Consider the regular expressions r_1 = (1+0)^*0(1+0)^* and r_2 = 11^* over the alphabet \Sigma=\{0,1\}, and let L_1 = L(r_1) and L_2 = L(r_2). Which one of the following is incorrect?

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BlackTom AI · Accepted Answer

We start by restating the problem to be clear about what is being evaluated.
Question: Consider the regular expressions r1 = (1+0)*0(1+0)* and r2 = 11^* over the alphabet {0,1}, with L1 = L(r1) and L2 = L(r2). Which one of the following is incorrect?
Answer options:
a. L2 ⊆ µL1, i.e., L2 is a subset of the complement of L1.
b. L1 ∪ L2 ≠ Σ*
c. L1^* = L1.
d. L1 ⊆ µL2, i.e., L1 is a subset of the complement of L2.

Now we analyze each option in turn, explaining why it could be true or false, with supporting reasoning.
Option a: L2 ⊆ complement of L1.
- L2 is described by 11^*, interpreted as 1 followed by zero or more 1s, i.e., the set of strings of the form 1^n with n ≥ 1. These are strings consisting entirely of 1s and of length at least 1.
- L1 consists of all strings over {0,1} that contain at least one 0 (since the pattern forces at least one 0 somewhere in the string). The complement of L1, over Σ*, consists of all strings with no 0, i.e., strings of only 1s, including the empty string.
- But L2 consists of strings 1^n with n ≥ 1, which are indeed strings with no 0 and of length at least 1. Therefore every string in L2 is in the complement of L1. Hence L2 ⊆ complement(L1) holds. This option is consistent.

Option b: L1 ∪ L2 ≠ Σ*.
- Σ* contains every possible string over {0,1}, including the empty string and all finite strings.
- L1 contains all nonempty strings that have at least one 0. L2 contains strings consisting solely of 1s with length at least 1.
- The only string not in L1 ∪ L2 would be the empty string ε, since ε has no 0 (so ε ∉ L1) and ε has length 0 (so ε ∉ L2). Therefore ε ∉ L1 ∪ L2, which means L1 ∪ L2 omits at least ε and thus cannot be all of Σ*. Hence L1 ∪ L2 ≠ Σ*. This option is true.

Option c: L1^* = L1.
- L1^* denotes the Kleene star of L1: all concatenations of zero or more strings from L1, including the empty string ε.
- Since L1 contains only strings that have at least one 0, any nonempty concatenation of members of L1 will also contain at least one 0, so such strings still lie in Σ* and, in particular, in L1 when they themselves satisfy the same condition. However, ε is included in L1^* but not in L1 itself because ε has no symbols and, crucially, does not contain a 0.
- Therefore ε ∈ L1^* but ε ∉ L1, so L1^* ≠ L1. This shows that this option claims an equality that is false. Hence this option is incorrect.

Option d: L1 ⊆ complement(L2).
- L2 consists of strings 1^n with n ≥ 1 (strings of only 1s, length at least 1).
- Any string in L1 contains at least one 0. Such a string cannot be of the form 1^n, since that would require the string to contain only 1s. Therefore, no string in L1 can be in L2.
- This implies L1 ∩ L2 = ∅ and, in particular, L1 is a subset of the complement of L2. So L1 ⊆ complement(L2) holds. This option is true.

Summary of reasoning about each option:
- a is true because every 11^* string is all 1s (at least one 1) and thus contains no 0, placing it in the complement of L1.
- b is true because ε is missing from L1 ∪ L2, so the union cannot cover Σ*.
- c is false because ε ∈ L1^* while ε ∉ L1, so the two sets cannot be equal.
- d is true because L1 strings all contain a 0, so none belong to L2, placing L1 entirely in the complement of L2.

The incorrect statement among the options is the one that asserts equality where it does not hold, namely option c. The rest of the options describe valid containments or non-coverages consistent with the definitions of L1 and L2.

Languages and Computation (COMP2049 UNNC) (SPC1 24-25) Lab Quiz 3 - Group A

Consider the regular expressions r_1 = (1+0)^0(1+0)^ and r_2 = 11^* over the alphabet \Sigma=\{0,1\}, and let L_1 = L(r_1) and L_2 = L(r_2). Which one of the following is incorrect?

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Languages and Computation (COMP2049 UNNC) (SPC1 24-25) Lab Quiz 3 - Group A

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