The following deals with Peano arithmetic I have trouble dealing with inequalities
Working assumptions :
6.22 Theorem Let m $\leqslant$ n denote the fact that m ∈ n or m = n. Then the relation is an order relation in ω.
Prove there is no natural number k such that m<k<m+.
I think by contradiction would be a good way
Attempted proof
m<k<m+ means m<k and k<m+
By 6.22 ,m<k implies m=k or m$\in$ k
and k<m+ implies k=m+ or k$\in$ m+
If m=k and k=m+ then m+=k+ and k+=m+
But k+ can’t be m’s successor and m simultaneously thus there is no natural number k satisfying the condition. Help