1. Context-Free Grammar (CFG)

A Context-Free Grammar (CFG) is a formal grammar that generates strings belonging to a context-free language.
It is defined as a 4-tuple:

G = (V, Σ, P, S)

where:

• V → Finite set of variables (nonterminals)
• Σ → Finite set of terminals, with V ∩ Σ = ∅
• P → Finite set of productions of the form A → α, where A ∈ V and α ∈ (V ∪ Σ)*
• S → Start symbol, S ∈ V

A string w ∈ Σ* belongs to the language L(G) if and only if

S ⇒* w
(i.e., w can be derived from S using the productions).

2. Pumping Lemma for Context-Free Languages (CFLs)

The Pumping Lemma for CFLs is used to prove that certain languages are not context-free.

Theorem (CFL Pumping Lemma):
If L is a context-free language, then there exists a pumping length p ≥ 1 such that every string s ∈ L with |s| ≥ p can be written as:

s = u v x y z

satisfying:

1. v y ≠ ε (at least one of v or y is nonempty),
2. |v x y| ≤ p,
3. For all i ≥ 0, the string u v^i x y^i z ∈ L.

Important clarifications:

• v and y are not required to be identical.
• x is not a “separator”; any of u, x, z may be empty.
• Only the constraints v y ≠ ε and |v x y| ≤ p are guaranteed.

3. Why the Pumping Lemma Holds (Proof Sketch)

• Convert the grammar of L into Chomsky Normal Form (CNF).

• Let the number of variables be k.

• Any parse tree deriving a sufficiently long string (|s| ≥ p) must contain a long path from root to leaf.

• By the pigeonhole principle, some variable A repeats along this path:

  S ⇒* u A z ⇒* u v A y z ⇒* u v x y z

• The repeated variable A allows pumping: replacing the subtree derived from the second A by more/fewer copies gives valid derivations.

• This proves u v^i x y^i z ∈ L for all i ≥ 0.

• The structure of the parse tree ensures that |v x y| ≤ p and v y ≠ ε.

4. Using the Pumping Lemma to Show a Language is Not Context-Free

To prove a language L is not context-free, proceed by contradiction:

1. Assume L is context-free.

2. Let p be the pumping length from the lemma.

3. Choose a specific string s ∈ L with |s| ≥ p.

4. For every valid decomposition s = u v x y z satisfying
   |v x y| ≤ p and v y ≠ ε,
   show there exists some i ≥ 0 (often i = 0 or i = 2) such that:

   u v^i x y^i z ∉ L

   which contradicts the lemma.

5. Therefore, L is not context-free.