theorem Th6:
    for a be Element of R,i be Element of n holds
    (1_1(i,R)).(UnitBag i) = 1.R & (a|(n,R)).(EmptyBag n) = a &
    (1_1(i,R)).(EmptyBag n) = 0.R & (a|(n,R)).(UnitBag i) = 0.R
    proof
      let a be Element of R, i be Element of n;
      set UBi = UnitBag i;
A1:   UBi in dom (0_(n,R)) & EmptyBag n in dom (0_(n,R)) by PRE_POLY:def 12;
A2:   (1_1(i,R)).(UBi) = (0_(n,R)+*(UBi,1_R)).(UBi) by HILBASIS:def 3
      .= 1.R by A1,FUNCT_7:31;
A3:   (a|(n,R)).(EmptyBag n)
       = (0_(n,R)+*(EmptyBag n,a)).(EmptyBag n) by POLYNOM7:def 8
      .= a by A1,FUNCT_7:31;
A4:   UBi <> EmptyBag n
      proof
        (UBi).i <> (EmptyBag n).i by HILBASIS:9;
        hence thesis;
      end;
A5:   (1_1(i,R)).(EmptyBag n)
      = (0_(n,R)+*(UBi,1_R)).(EmptyBag n) by HILBASIS:def 3
      .= (0_(n,R)).(EmptyBag n) by A4,FUNCT_7:32 .= 0.R;
      (a|(n,R)).(UBi) = (0_(n,R)+*(EmptyBag n,a)).(UBi) by POLYNOM7:def 8
      .=(0_(n,R)).(UBi) by A4,FUNCT_7:32 .= 0.R;
      hence thesis by A2,A3,A5;
    end;
