# Use the pumping lemma to prove that the following language is not context free

Can anyone help with the following problem ?

Let $$B = \{ a^{n}b^{m}c^{m}d^{2n} | n,m ≥ 0 \}$$, use the pumping lemma to prove B is not context-free

• Sorry, but $B$ is context-free. – Hendrik Jan Oct 6 '18 at 11:37
• Sorry, you are right, there is probably a mistake in the description of the exercise. – ElDon90 Oct 7 '18 at 8:26

Consider the following Context Free Grammar $$G$$:
$$S \rightarrow aSdd\ |\ X\ |\ \epsilon$$
$$X \rightarrow bXc\ |\ \epsilon$$
The language $$L(G)$$ generated by $$G$$ corresponds to the set of all the strings $$\{\epsilon, add, aadddd, ..., bc, bbcc,..., abcdd, abbccdd,...\}$$.
More generally, it corresponds to $$\{ a^{n}b^{m}c^{m}d^{2n} | n,m ≥ 0 \}$$.