Is Bob right?

Andrew and Bob are writing a computer program which gives the prime factorization of a number.

Andrew suggests the following structure for the program:

$\bullet\text{ User selects an integer, }x_0$ $\\ \bullet\text{ The integer is divided by an integer, }y \text{, in ascending order starting from 2 until } x_0=0\mod{y} \\ \text{ at which point the program prints } y \text{ and we find } x_1 = \dfrac{x_0}{y}.$ $\\ \bullet\text{ We repeat the instructions above using }x_1\text{ instead of } x_0\text{ and upon each iteration finding } x_n = \dfrac{x_{n-1}}{y} .$ $\\ \bullet\text{ The program stops executing when } x_n \text{ is itself prime.}$

For example,

$\text{Let, }x_0 = 20$

$20 = {0}\mod{2} \therefore x_1 = \dfrac{20}{2} = 10$

$\text{Repeating for } x_1,$

$10 = 0\mod{2} \therefore x_2 = \dfrac{10}{2} = 5$

$\text{Since } x_2 \text{ is prime the program stops executing. It has printed out } 20 = 2 \times 2 \times 5.$

Bob suggests that for some integers, the program would not work as $y$ would not be prime and so the program would not give the correct factorization.

Is Bob correct?