Minimal successor set - difficult

[Andrei1]

Prove that for all \(\displaystyle x,y\in\omega,\ \ x\subset y\vee y\subset x.\)

If I assume that the conclusion is false then I can prove that for some \(\displaystyle a\in x,\ b\in y\) we have \(\displaystyle a\notin b\) and \(\displaystyle b\notin a.\)

Also I am thinking that if assume the contrary then \(\displaystyle \omega\) minus \(\displaystyle \{x\}\) or minus \(\displaystyle \{y\}\) or both is a smaller successor set. Should I try to prove this?

I get stuck in trying to prove for sets from \(\displaystyle \omega\) the equivalence: \(\displaystyle a\subseteq b\wedge a\not=b\Leftrightarrow\exists c(a\cup c^+=b)\).

 

[Andrei1]

Here's the plan.
(1) Prove \(\displaystyle a\subset b^+\Rightarrow b\notin a\) by induction on \(\displaystyle a.\) Use also \(\displaystyle x=y\Rightarrow x^+=y^+.\)
(2) Prove \(\displaystyle a\subseteq b\Leftrightarrow a\subset b^+\). In proving (\(\displaystyle \Leftarrow\)) side use (1). In proving (\(\displaystyle \Rightarrow\)) side use \(\displaystyle x\subset x^+\), which follows from \(\displaystyle x\notin x.\)
(3) Prove \(\displaystyle b\subset a\Leftrightarrow b^+\subseteq a\). In proving (\(\displaystyle \Rightarrow\)) side use induction on \(\displaystyle a.\) Use also \(\displaystyle x=y\Rightarrow x^+=y^+\) and \(\displaystyle x\subseteq x^+.\) In (\(\displaystyle \Leftarrow\)) side you need \(\displaystyle x\notin x.\)
(4) Prove \(\displaystyle a\subset b\vee a=b\vee b\subset a\) by using (2) and (3). Use induction.