Proof Check Lemma 2.2.10 in Tao I aim to prove the following.
Lemma 2.2.10. Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b{+\!+} = a$.
I use the following.
Definition 2.2.1 (Addition of natural numbers). Let $m$ be a natural number. To add zero to $m$, we define $0 + m := m$. Now suppose inductively that we have defined how to add $n$ to $m$. Then we can add $n{+\!+}$ to $m$ by defining $(n{+\!+}) + m := (n + m){+\!+}$.
Proposition 2.2.6 (Cancellation law). Let $a, b, c$ be natural numbers such that $a + b = a + c$. Then we have $b = c$.
Axiom 2.4. Different natural numbers must have different successors; i.e., if $n, m$ are natural numbers and $n \neq m$, then $n{+\!+} \neq m{+\!+}$. Equivalently, if $n{+\!+} = m{+\!+}$, then we must have $n = m$.
Tao suggested the use of induction, so I am doubting the validity of my proof.
Proof:
Proceed by contradiction. Let us assume we have $2$ differing natural numbers $b$ and $c$, such that $b{+\!+} = a$ and $c{+\!+} = a$. 
Then we have $b{+\!+} = 0 + b{+\!+}$ and $c{+\!+} = 0 + c{+\!+}$ (definition of addition).
So we then have that $0 + b{+\!+} = 0 + c{+\!+}$, but then $b{+\!+} = c{+\!+}$ (Cancellation Law). This is a contradiction due to Axiom 2.4. 
I am self-studying real analysis, so I want to ensure that I am proceeding correctly. 
 A: You have proved the uniqueness. Also you need to show the existence of such $b$.
To do that, you need to consider the statement $$P(a)\equiv \text{ there exists a } b \text{ such that } b\!+\!\!+=a \text{ whenever } a\ne0$$
because $a$ is positive. Then induct on $a$.
Note that in certain step, the statement is vacuously true.

 We have to prove $$a\ne0\implies\exists b\in\mathbf N,\;\;b\!+\!\!+=a.$$ So, we induct on $a$. The base case $a=0$ is vacuously true. Now suppose inductively that the claim is true for $a$; we need to show the claim for $a\!+\!\!+$, i.e., $b'\!+\!\!+=a\!+\!\!+$ for some natural number $b'$. Thus, by induction hypothesis, we have $b\!+\!\!+=a$. Applying the increment (by Sustitution axiom of equality) we obtain $(b\!+\!\!+)\!+\!\!+=a\!+\!\!+$. Defining $b':=b\!+\!\!+$ the claim follows.

A: The proof of Cristhian is incomplete. The induction hypothesis does not allow you to assert immediately that there exists a $b$ such that $b{+\!+}=a$ (since the truth-value of $A\implies B$ by itself is not enough to know the truth-value of $B$). Instead, you should consider two different cases in the induction step: 1) $a=0$ and 2) $a\neq0$. 
In the first case, $a{+\!+}=0{+\!+}$ and so $b':=0$ is a unique natural number (by Axioms 2.1 and 2.4) such that $b'{+\!+}=a{+\!+}$. In the second case, both the antecedent $a\neq0$ and the conditional $P(a):=(a\neq0\implies\exists!b\in\mathbb N:b{+\!+}=a)$ are true by hypothesis. So, we know there exists a unique natural number $b$ such that $b{+\!+}=a$. This implies that $a{+\!+}=(b{+\!+}){+\!+}$. By Axioms 2.2 (existence) and 2.4 (uniqueness) there exists a unique natural number $b':=b{+\!+}$ such that $b'{+\!+}=a{+\!+}$.
The two cases allow us to state that $P(n)\implies P(n{+\!+})$. Since $P(0)$ is vacuously true, the Proposition follows by induction (Axiom 2.5).
A: To prove the existence, it will be easy if we consider the contrapositive of above statement

$\forall a \gt 0 \to \exists b \ (b++ = a)$
$\longleftrightarrow \forall b (b++ \ne a) \to \exists a (a = 0)$

the RHS implication follows from Axiom 2.3
