TPTP Problem File: MSC025^2.p

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% File     : MSC025^2 : TPTP v7.1.0. Released v5.5.0.
% Domain   : Syntactic
% Problem  : Nik's Challenge
% Version  : Especial.
%          : Theorem formulation : Bijection instantiated.
% English  : Force the $i type to only have two elements, then conjecture the
%            existence of a bijection between $o and $i.

% Refs     : [Sul13] Sultana (2013), Email to Geoff Sutcliffe
% Source   : [Sul13]
% Names    :

% Status   : Theorem
% Rating   : 0.38 v7.1.0, 0.62 v7.0.0, 0.57 v6.4.0, 0.67 v6.3.0, 0.60 v6.2.0, 0.57 v5.5.0
% Syntax   : Number of formulae    :    9 (   0 unit;   4 type;   0 defn)
%            Number of atoms       :   40 (  10 equality;  16 variable)
%            Maximal formula depth :    8 (   4 average)
%            Number of connectives :   20 (   5   ~;   2   |;   2   &;   6   @)
%                                         (   0 <=>;   5  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    3 (   3   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    6 (   4   :;   0   =)
%            Number of variables   :    5 (   0 sgn;   5   !;   0   ?;   0   ^)
%                                         (   5   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% SPC      : TH0_THM_EQU_NAR

% Comments : 
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thf(one,type,(
    one: $i )).

thf(two,type,(
    two: $i )).

thf(binarity_exhaust,axiom,(
    ! [X: $i] :
      ( ( X = one )
      | ( X = two ) ) )).

thf(binarity_distinc,axiom,(
    one != two )).

thf(b1_ty,type,(
    b1: $o > $i )).

thf(b1,axiom,(
    ! [X: $o] :
      ( ( X
       => ( ( b1 @ X )
          = one ) )
      & ( ~ ( X )
       => ( ( b1 @ X )
          = two ) ) ) )).

thf(b2_ty,type,(
    b2: $o > $i )).

thf(b2,axiom,(
    ! [X: $o] :
      ( ( X
       => ( ( b2 @ X )
          = two ) )
      & ( ~ ( X )
       => ( ( b2 @ X )
          = one ) ) ) )).

thf(goal,conjecture,(
    ! [F: $o > $i] :
      ( ! [X: $o] :
          ( ( F @ X )
         != ( F @ ~ ( X ) ) )
     => ( ( F = b1 )
        | ( F = b2 ) ) ) )).

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