TITLE: Proving something is a proper subset (Hints no solutions please)
QUESTION [7 upvotes]: Please DO NOT provide the full solution.
Prove that {$10k+7 | k \in\mathbb{Z}$} is a proper subset of {$5m-8|m\in\mathbb{Z}$}
I can see this is indeed true $\{\ldots, 7, 17, 27,\ldots\} \subset \{\ldots, 2, 7, 12, 17, 22, 27, \ldots\}.\,$ As for a proper subset, there is an element in the second set not contained in the first set.
How would I prove this? Unfortunately I have not progressed yet. Any small hints to start me off? I assume you have to manipulate the first one to arrive to the second one? 
Also, is there a standard approach to tackle these questions?

REPLY [5 votes]: You need to show that it is a subset, that is the element $10k+7$ with arbitrary $k\in \mathbb Z$ must be writable as an element $5m-8$, that is for each $k$ you must be able to specify an $m$ with $10k+7=5m-8$.
Also, you need to show that the subset is proper, that is you must exhibit at least one element $5m-8$ that cannot be written as $10k+7$.