CS 390 Solutions to Test I

1. T, T, T, T, F

2.

3 (a) a
(b) aa, ab
(c) aaa, aab, abb.
(d) By prefixing with a or by appendin b.
(e) By prefixing with a or by appendin b.
(f) (1) Basis:
(2) Induction: For any , and .
Extremal Clause: Nothing is in L unless it is obtained from (1) and (2).

4. Induction on the length of string.
Statement to prove: For any natural number n and any string x over alphabet , if , then { is regular.

Proof. Basis: n = 0. If , then . By the definition of regular language, { } is regular.
Induction: Suppose that for a natural number n and a string x, if , then {x} is regular.
If , then there is and such that x = ya. Hence {x} = {ya} = . But since , is regular. Also { is regular by the definition of regular language. Since a concatenation of regular languages is regular, is regular.

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