Prove $[(n, k)] = 7^{*} \iff n - k = 7$ given $7^{*} = [(8,1)] \in \mathbb{Z}$ and $k < n \in \mathbb{N}$ using any facts about addition in the Peano System.
$k < n \in \mathbb{N}$ is defined as $n = k + m$ for some $m \in \mathbb{N}$. For $k < n \in \mathbb{N}$, $n - k$ is defined as the $m$ such that $n = k + m$ (i.e. $n = k + (n - k)$. The equivalence relation $\sim$ on $\mathbb{N} \times \mathbb{N}$ whose equivalence classes are integers is defined as $(n, k) \sim (n', k')$ if $n + k' = n' + k$. Also, $(n, k) \sim (n', k') \iff [(n,k)] = [(n', k')]$.
For the $\leftarrow$ direction, we have to show that $7^{*} = [(8,1)] = [(n,k)]$, assuming $n - k = 7$. By the definition of subtraction in $\mathbb{N}$, we have $n = k + 7$. We can substitute this value of $n$ into $[(n, k)]$ so we have $[(k + 7, k)]$. If we can show $(k + 7, k) \sim (8,1)$ then we can use the fact that $(n, k) \sim (n', k') \iff [(n,k)] = [(n', k')]$ to show that $[(n,k)] = [(8,1)]$. So using the definition of $\sim$, we have $(k + 7) + 1 = k + 8$ so we have $k + 8 = k + 8$, and thus $\sim$, and thus $[(n,k)] = [(8,1)]$.
For the $\rightarrow$ direction, we suppose that $[(n, k)] = 7^{*} = [(8,1)]$, and want to prove that $n - k = 7$. By the definition of subtraction in $\mathbb{N}$, it's the same thing as proving that $n = k + 7$. We are also given in the problem that $k < n$, so we know by definition of subtraction that $n = k + m$ for some $m \in \mathbb{N}$. If we can prove $m = 7$, then we are done. We know that $n = 8$ and $k = 1$, so we have $8 = 1 + m$. This is the same as $1 + 7 = 1 + m$, so by cancellation we have $7 = m$. But this is incorrect (as per professor). What have I done incorrectly in the proof?