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
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