Kleene’s Theorem (Part-3) CSC312 Automata Theory Lecture # 14 Chapter # 7 by Cohen Kleene’s Theorem (Part-3)

Proof of Part-3 (Cont…) Rule 4: (Closure of an FA) Proof of Rule 4: If r is a RE and FA1 is a finite automaton that accepts exactly the language defined by r, then there is an FA called FA2 that will accepts exactly the language defined by RE r*. Proof of Rule 4: The language defined by r* must always contained the null word. The closure of an FA1 is same as concatenation of an FA with itself, excepts that the initial stat of the required FA is a final state as well.

Proof of Part-3 (Cont…) Proof of Rule 4 (Cont…) In order to decide initial and final states of new FA we have following rule - IF the start state of FA1 is also a final state OR the start state of FA1 is not re-enterable, THEN the start state of FA2 is z1 = x1 , which must be a final state of FA2 i.e. ±z1 = x1

Proof of Part-3 (Cont…) Proof of Rule 4 (Cont…) ELSE we make two states from x1 as z1 = x1 ; z1 as initial and final state i.e. ±z1 = x1 and z2 = x1 ; z2 as non-initial and non-final state Note: - whenever x1 appears under ‘new states’ columns of the transition table we treat it as z2. Other than the start state of FA2 (which is always also a final state), the final states of FA2 are the states zk = {xi’s} for which a number of the subset {xi’s} is a final state of FA1. (Solve exercise Q No. 6)