Regarding the language L2 where the #a's >= #b's I thought of the following:
S -> SaSbS
S -> SbSaS
S -> SaS
S -> ε
When it comes to L1 I thought of utilising L2 in the following way and give an infinite degree of freedom to the number of a's:
S -> L1 A L1
A -> aA'
A'-> aA' | ε
I know that there are similar questions answered on this topic but I wanted to make my own attempt since the answers provided are not the same and two distinct CFGs may produce the same language. Are the CFGs provided correct?
