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

theorem
  for a be Real, n be Nat, i be odd Nat holds
  ((a,-a) Subnomial (n+i)).i = a|^(n+i)
  proof
    let a be Real, n be Nat, i be odd Nat;
    A1: len ((a,a) Subnomial ((n+i)+1-1)) = n+i+1 &
     len ((a,-a) Subnomial ((n+i)+1-1)) = n+i+1;
    i >= 1 & i+(n+1) >= i + 0 by XREAL_1:6,NAT_1:14; then
    A2: i in dom ((a,a) Subnomial (i+n)) &
    i in dom ((a,-a) Subnomial (i+n)) by A1,FINSEQ_3:25; then
    ((a*1,(-1)*a) Subnomial (n+i)).i =
    ((a,a) Subnomial (n+i)).i*((1,-1) Subnomial (n+i)).i by STT
    .= a|^(n+i) by A2,CONST;
    hence thesis;
  end;
