Let r and s be regular expressions. Which of the following regular expression equivalences is false? Select one: (r+s)≡(s+r) (r∗s∗+s∗r∗)∗≡(r+r∗s)∗ (rs+sr)∗≡((r+s)(s+r))∗ (r+s)∗≡(r∗s∗)∗
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Let r and s be regular expressions. Which of the following regular expression equivalences is false?
(r+s)≡(s+r)
(r∗s∗+s∗r∗)∗≡(r+r∗s)∗
(rs+sr)∗≡((r+s)(s+r))∗
(r+s)∗≡(r∗s∗)∗
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- Consider the code C={00010, 01100,01011,11010,01010}. What is the minimum distance of C? For what k is the code C k-error detecting? Write the set of error patterns that C can detect?Construct a truth table for (p ∨ ¬ q) ∨ (¬ p ∧ q) Use the truth table that you constructed in part 1 to determine the truth value of (p ∨¬q) ∨ (¬ p ∧ q), given that p is true and q is false. Determine whether the given statement is a tautology, contradiction, or contingency. p V (~p V q) ~ (p ∧ q) ~p V ~qSuppose that you are given the following logical statement:¬ P ∧ Q → ¬ R ∨ SBased on our discussion of order of operations, which is the correct way to interpretthis?(a) ¬(P ∧ ((Q → (¬R)) ∨ S))(b) (¬(P ∧ (Q → (¬R)))) ∨ S(c) ((¬P) ∧ (Q → (¬R))) ∨ S(d) (¬P) ∧ ((Q → (¬R)) ∨ S)(e) (¬(P ∧ Q)) → (¬(R ∨ S))(f) ((¬P) ∧ Q) → ((¬R) ∨ S)(g) ((¬P) ∧ (Q → (¬R))) ∨ S(h) ((¬P) ∧ (Q → (¬R))) ∨ S(i) (¬P) ∧ ((Q → (¬R)) ∨ S)(j) ¬((P ∧ Q) → (¬(R ∨ S)))
- 1. Which of the following statements are true? (a) |N|=|Z| (b) |N|<|Z| (c) |Z|=א0 (d) |R|>|Z| 2. Which of the following statements are tautologies? (a) P → Q (b) P ∧ ¬P (c) P ∨ ¬P (d) P → P 3. The contrapositive equivalency of ¬Q → P is given by: (a) P → Q (b) Q → P (c) ¬P → Q (d) ¬Q→PReduce the proposition using laws. Simplify (q∨p)∧(¬p∨q) to q Select a law from below to apply to: (q∨p)∧(¬p∨q) Laws Distributive (a∧b)∨(a∧c)≡a∧(b∨c) (a∨b)∧(a∨c)≡a∨(b∧c) Commutative a∨b≡b∨a a∧b≡b∧a Complement a∨¬a≡T a∧¬a≡F Identity a∧T≡a a∨F≡a Double negation ¬¬a≡aThe binomial coefficient C(N,k) can be defined recursively as follows: C(N,0) = 1, C(N,N) = 1, and for 0 < k < N, C(N,k) = C(N-1,k) + C(N - 1,k - 1). Write a function and give an analysis of the running time to compute the binomial coefficients as follows: A. The function is written recursively.
- Describe the following sets by regular expressions, (a) {abba} (b) {a, ab} .Complete the truth table for the following compound statement.(p∨∼q)→(r∧∼p)(p∨∼q)→(r∧∼p) p q r ∼q∼q p∨∼qp∨∼q ∼p∼p r∧∼pr∧∼p (p∨∼q)→(r∧∼p)(p∨∼q)→(r∧∼p) T T T T T F T F T T F F F T T F T F F F T F F F Is the compound statement a tautology? No, the statement is not a tautology. Yes, the statement is a tautology.Exercise 1.A set W of strings of symbols is defined recursively by a, b, and d belong to W. If x belongs to W, so does a(x)d. Provide 3 strings made of 3 letters each belonging to W and show the procedure used to obtain them. Provide 3 strings made of at least 5 letters each belonging to W and show the procedure used to obtain them.
- if p and q are logical variables, which of the following is a tautology (i.e., always correct irrespective of specific value of variables) Select one: a. p → (q ∧ p) b. p ∨ (q → q) c. (p ∨ q) → q d. p ∨ (p → q)Describe the following in CFG: S --> sS | bC C --> cC | eGenerate a bunch of example, you will see that they are: Any number of a's followed, optionally followed one b and any number of c'sHence the answer is a*|a*bc* Calculation of power using recursion: 52^847 = 52 * 52^8461): ¬( p ∨ q ) ≡ ¬p ∧ ¬q The above law is called Group of answer choices De Morgans Law Absorption Law Complement Law Double Negation Law 2): To prove the this identity ¬p → (q → r) ≡ q → (p ∨ r) The following reasoning is valid: ¬p → (q → r)¬¬p ∨ (q → r) Conditional identity p ∨ (q → r) Double negation law p ∨ (¬q ∨ r) Conditional identity (p ∨ ¬q) ∨ r Associative law (¬q ∨p) ∨ r Commutative law ¬q ∨ (p ∨ r) Associative law q → (p ∨ r) Conditional identity Group of answer choices True False