There’s a number of methods to show a language is not context free. Going via some workouts, I am caught at a query that asks me to show {that a} language is certainly context free.

$L = {a^{(n+1)} b^{(m+2)}c^{(n+4)} | m, n ≥ 0 }$

I see that the langauge is equal to $L = {aa^nbbb^mccccc^n}$, however I am undecided if that helps in any respect.

Possibly breaking the language aside right into a concatenation of three languages like ${a^{n+1} | n>= 0}.{b^{m+2} | m >= 0}.{c^{n+4} | n>=0}$ may be useful, as a result of we will do a rule individually for every? The n needs to be equal for the left and proper aspect although, which I am having issue with.

Any assist can be a lot appreciated!