Question regarding the proof that quadratic probing always finds an empty slot if the table is less than half full
Suppose that $t=11$, $i=5$, $j=6$. Then $i+j \bmod t = 0$, and there is no contradiction. However, in this example $i &lt; t/2$ while $j &gt; t/2$. This sort of example cannot happen if $i,j &lt; t/2$, since then $i+j &lt; t$.