I've seen a few proofs for this lemma...but I've been playing around trying to put it in a form that makes the most intuitive sense to me. If anyone could verify the legitimacy of the proof, I would greatly appreciate it.
Let $a$ be a positive number in $\mathbb N$. Then there exists exactly one natural number $b$ such that $b\text{++} =a$.
The notation $++$ is from Tao's Analysis I book and denotes the successor function.
I am going to implement two consecutive contradiction-styled proofs to prove this statement.
Firstly, I am going to assume that the existence part is true and verify the uniqueness of the existence.
Suppose $b$ is not unique and there exists another element $c$ such that $c++ = a$. By transitivity of equality, $b++=a=c++ \implies b++=c++$. By one of the Peano Axioms, if $n++=m++$, then we must have $n=m$. Therefore, $b=c$ and we see that if there exists a $b$, then this $b$ must be unique.
Secondly, I want to prove that there simply exists a $b$ for all positive $a$. Consider the following induction where "predecessor" is the property of interest (for definition's sake, $b$ in the statement $b++=a$ is the predecessor):
Base Case: $1$ has a predecessor. Certainly. $0$ is the predecessor of $1$ because $0++=1$
$n$ assumption: assume $n$ has a predecessor.
prove $n++$ has a predecessor: of course $n++$ has a predecessor. It's predecessor is $n$ because $(n)++=n++$.
Therefore, $\forall n \geq 1$, where $n \in \mathbb N$, $n$ has a predecessor. Now by one of the Peano Axioms, $n++\neq0$ for all natural number $n$. This is equivalent to saying $0$ has no predecessor (based on my definition of predecessor).
So, let's assume by contradiction, that positive $a$ has no predecessor. Therefore, the only available element in $\mathbb N$ for $a$ to be is $0$. However, $a$ is positive, by assumption. Therefore, this is a contradiction so every positive $a$ must have a predecessor. And by the first contradiction (above), this predecessor must be unique.
Are there any issues with this proof? Thank you.