reserve G for Group,
  a,b for Element of G,
  m, n for Nat,
  p for Prime;

theorem Th1:
  (for r being Nat holds n <> p |^ r) implies
  ex s being Element of NAT st s is prime & s divides n & s <> p
proof
  assume
A1: for r being Nat holds n <> p |^ r;
  per cases;
  suppose
A2:  n = 0;
    per cases;
    suppose
A3:   p = 2;
      take s = 3;
      thus s is prime by PEPIN:41;
      thus s divides n by A2,INT_2:12;
      thus s <> p by A3;
    end;
    suppose
A4:   p <> 2;
      take s = 2;
      thus s is prime by INT_2:28;
      thus s divides n by A2,INT_2:12;
      thus s <> p by A4;
    end;
  end;
  suppose
A5: n <> 0;
A6: p > 1 by INT_2:def 4;
  reconsider r1 = p |-count n as Element of NAT;
A7: p |^ r1 divides n by A5,A6,NAT_3:def 7;
A8: not p |^ (r1+1) divides n by A5,A6,NAT_3:def 7;
  consider s1 be Nat such that
A9: n = p |^ r1 * s1 by A7,NAT_D:def 3;
  s1 >= 2
  proof
    assume not s1 >= 2;
    then s1 < 1 + 1;
    then s1 <= 1 by NAT_1:13;
    then s1 = 0 or s1 = 1 by NAT_1:25;
    hence contradiction by A5,A1,A9;
  end;
  then consider s be Element of NAT such that
A10: s is prime and
A11: s divides s1 by INT_2:31;
A12: s1 divides n by A9,NAT_D:def 3;
  take s;
  s <> p
  proof
    assume s = p;
    then consider s2 be Nat such that
A13: s1 = p * s2 by A11,NAT_D:def 3;
    n = p |^ r1 * p * s2 by A9,A13
     .= p |^ (r1+1) * s2 by NEWTON:6;
    hence contradiction by A8,NAT_D:def 3;
  end;
  hence thesis by A10,A11,A12,NAT_D:4;
end;
end;
