## TPTP Problem File: SEV062^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SEV062^5 : TPTP v7.5.0. Released v4.0.0.
% Domain   : Set Theory (Relations)
% Problem  : TPS problem T146A
% Version  : Especial.
% English  :

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_0338 [Bro09]
%          : T146A [TPS]

% Status   : Theorem
% Rating   : 0.08 v7.4.0, 0.00 v7.3.0, 0.10 v7.2.0, 0.00 v6.2.0, 0.17 v6.0.0, 0.00 v5.1.0, 0.25 v5.0.0, 0.00 v4.0.1, 0.33 v4.0.0
% Syntax   : Number of formulae    :    1 (   0 unit;   0 type;   0 defn)
%            Number of atoms       :   36 (   0 equality;  36 variable)
%            Maximal formula depth :   15 (  15 average)
%            Number of connectives :   35 (   0   ~;   0   |;   4   &;  24   @)
%                                         (   0 <=>;   7  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    6 (   6   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    2 (   0   :;   0   =)
%            Number of variables   :   15 (   0 sgn;  15   !;   0   ?;   0   ^)
%                                         (  15   :;   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
%          : Polymorphic definitions expanded.
%------------------------------------------------------------------------------
thf(cT146A_pme,conjecture,(
! [Xr: \$i > \$i > \$o,Xx: \$i,Xy: \$i] :
( ! [Xp: \$i > \$i > \$o] :
( ( ! [Xx0: \$i,Xy0: \$i] :
( ( Xr @ Xx0 @ Xy0 )
=> ( Xp @ Xx0 @ Xy0 ) )
& ! [Xu: \$i,Xv: \$i,Xw: \$i] :
( ( ( Xp @ Xu @ Xv )
& ( Xr @ Xv @ Xw ) )
=> ( Xp @ Xu @ Xw ) ) )
=> ( Xp @ Xx @ Xy ) )
=> ! [Xp: \$i > \$i > \$o] :
( ( ! [Xx0: \$i,Xy0: \$i] :
( ( Xr @ Xx0 @ Xy0 )
=> ( Xp @ Xx0 @ Xy0 ) )
& ! [Xx0: \$i,Xy0: \$i,Xz: \$i] :
( ( ( Xp @ Xx0 @ Xy0 )
& ( Xp @ Xy0 @ Xz ) )
=> ( Xp @ Xx0 @ Xz ) ) )
=> ( Xp @ Xx @ Xy ) ) ) )).

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