Is my proof of lemma 2.2.10 (Analysis 1 Tao) correct? I have recently begun self studying Real Analysis 1 by Tao. Such proofs are new to me and the solutions are not provided in the book. That is why I'm asking this question. Any feedback is welcome, thank you very much for your time!
There are two other questions about this lemma on here. However, they both used symbolic logic and hence remain inaccessible to me.
Axiom 2.4: Different natural numbers must have different successors; i.e., if $n,m$ are natural numbers and $n≠m,$ then $n++≠m++.$ Equivalently, if $n++=m++,$ then we must have $n=m.$
Axiom 2.5: Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number $n.$
I am trying to prove:
Lemma 2.2.10: Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b++=a.$
Proof: Let $P(n)$ be a property pertaining to a natural number $n.$ Assume that $P(0)$ is true and that $P(n)$ being true implies that $P(n++)$ is true. By axiom 2.5, $P(n)$ is true for all natural numbers $n.$ Assume that there exists a positive natural number $a$ such that $b++\not=a$ for any natural number $b.$ However, the existence of such a number would mean that $P(a)$ isn't necessarily true. This is a contradiction and hence there must exist a natural number $b$ such that $b++=a.$ Next, assume that $c$ is also a natural number such that $c++=a.$ Hence, $b++=c++.$ By axiom 2.4, this implies that $b=c.$ Consequently, there exists a unique natural number $b$ such that $b++=a$ for all positive natural numbers $a.$
$\square$
 A: Long comment

Let $P(n)$ be a property pertaining to a natural number $n$. [...] Assume that there exists a positive natural number a such that $b++≠a$ for any natural number $b$. However, the existence of such a number would mean that $P(a)$ isn't necessarily true.

But you have only restated Induction axiom, without defining the property $P(n)$ to which apply it.
In order to apply Induction, we have to define a suitable property $P(n)$:

"if $n$ is positive, there is a number $b$ such that $n=b++$".

Base step: $a=0$.
According to Definition 2.2.7 (Positive natural numbers) $0$ is not positive. Thus, using Ex Falso Quodlibet we have:

"if $a=0$, then (if $a$ is positive, there is a number $b$ such that $a=b++$)".

The result ($P(0)$) follows by Modus ponens.
Induction step: assume that the property $P(n)$ defined above holds for $a$ and consider $a++$.
We have that: $a++=a++$ (law of equality; see Appendix.7 Equality, page 329). Thus,

"there is a number $b$ such that $a++=b++$"

holds. But then it holds also: "if $a++$ is positive, there is a number $b$ such that $a++=b++$", i.e. $P(a++)$.
Having proved $P(a++)$, we have also that "if $P(a)$, then $P(a++)$" holds.
Now we have all the ingredients needed for applying Axiom 2.5 (Principle of mathematical induction): we have proved $P(0)$ and we have proved that whenever $P(a)$ is true, $P(a++)$ is also true. 
Then we can conclude that $P(a)$ is true for every natural number $a$, i.e. that:


"if $a$ is positive, there is a number $b$ such that $a=b++$".



The previous result grabs the key role of Induction: every natural number is "achievable" starting from $0$ and applying the successor operation.
