[TI] Compact: Clarifications & spelling fixes

semester3/ti-compact/parts/01_words-alphabets.tex

@@ -60,7 +60,7 @@ where the Program doesn't have to compile, i.e. we can describe processes inform
 \fancydef{Randomness} $x \in \wordbool$ random if $K(x) \geq |x|$, thus for $n \in \N$, $K(n) \geq \ceil{\log_2(n + 1)} - 1$
 
 \stepLabelNumber{theorem}
-\fancytheorem{Prime number} $\displaystyle \limni \frac{\text{Prime}(n)}{\frac{n}{\ln(n)}}$
+\fancytheorem{Prime number} $\displaystyle \limni \frac{\text{Prime}(n)}{\frac{n}{\ln(n)}} = 1$ with $\text{Prime}(n)$ the number of prime numbers on $[0, n] \subseteq \N$
 
 \fhlc{Cyan}{Proofs} Proofs in which we need to show a lower bound for Kolmogorov-Complexity (almost) always work as follows:
 Assume for contradiction that there are no words with $K(w) > f$ for all $w \in W$.

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

@@ -146,8 +146,8 @@ 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: $\delta: Q \times \Sigma \rightarrow \cP(Q)$.
-I.e., there can be any number of transitions for one symbol from $\Sigma$ from each state.
+The most notable differences between deterministic and non-deterministic FA is that the transition function is different: $\delta: Q \times \Sigma \rightarrow \cP(Q)$.
+I.e., there can be any number of transitions for one symbol of $\Sigma$ for each state.
 This is (in graphical notation) represented by arrows that have the same label going to different nodes.
 
 It is also possible for there to not be a transition function for a certain element of the input alphabet.