
theorem
  for a be Nat, b,n be non zero Nat holds
  a mod b = 1 implies a|^n mod b = 1
  proof
    let a be Nat, b,n be non zero Nat;
    assume a mod b = 1; then
    1 = ((a mod b)|^n) mod b;
    hence thesis by GR_CY_3:30;
  end;
