Language is not Regular
The language is not regular which can be showed via the pumping lemma for regular languages. In your proof you should have used a string that actually is in the language e.g.: w = xyz = 0^(2p) 1^(p). As using a string that is already not part of the language, would provide little information about the conclusions of the pumping lemma.
Then given the pumping lemma:
w = xyz = 0^(2p) 1^(p)|xy| <= p|y| >= 1q = xy^(i) z, q is in the language for all i.
As |xy| <= p, the substring |xy| exists of only zeroes. As |y| > 1, there is at least one zero in y, and at most p zeroes in y.
If we pumped y, then that would mean that the output string is of the form: #w(0) = #y(0) * i + #x(0) + p
The language asserts:
#w(0) = 2#w(1)#y(0) * i + #x(0) + p = 2p#y(0) * i + #x(0) = p
As we can pump i forever, we can always reach an inequality. Showing that there is no valid derivation among xyz, thus the pumping lemma for regular languages is not applicable, thus the language is not regular.
Language is Context-Free
The language is a context-free language, however, not by the grammar you have specified. The right context-free grammar for this language is:
S -> S0S0S1S | S0S1S0S | S1S0S0S | ε
Essentially, whenever a '1' is added we must ensure that 2 zeroes are added, the order is irrelevant, but they must be placed. Due to this assertion, we can recursively assume that for each '1' there are 2 zeroes. Which follows the 'w contains twice as many zeroes than ones'.
Language is Context-Sensitive, Decidable, or other language class
Showing that the language is context-sensitive, or some other complexity class is trivial. As we already showed that it is context-free, all language classes that contains the context-free language class as a subset also accepts this language.
Examples of language classes that have the context-free language class as subset:
- Context-Sensitive
- Recursive Enumerable
- Decidable
