# exercise 2.16

Show that the class of context-free languages is closed under the regular operations, union, concatenation, and star.

After the meeting with Calzavara, we'll rewrite the proofs using the grammars instead of PDAs

Indice dei contenuti

### Union

Let A, B be CFLs, then exist G, H CFGs when L(G) = A and L(H) = B. Let S1, S2 the start symbols for G and H.

We can build a new CFG with start symbol S where S → S1 | S2, including all the productions of G and H.
Before the union, we must ensure that the terminals in G and H are different, by replacing the common names.

### Concatenation

Let A, B be CFLs, then exist G, H CFGs when L(G) = A and L(H) = B. Let S1, S2 the start symbols for G and H.

We can build a new CFG with start symbol S where S → S1 S2, including all the productions of G and H.
As the union, we must ensure that the terminals in G and H are different.

### Star

Let A be CFL, then exist G CFG when L(G) = A. Let S1 the start symbols for G.

We can build a new CFG with start symbol S where S → S S1|ε, including all the productions of G.

ε warrantees that the recursion ends and that the new CFG accepts an empty string.

### Union

#### Proof idea

Union of two PDAs can be similar to the union of two NFA: a new start state points to both the old start states with an ε arrow. What about the stack? Since it's empty - by definition of PDA - at the new start state, the two arrows must not read or write anything from the stack, to be sure to keep the old starts states with an empty stack.

#### Proof

Let be A1, A2 two context-free languages. If A1 and A2 are context-free, there exists two CFG N1, N2 that recognize them. We define N1, N2 as follow:

N1 = {Q1, Σ, Γ1, δ1, q1-start, F1⊆Q1}

N2 = {Q2, Σ, Γ2, δ2, q2-start, F2⊆Q2}

Construct N that recognize A1 ∪ A2 as
N = {Q, Σ, Γ, δ, qstart, F⊆Q}
where

Q = Q1 ∪ Q2 ∪ qstart

Γ = Γ1 ∪ Γ2

qstart = "new" start state for N

F = F1 ∪ F2

δ = (for a ∈ Σ and b ∈ Γ)

• δ(q1, a, b) if q ∈ Q1
• δ(q2, a, b) if q ∈ Q2
• {q1, q2} if q = qstart and a = ε
• ∅ if q = qstart and a ≠ ε

Note that the stack is managed independently for each automaton N1 and N2

### Concatenation

#### Proof idea

As for the union, we can start our proof using the concatenation proof for the NFAs, that consists in linking all the final states of the "first" NFA to the start state of the "second".

Can we be sure that at the final state the stack is empty in every case? That's an open question.

If yes, the proof idea is completed.

#### Proof

Let be A1 a context-free language. If A1 is context-free, there exists a CFG N1 that recognize it. We define N1 as follow:

N1 = {Q1, Σ, Γ1, δ1, q1-start, F1⊆Q1}

Construct N that recognize A1* as
N = {Q, Σ, Γ, δ, qstart, F⊆Q}
where

Q = Q1∪ qstart

Γ = Γ1

qstart is the new start

F = F1∪ qstart

δ = (for a ∈ Σ and b ∈ Γ)

• δ(q1, a, b) if q ∈ Q1 and q ∉ F
• δ(q1, a, b) if q ∈ Q1 and q ∈ F and a ≠ ε
• δ(q1, a, b) ∪ qdiscard if q ∈ Q1 ∪ qdiscard and a = ε and b ≠ $• q2-start if q ∈ Q1∪ qdiscard and a = ε and b =$
• ∅ if q = qstart and a ≠ ε