[see also: moment, simultaneously]
an impulse acting at time $t=0$
This path stays in $B$ for all time.
This can be performed in $O(n)$ time.
The problem is to move all the discs to the third peg by moving only one at a time.
At times [= Occasionally] it will be useful to consider ......
We prove both lemmas at the same time.
At the time of writing [5], I was not aware of this reference.
It has been known for some time that ......
But this time boundedness on $U$ is enough; we do not need continuity on $V$.
Then $A$ has three times as many elements as $B$ has.
The integral of $F$ is $r$ times the sum of ......
Thus $X$ has $r$ times the length of $Y$.
The diameter of $A$ is three times that of $B$.
Clearly, $F$ is $r$ times as long as $G$.
Applying this argument $k$ more times, we obtain ......
......where $z^k=z... z$ ($k$ times)
Let $K$ be the number of times that $z$ returns to $B$.
Repeating this procedure enough times gives the desired triangulation.
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