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Is L = { $a^{n}b^{n}c^{2n}$ | n>=0 } a context free language?
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Push a onto the stack. Then for every a pop b. This way we get equal number of a and b. Now to push c which are twice of a or b, we don't have count as stack is empty. So it is not CFL. It can be done by LBA and hence CSL.

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