Over $\mathbb{R}$ is defined a relation $\le$ in the following way:
$$\forall a,b \in \mathbb{R} [a \le b \iff b - a \ge 0]$$
Demonstrate that $\le$ is a relation of total order.
For a relation to be of total order, it has to be reflexive, transitive, antisymmetric and total.
I already proved reflexive and transitive, but not so with the other two:
Antisymmetric
We have to prove that for arbitrary elements $a,b \in \mathbb{R}$,
$$(a \le b \land b \le a) \implies a = b$$
We start:
$$a \le b \land b \le a$$
$$b - a \ge 0 \land a - b \ge 0$$
$$b - a + a - b \ge 0$$
$$0 \ge 0$$
Umm... True. That's true. But I doubt this proves that $a = b$.
Total
We have to prove that $\forall a,b \in \mathbb{R}$, $(a \le b \lor b \le a)$:
$$a \le b \lor b \le a$$
$$b - a \ge 0 \lor a - b \ge 0$$
$$b - a + a - b \ge 0$$
$$0 \ge 0$$
Ok, clearly I don't see the bigger picture here.
As you can observe, I have the basic mindset of writing what I want to prove, then evaluate it with my premise and then continuously simplify it.
I fear that I'm missing the point of this kind of demonstration - I'm not particularly interested in this one exercise, but rather understand better the apparently abstract process that tends to be made for this kind of problem. Can you advice me here?