
theorem Th22:
  for p1,p2 being Prime, m being non zero Element of NAT st
  p1|^(p1 |-count m) = p2|^(p2 |-count m) & p1 |-count m > 0 holds p1 = p2
proof
  let p1,p2 be Prime;
  let m be non zero Element of NAT;
  assume
A1: p1|^(p1 |-count m) = p2|^(p2 |-count m);
A2: p1 > 1 by INT_2:def 4;
  assume p1 |-count m > 0;
  then p1 to_power (p1 |-count m) > 1 by A2,POWER:35;
  then
A3: p1|^(p1|-count m) > 1 by POWER:41;
  assume p1 <> p2;
  then p1 <> 1 & not p1 divides p2|^(p2 |-count m) by INT_2:def 4,NAT_3:6;
  then p1 |-count (p1|^(p1 |-count m)) = 0 by A1,NAT_3:27;
  then p1 |-count m = 0 by A2,NAT_3:25;
  hence contradiction by A3,NEWTON:4;
end;
