(* We need the simple form of Bezout above to show this *)
Fact is_rel_prime_div p q k : is_gcd p q 1 -> p div q*k -> p div k.
Fact is_rel_prime_div_r p q k : is_gcd p q 1 -> p div k*q -> p div k.
Hint Resolve divides_1 divides_mult_compat is_gcd_refl.
Fact is_gcd_0 p q : is_gcd p q 0 -> p = 0 /\ q = 0.
Fact is_gcd_rel_prime p q g : is_gcd p q g -> exists a b, p = a*g /\ q = b*g /\ is_gcd a b 1.
Fact is_lcm_mult p q l k : is_lcm p q l -> is_lcm (k*p) (k*q) (k*l).

Theorem is_gcd_lcm_mult p q g l : is_gcd p q g -> is_lcm p q l -> p*q = g*l.