reserve i,j,n,k for Nat,
  a for Element of COMPLEX,
  R1,R2 for Element of i-tuples_on COMPLEX;

theorem Th20:
  for F being FinSequence of COMPLEX st len (F*') >= 1 holds
  addcomplex $$ (F*') = (addcomplex $$ F)*'
proof
  let F be FinSequence of COMPLEX;
A1: len F = len (F*') by COMPLSP2:def 1;
  assume
A2: len (F*') >= 1;
  then consider f22 being sequence of COMPLEX such that
A3: f22.1 = (F*').1 and
A4: for n be Nat st 0 <> n & n < len (F*') holds f22.(n + 1)
  = addcomplex.(f22.n,(F*').(n + 1)) and
A5: addcomplex $$ (F*') = f22.(len (F*')) by FINSOP_1:1;
A6: (len (F*')) in Seg len (F*') by A2,FINSEQ_1:1;
  defpred P[set,set] means $2= f22.$1;
A7: for k be Nat st k in Seg len (F*') ex x being Element of COMPLEX st P[k,
  x];
  ex f2 being FinSequence of COMPLEX st dom f2 = Seg len (F*') & for k be
  Nat st k in Seg len (F*') holds P[k,f2.k] from FINSEQ_1:sch 5(A7);
  then consider f2 being FinSequence of COMPLEX such that
A8: dom f2 = Seg len (F*') and
A9: for k be Nat st k in Seg len (F*') holds P[k,f2.k];
  consider f9 being sequence of COMPLEX such that
A10: for n being Nat st 1<= n & n<=len (f2*') holds f9.n=(f2*').n by Th19;
A11: len (f2*')=len f2 by COMPLSP2:def 1;
  then 1<=len (f2*') by A2,A8,FINSEQ_1:def 3;
  then
A12: f9.1=(f2*').1 by A10;
A13: len f2 = len (F*') by A8,FINSEQ_1:def 3;
A14: for n st 0 <> n & n < len (F*') holds f2.(n + 1) = addcomplex.(f2.n,(F
  *').(n + 1))
  proof
    let n;
    assume that
A15: 0 <> n and
A16: n < len (F*');
A17: n+1<=len (F*') by A16,NAT_1:13;
A18: 0+1<=n by A15,NAT_1:13;
    then
A19: n in Seg len (F*') by A16,FINSEQ_1:1;
    1<=n+1 by A18,NAT_1:13;
    then (n+1) in Seg len (F*') by A17,FINSEQ_1:1;
    then f2.(n + 1) = f22.(n + 1) by A9
      .= addcomplex.(f22.n,(F*').(n + 1)) by A4,A15,A16
      .= addcomplex.(f2.n,(F*').(n + 1)) by A9,A19;
    hence thesis;
  end;
A20: for n being Nat st 0 <> n & n < len F holds (f2*').(n + 1) =
  addcomplex.((f2*').n,F.(n + 1))
  proof
    let n be Nat;
    assume that
A21: 0 <> n and
A22: n < len F;
A23: n<=len f2 by A1,A8,A22,FINSEQ_1:def 3;
A24: 0+1<=n by A21,NAT_1:13;
    then
A25: 1<=n+1 by NAT_1:13;
     reconsider c = (F.(n + 1))*' as Element of COMPLEX by XCMPLX_0:def 2;
A26: n+1<=len F by A22,NAT_1:13;
    then n+1<=len f2 by A1,A8,FINSEQ_1:def 3;
    then (f2*').(n + 1) = (f2.(n+1))*' by A25,COMPLSP2:def 1
      .= (addcomplex.(f2.n,(F*').(n + 1)))*' by A1,A14,A21,A22
      .= (addcomplex.(f2.n,c))*' by A25,A26,COMPLSP2:def 1
      .= (f2.n+c)*' by BINOP_2:def 3
      .= (f2.n)*'+c*' by COMPLEX1:32
      .= addcomplex.((f2.n)*',F.(n + 1)) by BINOP_2:def 3;
    hence thesis by A24,A23,COMPLSP2:def 1;
  end;
A27: for n be Nat st 0 <> n & n < len F holds f9.(n + 1) =
  addcomplex.(f9.n,F.(n + 1))
  proof
    let n be Nat;
    assume that
A28: 0 <> n and
A29: n < len F;
A30: 0+1<=n by A28,NAT_1:13;
    n+1<=len (f2*') by A1,A13,A11,A29,NAT_1:13;
    then
A31: f9.(n+1)=(f2*').(n+1) by A10,NAT_1:11;
    n<=len (f2*') by A1,A13,A29,COMPLSP2:def 1;
    then f9.n=(f2*').n by A10,A30;
    hence thesis by A20,A28,A29,A31;
  end;
A32: 1 in Seg len (F*') by A2,FINSEQ_1:1;
  set d=(addcomplex $$ (F*'))*';
A33: len F >=1 by A2,COMPLSP2:def 1;
  (f2*').(len F)=(f2*').(len (F*')) by COMPLSP2:def 1
    .= (f2.(len (F*')))*' by A2,A13,COMPLSP2:def 1
    .= d by A5,A9,A6;
  then
A34: d=f9.(len F) by A2,A1,A13,A10,A11;
  F.1=((F.1)*')*' .= ((F*').1)*' by A33,COMPLSP2:def 1
    .= (f2.1)*' by A3,A9,A32
    .= (f2*').1 by A2,A13,COMPLSP2:def 1;
  then d = addcomplex $$ F by A33,A12,A27,A34,FINSOP_1:2;
  hence thesis;
end;
