
theorem gcdp:
for p being Prime
for n being non zero Nat st n < p holds n gcd p = 1
proof
let p be Prime, n be non zero Nat;
assume A: n < p;
1 * n = n & 1 * p = p; then
B: 1 divides n & 1 divides p by NAT_D:def 3;
now let m be Nat;
  assume C: m divides n & m divides p; then
  per cases by INT_2:def 4;
  suppose m = 1;
    hence m divides 1;
    end;
  suppose m = p;
    hence m divides 1 by A,C,NAT_D:7;
    end;
  end;
hence thesis by B,NAT_D:def 5;
end;
