Question: Q 7 True / False 3 Points True / False: Every proper subset of a regular set is regular. True False True / False: Every

Q7 True/ False
3 Points
True/ False: Every proper subset of a regular set is regular.
True
False
True/ False: Every proper subset of a nonregular set is nonregular.
True
False
True/ False: The complement of a regular set is regular.
True
False
True/False: The complement of a nonregular set is nonregular
True
False
True/ False: The union of any two regular sets is regular.
True
False
True/ False: The union of two nonregular sets is nonregular.
True
() False
Q8 Pumping Lemma
2 Points
Select all and only true statements.
All regular languages have pumping lengths.
To prove that a language is regular, it's enough to show that it has a pumping length.
To prove that a language is nonregular, it's enough to show that it does not have any
pumping lengths.
To prove that a specific positive integer is not a pumping length for a given language, we
need to show that all strings are not "pumpable" relative to that length.
To prove that a specific positive integer is not a pumping length for a given language, we
need to show that there is a string in that language that is longer than that number and
that is not "pumpable" relative to that length in that language.
Q9 Pumping length
3 Points
Definition A positive integer p is a pumping length of a language L over means that, for each
string sin**, if |s|p and sinL, then there are strings x,y,z such that s=xyz and |y|>0,
for each i0,xyizinL, and |xy|p.
In particular, this means that a positive integer p is not a pumping length of a language L over
iff
EEs(|s|p??sinL??AAxAAyAAz((s=xyz??|y|>0??|xy|p)EEi(i0??xyiz!inL)))
True/ False: A pumping length for A1={1,01,001,0001,00001} is p=4
True
False
True/ False: A pumping length for A2={0j1|j0} is p=3
True
False
True/ False: For any language A, if p is a pumping length for A and p'>p, then p' is also a
pumping length for A.
True
False
Q 7 True / False 3 Points True / False: Every

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