
theorem Th30:
  for n being Ordinal, L being add-associative
  right_complementable right_zeroed distributive well-unital non trivial non
  empty doubleLoopStr, p being Series of n, L holds 1_(n,L)*'p = p
proof
  let n be Ordinal, L be add-associative right_complementable right_zeroed
distributive well-unital non trivial doubleLoopStr, p be Series of
  n, L;
  set O = 1_(n,L), cL = the carrier of L;
  now
    let b be Element of Bags n;
    consider s being FinSequence of cL such that
A1: (O*'p).b = Sum s and
A2: len s = len decomp b and
A3: for k being Element of NAT st k in dom s ex b1, b2 being bag of n
    st (decomp b)/.k = <*b1, b2*> & s/.k = O.b1*p.b2 by Def10;
    s is non empty by A2;
    then consider s1 being Element of cL, t being FinSequence of cL such that
A4: s1 = s.1 and
A5: s = <*s1*>^t by FINSEQ_3:102;
A6: Sum s = (Sum <*s1*>) + (Sum t) by A5,RLVECT_1:41;
A7: now
      per cases;
      suppose
        t = <*>(cL);
        hence (Sum t) = 0.L by RLVECT_1:43;
      end;
      suppose
A8:     t <> <*>(cL);
        now
          let k be Nat;
A9:       len s = len t + len <*s1*> by A5,FINSEQ_1:22
            .= len t +1 by FINSEQ_1:39;
          assume
A10:      k in dom t;
          then
A11:      t/.k = t.k by PARTFUN1:def 6
            .= s.(k+1) by A5,A10,FINSEQ_3:103;
          1 <= k by A10,FINSEQ_3:25;
          then
A12:      1 < k+1 by NAT_1:13;
          k <= len t by A10,FINSEQ_3:25;
          then
A13:      k+1 <= len s by A9,XREAL_1:6;
          then
A14:      k+1 in dom decomp b by A2,A12,FINSEQ_3:25;
A15:      dom s = dom decomp b by A2,FINSEQ_3:29;
          then
A16:      s/.(k+1) = s.(k+1) by A14,PARTFUN1:def 6;
          per cases by A13,XXREAL_0:1;
          suppose
A17:        k+1 < len s;
            consider b1, b2 being bag of n such that
A18:        (decomp b)/.(k+1) = <*b1, b2*> and
A19:        s/.(k+1) = O.b1*p.b2 by A3,A15,A14;
            b1 <> EmptyBag n by A2,A12,A17,A18,PRE_POLY:72;
            hence t/.k = 0.L*p.b2 by A11,A16,A19,Th25
              .= 0.L;
          end;
          suppose
A20:        k+1 = len s;
A21:        now
              assume b = EmptyBag n;
              then decomp b = <* <*EmptyBag n, EmptyBag n*> *> by PRE_POLY:73;
              then len t +1 = 0 qua Nat+1 by A2,A9,FINSEQ_1:39;
              hence contradiction by A8;
            end;
            consider b1, b2 being bag of n such that
A22:        (decomp b)/.(k+1) = <*b1, b2*> and
A23:        s/.(k+1) = O.b1*p.b2 by A3,A15,A14;
            (decomp b)/.len s = <*b,EmptyBag n*> by A2,PRE_POLY:71;
            then b2 = EmptyBag n & b1 = b by A20,A22,FINSEQ_1:77;
            then s.(k+1) = 0.L*(p.EmptyBag n) by A16,A23,A21,Th25
              .= 0.L;
            hence t/.k = 0.L by A11;
          end;
        end;
        hence Sum t = 0.L by MATRLIN:11;
      end;
    end;
A24: s is non empty by A2;
    then consider b1, b2 being bag of n such that
A25: (decomp b)/.1 = <*b1, b2*> and
A26: s/.1 = O.b1*p.b2 by A3,FINSEQ_5:6;
    1 in dom s by A24,FINSEQ_5:6;
    then
A27: s/.1 = s.1 by PARTFUN1:def 6;
    (decomp b)/.1 = <*EmptyBag n, b*> by PRE_POLY:71;
    then
A28: b2 = b & b1 = EmptyBag n by A25,FINSEQ_1:77;
    Sum <*s1*> = s1 by RLVECT_1:44
      .= 1.L*p.b by A4,A26,A28,A27,Th25
      .= p.b;
    hence (O*'p).b = p.b by A1,A6,A7,RLVECT_1:4;
  end;
  hence thesis;
end;
