Define $s:\mathbb{N}\rightarrow \mathbb{N} $ given by $s(n)=n+2$, with $n\in\mathbb{N}$. Prove that $\mathbb{N}$ and $s$ satisfy
- every $n\in\mathbb{N}$ has only one sucessor and $s$ is one-to-one.
- exists at least one element of $\mathbb{N}$, which is not sucessor of any $n\in\mathbb{N}$. But they don't satisfy the Finite induction principle
Here it is my progress: given $m, n\in\mathbb{N}$ such that $s(m)=s(n)$. So, $m+2=n+2$. From the cancelation, $m=n$.
From the definition of $s$, we have $s(\mathbb{N}) =\{3,4,\dots\}$, which means that 1 and 2 do not have successors.
Finally let's consider $X=\{1,3,5,\dots\}\neq\mathbb{N}$. Then, $s(1)=3\in X$ and if $p\in X$, so $$ s(p)= p+ 2 \in X $$.
Do you agree with me?