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Eliminate the variable B from the grammar S aSB|bB B aA|b. 2 Complete the proof of Theorem 6.1 by showing that (Given Below) S *G wS * w implies S...
1. Eliminate the variable B from the grammar
S → aSB|bB
B → aA|b.
2 Complete the proof of Theorem 6.1 by showing that (Given Below)
S ⇒*Gˆ wS ⇒*Ĝ w
implies
S ⇒*G w.S ⇒*G w
3. In Theorem 6.1, why is it necessary to assume that A and B are different variables?
THEOREM 6.1
Let G = (V, T, S, P) be a context-free grammar. Suppose that P contains a production of the form
A → x1Bx2.
Assume that A and B are different variables and that
B → y1 |y2|ꞏꞏꞏ|yn
is the set of all productions in P that have B as the left side. Let Ĝ = (V, T, S, PˆP̂ ) be the grammar in which PˆP̂ is
constructed by deleting
A → x1Bx2 (6.1)
from P, and adding to it
A → x1y1x2 |x1y2x2|ꞏꞏꞏ|x1ynx2.
Then
L (Ĝ) = L (G).
Proof: Suppose that w ∈ L (G), so that
S⇒*G w.S⇒*G w.
The subscript on the derivation sign ⇒ is used here to distinguish between derivations with different grammars. If this
derivation does not involve the production (6.1), then obviously
S⇒*Gˆ w.S⇒*Ĝ w.
If it does, then look at the derivation the first time (6.1) is used. The B so introduced eventually has to be replaced; we
lose nothing by assuming that this is done immediately (see Exercise 18 at the end of this section). Thus
S ⇒*G u1 Au2 ⇒ G u1x1Bx2u2 ⇒ G u1x1 yjx2u2.S ⇒*G u1 Au2 ⇒ G u1 x1 B x2 u2
⇒ G u1 x1 yj x2 u2.
But with grammar Ĝ we can get
S ⇒*Gˆ u1 Au2 ⇒ Gˆ u1 x1 yj x2u2.S ⇒*Ĝ u1 Au2 ⇒ Ĝ u1 x1 yj x2 u2.
Thus we can reach the same sentential form with G and Ĝ. If (6.1) is used again later, we can repeat the argument. It
follows then, by induction on the number of times the production is applied, that
S ⇒*Gˆ w.S ⇒*Ĝ w.
Therefore, if w ∈ L (G), then w ∈ L (Ĝ).
By similar reasoning, we can show that if w ∈ L (Ĝ), then w ∈ L (G), completing the proof.
4. Convert the grammar
S → AB|aB,
A → abb|λ,
B → bbA
into Chomsky normal form.
5.Convert the grammar
S → ab |aS| aaS|aSS
into Greibach normal form.