%--------------------------------------------------------------------------
% File     : SET771+4 : TPTP v8.2.0. Bugfixed v2.2.1.
% Domain   : Set Theory (Equivalence relations)
% Problem  : Equality of images defines a equivalence relation
% Version  : [Pas99] axioms.
% English  :

% Refs     : [Pas99] Pastre (1999), Email to G. Sutcliffe
% Source   : [Pas99]
% Names    :

% Status   : Theorem
% Rating   : 0.72 v8.2.0, 0.78 v8.1.0, 0.72 v7.5.0, 0.78 v7.4.0, 0.67 v7.3.0, 0.69 v7.1.0, 0.61 v7.0.0, 0.70 v6.4.0, 0.73 v6.3.0, 0.67 v6.2.0, 0.68 v6.1.0, 0.73 v6.0.0, 0.78 v5.5.0, 0.85 v5.4.0, 0.89 v5.2.0, 0.90 v5.1.0, 0.95 v5.0.0, 0.92 v4.1.0, 0.96 v4.0.1, 0.91 v4.0.0, 0.92 v3.7.0, 0.95 v3.3.0, 0.93 v3.2.0, 0.91 v3.1.0, 0.89 v2.7.0, 0.83 v2.6.0, 0.86 v2.5.0, 0.88 v2.4.0, 0.75 v2.3.0, 0.67 v2.2.1
% Syntax   : Number of formulae    :   34 (   1 unt;   0 def)
%            Number of atoms       :  175 (   7 equ)
%            Maximal formula atoms :   13 (   5 avg)
%            Number of connectives :  144 (   3   ~;   2   |;  71   &)
%                                         (  36 <=>;  32  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   19 (   9 avg)
%            Maximal term depth    :    2 (   1 avg)
%            Number of predicates  :   20 (  19 usr;   0 prp; 2-6 aty)
%            Number of functors    :   16 (  16 usr;   1 con; 0-5 aty)
%            Number of variables   :  169 ( 156   !;  13   ?)
% SPC      : FOF_THM_RFO_SEQ

% Comments :
% Bugfixes : v2.2.1 - Bugfixes in SET006+1.ax.
%--------------------------------------------------------------------------
%----Include set theory definitions
include('Axioms/SET006+0.ax').
%----Include mappings axioms
include('Axioms/SET006+1.ax').
%----Include equivalence relation axioms
include('Axioms/SET006+2.ax').
%--------------------------------------------------------------------------
fof(thIII07,conjecture,
    ! [F,A,B,R] :
      ( ( maps(F,A,B)
        & ! [X1,X2] :
            ( ( member(X1,A)
              & member(X2,A) )
           => ( apply(R,X1,X2)
            <=> ? [Y] :
                  ( member(Y,B)
                  & apply(F,X1,Y)
                  & apply(F,X2,Y) ) ) ) )
     => equivalence(R,A) ) ).

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