I am trying to prove part (e) of Proposition $2.2.12$ in Tao's analysis textbook that for natural numbers $a, b$, $a < b$ if and only if $a{+\!+} ≤ b$, where $a{+\!+}$ is the successor of $a$. I am having difficulty proving this, though it appears obvious. I am unsure of whether induction on $a$, holding $b$ constant, is ideal, or if this could be proved directly since it is provable from Tao's axioms that the successor of $a$ and $a + 1$, and we have access to most of the order axioms, including the fact that addition preserves any order.
I'd appreciate any hints on how to get started on this. Thanks.
We don't need induction. Assume $a < b$. From definition, there is some natural number $c$ s.t. $a + c = b$. $c ≠ 0$ (otherwise we would have $a = b$ from lemma 2.2.2), so $c$ is positive. From Lemma 2.2.10 we have there is $d$ s.t. $c = d{+\!+}$. Then from lemma 2.2.3 and proposition 2.2.4 we have $b = a + c = a + (d{+\!+}) = (a + d){+\!+} = (d + a){+\!+} = d + (a{+\!+}) = (a{+\!+}) + d$, so by definition $a{+\!+} ≤ b$.
For the other direction, if $a{+\!+} ≤ b$ then for some $d$ we have $(a{+\!+}) + d = b$, then $a + (d{+\!+}) = b$ so $a ≤ b$. And would $a = b$ we would have $a{+\!+} = a + (0{+\!+}) ≥ a = b$, and as we already have antisymmetry and $a{+\!+} ≤ b$, it would imply $a{+\!+} = b$ - but then $a + (0{+\!+}) = a + 0$, so from proposition 2.2.6 $0{+\!+} = 0$, but $0$ has no predecessor, and $0{+\!+}$ has one.