] :
( ( cQ @ Xx @ Xy )
=> ( cQ @ Xy @ Xx ) )
& ! [Xx: a,Xy: a,Xz: a] :
( ( ( cQ @ Xx @ Xy )
& ( cQ @ Xy @ Xz ) )
=> ( cQ @ Xx @ Xz ) ) ) )).

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