theorem
  r(#)(G + H) = r(#)G + r(#)H & r(#)(G - H) = r(#)G - r(#)H
  proof
    now
      let n be Element of NAT;
      thus (r(#)(G + H)).n = r(#)(G + H).n by Def1
      .= r(#)(G.n + H.n) by Def5
      .= r(#)(G.n) + r(#)(H.n) by VFUNCT_1:13
      .=(r(#)G).n + r(#)(H.n) by Def1
      .=(r(#)G).n + (r(#)H).n by Def1
      .=(r(#)G + r(#)H).n by Def5;
    end;
    hence r(#)(G + H) = r(#)G + r(#)H by FUNCT_2:63;
    now
      let n be Element of NAT;
      thus (r(#)(G - H)).n = r(#)(G - H).n by Def1
      .= r(#)(G.n - H.n) by Th3
      .= r(#)G.n - r(#)H.n by VFUNCT_1:15
      .= (r(#)G).n - r(#)H.n by Def1
      .= (r(#)G).n - (r(#)H).n by Def1
      .= (r(#)G - r(#)H).n by Th3;
    end;
    hence thesis by FUNCT_2:63;
  end;
