Let $A=(S_{1},\Sigma_{1},\delta_{1},I_{1},F_{1})$ and $B=(S_{2},\Sigma_{2},\delta_{2},I_{2},F_{2})$ be two automata. We define the juxtaposition of $A$ and $B$, written $AB$, as the sextuple $(S,\Sigma,\delta,I,F,\epsilon)$, as follows:

1. 1.

   $S:=S_{1}\stackrel{\cdot}{\cup}S_{1}$, where $\stackrel{\cdot}{\cup}$ denotes disjoint union,

2. 2.

   $\Sigma:=(\Sigma_{1}\cup\Sigma_{2})\stackrel{\cdot}{\cup}\{\epsilon\}$,

3. 3.

   $\delta:S\times\Sigma\to P(S)$ given by

   • $\delta(s,\epsilon):=I_{2}$ if $s\in F_{1}$, and $\delta(s,\epsilon):=\{s\}$ otherwise,

   • $\delta|(S_{1}\times\Sigma_{1}):=\delta_{1}$,

   • $\delta|(S_{2}\times\Sigma_{2}):=\delta_{2}$, and

   • $\delta(s,\alpha):=\varnothing$ otherwise (where $\alpha\neq\epsilon$).

4. 4.

   $I:=I_{1}$,

5. 5.

   $F:=F_{2}$.

Because $S_{1}$ and $S_{2}$ are considered as disjoint subsets of $S$, $I\cap F=\varnothing$. Also, from the definition above, we see that $AB$ is an automaton with $\epsilon$-transitions.

The way $AB$ works is as follows: a word $c=ab$, where $a\in\Sigma_{1}^{*}$ and $b\in\Sigma_{2}^{*}$, is fed into $AB$. $AB$ first reads $a$ as if it were read by $A$, via transition function $\delta_{1}$. If $a$ is accepted by $A$, then one of its accepting states will be used as the initial state for $B$ when it reads $b$. The word $c$ is accepted by $AB$ when $b$ is accepted by $B$.

Visually, the state diagram $G_{{A_{1}A_{2}}}$ of $A_{1}A_{2}$ combines the state diagram $G_{{A_{1}}}$ of $A_{1}$ with the state diagram $G_{{A_{2}}}$ of $A_{2}$ by adding an edge from each final node of $A_{1}$ to each of the start nodes of $A_{2}$ with label $\epsilon$ (the $\epsilon$-transition).