$7$ Divisibility and GCD Prove that $$a b c in Z$,($a $|$ c$)=($b $|$ c$)=1=$ gcd$($a$,$ b$)=($ab $|$ c$).$ following the instructions for an acceptable proof. The Coq statement is Lemma prob$7$ : forall a b c : Z$,($a $|$ c$)->($b $|$ c$)->$ Zis$\_$gcd a b $1->($a$*$b $|$ c$).$ The Zis gcd bezout lemma can be apply$$ed to get a Bezout statement, which can then be destructed, to get the Bezout relation from Zis gcd$.$