reserve m, n for Nat;

theorem Th23:
  for p being Prime holds p is square-free
proof
  let p be Prime;
  assume p is square-containing;
  then consider n being Prime such that
A1: n |^ 2 divides p;
A2: n divides n |^ 2 by NAT_3:3;
  then
A3: n divides p by A1,NAT_D:4;
  per cases by A3,INT_2:def 4;
  suppose
    n = 1;
    hence contradiction by INT_2:def 4;
  end;
  suppose
    n = p;
    then n = n |^ 2 by A1,A2,NAT_D:5;
    then n |^ 1 = n |^ 2;
    then n <= 1 by PEPIN:30;
    hence contradiction by INT_2:def 4;
  end;
end;
