Is my induction proof correct?

Question

Prove: $\forall n,m\in \mathbb{N_{0}}:n+m\in \mathbb{N_{0}}$.
Remark. By $\mathbb{N_{0}}$ we mean $0\in \mathbb{N}$.

Proof. Let $n,m\in \mathbb{N_{0}.}$

Providing $m=n=0,$

\begin{align*} n+m=n\tag*{because $n+0:=n$} \end{align*}

Thus $n\in \mathbb{N_{0}}$ given that $n=0$.

Now, assume \begin{align*}\forall k,s\in \mathbb{N}: k+s\in \mathbb{N}\end{align*}

And notice that $(k+1),(s+1)\in \mathbb{N}$. So, in particular, $(k+1)+(s+1)\in \mathbb{N}.$ Therefore $\forall n,m\in \mathbb{N}:n+m\in \mathbb{N}$ by the principle of mathematical induction.

Answer 1

I don't think it's okay to assume for $k$ and $s$ and show that it's true for $k+1$ and $s+1$. If you really want to use induction on two variables, then there's something called double induction which you need to use.

However, the proof can be done much more easily by using induction on one variable only.

Consider $m\in \mathbb{N}$.

Base Case: $m+1$ (or $0$ depending on which system you are using) belongs to $\mathbb{N}$ because by definition of $\mathbb{N}$ using Peano's axioms, the successor of a element in $\mathbb{N}$ belongs to $\mathbb{N}$.

Induction Case: Let $m+k\in \mathbb{N}$. Then using the same argument as above, $m+k+1\in \mathbb{N}$.

This completes the proof.