# Which of the following claims about languages are true, and which are false?

I have the following 4 claims about languages. I need to state whether each is true/false and provide a short justification.

1. $$\forall L : (L^+)^* = L^*$$
2. $$\forall L_1, L_2 : (L_1 = L_2 \iff L^*_1 = L^*_2)$$
3. $$\forall L_1, L_2 : L_1 L_1^* L_2 \subseteq L_1^* L_2$$
4. $$\forall L_1, L_2 : L_1^* L_2 \subseteq L_1 L_1^* L_2$$

Progress so far:

First I begin with a selection of definitions provided by my reference literature:

• Alphabet "$$\sum$$": A non empty finite set.
• Symbol: An element from an alphabet.
• Word: a finite sequence of symbols from an alphabet.
• Empty word ($$\varepsilon$$): A sequence consisting of 0 symbols.
• Language ($$L$$): A subset from $$\sum^*$$
• $$L^* = \bigcup_{i \geq 0} L^i$$
• $$L^+ = \bigcup_{i \geq 1} L^i$$

Attempts at solutions:

1. I have found a solution which claims true because $$L^* = L^+ \cup \{ \varepsilon \}$$. Perhaps someone could elaborate here? I interpret the question to mean that we first we take the union of all words $$\geq 1$$ and then the union of all worlds $$\geq 0$$, i.e I first compute the term inside the brackets before I compute the outer $$*$$. Am I making an error here?
2. I have a solution which claims that this is false by providing the counter example $$\forall w \in L_1 : w \in L_2 \land \forall w \in L_2 : w \in L_1, L_1 = \{ a , b \}, L_2 = \{ a , b, a, b \}$$. I thought that it was against the rules to define duplicate items in a set? I don't understand the idea here.

For the remaining solutions I have not made any progress.

$$L \subseteq L^+$$ so $$L^\ast \subseteq (L^+)^\ast$$.
The reverse is also true: if $$x$$ is a word in $$(L^+)^\ast$$, it is either equal to $$\varepsilon$$ which is also in $$L^\ast$$, or $$x= w_1w_2 \cdots w_n$$ where all $$w_i$$ are from $$L^+$$, so themselves concatenations of some finite number of words from $$L$$. So that makes $$x$$ of the same form, hence in $$L^\ast$$.
Check that $$L_1 = \{0,1,00\}$$ and $$L_2 = \{0,1,10\}$$ obey $$L_1^\ast = L_2^\ast$$.
3 is true as $$L_1L_1^\ast \subseteq L_1^\ast$$ already. Quite trivial to check.
1. $$L_1 =\{0\}$$ and $$L_2 = \{1\}$$ is a counterexample as $$1$$ is in the left language but not in the right one.