Question

Is this a context free language?

{a^(2k) b^n c^n : k >= 0 ∧ 0 <= n <= m} ∪
{a^(2k+1) b^n c^m :k >= 0 ∧ n >= m >= 0}

Answer 1

One way to prove a Language a Context-Free-Language is to write Context-Free-Grammar for the given language:(or either draw PDA)

The language below:

{a^(2k) bⁿ c^m : k >= 0 and 0 <= n <= m} U
{a^(2k+1) bⁿ c^m : k >= 0 and n >= m >= 0}

is Context Free Language

_{I think you have made mistake in writing question as I commented to you question, I am doing for above grammar}

We can write Context-Free-Grammar for this Language:

in Context-Free-Grammar productions of kind α --> β where α is a single variable.

S –> S₁ | S₂

S₁ generates this part {a^(2k) bⁿ c^m : k >= 0 and 0 <= n <= m} and S₂ generates {a^(2k+1) bⁿ c^m : k >= 0 and n >= m >= 0}

S₁ –> AB

A –> Aaa | ^

B –> bBc | ^

B –> Bc

And

S₂ –> AaC

C –> bCc | ^

C –> bC

In grammar S is start Variable and {S, S₁, S₂, A, B, C} all are variable.
So in above grammar every productions are in the form α --> β where α is a single variable hence given language is Context-Free-Language.

_{Let me know if you have other doubt or if your language I misunderstood}