Use the pumping lemma to prove that. L = {a^nb^2nc^3n|n epsilon N0} is not conte

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Use the pumping lemma to prove that. L = {a^nb^2nc^3n|n epsilon N0} is not context-free. Assume that the language L is context-free. Then for a large enough number p every string .s epsilon L with |s| > = p can be written as s = uvxyz, where |vxy| = 1, so that all strings of the form uv^2:xy^2 z with i > = 0 are in L. In particular, if we choose the string a^p^2pc^3p epsilon L for which we have |a^pb^2pc^3p| = 6p > = p, then setting a^pb^2c^3p = uvxyz: implies that a number of symbols, and at least one(!), wiIl be assigned to either u or v. By setting i = 0 those symbols are deleted from s. Let us denote the resulting string As s' ... complete the proof on your own.

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