 reserve n,i for Nat;
 reserve p for Prime;

theorem MoInv: :: MOEBIUS1:24 inverted
  n <> 0 & (for p being Prime holds p |-count n <= 1) implies
    n is square-free
  proof
    assume
A1: n <> 0 & for p being Prime holds p |-count n <= 1;
    assume n is square-containing; then
    consider p being Prime such that
A2: p |^ 2 divides n by MOEBIUS1:def 1;
    p |-count n >= 1 + 1 by A1,A2,Ciek;
    hence thesis by A1,NAT_1:13;
  end;
