The Correctness of LR(0) parsing

1. Lemma 2: For kernel items, [N -> ₁ . ₂ ] ( ₀, ) only if [ N ₁ . ₂ ] is valid for .

2. Now, assuming Lemma 1, we can prove Lemma 2 by induction on the length of . The basis step is so simple that we will look at the induction step first:

   induction

   Assume that we know that the theorem holds for all strings of length n and consider some string x such that is of length n and x is a single symbol.

   Suppose that [ N ₁ x . ₂ ] is an item in ( ₀, x ). The fact that this item is in this set implies that the item [ N ₁ . x ₂ ] must be in ( ₀, ). This, together with our inductive assumption implies that [ N ₁ . x ₂ ] must be valid for . Therefore, there exists a derivation:

   S' N ₁ x ₂

   with ₁ = . This, however implies that [ N ₁ x . ₂ ] is indeed valid for x.

   basis

   Similarly, when we consider strings of length 0, the only kernel item in ( ₀, eps) is [ S' . S]. The derivation S' S' S shows that this item is valid for eps.