Proof Check Lemma 2.2.10 in Tao

Solution 1:

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.

Solution 2:

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).