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First of all, note that your quest for "rules in place" is probably doomed. This looks pretty much like an exercise problem you would pose in class; there is not necessarily an intuitive rule or seman …

Note that the complement of $A \subseteq \Sigma^*$ is defined as $\Sigma^* \setminus A$; so yes: the complement of $\Sigma^*$ itself is the empty set. To be completely accurate, you need to state wi …

What we have with $\qquad X = AX + B$ is, quite literally, a recurrence of languages expressed in terms of the symbolic method. $B$ represents the base case of the recurrence. Arden's Lemma just tel …

Check your definitions: $*$ is the Kleene star. So, $\{a,b\}^*$ is the set of all finite strings over alphabet $\{a,b\}$.

Without knowing what kind of transformation you compute, it's unlikely that there's a useful word. Assuming that any two inputs are processed independently of each other, you're looking at (computable …

Here is what we (believe to) know: $\mathrm{HAL} \setminus \mathrm{CFL} \neq \emptyset$ (type-1, type-2) $\mathrm{CFL} \setminus \mathrm{HAL} \neq \emptyset$ (type-1) $\mathrm{CFL} \subseteq \mathrm …

There are plenty of possibilities. Others have already mentioned automata which offer a rich selection. Consider the following frameworks, too: Some languages can be defined directly by (co)inductiv …

Let $\mathrm{SDL}_k$ the shuffled Dyck language with $k$ types of parentheses, i.e. $\qquad \displaystyle \mathrm{SDL}_k = \mathrm{DL}([_1,]_1) \, ш\, \dots \, ш\, \mathrm{DL}([_k,]_k)$ with $ш$ the …

This is pretty standard notation for $\qquad \#_a(w) = |w|_a = $ number of occurrences of $a$ in $w$. Building a PDA for this language is straight-forward. Find a hint below.

Try to use closure properties instead. Closer hint:

You are mixing two things here: syntax and semantics. First, syntax. What is the language of strings $L_P \subseteq \Sigma^*$ with $\Sigma = \{\mathtt{a}, \dots, \mathtt{z}, \mathtt{)}, \mathtt{(}, \ …

You reasoning is not correct, and not only because you reach the wrong answer. You argued for one representation of the language that it can't be faithfully checked by one (idea for) a push-down auto …

The prefix and suffix operations can be expressed in terms of right- resp. left-quotients: $\qquad \operatorname{pref(L)} = L / \Sigma^*$ and $\qquad \operatorname{suff(L)} = L \backslash \Sigma^*$. …

An infinite number of potential identifiers results in an infinite language, but this is a "boring" feature with this consequence, because you can do without. We know that two variables are sufficient …

Hint: Try the usual approach to show set inclusion! That is, pick $w \in A^*$ and show that $w \in B$. Next hint: More elaborately: