
theorem Th19:
  for p being Prime, n being Nat st n < p holds not p divides n!
proof
  defpred P[Nat] means for p being Prime st $1<p holds not p divides $1!;
  let p be Prime;
  let n be Nat;
  assume
A1: n<p;
A2: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    assume
A3: P[n];
    for p being Prime st n+1<p holds not p divides ((n+1)!)
    proof
      let p be Prime;
      assume
A4:   n+1<p;
      assume p divides (n+1)!;
      then
A5:   p divides (n!)*(n+1) by NEWTON:15;
      n+0<n+1 by XREAL_1:6;
      then n<p by A4,XXREAL_0:2;
      then not p divides n! by A3;
      then p divides (n+1) by A5,NEWTON:80;
      hence contradiction by A4,NAT_D:7;
    end;
    hence thesis;
  end;
A6: P[0]
  by NAT_D:7,NEWTON:12,INT_2:def 4;
  for n being Nat holds P[n] from NAT_1:sch 2(A6,A2);
  hence thesis by A1;
end;
