## TPTP Problem File: SEV021^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV021^5 : 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_1068 [Bro09]

% Status   : Theorem
% Rating   : 0.82 v7.5.0, 1.00 v4.0.0
% Syntax   : Number of formulae    :    3 (   0 unit;   2 type;   0 defn)
%            Number of atoms       :   38 (   2 equality;  29 variable)
%            Maximal formula depth :   15 (   7 average)
%            Number of connectives :   33 (   0   ~;   0   |;   9   &;  17   @)
%                                         (   1 <=>;   6  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :   11 (  11   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    4 (   2   :;   0   =)
%            Number of variables   :   15 (   0 sgn;  10   !;   4   ?;   1   ^)
%                                         (  15   :;   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
%          : The conjecture is of the form A => B, where A is not needed to
%            prove B. A is an easily provable property of equality.
%------------------------------------------------------------------------------
thf(a_type,type,(
a: \$tType )).

thf(cP,type,(
cP: ( a > \$o ) > \$o )).

thf(cTHM262_D_EXT2_pme,conjecture,
( ! [Xq1: a > \$o,Xq2: a > \$o] :
( ( ( Xq1 = Xq2 )
& ( cP @ Xq1 ) )
=> ( cP @ Xq2 ) )
=> ( ( ! [Xp: a > \$o] :
( ( cP @ Xp )
=> ? [Xz: a] :
( Xp @ Xz ) )
& ! [Xx: a] :
? [Xp: a > \$o] :
( ( cP @ Xp )
& ( Xp @ Xx ) )
& ! [Xx: a,Xy: a,Xp: a > \$o,Xq: a > \$o] :
( ( ( cP @ Xp )
& ( cP @ Xq )
& ( Xp @ Xx )
& ( Xq @ Xx )
& ( Xp @ Xy ) )
=> ( Xq @ Xy ) ) )
=> ? [Q: a > a > \$o] :
( ( ^ [Xs: a > \$o] :
( ? [Xz: a] :
( Xs @ Xz )
& ! [Xx: a] :
( ( Xs @ Xx )
=> ! [Xy: a] :
( ( Xs @ Xy )
<=> ( Q @ Xx @ Xy ) ) ) ) )
= cP ) ) )).

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