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| In [[formal language]] theory, and in particular the theory of [[nondeterministic finite automata]], it is known that the '''union of two regular languages''' is a [[regular language]]. This article provides a proof of that statement.
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| ==Theorem==
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| For any regular languages <math>L_{1}</math> and <math>L_{2}</math>, language <math>L_{1}\cup L_{2}</math> is regular.''
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| ''Proof''
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| Since <math>L_{1}</math> and <math>L_{2}</math> are regular, there exist [[Nondeterministic finite automaton|NFAs]] <math>N_{1},\ N_{2}</math> that recognize <math>L_{1}</math> and <math>L_{2}</math>.
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| Let
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| ::<math> N_{1} = (Q_{1},\ \Sigma ,\ T_{1},\ q_{1},\ A_{1})</math>
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| ::<math> N_{2} = (Q_{2},\ \Sigma ,\ T_{2},\ q_{2},\ A_{2})</math>
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| Construct
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| ::<math>N = (Q,\ \Sigma ,\ T,\ q_{0},\ A_{1}\cup A_{2})</math>
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| where
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| ::<math>Q = Q_{1}\cup Q_{2}\cup\{q_{0}\}</math>
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| ::<math>T(q,x) = \left\{\begin{array}{lll}
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| T_{1}(q,x) & \mbox{if} & q\in Q_{1} \\
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| T_{2}(q,x) & \mbox{if} & q\in Q_{2} \\
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| \{q_{1}, q_{2}\} & \mbox{if} & q = q_{0}\ and\ x =\epsilon\\
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| \emptyset & \mbox{if} & q = q_{0}\ and\ x\neq\epsilon
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| \end{array}\right.
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| </math> | |
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| In the following, we shall use <math>p\stackrel{x,T}{\rightarrow}q</math> to denote <math>q\in E(T(p,x))</math>
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| Let <math>w</math> be a string from <math>L_{1}\cup L_{2}</math>. [[Without loss of generality]] assume <math>w\in L_{1}</math>.
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| Let <math>w = x_{1}x_{2}\cdots x_{m}</math> where <math>m\geq 0, x_{i}\in\Sigma</math>
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| Since <math>N_{1}</math> accepts <math>x_{1}x_{2}\cdots x_{m}</math>, there exist <math>r_{0}, r_{1},\cdots r_{m}\in Q_{1}</math> such that
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|
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| ::<math> q_{1}\stackrel{\epsilon , T_{1}}{\rightarrow}r_{0}\stackrel{x_{1} , T_{1}}{\rightarrow}r_{1}\stackrel{x_{2} , T_{1}}{\rightarrow}r_{2}\cdots r_{m-1}\stackrel{x_{m} , T_{1}}{\rightarrow}r_{m}, r_{m}\in A_{1}
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| </math>
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| Since <math>T_{1}(q,x) = T(q,x)\ \forall q\in Q_{1}\forall x\in\Sigma</math>
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| :: <math>r_{0}\in E(T_{1}(q_{1},\epsilon ))\Rightarrow r_{0}\in E(T(q_{1},\epsilon ))</math> | |
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| :: <math>r_{1}\in E(T_{1}(r_{0},x_{1} ))\Rightarrow r_{1}\in E(T(r_{0},x_{1} ))</math>
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| :::: <math>\vdots</math>
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| :: <math>r_{m}\in E(T_{1}(r_{m-1},x_{m} ))\Rightarrow r_{m}\in E(T(r_{m-1},x_{m} ))</math> | |
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| We can therefore substitute <math>T</math> for <math>T_{1}</math> and rewrite the above path as
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| <math>q_{1}\stackrel{\epsilon , T}{\rightarrow}r_{0}\stackrel{x_{1} , T}{\rightarrow}r_{1}\stackrel{x_{2} , T}{\rightarrow}r_{2}\cdots r_{m-1}\stackrel{x_{m} , T}{\rightarrow}r_{m}, r_{m}\in A_{1}\cup A_{2}, r_{0}, r_{1},\cdots r_{m}\in Q</math>
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| Furthermore,
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| :: <math>
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| \begin{array}{lcl}
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| T(q_{0}, \epsilon) = \{q_{1}, q_{2}\} & \Rightarrow & q_{1}\in T(q_{0}, \epsilon)\\
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| \\ & \Rightarrow & q_{1}\in E(T(q_{0}, \epsilon))\\
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| \\ & \Rightarrow & q_{0}\stackrel{\epsilon , T}{\rightarrow}q_{1}
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| \end{array}
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| </math>
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| and | |
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| :: <math>q_{0}\stackrel{\epsilon , T}{\rightarrow}q_{1}\stackrel{\epsilon , T}{\rightarrow}r_{0}\Rightarrow q_{0}\stackrel{\epsilon , T}{\rightarrow}r_{0}
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| </math>
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| The above path can be rewritten as
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| :<math>q_{0}\stackrel{\epsilon , T}{\rightarrow}r_{0}\stackrel{x_{1} , T}{\rightarrow}r_{1}\stackrel{x_{2} , T}{\rightarrow}r_{2}\cdots r_{m-1}\stackrel{x_{m} , T}{\rightarrow}r_{m}, r_{m}\in A_{1}\cup A_{2}, r_{0}, r_{1},\cdots r_{m}\in Q
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| </math>
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| Therefore, <math>N</math> accepts <math>x_{1}x_{2}\cdots x_{m}</math> and the proof is complete.
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| '''Note:''' The idea drawn from this mathematical proof for constructing a machine to recognize <math>L_{1}\cup L_{2}</math> is to create an initial state and connect it to the initial states of <math>L_{1}</math> and <math>L_{2}</math> using <math>\epsilon</math> arrows.
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| == References ==
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| * Michael Sipser, ''Introduction to the Theory of Computation'' ISBN 0-534-94728-X. ''(See . Theorem 1.22, section 1.2, pg. 59.)''
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| [[Category:Article proofs]]
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| [[Category:Formal languages]]
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| [[Category:Automata theory]]
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