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 $S(a)$, if $a$ is a positive number, then there exists exactly one natural number $b$ such that $b++=a$.
We want to show $S(a++)$, if $a++$ is a positive number, 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, we can 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, assume $x$ and $y$ are distinct natural numbers such that $x++=(a++)$ and $y++=(a++)$. By transitivity of equality we have $x++=y++$, and by Axiom 2.4, if $x++=y++$ then $x=y$. This contradicts $x$ and $y$ being distinct. And so there is at most one natural number $b$ 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 $b++=a++$.
By showing the base case and the inductive step, we have proved that for any positive number $a$, there exists only one natural number $b$ such that $b++=a$.
