The first thing to note is that $Y \\subseteq X$. You need to prove that for every $y \\in Y$ that $y \\in X$. Essentially you have to prove an infinite amount of stuff here, and we do this by taking a generic element of $Y$ and proving that it also belongs to $X$.
\nIf $y \\in Y$ then $y$ has the shape $y = \\var{c}k + \\var{d}$ for some $k \\in \\mathbb Z$. We need to show that it also has the shape $\\var{a}n + \\var{b}$ for some integer $n$ and so belongs to $X$ as well.
\n$\\begin{align*} y & = \\var{c} k + \\var{d} \\\\ & = \\var{a}\\times\\var{u} k + \\var{a} + \\var{b} \\\\ & = \\var{a}(\\var{u} k + 1) + \\var{b} \\\\ & = \\var{a}n + \\var{b}. \\end{align*}$
\nwhere $n = (\\var{u}k+1) \\in \\mathbb Z$. Because $y$ has the required shape, $y=\\var{a}n + \\var{b}$ for some integer $n$, it is also an element of $X$. So $y \\in X$.
\nSo for this part of this question you can choose any element of $Y$. All of them will automatically be included in $X$.
\nFor the second question you need to find an element of $X$ which which is not an element of $Y$. This does not require an infinite amount of work - all you need is one element of $X$ which is not in $Y$. The easiest way to proceede is to just try a few different values of $X$. A more systematic approach is to try the procedure above and see where it goes wrong.
\nBy definition the sets $X$ and $Y$ are equal when both $Y\\subseteq X$ and $X \\subseteq Y$ are true. We showed in the first question that the relation $Y\\subseteq X$ is true, but this does not matter much we showed in the second question that $X \\nsubseteq Y$. So the sets are not equal, instead we could say that $X$ is a proper subset of $Y$: $X \\subset Y$.
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"}, "rulesets": {}, "statement": "Consider the sets $X = \\left\\{ \\var{a}n + \\var{b} | n \\in \\mathbb Z\\right\\}$ and $Y = \\left\\{ \\var{c}k + \\var{d} | k \\in \\mathbb Z\\right\\}$.
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