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%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
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\usepackage{amsmath}            % Symbols
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%\usepackage[english,german,french,spanish,italian]{babel}              % Languages

% Document Info
\author{Torvak}
\pdfinfo{
  /Title (sets-languages-and-automata.pdf)
  /Creator (Cheatography)
  /Author (Torvak)
  /Subject (Sets | Languages \& Automata Cheat Sheet)
}

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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Sets | Languages \& Automata Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Torvak} via \textcolor{DarkBackground}{\uline{cheatography.com/32041/cs/12709/}}}
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\noindent
\begin{multicols}{3}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Torvak \\
  \uline{cheatography.com/torvak} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 16th September, 2017.\\
   Page {\thepage} of \pageref{LastPage}.
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{1.16956 cm} p{0.45947 cm} x{1.2531 cm} x{1.29487 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Set symbols}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Set}} & \{\} & a collection of elements & A = \{3,7,9,14\}, \{\{nl\}\} B = \{9,14,28\} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
{\bf{Such that}} & | & such that & A = \{x | x∈\textbackslash{}mathbb\{R\}, x\textless{}0\} \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
{\bf{Intersection}} & A∩B & belongs to both A AND B at & A ∩ B = \{9,14\} \tn 
% Row Count 9 (+ 3)
% Row 3
\SetRowColor{white}
{\bf{Union}} & A∪B & belongs to either A OR B & A ∪ B = \{3,7,9,14,28\} \tn 
% Row Count 11 (+ 2)
% Row 4
\SetRowColor{LightBackground}
{\bf{Subset}} & A⊆B & A is a subset of B. \{\{nl\}\}Set A is included in set B. & \{9,14,28\} ⊆ \{9,14,28\} \tn 
% Row Count 16 (+ 5)
% Row 5
\SetRowColor{white}
{\bf{Proper subset / strict subset}} & A⊂B & A is a subset of B, but A is not equal to B & \{9,14\} ⊂ \{9,14,28\} \tn 
% Row Count 20 (+ 4)
% Row 6
\SetRowColor{LightBackground}
{\bf{Not a subset}} & A⊄B & Set A is not a subset of set B & \{9,66\} ⊄ \{9,14,28\} \tn 
% Row Count 23 (+ 3)
% Row 7
\SetRowColor{white}
{\bf{Superset}} & A⊇B & A is a superset of B.\{\{nl\}\}Set A includes set B & \{9,14,28\} ⊇ \{9,14,28\} \tn 
% Row Count 27 (+ 4)
% Row 8
\SetRowColor{LightBackground}
{\bf{Proper superset / strict superset}} & A⊃B & A is a superset of B, but B is not equal to A & \{9,14,28\} ⊃ \{9,14\} \tn 
% Row Count 31 (+ 4)
\end{tabularx}
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\begin{tabularx}{5.377cm}{x{1.16956 cm} p{0.45947 cm} x{1.2531 cm} x{1.29487 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Set symbols (cont)}}  \tn
% Row 9
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{\bf{Not superset}} & A⊅B & Set A is not a superset of set B & \{9,14,28\} ⊅ \{9,66\} \tn 
% Row Count 3 (+ 3)
% Row 10
\SetRowColor{white}
{\bf{Power set}} & 2\textasciicircum{}A\textasciicircum{} & All subsets of A &  \tn 
% Row Count 5 (+ 2)
% Row 11
\SetRowColor{LightBackground}
{\bf{Power set}} & P(A) & All subsets of A &  \tn 
% Row Count 7 (+ 2)
% Row 12
\SetRowColor{white}
{\bf{Equality}} & A=B & Both sets have the same members & A=\{3,9,14\}, B=\{3,9,14\}, A=B \tn 
% Row Count 10 (+ 3)
% Row 13
\SetRowColor{LightBackground}
{\bf{Complement}} & A\textasciicircum{}c\textasciicircum{} & All the objects that do not belong to set A &  \tn 
% Row Count 14 (+ 4)
% Row 14
\SetRowColor{white}
{\bf{Relative complement}} & A\textbackslash{}B or A-B & Objects that belong to A and not to B & A = \{3,9,14\}, B = \{1,2,3\}, A \textbackslash{} B = \{9,14\} \tn 
% Row Count 18 (+ 4)
% Row 15
\SetRowColor{LightBackground}
{\bf{Symmetric difference}} & A∆B or A⊖B & Objects that belong to A or B but not to their \seqsplit{intersection} & A = \{3,9,14\}, B = \{1,2,3\}, A ∆ B = \{1,2,9,14\} \tn 
% Row Count 23 (+ 5)
% Row 16
\SetRowColor{white}
{\bf{Element of}} & a∈A & Set membership & A=\{3,9,14\}, 3	∈ A \tn 
% Row Count 25 (+ 2)
% Row 17
\SetRowColor{LightBackground}
{\bf{ Not element of}} & x∉A & No set membership & A=\{3,9,14\}, 1 ∉ A \tn 
% Row Count 27 (+ 2)
% Row 18
\SetRowColor{white}
{\bf{Ordered pair}} & (a,b) & Collection of 2 elements &  \tn 
% Row Count 29 (+ 2)
% Row 19
\SetRowColor{LightBackground}
{\bf{Cartesian product}} & A×B & Set of all ordered pairs from A and B &  \tn 
% Row Count 33 (+ 4)
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\begin{tabularx}{5.377cm}{x{1.16956 cm} p{0.45947 cm} x{1.2531 cm} x{1.29487 cm} }
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\mymulticolumn{4}{x{5.377cm}}{\bf\textcolor{white}{Set symbols (cont)}}  \tn
% Row 20
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\seqsplit{cardinality} & |A| or \#A & The number of elements of set A & A=\{3,9,14\}, |A|=3 \tn 
% Row Count 3 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
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\begin{tabularx}{5.377cm}{x{1.34379 cm} x{3.63321 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Set operations}}  \tn
% Row 0
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\seqsplit{Associativity} & (A∪B) ∪  C = A∪(B∪C)\{\{nl\}\}(A∩B)∩C = A∩(B∩C) \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\seqsplit{Commutativity} & A∪B = B∪A \{\{nl\}\} A∩B = B∩A \tn 
% Row Count 5 (+ 2)
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\seqsplit{Complementation} & A∪Not(A) = U\{\{nl\}\} A∩Not(A) =  ∅ \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
\seqsplit{Idempotence} & A∪A = A et A∩A \tn 
% Row Count 9 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Identité & A∪∅ = A\{\{nl\}\} A∩U = A \tn 
% Row Count 10 (+ 1)
% Row 5
\SetRowColor{white}
\seqsplit{Extrémité} & A∪U = U\{\{nl\}\}A∩∅ = ∅ \tn 
% Row Count 12 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\seqsplit{Involution} & Not(A) = A \tn 
% Row Count 13 (+ 1)
% Row 7
\SetRowColor{white}
Morgan's law & (Not(A∪B)) = Not(A)∩Not(B)\{\{nl\}\}(Not(A∩B)) = Not(A)∪Not(B) \tn 
% Row Count 16 (+ 3)
% Row 8
\SetRowColor{LightBackground}
\seqsplit{Distribuvité} & A∪(B∩C) = (A∪B) ∩ (A∪C)\{\{nl\}\}A∩(B∪C) = (A∩B) ∪ (A∩C) \tn 
% Row Count 19 (+ 3)
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\begin{tabularx}{5.377cm}{x{1.74195 cm} x{3.23505 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Languages - Definitions}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Formal language}} & vocabulary + grammar rules \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
{\bf{Monoid}} & Set having an internal binary operation (+,-,*/, U,...) and having a neutral element ∈. Ex: \textless{}E,+,∈\textgreater{} means all strings of E will be concaneted with operator + \tn 
% Row Count 9 (+ 7)
% Row 2
\SetRowColor{LightBackground}
{\bf{Grammar}} & (Vn, Vt, P, S) when:\{\{nl\}\}- Vn is a finite non-empty set whose elements are variables\{\{nl\}\}-  VT is a finite non-empy set of terminal states\{\{nl\}\}- Vn∩ε = ∅\{\{nl\}\} -P is a finite set whose elements are α -\textgreater{} β, known as production rules when α,  β,  ∈  (Vn∪ε)* but α should contain at least 1 symbol from Vn\{\{nl\}\}- S is a start symbol, where S ∈ Vn\{\{nl\}\} \tn 
% Row Count 24 (+ 15)
% Row 3
\SetRowColor{white}
{\bf{Symbol}} & Element of a set. Ex for set A = \{1, 2, 3\}, 2 is an element \tn 
% Row Count 27 (+ 3)
% Row 4
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{\bf{Alphabet}} & A finite and non empty set of symbols \tn 
% Row Count 29 (+ 2)
% Row 5
\SetRowColor{white}
{\bf{Length of a string}} & Number of symbols in a string. Ex: for w = aabbc, |w| = 5 \tn 
% Row Count 32 (+ 3)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
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\begin{tabularx}{5.377cm}{x{1.74195 cm} x{3.23505 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Languages - Definitions (cont)}}  \tn
% Row 6
\SetRowColor{LightBackground}
{\bf{Empty language}} & ∅ contains no string \tn 
% Row Count 2 (+ 2)
% Row 7
\SetRowColor{white}
{\bf{Empty string}} & ε where |ε| = 0 \tn 
% Row Count 4 (+ 2)
% Row 8
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{\bf{A\textasciicircum{}n\textasciicircum{}}} & Word of length {\bf{n}} from the alpabet {\bf{A}} \tn 
% Row Count 6 (+ 2)
% Row 9
\SetRowColor{white}
{\bf{A\textasciicircum{}*\textasciicircum{}}} & Words of finite length from the alphabet A or null (can be empty: ε) \tn 
% Row Count 9 (+ 3)
% Row 10
\SetRowColor{LightBackground}
{\bf{A\textasciicircum{}+\textasciicircum{}}} & Words of finite length from the alphabet A and NOT NULL (no ε) \tn 
% Row Count 12 (+ 3)
% Row 11
\SetRowColor{white}
{\bf{Regular expression}} & Rercursive language, the rules of the expression must be indepent.Ex:\{\{nl\}\}L1 = \{a\textasciicircum{}n\textasciicircum{}n\textasciicircum{}2n\textasciicircum{}|n\textgreater{}=0, m\textgreater{}= 0\} and\{\{nl\}\}L2 = \{a\textasciicircum{}n\textasciicircum{}b\textasciicircum{}m\textasciicircum{} | n = 2m\}\{\{nl\}\}{\emph{Here {\bf{L1 is regular and L2 is not}}. Because in L2 there is a dependency between n and m}} \tn 
% Row Count 21 (+ 9)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\begin{tabularx}{5.377cm}{x{1.69218 cm} x{3.28482 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Maths set reminders}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{N}} & \{0,1,2,3,...\} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
{\bf{Z}} & \{...,-3,-2,-1,0,1,2,3,...\} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
{\bf{D}} & \{d | d is a nb. having a finite number of decimals \} \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
{\bf{Q}} & \{ r | r is a rational nb. that can be written as the quotien a/b of a real integer number a by a whole integer (not null) b\}\} \tn 
% Row Count 9 (+ 5)
% Row 4
\SetRowColor{LightBackground}
{\bf{IR}} & \{...,-3,-2,-1,0,Ɣ, 1, √2, 2, e, 3, π,...\} \tn 
% Row Count 11 (+ 2)
% Row 5
\SetRowColor{white}
Successive inclusions & N ⊂ Z ⊂ D ⊂ Q ⊂ IR \tn 
% Row Count 13 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{5.377cm}{x{0.84609 cm} x{4.13091 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Automata definitions}}  \tn
% Row 0
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{\bf{DFA}} & Deterministic Finite Automata. A DFA has a is deterministic because from each state we are able to determine the next state. \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
{\bf{NDFA}} & Non Deterministic Finite Automata. A NDFA is non deterministic because we can't always determiner determine which will be the next state from the present state. \tn 
% Row Count 9 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{5.377cm}{x{1.89126 cm} x{3.08574 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Operations on Languages}}  \tn
% Row 0
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{\bf{Union}} & L∪M = \{x | x ϵ L or x ϵ M\} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
{\bf{Intersection}} & L∩M = \{x | x  ϵ L and  x ϵ M\} \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
{\bf{Difference(exclusion)}} & L\textbackslash{}M = \{x | x  ϵ L and x ∉ M\} \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
{\bf{Complement on A\textasciicircum{}*\textasciicircum{}}} & Comp(L) = A\textasciicircum{}{\emph{\textasciicircum{}\textbackslash{}L = \{x|x ϵ A\textasciicircum{}}}\textasciicircum{} and x ∉ L\} \tn 
% Row Count 8 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{L∪Ø = L \newline L∩Ø = L}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Identity of Regular Expressions}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{Ø + R = R} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Ø.R = R.Ø = Ø} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Λ.R = R.Λ = R} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Λ* = Λ and Ø* = Λ} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{R+R = R} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{R*R*=R*} \tn 
% Row Count 6 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{R.R* = R.*R} \tn 
% Row Count 7 (+ 1)
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\mymulticolumn{1}{x{5.377cm}}{(R*)=R*} \tn 
% Row Count 8 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Λ+R.R* = R* = Λ+R*.R} \tn 
% Row Count 9 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{(P.Q)*P=P(Q.P)*} \tn 
% Row Count 10 (+ 1)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{(P+Q)*=(P*.Q*)*=(P*+Q*)*} \tn 
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% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{(P+Q).R=P.R+Q.R and R(P+Q)=R.P+R.Q} \tn 
% Row Count 12 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.4931 cm} x{3.4839 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Graph types}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Reflexif}} & All states have a loop, meaning δ(state1, input) = state1 \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
{\bf{Symetric}} & All states have a direct way back , meaning:\{\{nl\}\}δ(A, x) = B and δ(B,y) = A \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
{\bf{Anti symetric}} & There never is a direct way back to the same state meaning, if :\{\{nl\}\}δ(A,x) = B then δ(B, y) ≠ A \tn 
% Row Count 10 (+ 4)
% Row 3
\SetRowColor{white}
{\bf{Transitive}} & There a shortcuts to a state, meaning if: \tn 
% Row Count 12 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Grammar construction}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Problem}} − {\emph{Suppose :}} \newline % Row Count 1 (+ 1)
{\bf{L(Gr) = \{a\textasciicircum{}n\textasciicircum{}b\textasciicircum{}m\textasciicircum{}c\textasciicircum{}k\textasciicircum{} | n \textgreater{}= 0, k \textgreater{} 1, m = n+k\}}} \newline % Row Count 3 (+ 2)
{\emph{We have to find out the grammar Gr which produces L(Gr).}} \newline % Row Count 5 (+ 2)
{\bf{Solution:}}  \newline % Row Count 6 (+ 1)
{\emph{Possible word}} {\bf{w = a\textasciicircum{}4\textasciicircum{}b\textasciicircum{}9\textasciicircum{}c\textasciicircum{}5\textasciicircum{}}} \newline % Row Count 7 (+ 1)
{\emph{We need to decompose b\textasciicircum{}9\textasciicircum{}:}} {\bf{w = a\textasciicircum{}4\textasciicircum{}b\textasciicircum{}4\textasciicircum{}b\textasciicircum{}5\textasciicircum{}c\textasciicircum{}5\textasciicircum{}}} \newline % Row Count 9 (+ 2)
{\emph{Now we can define :}} \newline % Row Count 10 (+ 1)
- {\bf{S1}} having the {\bf{same number of a}} followed by the {\bf{same number of b}}. \newline % Row Count 12 (+ 2)
- {\bf{S2}} having the {\bf{same number of b}} followed by the {\bf{same number of a}}. \newline % Row Count 14 (+ 2)
{\emph{So:}} \newline % Row Count 15 (+ 1)
S   = S1 S2 \newline % Row Count 16 (+ 1)
S1 = a S1 b | Λ \newline % Row Count 17 (+ 1)
S2 = b S2 a | bbcc \newline % Row Count 18 (+ 1)
{\emph{Why {\bf{bbcc}} as alternative to S2? Because in the case of a minimum word {\bf{w = b\textasciicircum{}2\textasciicircum{}c\textasciicircum{}2\textasciicircum{} = bbcc}} because k \textgreater{} 1 so in this case we do have m = n+k = 0+2}}% Row Count 22 (+ 4)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}