reserve k,n,m,l,p for Nat;
reserve n0,m0 for non zero Nat;
reserve f for FinSequence;
reserve x,X,Y for set;
reserve f1,f2,f3 for FinSequence of REAL;
reserve n1,n2,m1,m2 for Nat;

theorem Th16:
  n1 in NatDivisors n0 & m1 in NatDivisors m0 implies
  n1*m1 in NatDivisors(n0*m0)
proof
  reconsider b=n0*m0 as non zero Nat;
  assume
A1: n1 in NatDivisors n0;
  then
A2: 0 < n1;
A3: n1 divides n0 by A1,MOEBIUS1:39;
  assume
A4: m1 in NatDivisors m0;
  then
A5: 0 < m1;
  then reconsider a=n1*m1 as non zero Nat by A2;
A6: m1 divides m0 by A4,MOEBIUS1:39;
  for p being Element of NAT st p is prime holds p |-count a <= p |-count b
  proof
    reconsider n19=n1,m19=m1 as non zero Nat by A1,A4;
    let p be Element of NAT;
    assume p is prime;
    then reconsider p9=p as Prime;
A8: p9 |-count(n19*m19) = p9 |-count n19 + p9 |-count m19 & p9 |-count(n0*
    m0) = p9 |-count n0 + p9 |-count m0 by NAT_3:28;
    p9 |-count n19 <= p9 |-count n0 & p9 |-count m19 <= p9 |-count m0 by A3,A6,
NAT_3:30;
    hence p |-count a <= p |-count b by A8,XREAL_1:7;
  end;
  then ex c being Element of NAT st n0*m0=(n1*m1)*c by NAT_4:20;
  then n1*m1 divides n0*m0 by NAT_D:def 3;
  hence n1*m1 in NatDivisors(n0*m0) by A2,A5;
end;
