Over $\mathbb{Z}$, $aRb \iff a \le b \lor a = 3b$. Determine if it is total.
I think it is:
Have arbitrary elements $a,b \in \mathbb{Z}$.
We have to prove that $aRb \lor bRa$, which can be written as:
$$(a \le b \lor a = 3b) \lor (b \le a \lor b = 3a)$$
Further simplified:
$$(a \le b) \lor (b \le a)$$
Clearly, this should be a tautology. But I'm failing to further simplify this. What else can I do with it?
On
As nomen points out in a comment, the sentence $\forall a\, \forall b\, ((a\le b) \vee (b \le a))$ is not a tautology. Being a tautology would mean that it held for all structures of the language, or equivalently that it could be deduced using only the axioms of first order logic, and this is not the case.
The distinction between being a tautology and merely being true in a certain structure is important here because it affects what the next step in your argument will be. Rather than trying to further manipulate formulas using logical rules, you can simply notice that the structure $\mathbb{Z}$ (with the usual ordering) satisfies the formula $(a\le b) \vee (b \le a)$ for all choices of $a,b \in \mathbb{Z}$. In the context of this problem, this is probably something you may assert without further proof. However, it is important to realize that in doing so, you are saying something about the actual order relation $\le$ on $\mathbb{Z}$ and not merely about the predicate symbol that has the same name. So for example if $\le$ were interpreted as the relation of divisibility rather than the order relation, the formula $(a\le b) \vee (b \le a)$ would not be true of, say $a = 2$ and $b = 3$.
Your simplified $(a \le b) \lor (b \le a)$ is indeed a tautology, because its negation is the following contradiction $(x > y) \land (y > x)$.