
theorem Th21:
  for a,b being non zero Nat holds
  (for p being Element of NAT st p is prime holds p |-count a = p |-count b)
  implies a = b
proof
  let a,b be non zero Nat;
  assume
A1: for p being Element of NAT st p is prime holds p |-count a = p |-count b;
  then
  for p being Element of NAT st p is prime holds p |-count a <= p |-count b;
  then consider c being Element of NAT such that
A2: b=a*c by Th20;
  for p being Element of NAT st p is prime holds p |-count b <= p |-count
  a by A1;
  then consider d being Element of NAT such that
A3: a=b*d by Th20;
  a=a*(c*d) by A2,A3;
  then c = 1 by NAT_1:15,XCMPLX_1:7;
  hence thesis by A2;
end;
