Ok, here is what I have for the proof of this conjecture. Let me know if I'm on the right path? all input appreciated.
There exist integers $j$, $k$, and $m$, such that, $b = aj $ and $ c = ajk.$ Then $c = ajk $ (substituting $aj$ for $b$) let $m = jk$, then $c = ma, => a|c.$
Note that,
$$a|b \implies b=ma \,,$$
and
$$ b|c \implies c =nb \implies c = nma \implies c = q a \implies a|c \,,$$
for $m,n,q \in \mathbb{Z}$ and $q=nm.$
Since you given a/b, this means b=ka for some k-interger. Also b/c implies that c=lb for some interger l. now, substituting b-ka into c=lb gives
c=lka.....>>>c=lk(a)>>>>we conclude a can divide c
Hence a/c N:B_ the / is not a mathematical symbol for "divides" but rather use |...its just that i like it...lol
(In this answer all variables are integers, i.e., elements of $\mathbb{Z}$.)
By the definition of divisibility we are given that $\;n * a = b\;$ and $\;m * b = c\;$ for some $\;n\;$ and $\;m\;$. Now we are asked to find a $\;k\;$ which makes $\;k*a = c\;$:
\begin{align} & k*a = c \\ \equiv & \;\;\;\;\;\text{"use the only fact we know about $\;c\;$"} \\ & k*a = m*b \\ \equiv & \;\;\;\;\;\text{"use the other fact"} \\ & k*a = m*n*a \\ \Leftarrow & \;\;\;\;\;\text{"weaken using Leibniz' rule -- to achieve our goal"} \\ & k = m*n \\ \end{align}
Therefore we have found such a $\;k\;$, and hence proved $\;a|c\;$.
