theorem Th24:
  Y c= X implies chi(X,Y) = chi(Y,Y)
  proof
    assume
A1: Y c= X;
    now
      thus dom chi(X,Y) = Y by FUNCT_3:def 3
                       .= dom chi(Y,Y) by FUNCT_3:def 3;
      hereby
        let x be object;
        assume
A2:     x in dom chi(X,Y);
        then x in Y & x in X by A1;
        hence (chi(X,Y)).x = 1 by FUNCT_3:def 3
                           .= (chi(Y,Y)).x by A2,FUNCT_3:def 3;
      end;
    end;
    hence thesis;
  end;
