[TI] Compact: Add better explanation

semester3/ti-compact/parts/02_finite-automata.tex

@@ -20,7 +20,7 @@ We can note the automata using graphical notation similar to graphs or as a seri
         \item $\delta(q, a) = p$ transition from $q$ on reading $a$ to $p$
         \item $q_0$ initial state
         \item $F \subseteq Q$ accepting states
-        \item $\mathcal{L}_{EA}$ regular languages (accepted by FA)
+        \item $\cL_{EA}$ regular languages (accepted by FA)
     \end{itemize}
 \end{multicols}
 
@@ -144,7 +144,7 @@ Thus, all four words have to lay in pairwise distinct states and we thus need at
 
 
 \subsection{Non-determinism}
-The most notable differences between deterministic and non-deterministic FA is that the transition function maps is different: $\delta: Q \times \Sigma \rightarrow \mathcal{P}(Q)$.
+The most notable differences between deterministic and non-deterministic FA is that the transition function maps is different: $\delta: Q \times \Sigma \rightarrow \cP(Q)$.
 I.e., there can be any number of transitions for one symbol from $\Sigma$ from each state.
 This is (in graphical notation) represented by arrows that have the same label going to different nodes.
 
@@ -162,5 +162,5 @@ States are no now sets of states of the NFA in which the NFA could be in after p
 
 For each state, the set of states $P = \hdelta(q_0, z)$ for $|z| = n$ represents all possible states that the NFA could be in after doing the first $n$ calculations.
 
-Correspondingly, we add new states if there is no other state that is in the same branch of the calculation tree $\mathcal{B}_M(x)$.
+Correspondingly, we add new states if there is no other state that is in the same branch of the calculation tree $\cB_M(x)$.
 So, in other words, we execute BFS on the calculation tree.

semester3/ti-compact/parts/03_turing-machines.tex

@@ -26,14 +26,14 @@ As with normal TMs, the Turing Machine $M$ accepts $w$ if and only if $M$ reache
 Church's Thesis states that the Turing Machines are a formalization of the term ``Algorithm''.
 It is the only axiom specific to Computer Science.
 
-All the words that can be accepted by a Turing Machine are elements of $\mathcal{L}_{RE}$ and are called \bi{recursively enumerable}.
+All the words that can be accepted by a Turing Machine are elements of $\cL_{RE}$ and are called \bi{recursively enumerable}.
 
 
 \subsection{Non-Deterministic Turing Machines}
 The same ideas as with NFA apply here. The transition function also maps into the power set:
 \rmvspace
 \begin{align*}
-    \delta : (Q - \{ \qacc, \qrej \}) \times \Gamma \rightarrow \mathcal{P}(Q \times \Gamma \times \{ L, R, N \})
+    \delta : (Q - \{ \qacc, \qrej \}) \times \Gamma \rightarrow \cP(Q \times \Gamma \times \{ L, R, N \})
 \end{align*}
 
 Again, when constructing a normal TM from a NTM (which is not required at the Midterm, or any other exam for that matter in this course),

semester3/ti-compact/parts/04_computability.tex

@@ -18,19 +18,18 @@ Below is a list of countable objects. They all have corresponding Lemmas in the
 
 \rmvspace
 \drmvspace
-The following objects are uncountable: $[0, 1]$, $\R$, $\mathcal{P}(\wordbool)$
+The following objects are uncountable: $[0, 1]$, $\R$, $\cP(\wordbool)$
 
-\inlinecorollary $|\text{KodTM}| < |\mathcal{P}(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$
+\inlinecorollary $|\text{KodTM}| < |\cP(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$
 
 \fhlc{Cyan}{Proof of $L$ (not) recursively enumerable}
 
-Proving that a language \textit{is} recursively enumerable is as easy as providing a Turing Machine that accepts it.
+Proving that a language \textit{is} recursively enumerable is as difficult as providing a Turing Machine that accepts it.
 
-Proving that a language is \textit{not} recursively enumerable is a bit harder. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$.
+Proving that a language is \textit{not} recursively enumerable is likely easier. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$.
 
-As an example, we'll use the following language
+\inlineex Assume towards contradiction that $L_\text{diag} \in \cL_{RE}$. Let 
-
+\rmvspace
-Assume towards contradiction that $L_\text{diag} \in \mathcal{L}_{RE}$. Let 
 \begin{align*}
     L_{\text{diag}} & = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and $M_i$ does not accept } w_i \} \\
                     & = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and } d_{ii} = 0\}
@@ -40,7 +39,9 @@ Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines,
 
 This however leads to a contradiction, as $w_i \in L_\text{diag} \Longleftrightarrow d_{ii} = 0 \Longleftrightarrow w_i \notin L(M_i)$.
 
-In other words, $w_i$ is in $L_\text{diag}$ if and only if $w_i$ is not in $L(M_i)$, which contradicts our statement above.
+In other words, $w_i$ is in $L_\text{diag}$ if and only if $w_i$ is not in $L(M_i)$, which contradicts our statement above, in which we assumed that $L_\text{diag} \in \cL_{RE}$
+
+In other, more different, words, $w_i$ being in $L_\text{diag}$ implies (from the definition) that $d_{ii} = 0$, which from its definition implies that $w_i \notin L(M_i)$
 
 \setLabelNumber{theorem}{3}
-\inlinetheorem $L_\text{diag} \notin \mathcal{L}_{RE}$
+\inlinetheorem $L_\text{diag} \notin \cL_{RE}$