LL(1) cannot be ambiguous

Question

How can it be shown that no LL(1) grammar can be ambiguous?

I know what is ambiguous grammar but could not prove the above theorem/lemma.

Answer 1

Here's my first draft at a proof. It might need some fine tuning, but I think it covers all the cases. I think many solutions are possible. This is a direct proof.

(Side note: it is a pity SO doesn't support math, such as in LaTeX.)

Proof

Let T and N be the sets of terminal and non-terminal symbols.

Let the following hold

MaybeEmpty(s) = true <=> s ->* empty
First(s) = X containing all x for which there exists Y such that s ->* xY
Follow(A) = X containing all x for which there exists Y,Z such that S ->* YAxZ

Note that a grammar is LL(1) if the following holds for every pair of productions A -> B and A -> C:

1. (not MaybeEmpty(B)) or (not MaybeEmpty(C))
2. (First(B) intersect First(C)) = empty
3. MaybeEmpty(C) => (First(B) intersect Follow(A)) = empty

Consider a language with is LL(1), with A -> B and A -> C. That is to say there is some string of terminals TZ which admits multiple derivations by distinct parse trees.

Suppose that the left derivation reaches S ->* TAY ->* TZ. The next step may be either TAY -> TBY, or TAY -> TCY. Thus the language is ambiguous if both BY ->* Z and CY ->* Z. (Note that since A is an arbitrary non-terminal, if no such case exists, the language is non-ambiguous.)

Case 1: Z = empty

By rule 1 of LL(1) grammars, at most one of B and C can derive empty (non-ambiguous case).

Case 2: Z non-empty, and neither B nor C derive empty

By rule 2 of LL(1) grammars, at most one of B and C can permit further derivation because the leading terminal of Z cannot be in both First(B) and First(C) (non-ambiguous case).

Case 3: Z non-empty, and either MaybeEmpty(B) or MaybeEmpty(C)

Note the by rule 1 of LL(1) grammars, B and C cannot both derive empty. Suppose therefore that MaybeEmpty(C) is true.

This gives two sub-cases.

Case 3a: CY -> Y; and Case 3b: CY ->* DY, where D is not empty.

In 3a we must choose between BY ->* Z and CY -> Y ->* Z, but notice that First(Y) subset-of Follow(A). Since Follow(A) does not intersect First(B), only one derivation can proceed (non-ambiguous).

In 3b we must choose between BY ->* Z and CY ->* DY ->* Z, but notice that First(D) subset-of First(C). Since First(C) does not intersect First(B), only one derivation can proceed (non-ambiguous).

Thus in every case the derivation can only be expanded by one of the available productions. Therefore the grammar is not ambiguous.