Doubt

Question

L(G)={$a^mb^n$|m>n>=1} is the language context free?

Comments

@SHUBHAM SHASTRI

can you please share the image of PDA ...??

MY MOBILE IS NOT IN GOOD CONDITION :( :(

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It is definitely CFL. Since you map every b for a and remove corresponding number of a's and eventually if you are left with any number of a > = 1 then you accept it.

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It is a CFL. And I think this is the grammar for it.

$S\rightarrow aaABb$

$A\rightarrow aA|\epsilon$

$B\rightarrow aBb|\epsilon$

Answer 1

Yes it should he CFL please correct me of I am wrong