reserve m, n for Nat;

theorem
  for m, n being non zero Nat st not m, n are_coprime holds Moebius (m * n) = 0
proof
  let m, n be non zero Nat;
  assume not m, n are_coprime;
  then consider p being Prime such that
A1: p divides m & p divides n by PYTHTRIP:def 2;
  reconsider p as prime Element of NAT by ORDINAL1:def 12;
  p * p divides m * n by A1,NAT_3:1;
  then p |^ 2 divides m * n by WSIERP_1:1;
  then m * n is square-containing;
  hence thesis by Def3;
end;
