
theorem Th29:
  for n being Ordinal, L being add-associative
  right_complementable right_zeroed distributive right_unital non trivial non
  empty doubleLoopStr, p being Series of n, L holds p*'1_(n,L) = p
proof
  let n be Ordinal, L be add-associative right_complementable right_zeroed
distributive right_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: (p*'O).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 = p.b1*O.b2 by Def10;
    consider t being FinSequence of cL, s1 being Element of cL such that
A4: s = t^<*s1*> by A2,FINSEQ_2:19;
A5: now
      per cases;
      suppose
        t = <*>(cL);
        hence (Sum t) = 0.L by RLVECT_1:43;
      end;
      suppose
A6:     t <> <*>(cL);
        now
          let k be Nat;
A7:       len s = len t + len <*s1*> by A4,FINSEQ_1:22
            .= len t +1 by FINSEQ_1:39;
          assume
A8:       k in dom t;
          then
A9:       t/.k = t.k by PARTFUN1:def 6
            .= s.k by A4,A8,FINSEQ_1:def 7;
          k <= len t by A8,FINSEQ_3:25;
          then
A10:      k < len s by A7,NAT_1:13;
A11:      1 <= k by A8,FINSEQ_3:25;
          then
A12:      k in dom decomp b by A2,A10,FINSEQ_3:25;
A13:      dom s = dom decomp b by A2,FINSEQ_3:29;
          then
A14:      s/.k = s.k by A12,PARTFUN1:def 6;
          per cases by A11,XXREAL_0:1;
          suppose
A15:        1 < k;
            consider b1, b2 being bag of n such that
A16:        (decomp b)/.k = <*b1, b2*> and
A17:        s/.k = p.b1*O.b2 by A3,A13,A12;
            b2 <> EmptyBag n by A2,A10,A15,A16,PRE_POLY:72;
            hence t/.k = p.b1*0.L by A9,A14,A17,Th25
              .= 0.L;
          end;
          suppose
A18:        k = 1;
A19:        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,A7,FINSEQ_1:39;
              hence contradiction by A6;
            end;
            consider b1, b2 being bag of n such that
A20:        (decomp b)/.k = <*b1, b2*> and
A21:        s/.k = p.b1*O.b2 by A3,A13,A12;
            (decomp b)/.1 = <*EmptyBag n, b*> by PRE_POLY:71;
            then b1 = EmptyBag n & b2 = b by A18,A20,FINSEQ_1:77;
            then s.k = (p.EmptyBag n)*0.L by A14,A21,A19,Th25
              .= 0.L;
            hence t/.k = 0.L by A9;
          end;
        end;
        hence (Sum t) = 0.L by MATRLIN:11;
      end;
    end;
A22: s.len s = (t^<*s1*>).(len t +1) by A4,FINSEQ_2:16
      .= s1 by FINSEQ_1:42;
A23: Sum s = (Sum t) + (Sum <*s1*>) by A4,RLVECT_1:41;
    s is non empty by A2;
    then
A24: len s in dom s by FINSEQ_5:6;
    then consider b1, b2 being bag of n such that
A25: (decomp b)/.len s = <*b1, b2*> and
A26: s/.len s = p.b1*O.b2 by A3;
A27: s/.len s = s.len s by A24,PARTFUN1:def 6;
    (decomp b)/.len s = <*b, EmptyBag n*> by A2,PRE_POLY:71;
    then
A28: b1 = b & b2 = EmptyBag n by A25,FINSEQ_1:77;
    Sum <*s1*> = s1 by RLVECT_1:44
      .= p.b*1.L by A26,A28,A22,A27,Th25
      .= p.b;
    hence (p*'O).b = p.b by A1,A23,A5,RLVECT_1:4;
  end;
  hence thesis;
end;
