
theorem
  for a be Nat, b be non zero Nat, m,n be non zero Nat st m >= n holds
    a mod b|^m = 1 implies a mod b|^n = 1
  proof
    let a be Nat, b be non zero Nat, m,n be non zero Nat such that
    A1: m >= n;
    assume
    A2: a mod b|^m = 1;
    m - n >= n - n by A1,XREAL_1:9; then
    m - n in NAT by INT_1:3; then
    reconsider k = m - n as Nat;
    reconsider b as non trivial Nat by A2;
    1 = a mod b|^(n + k) by A2
    .= a mod (b|^n*b|^k) by NEWTON:8;
    hence thesis by MT;
  end;
