I'm having trouble with this question, for question one, do I prove that the statement satisfies reflexivity, antisymmetry, and transivity?
Would $m < m'$ satisfy reflexivity alone or do I have to include $n$ and $n'$? How would I prove antisymmetry, and trasivity?
I know that in a well founded order every nonempty subset has a minimal element, however I don't know how to prove that.
Let me call $m < m'$ condition 1 and $m = m' \land n \leq n'$ condition 2.
Reflexivity: For any $\langle m, n \rangle$, it can never be the case that $m < m$, so it does not satisfy condition 1. However, it must always be the case that $m = m$ and $n = n$ (and thus also $n \leq n$), and hence $\langle m,n \rangle \sqsubseteq \langle m,n \rangle$ by condition 2.
Antisymmetry: Assume $\langle m,n \rangle \sqsubseteq \langle x,y \rangle$ and $\langle x,y \rangle \sqsubseteq \langle m,n \rangle$. None of these two relations can satisfy condition 1. To see why, consider what happens when both satisfy it. Then we would have $m < x$ and $x < m$, which is impossible. So we know that at least one of them must satisfy condition 2, which means that $x = m$, which in turn means that none of them can satisfy condition 1. Hence, both must satisfy condition 2, so we get $x = m$ and $n \leq y \land y \leq n \implies y = n$, so $\langle m,n \rangle = \langle x,y \rangle$.
Transitivity: Try it yourself!
For wellfoundedness, try to look at the pairs componentwise. Assume you have some non-empty set of these pairs. By the wellfoundedness of the natural numbers, there must be some $\langle m,n \rangle$ with $m \leq m'$ for any other pair $\langle m',n' \rangle$. If $m < m'$, then this is your minimal element, otherwise $m = m'$. In that case, every pair in your set has the same first component. So we look at the second component. There must be some pair $\langle m,n \rangle$ with $n \leq n'$ for any other pair $\langle m',n' \rangle$. It can not be the case that $n = n'$, because then $\langle m,n \rangle = \langle m',n' \rangle$, so they are the same element. Thus $n < n'$, and this is your minimal element.