First we prove that two consecutive terms can never be equal. From $a_{n}=a_{n+1}=c$ it would follow immediately that $a_{n-1}=0$ and $a_{n-2}=a_{n-3}=c \quad(n>2)$. Finally we would have to have $a_{0}=a_{1}$ or $a_{0}=0$ resp. $a_{1}=0$, which is excluded. Hence $a_{n}>0$ also holds for all $n$.

Resolving the condition gives $a_{n+2}=a_{n+1}+a_{n}$ if $a_{n+2}>a_{n+1}$, and $a_{n+2}=a_{n+1}-a_{n}$ if $a_{n+2}<a_{n+1}$. From $a_{n+1}<a_{n}$ it therefore follows that $a_{n+2}>a_{n}$ and $a_{n+2}>a_{n+1}$. Hence the subsequence $b_{0}, b_{1}, b_{2}, \ldots$ that arises by omitting all terms that are smaller than both their predecessor and their successor is strictly monotonically increasing.

If we now show that $b_{m+1}-b_{m} \geq b_{m}-b_{m-1}$ holds for all $m \geq 2$, then we have for this subsequence an arithmetic sequence with positive common difference as a lower bound, from which the unboundedness follows directly. For this we set $b_{m+1}=a_{n+2}$, where $a_{n+2}>a_{n+1}$ is to hold. For $a_{n+1}>a_{n}$ we have $b_{m}=a_{n+1}$ and $b_{m-1} \geq a_{n-1}$ (because either $b_{m-1}=a_{n-1}$ or $b_{m-1}=a_{n}>a_{n-1}$ holds). Thus $b_{m+1}-b_{m}=a_{n}=a_{n+1}-a_{n-1} \geq b_{m}-b_{m-1}$. For $a_{n+1}<a_{n}$, on the other hand, we have $b_{m}=a_{n}$ and $b_{m-1} \geq a_{n-1}$ (because either $b_{m-1}=a_{n-1}$ or $b_{m-1}=a_{n-2}>a_{n-1}$ holds). So here $b_{m+1}-b_{m}=a_{n+1}=a_{n}-a_{n-1} \geq b_{m}-b_{m-1}$.