## TPTP Problem File: SEV198^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV198^5 : 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_1117 [Bro09]

% Status   : Unknown
% Rating   : 1.00 v4.0.0
% Syntax   : Number of formulae    :    5 (   0 unit;   4 type;   0 defn)
%            Number of atoms       :   65 (  12 equality;  38 variable)
%            Maximal formula depth :   25 (   7 average)
%            Number of connectives :   41 (   1   ~;   2   |;   9   &;  24   @)
%                                         (   0 <=>;   5  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    5 (   5   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    7 (   4   :;   0   =)
%            Number of variables   :   16 (   0 sgn;  10   !;   6   ?;   0   ^)
%                                         (  16   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% SPC      : TH0_UNK_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(c0,type,(
c0: a )).

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

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

thf(cS_INCL_LEM6_pme,conjecture,
( ( ! [Xx0: a,Xy: a,Xu: a,Xv: a] :
( ( ( cP @ Xx0 @ Xu )
= ( cP @ Xy @ Xv ) )
=> ( ( Xx0 = Xy )
& ( Xu = Xv ) ) )
& ! [Xx0: a,Xy: a] :
( ( cP @ Xx0 @ Xy )
!= c0 ) )
=> ( ! [R: a > a > a > \$o] :
( ( \$true
& ! [Xa: a,Xb: a,Xc: a] :
( ( ( ( Xa = c0 )
& ( Xb = Xc ) )
| ( ( Xb = c0 )
& ( Xa = Xc ) )
| ? [Xx1: a,Xx2: a,Xy1: a,Xy2: a,Xz1: a,Xz2: a] :
( ( Xa
= ( cP @ Xx1 @ Xx2 ) )
& ( Xb
= ( cP @ Xy1 @ Xy2 ) )
& ( Xc
= ( cP @ Xz1 @ Xz2 ) )
& ( R @ Xx1 @ Xy1 @ Xz1 )
& ( R @ Xx2 @ Xy2 @ Xz2 ) ) )
=> ( R @ Xa @ Xb @ Xc ) ) )
=> ( R @ x @ c0 @ c0 ) )
=> ( x = c0 ) ) )).

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