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The canonical residual automaton of a regular language has its distinct left quotients of a formal language as states, initial state $L$, accepting states those containing the empty word, and letter transition $R\mapsto a^{-1}R$. It is a complete minimal deterministic finite automaton; every other complete minimal deterministic finite automaton for that language is isomorphic to it.

An irreducible deterministic automaton has the fewest states among deterministic automata for its language and is unique up to isomorphism. Its states are the Myhill--Nerode classes.

The minimal deterministic finite automaton is unique up to an isomorphism preserving the initial state, transitions and accepting states. We use complete deterministic finite automata over the fixed alphabet; the PDF's abbreviation FDA has this meaning. State names themselves cannot be unique.

For words $u,v\in {\Sigma }^{\ast }$, introduce the Myhill-Nerode equivalenceIt is an equivalence relation, and appending the same letter to equivalent words preserves it: a suffix $w$ after $ua$ is the suffix $aw$ after $u$. Since $L$ is a regular language, take any recognizing deterministic finite automaton. Words reaching the same state are equivalent, since every further suffix gives the same computation. Thus ${\sim }_L$ has finite index.

The canonical residual automaton has one state $[u]$ for each class, initial state $[ϵ]$, transition $[u]\stackrel{a}{\to }[ua]$, and accepting states those with $u\in L$. These choices are well-defined by Myhill-Nerode equivalence. Induction on the input length shows that reading $u$ reaches $[u]$, so the deterministic finite automaton recognizes $L$, and all its states are accessible states of a deterministic finite automaton. The same state can be described by the left quotient of a formal languageTwo states are different precisely when some suffix distinguishes their acceptance behavior.

Every recognizing deterministic finite automaton has at least as many states as there are classes: pick a representative from each class; two different representatives cannot reach the same state. The canonical residual automaton attains this bound, and therefore is a minimal deterministic finite automaton.

For uniqueness, let $\mathcal{A}$ be any minimal deterministic finite automaton. All its states are accessible, since removing inaccessible states leaves a complete recognizing deterministic finite automaton with fewer states. Map a state reached by $u$ to $[u]$. This is well-defined because two words reaching the same state are equivalent. It is surjective because every class has a representative. Both sets have the minimal number of states, so it is a bijection. It preserves the initial state and transitions by construction, and preserves accepting states by taking the empty word as the suffix. This is the required isomorphism with the canonical residual automaton, provingThe empty word is included throughout, and empty or universal regular languages have the corresponding one-state complete deterministic finite automata.

Suppose a path took $p$ to a state $q$. Part (i) implies that no accept state is reachable from either $p$ or $q$. Thus every possible continuation is rejected from both states, so $p$ and $q$ are indistinguishable states of a deterministic finite automaton. Distinct indistinguishable states cannot occur in a minimal deterministic finite automaton. Therefore $q=p$, and there is no path from $p$ to any other state.

Equivalently, a cycle state in a minimal deterministic finite automaton for a finite language is the unique rejecting sink.

Let $\hat{\delta }$ be the extended transition function of a deterministic finite automaton. States $p,q\in Q$ are equivalent, or indistinguishable, whenThe quotient deterministic finite automaton by indistinguishable states $D/\sim$ has state set $Q/\sim$, initial state $[q_0]$, accepting states $\{[q]:q\in F\}$, and transitionRight invariance of $\sim$ makes this well-defined, and the quotient accepts $L(D)$.

To prove minimality, choose for each state $q$ of the accessible automaton $D$ a word $w_q$ with ${\hat{\delta }}_D(q_0,w_q)=q$. Run the same word in any DFA $A$ accepting $L(D)$, and assign the class $[q]$ the reached state of $A$. If two distinct classes $[p]\ne [q]$ reached the same state of $A$, every continuation $u$ would have the same acceptance result after $w_p$ and $w_q$. Since $A$ and $D$ accept the same language, this would make $p$ and $q$ indistinguishable in $D$, a contradiction. Thus $A$ has at least one distinct state for every class in $D/\sim$, proving that the quotient is the minimal deterministic finite automaton.

For divisibility by seven, take states $Q=\{0,1,\dots ,6\}$, initial state $0$, accepting set $F=\{0\}$, and transitionsAfter reading a word, the state is its binary value modulo seven, so this binary divisibility automaton accepts exactly the multiples of seven, with leading zeros allowed. Every state is accessible: the three-bit words representing $0,1,\dots ,6$ reach the corresponding residues.

For distinct residues $p,q$, let $w_p$ be the three-bit representation of $-p(\mathrm{mod}7)$, chosen in $\{0,\dots ,6\}$. Since $2^3\equiv 1(\mathrm{mod}7)$, reading $w_p$ from state $p$ ends atwhereas reading it from $q$ ends at $q-p\ne 0$. Thus $w_p$ distinguishes $p$ from $q$. All seven states are pairwise distinguishable, so the DFA is minimal by the Myhill-Nerode theorem.

The Knaster-Tarski theorem states that the fixed points of a monotone map $f:L\to L$ on a complete lattice form a complete lattice. LetFor every $x\in P$, monotonicity gives $f(m)\le f(x)\le x$, hence $f(m)\le m$. Applying $f$ once more gives $f(f(m))\le f(m)$, so $f(m)\in P$. The definition of $m$ then gives $m\le f(m)$, and therefore $f(m)=m$. Thusis the least fixed point. The order-dual argument shows thatis the greatest fixed point.

More generally, for a family $S$ of fixed points, let $P_S$ be the set of prefixed points $x$ satisfying $f(x)\le x$ and $s\le x$ for every $s\in S$. Its meet $m_S$ is again prefixed. Since $s=f(s)\le f(m_S)$ for every $s\in S$, the element $f(m_S)$ also lies in $P_S$, so the preceding argument gives $f(m_S)=m_S$. It is the join of $S$ within the fixed-point order. The dual construction gives the meet of $S$ within that order, proving completeness.

For the Myhill-Nerode theorem, let $A\subseteq {\Sigma }^{\ast }$ and define the Myhill-Nerode equivalenceThis is an equivalence relation and a right congruence: $u{\sim }_{A}v$ implies $ua{\sim }_{A}va$ for every letter $a$.

If a deterministic finite automaton accepts $A$, any two words reaching the same state are equivalent, because every continuation has the same subsequent run. Hence ${\sim }_A$ has at most as many classes as the automaton has states. Conversely, if ${\sim }_A$ has finitely many classes, define an automaton with state set ${\Sigma }^{\ast }/{\sim }_A$, initial state $[ϵ]$, transitionand accepting states $[u]$ with $u\in A$. Right congruence makes the transition well defined, and induction on word length shows that the state reached by $u$ is $[u]$, so the automaton accepts exactly $A$. Therefore $A$ is a regular language exactly when ${\sim }_A$ has finite index. Moreover, every automaton for $A$ has at least one state for each equivalence class, so this quotient is the minimal deterministic finite automaton.

For $0\le j\le 9$, consider the ten prefixesIf $0\le i<j\le 9$, appendThenbecause $i+11-j\le 10$, whereasIndeed, in a word ending in one $a$, the prefix before the final positive run of $a$'s must have length at most ten. Hence every pair $u_i,u_j$ is distinguished by some suffix. They occupy ten distinct Myhill-Nerode equivalence classes, so the minimal deterministic finite automaton for $L$ has at least ten states.