## TPTP Problem File: SEV008^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV008^5 : TPTP v7.5.0. Released v4.0.0.
% Domain   : Set Theory (Relations)
% Problem  : TPS problem THM261
% Version  : Especial.
% English  : A partition defines an equivalence relation.

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_0491 [Bro09]
%          : THM261 [TPS]

% Status   : Theorem
% Rating   : 0.09 v7.5.0, 0.00 v7.4.0, 0.11 v7.2.0, 0.00 v7.1.0, 0.25 v7.0.0, 0.14 v6.4.0, 0.17 v6.3.0, 0.20 v6.2.0, 0.14 v5.5.0, 0.17 v5.4.0, 0.20 v5.1.0, 0.40 v5.0.0, 0.20 v4.1.0, 0.00 v4.0.1, 0.67 v4.0.0
% Syntax   : Number of formulae    :    2 (   0 unit;   1 type;   0 defn)
%            Number of atoms       :   51 (   1 equality;  50 variable)
%            Maximal formula depth :   14 (   8 average)
%            Number of connectives :   48 (   0   ~;   0   |;  19   &;  24   @)
%                                         (   0 <=>;   5  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :   11 (  11   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    3 (   1   :;   0   =)
%            Number of variables   :   18 (   0 sgn;  10   !;   8   ?;   0   ^)
%                                         (  18   :;   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
%          : Polymorphic definitions expanded.
%------------------------------------------------------------------------------
thf(a_type,type,(
a: \$tType )).

thf(cTHM261_pme,conjecture,(
! [P: ( a > \$o ) > \$o] :
( ( ! [Xp: a > \$o] :
( ( P @ Xp )
=> ? [Xz: a] :
( Xp @ Xz ) )
& ! [Xx: a] :
? [Xp: a > \$o] :
( ( P @ Xp )
& ( Xp @ Xx )
& ! [Xq: a > \$o] :
( ( ( P @ Xq )
& ( Xq @ Xx ) )
=> ( Xq = Xp ) ) ) )
=> ( ! [Xx: a] :
? [S: a > \$o] :
( ( P @ S )
& ( S @ Xx )
& ( S @ Xx ) )
& ! [Xx: a,Xy: a] :
( ? [S: a > \$o] :
( ( P @ S )
& ( S @ Xx )
& ( S @ Xy ) )
=> ? [S: a > \$o] :
( ( P @ S )
& ( S @ Xy )
& ( S @ Xx ) ) )
& ! [Xx: a,Xy: a,Xz: a] :
( ( ? [S: a > \$o] :
( ( P @ S )
& ( S @ Xx )
& ( S @ Xy ) )
& ? [S: a > \$o] :
( ( P @ S )
& ( S @ Xy )
& ( S @ Xz ) ) )
=> ? [S: a > \$o] :
( ( P @ S )
& ( S @ Xx )
& ( S @ Xz ) ) ) ) ) )).

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