
theorem CF1:
for n being non zero Nat
for p being Prime ex k,m being Nat st n = m * p|^k & not p divides m
proof
let n be non zero Nat, p be Prime;
per cases;
suppose AS: p divides n;
  defpred P[Nat] means p|^($1) divides n;
  A: now let k be Nat;
     assume A0: P[k];
     A2: p|^k <= n by A0,NAT_D:7;
     now assume k > n; then
       A3: p|^k > p|^n by NAT_6:2;
       p > 1 by INT_2:def 4; then
       p >= 1 + 1 by INT_1:7; then
       p|^n > n by NEWTON:86;
       hence contradiction by A2,A3,XXREAL_0:2;
       end;
     hence k <= n;
     end;
     p|^1 = p; then
  B: ex k being Nat st P[k] by AS;
  consider k being Nat such that
  C: P[k] & for m being Nat st P[m] holds m <= k from NAT_1:sch 6(A,B);
  p|^k divides n & p|^k is non zero by C,NEWTON:87; then
  n/(p|^k) is Nat by Lm3b; then
  consider m being Nat such that D: m = n/(p|^k);
  take k, m;
  E: p|^k > 0 by NEWTON:83;
  thus F: n = n * 1 .= n * ((p|^k) / (p|^k)) by E,XCMPLX_1:60 .= m * p|^k by D;
  now assume p divides m; then
    consider l being Nat such that G: p * l = m by NAT_D:def 3;
    H: k + 0 < k + 1 by XREAL_1:8;
    n = l * (p * p|^k) by G,F .= l * p|^(k+1) by NEWTON:6;
    hence contradiction by C,H,NAT_D:def 3;
    end;
  hence thesis;
  end;
suppose AS: not p divides n;
  take 0, n;
  thus n = n * 1 .= n * p|^0 by NEWTON:4;
  thus not p divides n by AS;
  end;
end;
