Proof for : if $((a \mid b),$ & $(a \nmid c))$, then $b\nmid c.$ The reason for posting this question is a slightly different approach being used for the proof.
If $a\mid b$, & $a\nmid c$, then $b\nmid c$. I have used the contradiction approach: 
$\exists k \in \mathbb{Z}, b=ak$. If assume that $b \mid c$, then $\exists m \in \mathbb {Z}, c=bm$. This means that $c=akm$. But if $\forall l \in \mathbb{Z}, c \ne al$, then this is proved wrong, as $l=km$. Hence, there can be no integer value of $l=km$ possible.
 A: Yes, you've successfully proven the implication, by using proof by contradiction.
You are assuming the premise, and the negation of the consequent.  
If we call $P: a\mid b$, and $Q: a\nmid c$, and $R: b\nmid c$,
You've assumed $P\land Q$ and $\lnot R = b\mid c$
$P\land Q\tag 1$  
$P\;\;\tag{from (1)}$
$Q\;\;\tag {from(1)}$
$\quad|\lnot R\tag{(2): Assumption }$ 
$\qquad||\lnot Q\;\;\tag{as given by asker}$
$\qquad|| \lnot P \lor \lnot Q\tag{disjunction intro}$
$\qquad|| \lnot (P \land Q)\tag {DeMorgan's}$
$\qquad|| (P\land Q) \land \lnot (P\land Q)\tag{conjunction intro}$
$\lnot (\lnot R)\tag {follows from contradiction}$
$R\tag{double negation}$
So we have proven $$\big((P\land Q)\land \lnot R\big) \to \lnot(P\land Q)$$

We've reached a contradition.  We conclude $(P\land Q) \to R$.
But note, we can are essentially done when we arrive at $\lnot Q$, because we reach $Q\land \lnot Q \equiv \bot$, which is where you stopped (having obtained a contradiction). Logically, we have proven  $\lnot\lnot R$, or $R$, and hence $(P\land Q) \to R$. 
