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lfa:reg [2018/07/23 13:25]
pdmatei [Definition] 		lfa:reg [2018/07/23 14:47] (current)
pdmatei
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 Therefore, we have shown that **NFAs and DFAs accept the same languages**,​ i.e. $math[L(NFA) = L(DFA)]. In other words, if a language $math[L] is accepted by some DFA $math[M] ($math[L=L(M)]),​ then it can also be accepted by some NFA, and vice-versa. Therefore, we have shown that **NFAs and DFAs accept the same languages**,​ i.e. $math[L(NFA) = L(DFA)]. In other words, if a language $math[L] is accepted by some DFA $math[M] ($math[L=L(M)]),​ then it can also be accepted by some NFA, and vice-versa.
  
-It remains to establish the relationshi +It remains to establish the relationship ​between $math[LR] and $math[L(DFA)] (or equivalently $math[L(NFA)]).
-p between $math[LR] and $math[L(DFA)] (or equivalently $math[LNFA]).+
  
 ===== Equivalence between Regular Expressions and Automata ===== ===== Equivalence between Regular Expressions and Automata =====
  
 $justtheorem $justtheorem
-Let $math[M] be a DFA. There exists a **regular expression** $math[E], such that $math[L(E)=L(A)]. +Let $math[M] be a DFA. There exists a **regular expression** $math[E], such that $math[L(E)=L(M)]. 
 $end $end
  
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 If $math[i = j], then: If $math[i = j], then:
- * $math[R^{(0)}_{ii}] may contain **zero** transitions,​ hence $math[R^{(0)}_{ij} = \epsilon] +  ​* $math[R^{(0)}_{ii}] may contain **zero** transitions,​ hence $math[R^{(0)}_{ij} = \epsilon] 
- * $math[R^{(0)}_{ii}] may contain one transition, and the construction follows the above rules, yielding some regular expression $math[E_0]. ​+  * $math[R^{(0)}_{ii}] may contain one transition, and the construction follows the above rules, yielding some regular expression $math[E_0]. ​
 We combine the two situations in a single one: $math[R^{(0)}_{ii} = \epsilon \cup E_0], where $math[E_0] is constructed as above. We combine the two situations in a single one: $math[R^{(0)}_{ii} = \epsilon \cup E_0], where $math[E_0] is constructed as above.