theorem Th56:
  hom?-(a) = (curry (hom??(C))).(id a) & hom-?(a) = (curry' (hom?? (C))).(id a)
proof
  reconsider T = hom??(C) as Function of [:the carrier' of C,the carrier' of C
  :],Maps(Hom(C));
  now
    let f;
    thus ((curry T).(id a)).f = T.(id a,f) by FUNCT_5:69
      .= [[Hom(cod id a,dom f),Hom(dom id a,cod f)],hom(id a,f)] by Def23
      .= [[Hom(cod id a,dom f),Hom(dom id a,cod f)],hom(a,f)] by Th52
      .= [[Hom(a,dom f),Hom(dom id a,cod f)],hom(a,f)]
      .= [[Hom(a,dom f),Hom(a,cod f)],hom(a,f)]
      .= (hom?-(a)).f by Def20;
  end;
  hence hom?-(a) = (curry (hom??(C))).(id a) by FUNCT_2:63;
  now
    let f;
    thus ((curry' T).(id a)).f = T.(f,id a) by FUNCT_5:70
      .= [[Hom(cod f,dom id a),Hom(dom f,cod id a)],hom(f,id a)] by Def23
      .= [[Hom(cod f,dom id a),Hom(dom f,cod id a)],hom(f,a)] by Th52
      .= [[Hom(cod f,a),Hom(dom f,cod id a)],hom(f,a)]
      .= [[Hom(cod f,a),Hom(dom f,a)],hom(f,a)]
      .= (hom-?(a)).f by Def21;
  end;
  hence thesis by FUNCT_2:63;
end;
