Prove Proposition 2.2.10.
Lemma 2.2.10::
Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b++=a$.
Let $S(a)$ be the statement "if $a$ is a positive number then there exists exactly one natural number $b$ such that $b++=a$". We will use induction on $a$.
We need to show the base case $S(0)$, and the inductive step $S(a)\implies S(a++)$.
Base Case
We need to show that if $0$ is a positive number then there exists exactly one natural number $b$ such that $b++=0$.
However, 0 is not a positive number, and so the base case is vacuously true.
Inductive Step
We assume $S(a)$ and aim to show $S(a++)$, if $a++$ is a positive number then there exists exactly one natural number $b$ such that $b++=(a++)$. We need to show both the existence and uniqueness of $b$.
Since $a++$ exists so does $a$. We set $b=a$. By the substitution axiom of equality, we have $b++=a++$. And so at least one $b$ exists such that $b++=a++$.
For the purpose of contradiction, let's assume a natural number $b'$, distinct from $b$, also satisfies $b'++=a++$. By Axiom 2.4 this would mean $b'=a$, contradicting that $b'$ is distinct from $b$. This means there is at most one natural number $b$, specifically $b=a$, such that $b++=(a++)$.
By showing there is at least one $b$, and at most one $b$, that satisfies $b++=a++$, we have shown there is exactly one $b$ such that $b++=a++$, proving the inductive step.
Note that we didn't require the induction hypothesis, $S(a)$.
By showing the base case and the inductive step, we have proved thatfor any positive number $a$, there exists only one natural number $b$ such that $b++=a$.
