Assume that there are two integers $a$ and $b$ such that $0<a<b$, yet they both have the same Zeckendorf representation $Z$ with $k$ elements all 0s or 1s. We compute

where $F_{x}$ is the $x$th Fibonacci number. We can be assured that there is only one possible value for $c$ since all $F_{i}$ are distinct for $i>0$ and each $F_{i}$ was added only once or not at all, since each $Z_{i}$ is limited by definition to 0 or 1. Now $c$ holds the value of the Zeckendorf representation $Z$. If $c=a$, it follows that $c<b$, but that would mean that $Z$ is not the Zeckendorf representation of $b$ after all, hence this results in a contradiction of our initial assumption. And if on the other hand $c=b$ and $c>a$, then this leads to a similar contradiction as to what is the Zeckendorf representation of $a$ really is. ∎