This question arose from seeing two different answers to Exercise $2.2.2$ on Terence Tao's Analysis I; which asks us to prove Lemma $2.2.10$ which states:
Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b++=a$.
Now, this exercise has been talked about before on this site: 1, 2, 3, 4, 5
But none of the proofs do what the author of these proposed answers has done for their base case:
"Base case for $a=1$: we know that $b=0$ matches this property, since $0++=1$ by
Definition $2.1.3$. Furthermore, there is only one solution. Suppose that is another
natural number b such that $b++=1$. Then, we would have $b++=0++$, which would
imply $b=0$ by Axiom $2.4$. The base case is demonstrated."
Is there any way to justify that this is correct? Because every other proof of this I've seen says that the base case $a=0$ is vacuously true. Induction via the Peano Axioms isn't even defined to work for $P(1)$ and strong induction is proved in Exercise $2.2.5$ so using it now, even though it doesn't cause any circularity problems as far as I can see, doesn't seem ideal. An even more closely related question 6 asks this same thing but doesn't answer whether this proof without any further justification is correct.
Best Answer
Even though this response isn't the same as the answer presented by the author, since the strong principle of induction isn't mentioned until later, as you said, I think this is the more rigorous solution. As you already brought up that post, however, I'm going to assume you've already read that, so I've tried rewriting the base case to make the argument more explicit.
Base case: $a = 0$
By the lemma we're trying to prove, we must let $a$ be a positive number.
Consider Definition 2.27:
If we consider $n=a=0$, since $n$ is equal to $0$, by definition, n is not positive, so $a$ is not positive.
This is a contradiction, so the base case holds.