## TPTP Problem File: SEV021^6.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV021^6 : TPTP v8.0.0. Released v5.5.0.
% Domain   : Set Theory (Relations)
% Problem  : TPS problem from EQUIVALENCE-RELATIONS-THMS
% Version  : Especial.
% English  :

% Refs     : [Sul12] Sultana (2012), Email to Geoff Sutcliffe
% Source   : [Sul12]
% Names    :

% Status   : Theorem
% Rating   : 0.55 v7.5.0, 0.57 v7.4.0, 0.89 v7.2.0, 0.88 v7.0.0, 1.00 v6.3.0, 0.80 v6.2.0, 0.86 v6.1.0, 1.00 v5.5.0
% Syntax   : Number of formulae    :    5 (   1 unt;   3 typ;   1 def)
%            Number of atoms       :   13 (   3 equ;   0 cnn)
%            Maximal formula atoms :    8 (   6 avg)
%            Number of connectives :   38 (   0   ~;   0   |;  11   &;  20   @)
%                                         (   1 <=>;   6  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   15 (   8 avg)
%            Number of types       :    2 (   1 usr)
%            Number of type conns  :   12 (  12   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    3 (   2 usr;   0 con; 1-2 aty)
%            Number of variables   :   17 (   3   ^  10   !;   4   ?;  17   :)
% SPC      : TH0_THM_EQU_NAR

%          : The conjecture is of the form A => B, where A is not needed to
%            prove B. A is an easily provable property of equality.
%          : This version has the relation Q instantiated.
%------------------------------------------------------------------------------
thf(a_type,type,
a: \$tType ).

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

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

thf(cQ_def,definition,
( cQ
= ( ^ [X: a,Y: a] :
? [S: a > \$o] :
( ( cP @ S )
& ( S @ X )
& ( S @ Y ) ) ) ) ).

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 ) ) )
=> ( ( ^ [Xs: a > \$o] :
( ? [Xz: a] : ( Xs @ Xz )
& ! [Xx: a] :
( ( Xs @ Xx )
=> ! [Xy: a] :
( ( Xs @ Xy )
<=> ( cQ @ Xx @ Xy ) ) ) ) )
= cP ) ) ) ).

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