Defn: Let Leq be the set of all strings over
{0, 1} having an equal number 0s and 1s.
Question: Is Leq regular?
Theorem 4.1: (The pumping lemma for
regular languages) Let L be a regular
language. Then there is an n ∈ N such that for
every w ∈ L for which |w| ≥ n, we can write
w = xyz such that:
1. y 6= ²;
2. |xy| ≤ n; and
3. ∀k ≥ 0 : xy k z ∈ L.
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Claim: Leq is not regular.
Proof: By contradiction. Assume Leq is
regular. Let n be the constant guaranteed by
the pumping lemma. Let w = 0n 1n . Clearly,
w ∈ Leq . By the pumping lemma, we can write
w = xyz such that y 6= ², |xy| ≤ n, and
∀k ∈ N : xy k z ∈ Leq . Because |xy| ≤ n, y = 0i
such that 0 < i ≤ n. Then
xy 0 z = 0n−i 1n 6∈ Leq — a contradiction.
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Proof of Pumping Lemma (sketch):
• Let A = (Q, Σ, δ, q0 , F ) be a DFA s.t.
L(A) = L.
• Let n = |Q|, w ∈ L, |w| ≥ n.
• For 0 ≤ i ≤ n, let wi denote the prefix of
w with length i.
• By the Pigeonhole Principle,
∃ 0 ≤ i < j ≤ n : δ̂(q0 , wi ) = δ̂(q0 , wj ).
• Let x = wi , xy = wj , xyz = w (note that
|xy| ≤ n, y 6= ²).
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Proof continued:
• By induction on k, δ̂(q0 , xy k ) = δ̂(q0 x) for
all k ∈ N.
δ̂(q0 , xy k z) = δ̂(δ̂(q0 , xy k ), z)
= δ̂(δ̂(q0 , x), z)
= δ̂(δ̂(q0 , xy), z)
= δ̂(q0 , xyz)
∈ F.
• Therefore, xy k z ∈ L(A) = L.
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Theorem 4.5: The set of regular languages
over Σ is closed under complementation; i.e., if
L is a regular language over Σ, then Σ∗ − L is
regular.
Proof: Let A = (Q, Σ, δ, q0 , F ) be a DFA s.t.
L(A) = L. Let A0 = (Q, Σ, δ, q0 , Q − F ). Then
Σ∗ − L = Σ∗ − {w ∈ Σ∗ | δ̂(q0 , w) ∈ F }
= {w ∈ Σ∗ | δ̂(q0 , w) ∈ Q − F }
= L(A0 ).
Thus, Σ∗ − L is regular.
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Theorem 4.8: The set of regular languages
over Σ is closed under intersection.
Proof: Let L1 and L2 be regular languages
over Σ. By DeMorgan’s laws,
L1 ∩ L 2 = L 1 ∪ L 2 .
Because the set of regular languages over Σ is
closed under union and complementation,
L1 ∩ L2 is regular.
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Defn: Let Σ and ∆ be two alphabets. A
homomorphism from Σ∗ to ∆∗ is any function
h : Σ∗ → ∆∗ such that ∀x, y ∈ Σ∗ ,
h(xy) = h(x)h(y).
Defn: Let Σ and ∆ be two alphabets, and let
∗
∗
R ⊆ Σ∗ × ∆∗ . We define R̃ : 2Σ → 2∆ such
that
R̃(L) = {y ∈ ∆∗ |∃x ∈ L : hx, yi ∈ R}
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Example: Let h : {a, b}∗ → {0, 1}∗ be the
homomorphism defined by
h(a) = 01
h(b) = 10.
Then h̃(a∗ b∗ ) = (01)∗ (10)∗ .
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For a string w ∈ {a, b}∗ , let w denote the
homomorphism defined by
a=b
b = a.
Let L = {xxR w | x, w ∈ {0, 1}∗ , x 6= ²}. Then
−1 (L) = {x ∈ {a, b}∗ | ∃y ∈ L : h(x) = y}
hg
= {xxR w ∈ {a, b}∗ | x 6= ²}
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Theorem 4.14: Let h : Σ∗ → ∆∗ be a
homomorphism. If L ⊆ Σ∗ is regular, then
h̃(L) is also regular.
Theorem 4.16: If h : Σ∗ → ∆∗ is a
homomorphism and L ⊆ ∆∗ is regular, then
−1 (L) is regular.
hg
Claim: The language
L = {xxR w | x, w ∈ {0, 1}∗ , x 6= ²} is not
regular.
Proof idea:
−1 (L) ∩ a∗ b∗ = {ai bj | 0 < i ≤ j}.
hg
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Proof of Theorem 4.14 (sketch): By
induction on L ∈ R(Σ).
Base Case 1: h̃(∅) = ∅.
Base Case 2: h̃({²}) = {²}.
Base Case 3: For a ∈ Σ, h̃({a}) = {h(a)}.
IH: Assume that for L1 , L2 ∈ R(Σ), h̃(L1 ) and
h̃(L2 ) are regular.
IS 1: h̃(L1 ∪ L2 ) = h̃(L1 ) ∪ h̃(L2 ).
IS 2: h̃(L1 L2 ) = h̃(L1 )h̃(L2 ).
IS 3: h̃(L∗1 ) = h̃(L1 )∗ .
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Proof of Theorem 4.16 (sketch):
• Let A = (Q, ∆, δA , q0 , F ) be a DFA,
L(A) = L.
• Let B = (Q, Σ, δB , q0 , F ), where
δB (q, a) = δ̂(q, h(a)).
• By right induction on w ∈ Σ∗ ,
δ̂B (q0 , w) = δ̂A (q0 , h(w)).
−1 (L) = L(B).
• Therefore, hg
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