# Check my work on the the following theorem is correct

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

$$(m+n)+k=m+(n+k)$$.

$$6.11$$ 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

https://math.stackexchange.com/a/3796851/837396

Help🥶

• You should clearly define what is $\omega$. Nov 6, 2020 at 16:56
• Ok,I did that. See main post
– user837396
Nov 6, 2020 at 17:27
• 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. Nov 6, 2020 at 17:34
• 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
– user837396
Nov 6, 2020 at 18:12
• 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 Nov 7, 2020 at 1:14