Exercise 2.2.2 (Existence and uniqueness of predecessor)

How to think about the exercise. This exercise asks us to show that it makes sense to talk about the predecessor of a positive number. For instance, the predecessor of $3$ is $2$. Why is there a hypothesis that $a$ must be a positive number? Because by axiom 2.3 the number $0$ does not have a predecessor (within the set of natural numbers).

This exercise is almost a standard induction exercise, but not quite. The one subtlety is that the inductive hypothesis is never used. This might seem suspicious to you, but it’s actually harmless.

Model solution. Let $P(a)$ be the statement "If $a$ is positive, then there is a unique natural number $b$ such that $b++=a$." We shall induct on $a$. The base case is vacuously true, since $0$ is not a positive number (definition 2.2.7).

Now suppose inductively that $P(a)$ is true. We must show that $P(a++)$ is true. To show that $P(a++)$ is true, it suffices to show that there is a unique number $b$ such that $b++=a++$ (this is because given statements $p,q$, if $q$ is true, then $p⟹q$ is also true no matter what $p$ says). To do this, we must show two things: (1) there is a natural number $b$ such that $b++=a++$; (2) if $c$ is a natural number such that $c++=a++$, then $c=b$. To show (1), let $b:=a$. Then indeed we have $b++=a++$. To show (2), let $c$ be a natural number such that $c++=a++$. Then by axiom 2.4, we see that $c=a$. But we also know $b=a$, so $c=b$ as required. This closes the induction.

Source: Issa Rice’s blog.