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An abstract family of languages is a family closed under specified operations, including union, concatenation, iteration, appropriate homomorphisms, inverse homomorphisms and intersection with regular languages. These axioms organize closure properties of formal languages.

An automatic group admits an automatic structure for a group: a regular language of representatives and synchronous group multiplier automata recognizing right multiplication by each generator and equality of representatives. Representatives need not be unique. This finite-state description yields effective algorithms for the word problem for a group.

An automatic structure for a group consists of a finite generating alphabet $A$, a surjective representative regular language $L\subseteq A^{\ast }$, and finite-state automata accepting the padded convolutions of words $(u,v)$ in $L$ satisfying $\overline{u}x=\overline{v}$ for each $x\in A\cup \{\epsilon \}$. Symmetry of $A$ and uniqueness of representatives are additional conditions, not part of this definition.

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.

Choose a finite symmetric generating alphabet $\Sigma$ and one word $\sigma (g)$ representing each group element, with $\sigma (1)$ the empty word. The associated paths in the Cayley graph start at the identity and traverse one edge per unit time, remaining at their endpoint after completion. A synchronous combing has a constant $K$ such that paths to adjacent endpoints stay within $K$ at equal times. An asynchronous version permits increasing reparametrizations. The formal language need not be a regular language; regular synchronous representative languages define automatic groups.

A formal language with finitely many words. Every finite language is a regular language: take a finite union of the regular expressions spelling its members, with the empty language represented separately when there are no members.

A finite-state machine processes inputs using a finite set of automaton states and input-dependent automaton transitions. An automaton state records everything the machine remembers; an automaton transition says how that memory changes on the next symbol. A finite-state machine may also emit outputs. Here the emphasis is on recognizing a formal language: start in a designated automaton state, read a word from an alphabet, and accept exactly when the final automaton state is designated accepting.

Finite memory does not impose a maximum input length. A finite-state machine can read arbitrarily long words, revisiting its automaton states indefinitely. The restriction is that the part already read must be compressed into one of finitely many summaries. For divisibility by three, a word of binary digits can describe an enormous integer, yet its congruence class modulo three is all that the machine needs to retain. Two already-read words giving the same automaton state behave identically under every subsequent continuation in a DFA.

This makes finite-state machines useful whenever past details can be forgotten safely: checking whether a sequence of commands follows an allowed order, or recognizing a pattern with a regular expression. The beautiful design question is which distinctions about the past matter to the future. DFA minimization makes that question precise by merging indistinguishable states of a deterministic finite automaton. Both DFAs and NFAs recognize exactly the regular languages.

Regular languages are closed under finite intersection, but not countably infinite intersection. Indeed, for any formal language $K\subseteq {\Sigma }^{\ast }$ over a finite alphabet, let $L_i$ contain every word of length at least $i$, together with the members of $K$ of length less than $i$. Each $L_i$ is a regular language, since its exceptional short part is a finite language, and ${\bigcap }_{i\ge 1}{L}_i=K$. Thus every formal language, including one that is not a regular language, can arise in this way.

This theorem characterizes a regular language by regular expressions and acceptance by a finite-state automaton. Kleene's theorem states that a language is denoted by a regular expression exactly when it is accepted by a finite automaton.

For a word $u$ and a formal language $L$, the left quotient $u^{-1}L=\{w:uw\in L\}$ is the possible continuation language after the prefix $u$. Two prefixes have the same left quotient exactly when they are Myhill-Nerode equivalent. For a regular language only finitely many distinct left quotients occur.

Exactly the regular languages. A read-only-from-left Turing acceptor has only its current control state as usable memory: it never changes a tape symbol, and its head never returns to an earlier cell. This is the behaviour of a deterministic finite automaton while it reads the input.

To make the correspondence precise, retain the control states and the transitions on the input alphabet. Replace accepting and rejecting halts by absorbing states, so that halting before the end of a word over an alphabet has the same effect on every continuation. For each remaining state $q$, follow the transitions on blank symbols and mark $q$ accepting in the deterministic finite automaton exactly when this blank-only continuation eventually accepts. This marking is a finite calculation: either a halting state is reached or a state repeats, after which the blank-only run loops. The resulting deterministic finite automaton accepts exactly the same formal language as the read-only-from-left Turing acceptor. If the latter is a decider, the blank-only run from every reachable end-of-input state halts.

Conversely, simulate any deterministic finite automaton by a read-only-from-left Turing acceptor, using its state transitions while reading the input. At the first blank symbol, halt and accept or reject according to the current deterministic finite automaton state. This machine halts on every input and decides the required regular language.

Let $A=(Q,\{0,1\},\delta ,q_0,F)$ be a deterministic finite automaton recognizing $L$. Construct a nondeterministic finite automaton $B$ with automaton state set $Q\times \{0,1\}$. Its second coordinate records whether the extra zero has already been consumed. Start at $(q_0,0)$ and make exactly the automaton states in $F\times \{1\}$ accepting.

For each $q\in Q$, $i\in \{0,1\}$, and $a\in \{0,1\}$, include the ordinary automaton transitionAlso include the extra automaton transitionThis consumes the inserted zero without changing the simulated automaton state of $A$. At a zero in layer zero, both the ordinary and extra choices are available. There are no automaton transitions from layer one back to layer zero.

If $xy\in L$, run $A$ on $x$ in layer zero, consume the extra zero by the displayed automaton transition, and run $A$ on $y$ in layer one. The resulting run accepts $x0y$ in $B$.

Conversely, every accepting run of $B$ starts in layer zero and finishes in layer one, so it uses exactly one extra automaton transition. Split its input as $x0y$ at that move. Removing that consumed zero leaves precisely a run of $A$ on $xy$, ending in $F$. Hence $xy\in L$. The construction also permits an insertion at either end, including insertion into the empty word.

Therefore $L(B)=L^{\prime }$. Since $B$ has $2|Q|$ automaton states, $L^{\prime }$ is a regular language. This is a single-symbol insertion into a regular language.

No: regular languages are not closed under countably infinite intersection. Let $\Sigma =\{0,1\}$ and putThe first term in $L_i$ has regular expression $(0\cup 1{)}^i(0\cup 1{)}^{\ast }$, where the fixed power $i$ means $i$ copies. The second term is a finite language, a finite union of single words. Their union is therefore a regular language.

If $w\in K$, then for each $i$ either $|w|\ge i$ or $w$ belongs to the second term, so $w\in L_i$. If $w\notin K$, choose $i>|w|$. It belongs to neither term of $L_i$. ConsequentlyThe formal language $K$ is not a regular language. Indeed, for a proposed pumping length $p\ge 1$, the pumping lemma for regular languages applied to $0^p{1}^p$ must pump a nonempty block consisting only of zeros. Repeating that block increases the number of zeros without increasing the number of ones, leaving $K$. This contradiction proves the counterexample. Finite intersection closure does not extend to this infinite intersection of regular languages.

Every generated word satisfieswhere ${\#}_a,{\#}_b$ count occurrences of the indicated terminal symbols. To see this, follow the unique chain of $S$ occurrences in a parse tree. Each use of $S\to UST$ introduces one $a$ from $U$, at least one $b$ from its rightmost $T$, and possibly one further $b$ from $U\to aT$. Each use of $S\to ST$ adds one $b$ and no $a$. The terminating $S\to \epsilon$ contributes neither. Hence no generated word has more $a$ symbols than $b$ symbols.

Also, for every $n\ge 0$, the word $a^n{b}^n$ is generated: apply $S\to UST$ exactly $n$ times, choose $U\to a$ each time, then use $S\to \epsilon$ and all remaining $T\to b$.

Suppose $L(G)$ were a regular language, and let $p\ge 1$ be a pumping length from the pumping lemma for regular languages. Apply it to $w=a^p{b}^p$. Every allowed decomposition $w=xyz$ with $|xy|\le p$ and $|y|>0$ has $y=a^r$ for some $r\ge 1$. Pumping twice giveswhich violates the proved counting inequality and hence is not generated. This contradicts the pumping lemma for regular languages. Therefore $L(G)$ is not a regular language. Ambiguity was not used: the obstruction is the unbounded constraint on symbol counts.

The formal language is not a regular language. Let $K$ denote it and consider the infinitely many prefixesFor any $i\ne j$, append the word $z_i=(bc{)}^i$. Then $u_i{z}_i\in K$, whereas $u_j{z}_i\notin K$: the latter word has $j$ occurrences of $a$ and $i$ occurrences of $c$, and every word of $K$ has equal numbers of those two symbols.

Thus $u_i$ and $u_j$ are not Myhill-Nerode equivalent. There are infinitely many Myhill-Nerode equivalence classes, so the Myhill-Nerode theorem rules out a finite-state automaton recognizing $K$. Equivalently, if two such prefixes left a deterministic finite automaton in the same automaton state, appending $z_i$ could not distinguish their acceptance, giving the same contradiction.

The formal language is a regular language. A regular expression for it isThe first Kleene star independently chooses $m$ copies of the word $ab$, and the second independently chooses $n$ copies of $bc$. Their concatenation produces exactly the required words. Either count may be zero, including both counts zero, which gives the empty word. By Kleene theorem, the regular expression establishes regularity.

First construct the bounded-difference automaton. Fix $K$ as above, enlarging it if necessary. Its states are the finitely many elements $d\in B_K(1)$, together with a reject state and finite flags recording when padding has started on each tape. The initial state is $1$. Read the pair of words synchronously, padding their ends with a symbol $\#$ representing the identity. If the current prefixes represent $g,h$, the difference state is $d=g^{-1}h$. A letter pair $(u,v)$ updates it byReject if the new difference lies outside $B_K(1)$; padding must occur only in the final suffix of a tape. The accepting state for the generator $a$ is $d=a$. Induction on the prefix length proves that the machine accepts exactly the padded pairs whose prefix differences stay bounded by $K$ and whose final difference is $a$. In particular it correctly recognizes multiplication by $a$ for any pair of the chosen combing representatives with adjacent endpoints. Taking accepting state $1$ gives the equality machine.

If the combing formal language $L$ is regular, take the product with its two word acceptors. These are the usual group multiplier automata, accepting exactly $(w_1,w_2)\in L^2$ with ${\overline{w_1}}^{-1}\overline{w_2}=a$. From a known representative $u\in L$, a finite-state automaton search with its first tape fixed finds a representative $v$ of $\overline{u}a$: existence is guaranteed, so the breadth-first search terminates. Equality is tested by the identity multiplier. Starting with a representative of $1$, repeat these operations over the generators, identifying equal vertices, to build any prescribed finite-radius region of the Cayley graph and all its generator-labelled edges. Iterating multiplication over an input word and testing equality with the identity gives a word problem algorithm.

A bare geometric combing does not imply that $L$ is a regular language. Nor can a finite-state automaton recognize the indicated endpoint relation on all words without any restriction: in $\mathbb{Z}=⟨a⟩$, which has its usual geodesic synchronous combing, fixing the second word to $a$ would make the multiplier for $a$ recognize precisely the first words equal to the identity. Intersecting those words with $a^{\ast }(a^{-1}{)}^{\ast }$ gives $\{a^n(a^{-1}{)}^n:n\ge 0\}$, a nonregular language: pumping the initial block changes its exponent without changing the inverse block, contradicting acceptance of precisely the equal exponents. Thus the literal unrestricted-pair request is false; the machines above recognize bounded fellow-travelling pairs, or full representative-pair relations under the additional automatic-language hypothesis.

Nevertheless the final decidability assertion is true for a synchronous combable group, without a regular-language assumption. Here is a terminating filling bound, avoiding any assumption that the given combing is computable. Let $D$ be the number of words of length at most $K$ over $\Sigma$. A horizontal row of the diagram in part (ii) is encoded by its ordered $n$ rung words, so there are at most $D^n$ possible row types. If two rows have the same word tuple, delete the band between them and identify corresponding labelled rungs. This preserves the outside boundary and leaves a valid diagram of the same short relators. Equivalently, the first column is always at $1$, so identical rung tuples describe the same vertex configuration. Repeatedly remove such bands. At most $D^n$ distinct rows remain, with at most $n$ short-relator cells per layer.

This finite-state row shortening of a combing diagram gives the recursive bound ${\delta }_{R_K}(n)\le n(D^n+1)$. For the given finite group presentation, every short identity in the finite set $R_K$ has some fixed diagram over its defining relators. Let $C$ bound their areas. Replacing each short face by its fixed diagram givesPart 3(ii) now supplies a terminating word decider. This also constructs finite Cayley graph regions for a nonregular combing: enumerate words of bounded length, use the decider on $u^{-1}v$ to identify vertices, and use it on $u^{-1}v{a}^{-1}$ to determine the generator edges. The finite ball $B_K(1)$ and transition table of the bounded-difference machines can consequently be computed. The constants depend on a fixed combing and are not a uniform procedure certifying combability of an arbitrary input group presentation. Every finite group presentation of this synchronous combable group has solvable word problem, by the presentation-invariance proof in Q3.

Work with a finite symmetric alphabet $\Sigma =A^{\pm 1}$. If $w=a_1\cdots a_m$, its path $\hat{w}$ starts at $1$, visits $a_1\cdots a_j$ at integer time $j$, travels at unit speed along the intervening generator edges, and stays at its endpoint for $t\ge m$.

Two such paths synchronously $K$-fellow travel ifAn asynchronous version permits increasing time reparametrizations. Checking all integer times suffices after a bounded enlargement of $K$ for the edge interiors.

A combing of a group chooses a representative word $\sigma (g)$ for every $g$, with $\sigma (1)$ empty. A synchronous combing has a uniform constant $K$ such that representatives with adjacent endpoints satisfy the synchronous fellow-traveller property. A group admitting this is a synchronous combable group. The formal language need not be regular or computable. A regular language of representatives satisfying the synchronous condition instead gives an automatic group. This distinction is relevant to the finite-state automata request in part (iii); the present argument uses the synchronous meaning throughout. The definitions and distinction are also made explicitly in Combings of Groups and the Grammar of Reparameterization, Section 1.

Yes: regular languages are closed under the shuffle of formal languages. The construction must allow letters common to both alphabets to be assigned to either input word.

Take complete deterministic finite automata ${\mathcal{A}}_i=(Q_i,{\Sigma }_i,{\delta }_i,s_i,F_i)$ recognizing $L_i$. Construct a nondeterministic finite automaton on $Q_1\times Q_2$ over $\Sigma ={\Sigma }_1\cup {\Sigma }_2$. Its initial state is $(s_1,s_2)$ and its accepting states form $F_1\times F_2$. On a letter $a$, its possible moves from $(p,q)$ areBoth moves are allowed when $a$ belongs to both alphabets. They may coincide; that causes no difficulty.

For every run, record whether each move updated the first or the second coordinate. The letters assigned to each coordinate, in their original order, form two words $u_1,u_2$. The final coordinate states are exactly the states reached by reading $u_i$ in ${\mathcal{A}}_i$. Thus an accepting run expresses the input as an interleaving of a word of $L_1$ and a word of $L_2$.

Conversely, given such an interleaving, assign each input position to the word from which it came. The corresponding choices of transitions form a run ending in $F_1\times F_2$. This proves equality between the recognized formal language and the shuffle of formal languages, in both directions. No assumption of disjoint alphabets is needed. If one contributing word is the empty word, no move need update that coordinate; the initial pair is accepting exactly when both empty words are accepted.

Apply the powerset construction to this nondeterministic finite automaton to obtain a deterministic finite automaton. In particular,

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.