I am independently working through [Tao's Analysis I](http://terrytao.wordpress.com/books/analysis-i/), and one of the exercises is to prove that every positive natural number has a unique predecessor. The actual lemma is (where $n++$ denotes the successor of $n$):

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> Let $a$ be a positive [natural] number. Then there exists exactly one natural number $b$ such that $b++=a$.
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The hint given is to use induction, and the book uses the Peano axioms with $0\in\mathbb{N}$.

Here is my attempt at a proof, but I am not sure if I applied induction properly (I didn't seem to need the inductive hypothesis in proving the $k+1$ case), and I am even more unsure if uniqueness is proven:

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> The statement to prove is $(\forall a\not=0)(\exists!b)(b++=a)$, which is the same as $\forall a\exists!b(a\not=0\rightarrow (b++=a))$. Induction is used on $a$.
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> The basis step is to prove for $a=0$, which is $\exists!b(0\not=0\rightarrow (b++=0))$. Since the antecedent of the conditional is false, the statement is vacuously true. [But is $b$ unique...?]
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> For the inductive step, it is assumed that for an arbitrary $k$, $\exists!b(k\not=0\rightarrow (b++=k))$. Then $\exists!b((k++\not=0)\rightarrow (b++=k++))$ is to be derived. The second of these (the one being derived) can be rewritten as $(k++\not=0)\rightarrow (c++=k++)$, for some $c$.
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> $k++\not=0$ is true even without assuming it, since $0$ does not have a predecessor. But $c++=k++$ implies $c=k$ (because the successor function is injective), which shows that a [unique?] $c$ can be chosen such that it is equal to $k$. This closes the induction and thus every positive natural number has a unique successor.
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My questions are: (1) Is this proof correct? (2) If the answer to the first question is 'no', is the general idea for the proof correct? (3) If the answers to both of the preceding questions are 'no', could someone point me in the right direction for a correct proof?

Any help would be appreciated. Thanks.