In each of the following scenarios use the Well Ordering Principle to answer the questions:
a) Is there a process that is capable of generating an infinitely long sequence of numbers $a_1,a_2,a_3\dots$, such that for all $n\ge1, a_n>a_{n+1}$? Present a formal proof.
b) given a natural number n, we keep taking away multiples of $5$ from $n$ as often as possible. Directly apply the WOP to a suitably designed set $S$, to prove that this process is going to end and it will leave one of the numbers $\{0,1,2,3,4\}$. Also make sure to prove that this number does not change when we repeat the process; that is, this remaining number is $\underline{\text{unique}}$ (it is always the same.) Use the uniqueness of the remainder to prove that the number of times that this process could run,$t≥0$ must also be unique.
$a)$ The question didn't say what kind of number $a_n$ is, but since it's asking a WOP-proof, I suppose $a_n\in\mathbb{N}$, where $\mathbb{N}=\mathbb{Z^+}$, and it also didn't define what is a 'process' formally, but since we are dealing with natural numbers, I suppose it's talking about function with domain and codomain $\mathbb{N}$, that I'll prove there is no such process capable to do it, formally we can write as the following:
$$\text{WTS }\neg(\underbrace{\exists p:\mathbb{N}\to\mathbb{N}}_{\text{Exists such process}},s.t.\underbrace{\text{range}(p)=\{a_n\in\mathbb{N}:\forall n\ge1,a_n>a_{n+1}\}}_{\text{Infinitely long strictly decreasing sequence of natural number}}\wedge\underbrace{\text{range}(p)\neq\varnothing}_{\text{Capable of generating}})$$
By WOP knows that:$$\forall S\subseteq\mathbb{N}, S\neq\varnothing\rightarrow\exists x\in S,s.t.\forall y\in S,x\le y$$
Assume the negation is True, that we have $$\text{range}(p)\neq\varnothing$$
Which implies $$\exists x\in \text{range}(p),s.t.\forall y\in \text{range}(p),x\le y$$
But we also have $$\text{range}(p)=\{a_n\in\mathbb{N}:\forall n\ge1,a_n>a_{n+1}\}$$
Let $x$ be $a_n$ then $\exists a_{n+1}\in\text{range}(p),s.t. a_n>a_{n+1}$,
Which contradict with $x$ is the smallest element in $\text{range}(p)$ $\tag*{$\square$}$
$b)$ First we formalize the question for a formal proof $\dots$
$$\text{WTS }(\exists p:\mathbb{N}\to\mathbb{N},s.t.\forall n\in\mathbb{N},p(n) = n-5)$$
$$\rightarrow\exists!t\ge0,s.t.\underbrace{p\circ\dots\circ p(n)}_{t \text{ times for $p$}}\in\{0,1,2,3,4\}$$
Proof.
$\vdots$
Maybe it's not too hard to write a proof for this $\dots$ but how does this related to a "suitably designed set S" and how do I prove it with Well ordering principle?
Any help or hint or suggestion would be appreciated.
Your answer for (a) looks good to me!
For (b), consider the set $S = \{x_n ~|~ x_{n+1} = x_n - 5 \}$ (with $x_0$ the starting number). By WOP, $S$ has a least element, and so the process stops.
Now, let $x$ be the least element of $S$ (i.e., the number we stop at). If $x \geq 5$, then we could subtract $5$ from $x$, but then $x - 5 \in S$, by definition. This contradicts the minimality of $x$.
As for the uniqueness of this solution, say instead of subtracting $5$ on each step, we subtract some multiple of $5$ (as the problem states). Then the resulting sequence $y_n$ is a subsequence of the $x_n$ above. Can you use this to show that the least element of $y_n$ must agree with the least element of $x_n$?
I hope this helps ^_^