$\newcommand{\naturals}{\mathbb{N}}$
Use the Well-Ordering Principle to prove that every natural number greater than or equal to 11 can be written in the form $2s+5t$, for some natural numbers $s$ and $t$.
We are trying to prove $\forall n \in \naturals, n \geq 11, n = 2s+5t$, where $s, t \in \naturals$.
First, check $n = 11$. Then, $11=2s+5t$, $s=3,t=1$, where $s,t \in \naturals$. So, this case holds.
Next, check $n = 12$. Then, $12=2s+5t$, $s=1,t=2$, where $s,t \in \naturals$. So, this case holds.
Now, let us define set $A = \{n \in \naturals : n > 11, n \neq 2s+5t, s,t \in \naturals\}$. In English, this is the set of all natural numbers larger than 11 which cannot be written in the form $2s+5t$, with $s,t \in \naturals$. Seeking a contradiction, let us assume $A \neq \emptyset$. Then, the WOP guarantees that set $A$ has some smallest element $m$ where $m > 11$ but which cannot be written $m=2s+5t$.
We have shown $n=11$ and $n=12$ cannot be in $A$, so we will say $m \geq 13$.
Since $m \neq 2s+5t$, (not sure how to continue from here).
I feel as though I am missing how to apply the WOP in order to cause a contradiction. I hypothesize I need to somehow manipulate $n$ (like $n-2+2$ so that it somehow becomes the smallest element of $A$, smaller than $m$? But I am not sure how to do this. I am aware that this is not the only way to prove this statement, however I have been asked to use the WOP specifically seeking a contradiction.
We are trying to prove $\forall n \in \mathbb{N}, n \geq 11, n = 2s+5t$, where $s, t \in \mathbb{N}$.
First, check $n = 11$. Then, $11=2s+5t$, $s=3,t=1$, where $s,t \in \mathbb{N}$. So, this case holds.
Next, check $n = 12$. Then, $12=2s+5t$, $s=1,t=2$, where $s,t \in \mathbb{N}$. So, this case holds.
Now, let us define set $A = \{n \in \mathbb{N} : n \geq 11, n \neq 2s+5t, s,t \in \mathbb{N}\}$. In English, this is the set of all natural numbers larger than 11 which cannot be written in the form $2s+5t$, with $s,t \in \mathbb{N}$.
Seeking a contradiction, let us assume $A \neq \emptyset$. Then, the WOP guarantees that set $A$ has some smallest element $m$ where $m \geq 11$ but which cannot be written $m=2s+5t$.
We have shown $n=11$ and $n=12$ cannot be in $A$, so we will say $m \geq 13$.
Since $m \geq 13$, then $m-2 = 2s+5t$. However, this would mean that $m=2(s+1)+5t$, but we said that $m \neq 2s+5t$. Thus we have reached a contradiction, so it must be true that $\forall n \in \mathbb{N}, n \geq 11, n = 2s+5t$.