Previous Exercise:
This is a learning note of a course in CISPA, UdS. Taught by Bernd Finkbeiner
This is an example exercise to express LTL formula into Alternating Büchi Automata. For further definitions, you may check Section 7.2.
Question
Use the construction from the lecture to construct an alternating Büchi automaton $\mathcal{A}$ such that $$\mathcal{L(A)=L}((\Diamond p)\ \mathcal{U}\ (\square q))$$
Solution
we build the following alternating Büchi automaton $\mathcal{A} = (2^{p,q},Q,\varphi,\delta,\small\text{BÜCHI} \normalsize(F))$ which recognizes the models of $\varphi=(\Diamond p)\ \mathcal{U}\ (\square q)$. First we consider the transition function $\delta$ for an arbitrary symbol $a\in2^{p,q}:$
$\begin{array}{lll}
\hspace{1cm}\bullet &&\delta(\square q,a)\newline
&=&\delta(\neg(\textit{true}\ \mathcal{U}\ \neg q) ,a)\newline
&=&\overline{\delta(\textit{true}\ \mathcal{U}\ \neg q ,a)}\newline
&=&\overline{\delta(\neg q,a)\vee(\delta(\textit{true},a)\wedge(\textit{true}\ \mathcal{U}\ \neg q))}\newline
&=&\overline{\delta(\neg q,a)}\wedge\overline{(\delta(\textit{true},a)\wedge (\textit{true}\ \mathcal{U}\ \neg q))}&(\textit{true}\wedge \psi=\psi)\newline
&=&\overline{\delta(\neg q,a)}\wedge\overline{(\textit{true}\ \mathcal{U}\ \neg q)}&(\text{Using line 1})\newline
&=&\overline{\delta(\neg q,a)}\wedge\square q\newline
&=&\left\lbrace \begin{array}{lll}
\square q&\text{if }q\in a\newline
\textit{false}&\text{if }q\notin a\newline
\end{array}\right.
\end{array}
\ \newline \ \newline
\begin{array}{lll}
\hspace{1cm}\bullet &&\delta(\Diamond p,a)\newline
&=&\delta(\textit{true}\ \mathcal{U}\ p ,a)\newline
&=&\delta(p,a)\vee(\delta(\textit{true},a)\wedge\Diamond p)&(\textit{true}\wedge \psi=\psi)\newline
&=&\delta(p,a)\vee\Diamond p\newline
&=&\left\lbrace \begin{array}{lll}
\Diamond p&\text{if }p\notin a\newline
\textit{true}&\text{if }p\in a\newline
\end{array}\right.
\end{array}
$
By Substitution above result into $\delta((\Diamond p)\ \mathcal{U}\ (\square q),a)=\delta(\square q,a)\vee(\delta(\Diamond p,a)\wedge(\Diamond p)\ \mathcal{U}\ (\square q))$, we have:
$$\delta((\Diamond p)\ \mathcal{U}\ (\square q),a)=\left\lbrace
\begin{array}{lll}
\Diamond p\wedge((\Diamond p)\ \mathcal{U}\ (\square q))&\text{if }a =\varnothing &(\vee\ \textit{false}\text{ is omitted.})\newline
(\Diamond p)\ \mathcal{U}\ (\square q)&\text{if }a =\lbrace p\rbrace&(\wedge\ \textit{true}\text{ is omitted.})\newline
\square q\vee(\Diamond p\wedge((\Diamond p)\ \mathcal{U}\ (\square q)))&\text{if }a =\lbrace q\rbrace\newline
\square q\vee((\Diamond p)\ \mathcal{U}\ (\square q))&\text{if }a =\lbrace p,q\rbrace&(\wedge\ \textit{true}\text{ is omitted.})\newline
\end{array}\right.$$