reserve a, b, n for Nat,
  r for Real,
  f for FinSequence of REAL;
reserve p for Prime;

theorem Th5:
  p divides a |^ b implies p divides a
proof
  assume that
A1: p divides a |^ b and
A2: not p divides a;
  reconsider p,a as Element of NAT by ORDINAL1:def 12;
  defpred P[Nat] means p divides a|^($1+1);
A3: for k being Nat st k <> 0 & P[k] ex n being Nat st n < k & P[n]
  proof
    let k be Nat such that
A4: k <> 0 and
A5: P[k];
A6: p divides a |^ k * a by A5,NEWTON:6;
    take k-'1;
A7: k >= 0+1 by A4,NAT_1:13;
    then k-1 >= 0+1-1 by XREAL_1:13;
    then k-'1 = k-1 by XREAL_0:def 2;
    then k-'1 < k-0 by XREAL_1:15;
    hence k-'1 < k;
    k-'1+1 = k by A7,XREAL_1:235;
    hence thesis by A2,A6,NEWTON:80;
  end;
  now
    assume 0+1 > b;
    then b = 0 by NAT_1:13;
    then p divides 1 by A1,NEWTON:4;
    then p = 1 by WSIERP_1:15;
    hence contradiction by INT_2:def 4;
  end;
  then b = b-'1+1 by XREAL_1:235;
  then
A8: ex k being Nat st P[k] by A1;
  P[ 0 ] from NAT_1:sch 7(A8,A3);
  hence thesis by A2;
end;
