We have to check each and every option.
Option $A:$
- $A \rightarrow B \mid a \mid-CBD$
- $B \rightarrow C\mid b$
- $C \rightarrow A \mid c$
- $D \rightarrow d$
This grammar is not immediately left-recursive, because there is no single production $X \rightarrow X \alpha.$ However, it is left recursive because there are valid derivations of the form $A\overset{\ast}{\rightarrow} A\alpha\:(\text{and}\: B\overset{\ast}{\rightarrow} B\beta\:\text{and}\:C\overset{\ast}{\rightarrow} C\gamma).$
For example$:A \rightarrow B \rightarrow C \rightarrow A,$ so $A\overset{\ast}{\rightarrow} A.$
Option $B:$
- $A \rightarrow Ba \mid C$
- $B \rightarrow AA$
- $C \rightarrow B \mid b$
For example$:A \rightarrow C \rightarrow B \rightarrow AA,$ so $A\overset{\ast}{\rightarrow} A.$
Option $C:$
- $Q \rightarrow QED \mid q$
- $E \rightarrow e$
- $D \rightarrow NFA \mid d$
- $N \rightarrow DFA \mid n$
- $F \rightarrow f$
- $A \rightarrow a$
For example$:D \rightarrow NFA \rightarrow DFAFA.$ so $D\overset{\ast}{\rightarrow} D.$
Option $D:$
- $S \rightarrow (L) \mid x$
- $L \rightarrow SL'$
- $L' \rightarrow \epsilon \mid ,SL'$
In this grammar, there is no direct or indirect left recursion.
Actually, the above grammar is free from left recursion, when we remove left recursion from the following grammar.
- $S \rightarrow (L) \mid x$
- $L \rightarrow L,S \mid S$
References:
So, the correct answer is $A; B; C.$
