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

theorem ES:
  for a be Real, n be even Nat holds
    Sum ((a,-a) Subnomial n) = a|^n
  proof
    let a be Real, n be even Nat;
    per cases;
    suppose
      A1: a is zero;
      per cases;
      suppose
        B1: n = 0;
        Sum ((a,-a)Subnomial n) = (0+1)*a|^n by A1,B1,SAA;
        hence thesis;
      end;
      suppose
        B1: n > 0;
        (a,-a) Subnomial n = len ((0,0) Subnomial (n+1-1))|-> 0 by B1,A1;
        hence thesis by A1,B1,NAT_1:14,NEWTON:11;
      end;
    end;
    suppose
      A1: not a is zero;
      (a-(-a))*Sum ((a,-a) Subnomial n) = a|^(n+1)-((-1)*a)|^(n+1) by SumS
      .= a|^(n+1)-(-1)|^(n+1)*a|^(n+1) by NEWTON:7
      .= 2*a|^(n+1)
      .= 2*(a*a|^n) by NEWTON:6
      .= (2*a)*a|^n;
      hence thesis by A1,XCMPLX_1:5;
    end;
  end;
