%------------------------------------------------------------------------------
% File     : SYO558^1 : TPTP v9.0.0. Bugfixed v7.5.0.
% Domain   : Syntactic
% Problem  : Forall on $i can be expressed in terms of choice on $
% Version  : Especial.
% English  :

% Refs     : [Bro11] Brown (2011), Email to Geoff Sutcliffe
% Source   : [Bro11]
% Names    : CHOICE30 [Bro11]

% Status   : Theorem
% Rating   : 0.33 v8.1.0, 0.50 v7.5.0
% Syntax   : Number of formulae    :    3 (   1 unt;   1 typ;   0 def)
%            Number of atoms       :    2 (   1 equ;   0 cnn)
%            Maximal formula atoms :    1 (   1 avg)
%            Number of connectives :   11 (   1   ~;   0   |;   0   &;   8   @)
%                                         (   0 <=>;   1  =>;   0  <=;   0 <~>)
%            Maximal formula depth :    5 (   3 avg)
%            Number of types       :    2 (   0 usr)
%            Number of type conns  :    5 (   5   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    3 (   1 usr;   1 con; 0-2 aty)
%                                         (   1  !!;   0  ??;   0 @@+;   0 @@-)
%            Number of variables   :    5 (   3   ^;   1   !;   1   ?;   5   :)
% SPC      : TH1_THM_EQU_NAR

% Comments : (eps (complement P)) can only satisfy P if everything satisfies P.
%            Hence (^P.P(eps (^x.~Px))) = (^P.!! P).
%          : Note that this problem uses !! for forall instead of the ! 
%            quantifier.
% Bugfixes : Added type argument to !!.
%------------------------------------------------------------------------------
thf(eps,type,
    eps: ( $i > $o ) > $i ).

thf(choiceax,axiom,
    ! [P: $i > $o] :
      ( ? [X: $i] : ( P @ X )
     => ( P @ ( eps @ P ) ) ) ).

thf(conj,conjecture,
    ( ( ^ [P: $i > $o] :
          ( P
          @ ( eps
            @ ^ [X: $i] :
                ~ ( P @ X ) ) ) )
    = ( ^ [P: $i > $o] : ( !! @ $i @ P ) ) ) ).

%------------------------------------------------------------------------------