
theorem
  for a be Nat, b,n be non zero Nat holds
  (a mod b)|^n >= a|^n mod b
  proof
    let a be Nat, b,n be non zero Nat;
    reconsider m = n-1  as Nat;
    A1: (a mod b)*(a mod b)|^m = (a mod b)|^(m+1) by NEWTON:6;
    ((a mod b)*(a mod b)|^m) mod b <= (a mod b)*(a mod b)|^m by AMB;
    hence thesis by A1,GR_CY_3:30;
  end;
