Does $(p(0) \wedge (P(n) \implies P(n-1))) \implies P(n) \forall n\leq 0$?
In other words, what I'm asking is, can I use the axiom of induction for negative numbers? Why/why not?
E: This is not a duplicate from that other post, I ask nowhere about having a negative argument for the base case.
On
Yes, of course. Just define $Q(n) = P(-n)$. Then $P(n) \implies P(n-1) \ \forall n \le 0$ is equivalent to $Q(-n) \implies Q(1-n) \ \forall n \le 0$ which is the usual induction step (because $-n \ge 0$).
On
Broadly speaking, you can use induction for any set of objects that can be put into 1-to-1 correspondence with $\mathbb{N}$, and has the property that any subset has a minimal element.
On
Yes, you can.
Essential for that is that relation $<$ is a well-order relation on $\mathbb Z_{\geq0}$.
Likewise $>$ is a well-order relation on $\mathbb Z_{\leq0}$ and moreover the map $n\mapsto-n$ is an isomorphism between the structures $(\mathbb Z_{\geq0},<)$ and $(\mathbb Z_{\leq0},>)$.
yes, you can do that. The best way to see that it works, is, to define $n=-m$, and what you end up with is the usual axiom of induction