Proof

• \(L(G) \subseteq L(D)\):
  • For any \(x \in L(G)\), there is a derivation \[ S \yields u_1 \yields u_2 \yields \ldots \yields u_n \yields x \]
  • Due to \(G\)’s special production rule structure,
    each \(u_i = x_i R_i\), where \(x_i \sqsubseteq x\) is a prefix of \(x\).
  • The state \(R_i\) is the state of \(D\) after reading \(x_i\).
  • State \(R_n\) must be an accept state for the last nonterminal to be erased via \(R_n \to \emptystring\) in \(u_n = x R_n\).
  • Thus the computation sequence \(S, R_1, R_2, \ldots, R_n\) accepts \(x\), showing \(L(G) \subseteq L(D)\).

Proof

• \(L(D) \subseteq L(G)\):
  • For any \(x \in L(D)\), there is an accepting
    computation sequence (i.e., \(R_n \in F\)): \(\quad S, R_1, R_2, \dots, R_n\)
  • This can be translated into a derivation of \(x\) by applying the corresponding production rules in order: \[\begin{equation*} \fragment{ \begin{aligned} S &\to x[1] R_1 \\ R_1 &\to x[2] R_2 \\ R_2 &\to x[3] R_3 \\ &\cdots \\ R_{n-1} &\to x[n] R_n \\ R_{n} &\to \emptystring \\ \end{aligned} \quad \quad \quad \begin{aligned} S & \yields x[1] R_1 \\& \yields x[1] x[2] R_2 \\& \yields x[1] x[2] x[3] R_3 \\& \yields \ldots \\& \yields x[1] x[2] \cdots x[n] R_n \\& \yields x[1] x[2] \cdots x[n] = x \end{aligned} } \end{equation*}\]

• Left Regular Grammars are also equivalent in power to DFAs/NFAs.
• We could prove this using similar arguments to the RRG case.
• Or we can notice:
  • Consider converting an RRG \(G\) to an LRG \(G'\) by replacing each rule \(X \to aY\) with \(X \to Ya\).
  • What is the relationship between \(L(G)\) and \(L(G')\)? \(\quad L(G') = \reverse{L(G)} = \setbuild{\reverse{w}}{w \in L(G)}\)
  • NFA-decidable languages are closed under reversal! (exercise at home)
  • Thus \(L(G')\) is NFA-decidable if and only if \(L(G)\) is.

• \(\stackrel{\Large \nwarrow}{\large \leftarrow}\) base cases
• 
• Note that this is equivalent to the NFA

• The final simplification step operates on an EA that looks like:

  

• We can rip state “i” out of the diagram but still accept strings whose computation sequences go through \(i\), by changing the transition between \(s\) and \(a\):

  

• Both EAs accept exactly strings of the form \(w \in L(W)\), or \(x y_1 y_2 \dots y_k z\) for some \(k \in \N\) where \(x \in L(X)\), each \(y_i \in L(Y)\), and \(z \in L(Z)\).

  • In general, some of the transitions may not exist, e.g., if there is no \(s \stackrel{W}{\to} a\) transition, use \(X Y^* Z\); if no self-loop on \(i\), use \(W \cup XZ\); if neither, then use \(XZ\).

When more than three states remain:

• Select any state other than \(s\) or \(a\) and call it \(i\).
• Iterate over pairs of states \(q\) and \(r\) with transitions \(q \stackrel{X}{\to} i\) and \(i \stackrel{Z}{\to} r\) (even if \(q = r\))
• For each of these pairs, transform:
  into:
• Repeat until only 2 states remain!
• More precisely, iterate over pairs of transitions in/out of \(i\) (more on this in a bit).

• How to get from \(q\) to \(t\) going through \(i\)? \(0^*1\) (go from \(q\) to \(i\)) \((00)^*\) (loop on \(i\)) \(1^+\) (go from \(i\) to \(t\))
  • add new transition: \(q \stackrel{0^*1 (00)^* 1^+}{\to} t\)
• How to get from \(q\) to \(r\) going through \(i\)? \(\fragment{0^*1} \fragment{(00)^*} \fragment{\emptystring} \fragment{= 0^*1 (00)^*}\)
  • replace existing \(q \stackrel{110}{\to} r\) transition with: \(q \stackrel{110\ \ \cup\ \ 0^*1(00)^*}{\to} r\)
• How to get from \(t\) to \(r\) going through \(i\)? \(\fragment{01 (00)^* \emptystring = 01 (00)^*}\)
  • add new transition: \(t \stackrel{01(00)^*}{\to} r\)
• How to get from \(t\) to \(t\) going through \(i\)? \(\fragment{01 (00)^* 1^+}\)
  • add new self-transition: \(t \stackrel{01(00)^*1^+}{\to} t\)

• More generally, when removing a state \(i\), we must consider all pairs of transitions in/out of \(i\), along with the states they come from/go to.
• We often write two transitions with the same from/to states with a single arrow, for example, \(q \stackrel{0,1}{\to} i\)… Technically these are two different transitions: \(q \stackrel{0}{\to} i\) and \(q \stackrel{1}{\to} i\).
• Alternatively (and more conveniently when transforming an NFA to a regex by hand), one could think of the two NFA transitions represented by \(q \stackrel{0,1}{\to} i\) as a single EA transition \(q \stackrel{0 \cup 1}{\to} i\).