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

Proof. For any natural numbers $m$ and $n$, let $S(m,n)$ denote the statement $$m+n\in \mathbb{N_{0}}$$

Base step ($m=n=0$): The statement $S(0,0)$ says $0+0\in \mathbb{N_{0}}$ which is true because \begin{align*} m+n=m\in \mathbb{N_{0}}\tag{$\forall k\in \mathbb{N_{0}}:k+0:=0$ and $m=0$}\\ \end{align*}

Induction on $m$: We are to prove that $\forall m\in \mathbb{N_{0}}:m+0\in \mathbb{N_{0}}$ yet this follows immediately from two facts, namely: $m+0=m\in \mathbb{N_{0}}$ and $p\implies p$ is a tautology.

Thus far \begin{align*}\forall m\in \mathbb{N_{0}}:m+0\in \mathbb{N_{0}}\tag{1}\end{align*}

Fix an arbitrary $p\in \mathbb{N_{0}}$.

Base Step($n=0$): The statement $S(p,0)$ is true given that $p+0=p\in \mathbb{N_{0}}$.

Induction on $n$: The statement $\forall q\in \mathbb{N_{0}}: p+q\in \mathbb{N_{0}}\implies p+(q+1)\in \mathbb{N_{0}}$ is trivial. Let's observe that $p+1\in \mathbb{N_{0}}$,

\begin{align*} p+(q+1)&=(p+q)+1\tag{$\forall k,l\in \mathbb{N_{0}}:k+(l+1):=(k+l)+1$}\\ \end{align*}

And $(p+q)+1\in \mathbb{N_{0}}$ for $(p+q)+1$ is the succesor of $p+q\in \mathbb{N_{0}}$. So,\begin{align*}\forall n\in \mathbb{N_{0}}: p+n\in \mathbb{N_{0}}\ \text{for a fixed}\ p\in \mathbb{N_{0}} \tag{2}\end{align*}

Thus, along with $(1)$ and $(2)$, by mathematical induction it is proved that $\forall m,n\in \mathbb{N_{0}}:S(m,n).$

RobPratt
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    Is this for pedagogical purposes only? I am sure the induction theorem is strongly based upon the structure of the natural numbers, so it's at least fishy to me to prove that the natural numbers are closedunder addition using induction. – William M. Jun 27 '21 at 14:52
  • In general, if you have a family of propositions $P(n, m)$ and you want to use induction, the typical way to go is to consider an "arbitrary but fixed $m$ and use induction on $n \mapsto P(n,m)$," having done this, you have $P(n,m)$ for all pairs $(n,m).$ – William M. Jun 27 '21 at 14:55
  • What you are doing is essentially correct (assuming $\forall n \in \mathbb N_0: n + 1 \in \mathbb N_0$ is given.). What I must ask is, what foundational assumptions are you making here (about the set $\mathbb N_0$ and operator $+$)? By your own work you admit there is some binary function $+$ which takes natural numbers as inputs. Generally speaking, when $+$ is defined it is done so inductively, in much the way you have done your induction, meaning $m + n \in \mathbb N_0$ for $m,n \in \mathbb N_0$ by construction and thus the theorem you want to prove is vacuous. – user2628206 Jun 27 '21 at 14:56
  • @WillM. I promise it's just for pedagofical purposes. –  Jun 27 '21 at 17:05
  • @user2628206 Is the proof wrong? I'm a newbie. –  Jun 27 '21 at 17:18
  • Does this answer your question? Is my induction proof correct? – user2628206 Jun 27 '21 at 18:05

1 Answers1

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Seems to me like your proof is fine.

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.

Sayan Dutta
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  • I'm bearly learning how to do proofs, May I ask you a question? What does it mean for an arbitrary $n$ (let's say in $\mathbb{N}$) to be fixed? Does it mean it doesn't vary or something? –  Jun 27 '21 at 20:42
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    @Hanny Boy Yeah, kind of. It doesn't vary over the proof. – Sayan Dutta Jun 27 '21 at 20:57
  • Cheers, mate! Just picked up on that. –  Jun 27 '21 at 21:13