# Prove or disprove if L is CFL? [duplicate]

Given $$L=\{a^ib^jc^k | i\neq j \space and \space j=k\}$$. Is this CFL? How do I write CFG for it or prove it with pumping lemma? Thanks.

• Have you tried proving that this is not context-free using the pumping lemma? – Yuval Filmus Oct 25 '18 at 22:05
• Since the fastest and accuratest universal gun, Yuval's is kicking in, I am leaving. (This comment may be deleted later.) – Apass.Jack Oct 25 '18 at 22:07
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Suppose that $$L$$ were context-free. According to Ogden's lemma, there is a constant $$p$$ such that each word in $$L$$ with at least $$p$$ marked positions satisfies the constraints of the lemma. Consider the word $$s = \underline{a^p}b^{p+p!}c^{p+p!}$$, in which the underlined part is marked. According to Ogden's lemma, there is a decomposition $$s = uvwxy$$ in which $$vx$$ contains at least one $$a$$, and $$uv^iwx^iy \in L$$ for all $$i \geq 0$$. We now consider several cases:
1. $$x$$ contains $$b$$s but not $$c$$s, or $$c$$s but not $$b$$s. Choosing $$i = 0$$, we obtain a word in which the number of $$b$$s differs from the number of $$c$$s, and so does not belong to $$L$$.
2. $$x$$ contains both $$b$$s and $$c$$s. Choosing $$i = 2$$, we obtain a word not belonging to $$a^*b^*c^*$$, and so not belonging to $$L$$.
3. $$x$$ contains no $$b$$s nor $$c$$s, and $$v \notin a^*$$. In this case $$x = \epsilon$$, and so $$v$$ must contain at least two different characters. Choosing $$i = 2$$, we again obtain a word not belonging to $$a^*b^*c^*$$, and so not belonging to $$L$$.
4. $$vx \in a^+$$, say $$vx = a^q$$. Let $$i = p!/q+1$$. Then $$uv^iwx^iy = a^{p+p!}b^{p+p!}c^{p+p!} \notin L$$.
We have obtained a contradiction, and so $$L$$ is not context-free.