reserve m, n for Nat;

theorem
  for p being Prime, m being Nat st m is square-free & p divides m holds
  { d where d is Element of NAT : 0 < d & d divides m & not p divides d } =
  { d where d is Element of NAT : 0 < d & d divides (m div p) }
proof
  let p be Prime, m be Nat;
  assume that
A1: m is square-free and
A2: p divides m;
  set B = { d where d is Element of NAT : 0 < d & d divides (m div p) };
  set A = { d where d is Element of NAT : 0 < d & d divides m & not p divides
  d };
  thus A c= B
  proof
    let x be object;
    assume x in A;
    then consider d being Element of NAT such that
A3: d = x & 0 < d and
A4: d divides m & not p divides d;
    d divides (m div p) by A2,A4,Th28;
    hence thesis by A3;
  end;
  let x be object;
  assume x in B;
  then consider d being Element of NAT such that
A5: d = x & 0 < d and
A6: d divides (m div p);
  d divides m & not p divides d by A1,A2,A6,Th27;
  hence thesis by A5;
end;
