2 deleted 2 characters in body edited Jan 6 at 19:55 MrGeek 11655 bronze badges AA -> ωα_1ωα1 | ωα_2ωα2 | ... | ωα_n (where _i is an index, not the terminal _ and i)ωαn (where αi is the i-th α, not the symbols α and i), with α_1α1 ≠ α_2α2 ≠ ... ≠ α_nαn. We can then easily show that:∩∩(i=1,..,n) FIRST(α_iωαi) ≠ ØØ SS -> Sα | ββ  A -> ωα_1 | ωα_2 | ... | ωα_n (where _i is an index, not the terminal _ and i)  with α_1 ≠ α_2 ≠ ... ≠ α_n. We can then easily show that:∩(i=1,..,n) FIRST(α_i) ≠ Ø S -> Sα | β  A -> ωα1 | ωα2 | ... | ωαn (where αi is the i-th α, not the symbols α and i), with α1 ≠ α2 ≠ ... ≠ αn. We can then easily show that:∩(i=1,..,n) FIRST(ωαi) ≠ Ø S -> Sα | β  1 answered Jan 6 at 19:23 MrGeek 11655 bronze badges I have done some more research, and I think I've found a solution for the 1st and 2nd questions, as for the 3rd one, I found an existing solution on StackOverflow for it, the proof attempts are written below: We start by writing the three rules of the definition of an LL(1) grammar: For every production A -> α | β with α ≠ β: FIRST(α) ∩ FIRST(β) = Ø. If β =>* ε then FIRST(α) ∩ FOLLOW(A) = Ø (also, if α =>* ε then FIRST(β) ∩ FOLLOW(A) = Ø). Including ε in rule (1) implies that at most one of α and β can derive ε. Proposition 1: A non-factored grammar is not LL(1). Proof: If a grammar G is non-factored then there exists a production in G of the form: A -> ωα_1 | ωα_2 | ... | ωα_n (where _i is an index, not the terminal _ and i)  with α_1 ≠ α_2 ≠ ... ≠ α_n. We can then easily show that: ∩(i=1,..,n) FIRST(α_i) ≠ Ø  which contradicts rule (1) of the definition, thus, a non-factored grammar is not LL(1). ∎ Proposition 2: A left-recursive grammar is not LL(1). Proof: If a grammar is left-recursive then there exists a production in G of the form: S -> Sα | β  Three cases arise here: If FIRST(β) ≠ {ε} then:     FIRST(β) ⊆ FIRST(S) =>  FIRST(β) ∩ FIRST(Sα) ≠ Ø which contradicts rule (1) of the definition. If FIRST(β) = {ε} then: 2.1. If ε ∈ FIRST(α) then: ε ∈ FIRST(Sα) which contradicts rule (3) of the definition. 2.2. If ε ∉ FIRST(α) then:     FIRST(α) ⊆ FIRST(S) (because β =>* ε) =>  FIRST(α) ⊆ FIRST(Sα) ........ (I) we also know that: FIRST(α) ⊆ FOLLOW(S) ........ (II) by (I) and (II), we have: FIRST(Sα) ∩ FOLLOW(S) ≠ Ø and since β =>* ε, this contradicts rule (2) of the definition. In every case we arrive at a contradiction, hence, a left-recursive grammar is not LL(1). ∎ Proposition 3: An ambiguous grammar is not LL(1). Proof: While the above proofs are mine, this one is not, it's by Kevin A. Naudé which I got from his answer that is linked below: https://stackoverflow.com/a/18969767/6275103