Consider the grammar below A → BB B → bB | c B → c

Grammar: A → B B , B → bB | c

• Compute FIRST sets:

  FIRST(B) = {b, c}. There is no ε-production for B.

• LL(1) check:

  The two alternatives for B start with different terminals (b and c), so their FIRST sets are disjoint. A has a single production A → B B, so there are no competing productions for A.

• Conclusion about LL(1):

  Because alternatives for each nonterminal have disjoint FIRST sets and no ε is involved, the grammar is LL(1).

• LR(1) check:

  LR(1) parsers are more general than LL(1); any LL(1) grammar will have LR(1) item sets with no shift/reduce or reduce/reduce conflicts. Therefore this grammar is also LR(1).

Final answer: the grammar is both LL(1) and LR(1).