# Strongly connected and completely specified Moore equivalent of a Mealy Machine

Problem: Prove that if a Mealy machine is strongly connected and completely specified, the corresponding Moore machine will also be strongly connected and completely specified.

My approach so far:

As the Mealy Machine is completely specified, we have 2 cases for each state:

1. All transitions coming into this state have the same output.

2. At least 2 transitions coming into this state have different outputs.

For case 1: The equivalent for this state would simply be changing the notation, i.e. mark all transitions into this state simply by the input symbols and just mention the output with the state as is.

Now, as the machine is completely specified, each such input transition and it's corresponding output will be defined, and the states resulting due to such states will also have completely specified transitions.

For case 2: The equivalent for such a state requires the 'splitting'/addition of new states such that each 'split' state corresponds to the different outputs that can be generated. We split a state into a 2 corresponding states.

Now, as the machine is completely specified, each such input transition and it's corresponding output will be defined, and the states resulting due to such states will also have completely specified transitions.

Strongly Connected Machine: If for any two states $S_i, S_j$ of the machine, there exists a finite sequence of input characters that can make the machine transition from $S_i$ to $S_j$, the machine is strongly connected.

Completely Specified Machine: Consider $M = (I,O,S,F,Z)$

• $I:$ Input Symbols
• $O:$ Output Symbols
• $S:$ Set of States
• $F:$ Transition function $F: S\times I \rightarrow S$
• $Z:$ Output function
• $Z: S \rightarrow O$ (Moore)
• $Z:$ $S\times I \rightarrow O$ (Mealy)

Now, if for every $(p,x)$ such that $(p \in S, x \in I)$ there exist some $q \in S$ and $y \in O$ such that $F(p,x) = q$ and $Z(p,x) = y$ (Mealy), $Z(p) = y$ (Moore)

My Question:

I'm having a hard time trying to formally prove the above while I can see it is true with the above intuition. I can similarly say that if the mealy machine was strongly connected, even after replacing it with 1(or 2) states, it would most definitely still be strongly connected. I was looking for help on formalising this above approach or a more solid proof on this.

## 1 Answer

Let $M_1 = \langle Q, \Sigma, \Delta, \delta, \lambda, q_0 \rangle$ be a Mealy machine where $\Delta$ is the output alphabet and $\lambda$ is the output function.

Let us denote states of the corresponding Moore machine by $q_{ib}$ if there is input $a$ and output $b$ such that $\delta(q_j, a) = (q_i, b)$. This means if the Mealy machine enters a state $q$ by emitting more than two different outputs then the state $q_i$ "splits". Also, denote corresponding Moore machine's transition function by $\delta'$.

Now consider two states $q_{ib_1}$ and $q_{jb_2}$ (of corresponding Moore machine). Existence of the state $q_{jb_2}$ implies that there is a state $q_k$ and input symbol $a$ such that $\delta(q_k, a) = (q_j, b_2)$. Since we made assumption that the Mealy machine is strongly connected, there is a string $w$ such that $\delta(q_i, w) = (q_k, b_k)$ for some output symbol $b_k$. So, we have $$\delta'(q_{ib_1}, wa) = \delta'(\delta'(q_{ib_1}, w),a) = \delta'(q_{kb_k}, a) = q_{jb_2}$$ This shows that the corresponding Moore machine is also strongly connected.

As for the completely specification, let $q_{ib_1}$ be a state of the Moore machine, and $a$ is an input symbol. Then $\delta'(q_{ib_1}, a)= (q_{jb_2}, b_2)$, where $\delta(q_i, a) = (q_j, b_2)$, which shows it is completely specified.