TPTP Problem File: SEV427^1.p

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%------------------------------------------------------------------------------
% File     : SEV427^1 : TPTP v7.5.0. Released v5.2.0.
% Domain   : Set Theory
% Problem  : If two sets cover a type, a choice function must give an element 
% Version  : Especial.
% English  :

% Refs     : [Bro11] Brown (2011), Email to Geoff Sutcliffe
% Source   : [Bro11]
% Names    : CHOICE33 [Bro11]

% Status   : Theorem
% Rating   : 0.08 v7.4.0, 0.00 v6.2.0, 0.17 v6.0.0, 0.00 v5.2.0
% Syntax   : Number of formulae    :    6 (   0 unit;   3 type;   0 defn)
%            Number of atoms       :   15 (   0 equality;   6 variable)
%            Maximal formula depth :    5 (   4 average)
%            Number of connectives :   12 (   0   ~;   2   |;   0   &;   9   @)
%                                         (   0 <=>;   1  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    5 (   5   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    5 (   3   :;   0   =)
%            Number of variables   :    3 (   0 sgn;   2   !;   1   ?;   0   ^)
%                                         (   3   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% SPC      : TH0_THM_NEQ_NAR

% Comments : Assume eps is a choice function on $i and p and q are sets 
%            covering $i. Since $i is nonempty, either p or q is nonempty and 
%            so (eps @ p) is in p or (eps @ q) is in q.
%------------------------------------------------------------------------------
thf(eps,type,(
    eps: ( $i > $o ) > $i )).

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

thf(p,type,(
    p: $i > $o )).

thf(q,type,(
    q: $i > $o )).

thf(pq,axiom,(
    ! [X: $i] :
      ( ( p @ X )
      | ( q @ X ) ) )).

thf(conj,conjecture,
    ( ( p @ ( eps @ p ) )
    | ( q @ ( eps @ q ) ) )).

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