reserve a,b,i,j,k,l,m,n for Nat;

theorem RS:
  for a,b be Real holds
    (a,b) Subnomial (n+1) = (a*((a,b)Subnomial n))^<*b|^(n+1)*>
proof
  let a,b be Real;
  A0: (a,b)Subnomial n is FinSequence of COMPLEX by RVSUM_1:146;
  A0a: len ((a,b) Subnomial n) = len (a*(a,b) Subnomial n) by COMPLSP2:3;
  A2: len ((a,b) Subnomial ((n+1)+1-1)) = (n+1) + 1 &
  len ((a,b) Subnomial (n+1-1)) = n+1; then
  A3: len ((a,b) Subnomial (n+1))
  = len ((a,b) Subnomial n) + len <*b|^(n+1)*> by FINSEQ_1:39
  .= len (a*((a,b)Subnomial n )) + len <*b|^(n+1)*> by COMPLSP2:3
  .= len (a*((a,b)Subnomial n)^<*b|^(n+1)*>) by FINSEQ_1:22;
 for k be Nat st 1 <= k <= len ((a,b) Subnomial (n+1)) holds
 ((a,b) Subnomial (n+1)).k = (a*((a,b)Subnomial n)^<*b|^(n+1)*>).k
 proof
   let k be Nat such that
   B0: 1 <= k <= len ((a,b) Subnomial (n+1));
   reconsider m = k-1 as Nat by B0;
   per cases;
   suppose
     C1: k in dom ((a,b) Subnomial n); then
     C2: 1 <= k <= len ((a,b) Subnomial n) by FINSEQ_3:25; then
     m+1 <= n+1 by Def2; then
     reconsider l = n-m as Element of NAT by XREAL_1:6,NAT_1:21;
     C3: l+1 = (n+1) - m;
     k in dom ((a,b) Subnomial (n+1)) by B0,FINSEQ_3:25; then
     ((a,b) Subnomial (n+1)).k = a|^(l+1)*b|^m by Def2,C3
     .= a*a|^l*b|^m by NEWTON:6
     .= a*(a|^l*b|^m)
     .= a*((a,b) Subnomial n).k by C1,Def2
     .= (a*(a,b) Subnomial n).k by A0,COMPLSP2:16
     .= ((a*(a,b) Subnomial n)^<*b|^(n+1)*>).k by A0a,C2,FINSEQ_1:64;
     hence thesis;
   end;
   suppose
     not k in dom ((a,b) Subnomial n); then
     not k <= n+1 by B0,A2,FINSEQ_3:25; then
     k >= (n+1)+1 by INT_1:7; then
     C1: k = (n+1)+1 by B0,A2,XXREAL_0:1;
     len (a*(a,b) Subnomial n) = n+1 by A2,NEWTON:2; then
     ((a*(a,b) Subnomial n)^<*b|^(n+1)*>).((n+1)+1)
     = ((a,b) In_Power (n+1)).((n+1)+1) by NEWTON:29;
     hence thesis by NS,C1;
   end;
 end;
  hence thesis by A3;
end;
