theorem Th24:
  r (*) (LR1+LR2) = (r(*)LR1) + (r(*)LR2)
  proof
    per cases;
    suppose A1: r = 0;
      set Z=ZeroLC(R);
      A2: now let v be Element of R;
            thus (Z+Z).v = Z.v+Z.v by RLVECT_2:def 10
                        .= Z.v+0 by RLVECT_2:20
                        .= Z.v;
          end;
      thus r(*)(LR1+LR2) = Z by A1,Def2
                        .= Z+Z by A2
                        .= (r(*)LR1)+Z by A1,Def2
                        .= (r(*)LR1)+(r(*)LR2) by A1,Def2;
    end;
    suppose A3: r<>0;
      now let v be Element of R;
        thus(r(*)(LR1+LR2)).v = (LR1+LR2).(r"*v) by A3,Def2
                             .= LR1.(r"*v)+LR2.(r"*v) by RLVECT_2:def 10
                             .= (r(*)LR1).v+LR2.(r"*v) by A3,Def2
                             .= (r(*)LR1).v+(r(*)LR2).v by A3,Def2
                             .= ((r(*)LR1)+(r(*)LR2)).v by RLVECT_2:def 10;
      end;
      hence thesis;
    end;
  end;
