Prove that if $m \leqslant n$ then $\exists!\ p\in\omega$. such that $m+p=n$

Correction:induction on $n$ thanks to @Brian M. Scott

Working assumptions:

6.1 Definition By the set of the natural numbers we mean the intersection of all the successor sets. The set of the natural numbers is designated by the symbol $\omega$; every element of $\omega$ is called a natural number.

$\omega$={0,1,2,3,…,n,…:n$\in$ $Z^{+}$ }

$6.22$ Theorem

Let $m \leqslant n$ denote the fact that $m \in n$ or $m = n$. Then the relation is an order relation in $\omega$.

$6.13$ Theorem


$6.11$ Lemma

$n+ =1+n$, where $1$ is defined to be $0+$

Attempted proof

Let $m,n \in\omega $ be arbitrary.

Define $L_{pn} = \{m\in\omega : \text{ If } m\leqslant n \text{ then }\exists!\ p \text{ such that } n+p=m \}$

Base case: $m=0, p=n$ so $0 \in L_{pn}$

Inductive case:

Assume $m\in L_{pn}$ We will show $m+\in L_{np}$

If $m \leqslant n$ then $m=n$ or $m \in n$ If $m=n$ we are done. Otherwise if $m \in n$ ; $m \subset n$ so that $n= m^{+} +p$

Then $$\begin{align} n &= m^{+} +p\\ &= (m+p)+ \qquad\text{ Ind.hyp.}\\ &= 1+(m+p) \qquad\text{by 6.11}\\ &= (m+1)+p \qquad\text{by 6.13}\\ &= p+(m+)\qquad\quad\text{by 6.11}\end{align}$$ Thus $m+ \in L_{pn}$

It seems to me some of my proof deals with this post



  • $\begingroup$ You should clearly define what is $\omega$. $\endgroup$
    – Sigma
    Nov 6, 2020 at 16:56
  • $\begingroup$ Ok,I did that. See main post $\endgroup$
    – user837396
    Nov 6, 2020 at 17:27
  • $\begingroup$ Wow. The set of natural numbers is more commonly denoted by $\mathbb{N}$. One more thing, add the "elementary number theory" tag. I cannot solve your problem because I don't know number theory. $\endgroup$
    – Sigma
    Nov 6, 2020 at 17:34
  • $\begingroup$ Problem deals with Peano’s arithmetic. The text I work from is A Book of Set theory by Charles Pinter . I have the PDF version which has typos $\endgroup$
    – user837396
    Nov 6, 2020 at 18:12
  • 1
    $\begingroup$ I fixed up your formatting problem. You can see how it is done by checking the edit (though arrays are a better way of handling the offset step explanations than I used). You can find out more on Latex via Mathjax here $\endgroup$ Nov 7, 2020 at 1:14


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