## TPTP Problem File: SYN036^5.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : SYN036^5 : TPTP v7.1.0. Released v4.0.0.
% Domain   : Syntactic
% Problem  : TPS problem X2129
% Version  : Especial.
% English  :

% Refs     : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source   : [Bro09]
% Names    : tps_0554 [Bro09]
%          : THM57 [TPS]
%          : X2129 [TPS]

% Status   : Theorem
% Rating   : 0.38 v7.1.0, 0.43 v7.0.0, 0.38 v6.4.0, 0.43 v6.3.0, 0.50 v6.2.0, 0.67 v6.1.0, 0.50 v6.0.0, 0.33 v5.5.0, 0.20 v5.4.0, 0.25 v5.2.0, 0.00 v5.1.0, 0.25 v5.0.0, 0.00 v4.0.1, 0.33 v4.0.0
% Syntax   : Number of formulae    :    3 (   0 unit;   2 type;   0 defn)
%            Number of atoms       :   16 (   0 equality;   8 variable)
%            Maximal formula depth :    7 (   4 average)
%            Number of connectives :   15 (   0   ~;   0   |;   0   &;   8   @)
%                                         (   7 <=>;   0  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    2 (   2   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    4 (   2   :;   0   =)
%            Number of variables   :    8 (   0 sgn;   4   !;   4   ?;   0   ^)
%                                         (   8   :;   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
%          : This was an early Challenge Problem given by Peter Andrews in
%            1979 or 1980.
%          :
%------------------------------------------------------------------------------
thf(cQ,type,(
cQ: \$i > \$o )).

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

thf(cX2129,conjecture,
( ( ? [Xx: \$i] :
! [Xy: \$i] :
( ( cP @ Xx )
<=> ( cP @ Xy ) )
<=> ( ? [Xx: \$i] :
( cQ @ Xx )
<=> ! [Xy: \$i] :
( cP @ Xy ) ) )
<=> ( ? [Xx: \$i] :
! [Xy: \$i] :
( ( cQ @ Xx )
<=> ( cQ @ Xy ) )
<=> ( ? [Xx: \$i] :
( cP @ Xx )
<=> ! [Xy: \$i] :
( cQ @ Xy ) ) ) )).

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