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Consider an $\varepsilon$-tree CFG. If for every pair of productions $A\rightarrow u$ and $A\rightarrow v$

  1. If $\text{FIRST(u)} \cap \text{FIRST(v)}$ is empty then the CFG has to be $LL(1).$
  2. If the CFG is $LL(1)$ then $\text{FIRST(u)} \cap \text{FIRST(v)}$ has to be empty.
  3. Both $(A)$ and $(B)$
  4. None of the above
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FIRST(u) ∩ FIRST(v) is empty iff the CFG has to be LL(1).

Hence Option C should be correct.
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