
theorem lemp:
for p1,p2 being Prime
for n1,n2 being Nat st (n1 <> 0 or n2 <> 0) &
p1|^n1 = p2|^n2 holds p1 = p2 & n1 = n2
proof
let p1,p2 be Prime, n1,n2 be Nat;
assume that
AA: n1 <> 0 or n2 <> 0 and
AS: p1|^n1 = p2|^n2;
per cases by AA;
suppose that
S1: n1 is zero and
S2: n2 is non zero;
A1: p1|^n1 = 1 by S1,NEWTON:4;
  p2 divides p2|^n2 by S2,NAT_3:3;
  then p2 = 1 by AS,A1,INT_2:13;
  hence thesis by INT_2:def 4;
end;
suppose n1 is non zero;
then reconsider m = n1 - 1 as Nat;
n1 = m + 1; then
p1|^n1 = p1|^m * p1 by NEWTON:6;
hence p1 = p2 by AS,NAT_D:def 3,NAT_3:6;
then n1 <= n2 & n2 <= n1 by AS,NAT_6:2;
hence thesis by XXREAL_0:1;
end;
end;
