Is it possible to generate all sequences of ‘0’ and ‘1’ using only these 4 rules? I want to show you, a problem I was thinking about and it’s interesting for me, but I have no idea how to solve it. The question is:
There are 4 rules of replacement (the order of numbers matters):

*

*‘00’ --> ‘000’


*‘01’ --> ‘001’ or ‘010’ or ‘100’


*‘10’ --> ‘011’ or ‘110’ or ‘101’


*‘11’ --> ‘111’
For example:
‘1001’ = ‘10|01’ --> ‘011|010’ = ‘011010’
‘1001’ --> ‘011010’
Can you generate every finite sequence of ‘1’ and ‘0’ starting from ‘010’ or just ‘10’?
If yes show how or show the proof. If no say why and what conditions must be true (including e. g. how starting sequence should look like) to make it possible.
I think the division of sequence we generate can matters because sequence ‘1010’ can be:
‘10|10’ --> ‘011|110’
‘1010’ --> ‘011110’
or using different division:
‘1|01|0’ --> ‘1|010|0’
‘1010’ --> ‘10100’
so another thing about problem is show how this fact is related to the solution.
This question was invented by me, so I don’t even know if it's the best wording of a problem but you can write what you think about it :)
PS
Preventively sorry for my English. I’m not from England or United States. And I'm not sure it's good question to Mathematics Stack Exchange. I'm new here so you can write me whether this question match this site or not.
 A: Would you not start with the sequence you want to end with (of length $\ge 3$) and go backwards? To go backwards, you can always chop off the first three digits and map them back to two digits - until you reach a sequence with only two digits. For example:
$$100|100101\to 011|00101\to 100|0101\to 010|101\to 011|01\to 100|1\to 011\to 10$$
which corresponds to the expansion:
$$10\to 01|1\to 10|01\to 01|101\to 01|0101\to 10|00101\to 01|100101\to 100100101$$
Every use of an expansion rule extends the sequence by $1$, so starting from any sequence of length $2$ you can reach a sequence of length $n$ in exactly $n-2$ expansions.
Now if you count multiple expansions done on the same string as one step, there will, of course, be fewer steps. My guess is that the number of steps will behave $\sim\log_{3/2}n$ in the best case because, when going "backwards", for big $n$, you can replace approximately $n/3$ three-element subsequences by two-element subsequences, reducing the size by approximately the factor $2/3$. This would, however, warrant a rigorous proof (which I don't have at the moment but could give it a try - or you could also try yourselves!).
