SLD resolution

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.

Theorem

For any regular languages $L_{1}$ and $L_{2}$, language $L_{1}\cup L_{2}$ is regular.

Proof

Since $L_{1}$ and $L_{2}$ are regular, there exist NFAs $N_{1},\ N_{2}$ that recognize $L_{1}$ and $L_{2}$.

Let

$N_{1}=(Q_{1},\ \Sigma ,\ T_{1},\ q_{1},\ A_{1})$

$N_{2}=(Q_{2},\ \Sigma ,\ T_{2},\ q_{2},\ A_{2})$

Construct

$N=(Q,\ \Sigma ,\ T,\ q_{0},\ A_{1}\cup A_{2})$

where

$Q=Q_{1}\cup Q_{2}\cup \{q_{0}\}$

$T(q,x)=\left\{{\begin{array}{lll}T_{1}(q,x)&{\mbox{if}}&q\in Q_{1}\\T_{2}(q,x)&{\mbox{if}}&q\in Q_{2}\\\{q_{1},q_{2}\}&{\mbox{if}}&q=q_{0}\ and\ x=\epsilon \\\emptyset &{\mbox{if}}&q=q_{0}\ and\ x\neq \epsilon \end{array}}\right.$

In the following, we shall use $p{\stackrel {x,T}{\rightarrow }}q$ to denote $q\in E(T(p,x))$

Let $w$ be a string from $L_{1}\cup L_{2}$. Without loss of generality assume $w\in L_{1}$.

Let $w=x_{1}x_{2}\cdots x_{m}$ where $m\geq 0,x_{i}\in \Sigma$

Since $N_{1}$ accepts $x_{1}x_{2}\cdots x_{m}$, there exist $r_{0},r_{1},\cdots r_{m}\in Q_{1}$ such that

$q_{1}{\stackrel {\epsilon ,T_{1}}{\rightarrow }}r_{0}{\stackrel {x_{1},T_{1}}{\rightarrow }}r_{1}{\stackrel {x_{2},T_{1}}{\rightarrow }}r_{2}\cdots r_{m-1}{\stackrel {x_{m},T_{1}}{\rightarrow }}r_{m},r_{m}\in A_{1}$

Since $T_{1}(q,x)=T(q,x)\ \forall q\in Q_{1}\forall x\in \Sigma$

$r_{0}\in E(T_{1}(q_{1},\epsilon ))\Rightarrow r_{0}\in E(T(q_{1},\epsilon ))$

$r_{1}\in E(T_{1}(r_{0},x_{1}))\Rightarrow r_{1}\in E(T(r_{0},x_{1}))$

$\vdots$

$r_{m}\in E(T_{1}(r_{m-1},x_{m}))\Rightarrow r_{m}\in E(T(r_{m-1},x_{m}))$

We can therefore substitute $T$ for $T_{1}$ and rewrite the above path as

$q_{1}{\stackrel {\epsilon ,T}{\rightarrow }}r_{0}{\stackrel {x_{1},T}{\rightarrow }}r_{1}{\stackrel {x_{2},T}{\rightarrow }}r_{2}\cdots r_{m-1}{\stackrel {x_{m},T}{\rightarrow }}r_{m},r_{m}\in A_{1}\cup A_{2},r_{0},r_{1},\cdots r_{m}\in Q$

Furthermore,

${\begin{array}{lcl}T(q_{0},\epsilon )=\{q_{1},q_{2}\}&\Rightarrow &q_{1}\in T(q_{0},\epsilon )\\\\&\Rightarrow &q_{1}\in E(T(q_{0},\epsilon ))\\\\&\Rightarrow &q_{0}{\stackrel {\epsilon ,T}{\rightarrow }}q_{1}\end{array}}$

and

$q_{0}{\stackrel {\epsilon ,T}{\rightarrow }}q_{1}{\stackrel {\epsilon ,T}{\rightarrow }}r_{0}\Rightarrow q_{0}{\stackrel {\epsilon ,T}{\rightarrow }}r_{0}$

The above path can be rewritten as

$q_{0}{\stackrel {\epsilon ,T}{\rightarrow }}r_{0}{\stackrel {x_{1},T}{\rightarrow }}r_{1}{\stackrel {x_{2},T}{\rightarrow }}r_{2}\cdots r_{m-1}{\stackrel {x_{m},T}{\rightarrow }}r_{m},r_{m}\in A_{1}\cup A_{2},r_{0},r_{1},\cdots r_{m}\in Q$

Therefore, $N$ accepts $x_{1}x_{2}\cdots x_{m}$ and the proof is complete.

Note: The idea drawn from this mathematical proof for constructing a machine to recognize $L_{1}\cup L_{2}$ is to create an initial state and connect it to the initial states of $L_{1}$ and $L_{2}$ using $\epsilon$ arrows.

References

• Michael Sipser, Introduction to the Theory of Computation ISBN 0-534-94728-X. (See . Theorem 1.22, section 1.2, pg. 59.)