# prove that the language is not context free language

Prove that the language $$\{a^i b^j c^k$$ $$|$$ $$i>j>k \geq 0\}$$ is non context free language without using the pumping lemma.

we could use some helpful facts to solve the question:

1. $$A$$ is a context free language and $$B$$ is regular, then $$A \cap B$$ is context free.
2. $$L_1$$, $$L_2$$ are not context free languages, the intersection $$L_1 \cap L_2$$ is not context free as well.
• The second helpful fact is wrong. Take two disjoint non-context-free languages. – Yuval Filmus Apr 11 at 17:38
• if L1 is non context free language {a^i b^j c^k| i >=j>=k>=0} and L2 is non context free language {a^i b^j c^k| k>=j>=i >=0} the intersection of them is not context free – user102789 Apr 11 at 17:56
• If $L_1 = L_2$ the helpful fact is also true. It's true for some pairs $L_1,L_2$, and false for others. I'd say that's not so helpful. – Yuval Filmus Apr 11 at 23:26
• May I ask why you are interested in "without the pumping lemma"? Please explain, especially if you created this problem by yourself. Otherwise, can you provide a reference? – Apass.Jack Apr 12 at 3:56
• there is a question in "Sipser" book about that, the question asked to prove the language is non-context-free language by using fact ( a context free language intersection regular language is context free language ). the question in chapter 2 question # 2.30 – user102789 Apr 12 at 4:29