theorem
  (a1+b1)|^(n+1) = a1|^(n+1) + b1|^(n+1) + a1*b1*c1 implies
  (a1+b1)|^(n+2) = a1|^(n+2) + b1|^(n+2) + a1*b1*(a1|^n + b1|^n + c1*(a1+b1))
  proof
    assume (a1+b1)|^(n+1) = a1|^(n+1) + b1|^(n+1) + a1*b1*c1;
    then (a1+b1)|^(n+1+1) =
    (a1+b1)*(a1|^(n+1) + b1|^(n+1) + a1*b1*c1) by NEWTON:6
    .= a1*a1|^(n+1) + b1*a1|^(n+1) +
     a1*b1|^(n+1) + b1*b1|^(n+1) + (a1+b1)*a1*b1*c1
    .= a1|^(n+1+1)+ b1*a1|^(n+1) +
     a1*b1|^(n+1) + b1*b1|^(n+1) + a1*b1*(c1*(a1+b1)) by NEWTON:6
    .= a1|^(n+2) + b1*(a1*a1|^n) +
    a1*b1|^(n+1) + b1*b1|^(n+1) + a1*b1*c1*(a1+b1) by NEWTON:6
    .= a1|^(n+2) + b1*a1*a1|^n + a1*(b1*b1|^n)
    + b1*b1|^(n+1) + a1*b1*c1*(a1+b1) by NEWTON:6
    .= a1|^(n+2) + b1*a1*a1|^n + a1*b1*b1|^n
    + b1|^(n+1+1) + a1*b1*(c1*a1 + c1*b1) by NEWTON:6
    .= a1|^(n+2) + b1|^(n+2) + a1*b1*(a1|^n + b1|^n + c1*(a1 + b1));
    hence thesis;
  end;
