====== Deterministic automata ======

$def[DFA]
A **Deterministic Finite Automata** is a tuple $math[(K,\Sigma,\delta,q_0,F)], where $math[K] is a set of states, $math[\Sigma] is the alphabet, $math[\Delta : K \times \Sigma \rightarrow K] is a **total** transition function. 
$end

$def[Configuration]
The **configuration** of a DFA is an element of $math[K\times\Sigma^*].
$end

$def[One-step move]
We call $math[\vdash_M \subseteq (K\times \Sigma^*) \times (K\times \Sigma^*)] a **one-step** move relation, over configurations. The relation is defined as follows:
  * $math[(q,w) \vdash_M (q',w')] iff $math[w=cw'] for some symbol $math[c] of the alphabet, and $math[\delta(q,c)=q']

We write $math[\vdash_M^*] to refer to the **reflexive and transitive closure** of $math[\vdash_M]. Hence, $math[(q,w)\vdash_M^*(q',w')] means that DFA $math[M] can reach configuration $math[(q',w')] from configuration $math[(q,w)] is zero or more steps.
$end

$def[Acceptance]
We say a word $math[w] is **accepted** by a DFA $math[M] iff $math[(q_0,w)\vdash_M^*(q,\epsilon)] and $math[q\in F] ($math[q] is a final state).
$end


===== NFA to DFA transformation =====

Let $math[M=(K,\Sigma, \Delta, q_0, F)] be an NFA. We assume $math[M] **does not contain transitions on words of length larger than 1**. If $math[(q,w,q')\in\Delta] for some $math[w=c_1\ldots c_n] of size 2 or more, we construct intermediary states $math[q^1, \ldots, q^{n+1}] as well as transitions $math[(q^1,c_1,q^2), \ldots (q^n,c_n,q^{n+1})], where $math[q=q^1] and $math[q'=q^n]

We denote by $math[E_M(q) = \{p\in K\mid (q,\epsilon)\vdash^*_M (p,\epsilon)\}], the **$math[\epsilon]-closure** of state $math[q]. In effect $math[E_M(q)] contains **all states reachable from $math[q] via $math[\epsilon]-transitions**. When the automaton $math[M] is understood from the context, we omit the subscript and simply write $math[E(q)].

We build the DFA $math[M'=(K',\Sigma, \delta, q_0', F')] as follows:
  * $math[K'=2^{K}] - each state of $math[M'] is a **subset** of states from the NFA. It may be the case that some such states are not reachable, hence we shall ignore them from our construction;
  * $math[q_0' = E(q_0)]
  * $math[\delta(Q,c) = \displaystyle \cup_{q\in Q, (q,c,q')\in\Delta} E(q')] - a transition from $math[Q] on symbol $math[c] ends in the reunion of all $math[\epsilon]-closures of states in $math[M] reachable from some member of $math[Q] on symbol $math[c].
  * $math[F'=\{Q\subseteq K'\mid Q \cap F \neq\emptyset\}] - a state is final in $math[M'] iff it contains
 some
 final state in $math[M].

===== Correctness of the transformation =====

$prop[1]
For all $math[q,p\in K], $math[(q,w)\vdash^*_M (p,\epsilon)] iff $math[(E(q),w)\vdash^*_{M'}(P,\epsilon)], for some $math[P] such that $math[p\in P].
$end

The proposition states that, for each path in NFA $math[M] starting on $math[q] which //consumes// word $math[w] (hence ends up in configuration $math[(p,\epsilon)]), there is an //equivalent// path in the DFA $math[M'], which starts in the $math[\epsilon]-closure of $math[q] and ends in some state $math[P] which contains $math[p] --- and vice-versa.

Proposition 1 is essential for proving the following result:

$justtheorem
Let $math[M'] be a DFA constructed from NFA $math[M] according to the above rules. Then $math[L(M)=L(M')].
$end

$proof
The languages of the two machines coincide if, for all words: $math[w\in L(M)] iff $math[w\in L(M')], thus:
  * $math[(q_0,w)\vdash^*_M (p,\epsilon)] with $math[p\in F] iff $math[(E(q_0),w)\vdash^*_{M'}(P,\epsilon)] with $math[p\in P].

The above statement follows immediately from Proposition 1, where:
  * $math[q] is the initial state of $math[M]
  * $math[p] is some final state of $math[M]
as well as from the definition of $math[F'].
$end

We now turn to the proof of Proposition 1:

$proof
The proof is by **induction** over the length of the word $math[w].

** Basis step**: $math[\mid w\mid=0] that is $math[w=\epsilon]
  * //direction $math[\implies]:// 
    - Suppose $math[(q,\epsilon) \vdash^*_M(p,\epsilon)].
    - From the definition of $math[E] and 1., we have that $math[p\in E(q)].
    - Since $math[\vdash^*_{M'}] is **reflexive**, we have $math[(E(q),\epsilon)\vdash^*_{M'}(E(q),\epsilon)].
    - Therefore, we have $math[E(q),\epsilon)\vdash^*_{M'}(P,\epsilon)] with $math[p\in P]: $math[P] is actually $math[E(q)].

  * //direction $math[\impliedby]://
    - Suppose $math[(E(q),\epsilon) \vdash^*_{M'} (P,\epsilon)]
    - Since $math[\delta] does not allow $math[\epsilon]-transitions (and $math[\vdash^*_{M'}] is reflexive), it follows that $math[E(q)=P]. 
    - By the definition of $math[E], we have that $math[(q,\epsilon)\vdash^*_{M}(p,\epsilon)] for any $math[p\in E(q)].

** Induction hypothesis**: suppose that the claim is true for all strings w such that $math[\mid w\mid\leq k] for $math[k\geq0]

** Induction step**: we prove for any string $math[w] of length $math[k+1]; let $math[w'=wa] (hence $math[a] is the last symbol of $math[w']).
  * //direction $math[\implies]:// 
    - Suppose $math[(q,wa)\vdash^*_{M} (p,\epsilon)]. 
    - By the definition of $math[\vdash^*], we have: $math[(q,w)\vdash^*_{M} (r_1,a) \vdash_{M} (r_2,\epsilon)\vdash^*_{M} (p,\epsilon)]. In other words, there is a //path// from $math[q] which takes us to $math[r_1] by consuming $math[w], then to $math[r_2] via a **one-step** transition, then to $math[p] in zero or more $math[\epsilon]-transitions. Notice that $math[p] may be equal to $math[r_2], which is taken into account since $math[\vdash^*_{M}] is reflexive.
    - By the construction of $math[\vdash^*_{M}], we also have $math[(q,w)\vdash^*_{M}(r_1,\epsilon)]
    - From 3. by **induction hypothesis**, we have $math[(E(q),w)\vdash^*_{M'}(R_1,\epsilon)] with $math[r_1 \in R_1]
    - By construction of $math[\vdash^*_{M'}], we have $math[(E(q),wa)\vdash^*_{M'}(R_1,a)]
    - Since $math[(r_1,a) \vdash_{M} (r_2,\epsilon)], by the definition of $math[\delta], we have $math[E(R_2) \subseteq \delta(R_1,a)]. 
    - Since $math[(r_2,\epsilon) \vdash_{M} (p,\epsilon)] it follows that $math[p \in E(r_2)], and therefore $math[p \in \delta(R_1,a)]. 
    - In effect, from 5. and 7. we have shown that $math[(E(q),wa)\vdash^*_{M'}(R_1,a)\vdash_{M'}(R_2,\epsilon)] with $math[p\in R_2], which concludes our proof.

  * //direction $math[\impliedby]://
    - Suppose $math[(E(q),wa)\vdash^*_{M'}(P,\epsilon)]. 
    - Since no $math[\epsilon]-transitions are allowed in a DFA, we must have: $math[(E(q),wa)\vdash^*_{M'}(R,a)\vdash_{M'}(P,\epsilon)].
    - By construction of $math[\vdash^*_{M'}]: $math[(E(q),w)\vdash^*_{M'}(R,\epsilon)].
    - By **induction hypothesis**, $math[(q,w)\vdash^*_{M}(r,\epsilon)], with $math[r\in R].
    - By construction of $math[\vdash^*_{M}]: $math[(q,wa)\vdash^*_{M}(r,a)].
    - Since $math[(R,a)\vdash_{M'}(P,\epsilon)], by the definition of $math[\delta], for some $math[r\in R], $math[(r,a,x)\in\Delta] and $math[E(x)\subseteq P]. 
    - From 6. $math[(q,wa)\vdash^*_{M}(r,a)\vdash_{M}(x,\epsilon)] with $math[x\in E(x)\subseteq P] which completes our proof.
$end