# Taking a natural number different from zero as a base case in an inductive proof

Principle of mathematical induction states that if a subset $$S$$ of a successor set $$\omega$$ is also a successor set, then $$S=\omega$$. In primitive terms, it is formulated as:

if $$S \subset \omega$$, if $$0 \in S$$, and if $$n^+ \in S$$ whenever $$n \in S$$, then $$S=\omega$$.

Now, I wonder what if a statement that I want to prove holds for numbers starting from $$b \in \omega$$ for example. Would it be correct to introduce

1. a set $$L$$ that contains all natural numbers that satisfy the statement

2. a function $$s: \omega \to \omega, s(n) = n^+$$

3. a function $$r: \omega \to \omega, \begin{cases} r(0) = b\\ r(n^+) = s(r(n))\\ \end{cases}$$

4. a set $$S$$ that contains all natural numbers for which $$r(n)\in L$$

and show that $$0$$, $$n$$ and $$n^+$$ are in $$S$$? Because I can't just start proving inductively from $$b$$ and say that it holds for all numbers starting from $$b$$, as the principle of induction clearly states that $$0$$ has to be in $$S$$ too. So, I introduce a trick, i.e. mapping to perform induction.

Is it correct to do so? If not, then how to formally show that I can start from any number?

"For all $$b \in \omega$$: if $$S \subset \omega$$, if $$b \in S$$, and if $$n^+ \in S$$ whenever $$n \in S$$, then $$S=\{ n \in \omega | n \geq b \}$$."
Or, you can use the principle as stated and prove the base and step for the property $$P(n)$$ defined as $$n \geq b \to P'(n)$$, where $$P'(n)$$ is the 'real' property you are interested in, i.e. the property you want to prove all numbers $$n \geq b$$ to have. This works, since $$P(n)$$ will be trivially true for all $$n < b$$.