Construct a CFG for the regular expression (a+b)(aa+bb+(ab+ba)(aa+bb)*(ab+ba))*
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Construct a CFG for the regular expression
(a+b)(aa+bb+(ab+ba)(aa+bb)*(ab+ba))*
please answer with detail
Step by step
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- Use structural induction to show that no matter howcomplex structure a CTL formula Φ may contain, e.g., AFAGEGA[AXp U q] and etc., M, s ⊨ Φis always a Boolean expression.Computer Science Use structural induction to show that no matter how complex structure a CTL formula Φ may contain, e.g., AFAGEGA[AXp U q] and etc., M, s ⊨ Φis always a Boolean expression.Simplify the following expression (Boolean Algebra) such that WXY will be the only remaining terms and Z will be cancelled in the expression. A = W'XYZ + WX'YZ + WXY'Z + WXYZ' + WXYZ
- Generate a -NFA describing the same language as the regular expression ((11+00)(101)∗ (01∗+ (00)∗0))∗ . Show its transition diagram.Sub:-Flat Construct a NFA in which triple '1' is followed by triple '0' over Σ = {0, 1).If you knew the order of precedence for operators in boolean algebra is () then, ' then, ⊕ then, . then, +. with terminals a, b , c. 1-Design a CFG to generate its expressions unambiguosly. 2-Give the PDA equivalent to this CFG
- Simplify the below expressions a) By using Boolean algebra b) By using Karnaugh maps. i) z=xy+xy' ii) z= x+xyactoring is a powerful simplification technique in Boolean algebra, just as it is in real-number algebra. Show how you can use factoring to help simplify the following Boolean expressions: A. F = Y + YZB. F = XY’Z + XY’Z’Give a regular expression for LR, where L is the language L ((ab + b)* b (a + ab)*)
- What is (a --> b) and (¬b --> c) and (¬ c↔¬b) logically equivalent to? Use logic laws and label themComputer Science Show the language L= (w: each occurrence of the substring bb is followed immediately by aa)is regular in two ways by providing both a DFA acceptor and a regular CFG generator.Simplify Boolean function using K map. Variables are marked in a,b,c,d order 0 0 1 1 0 1 1 1 1 1 1 0 1 1 0 0 Question 37 options: ac’ + ad + bd + a’c ac’ + bd’ + a’c ac’ + bd + a’c None of above c’a + b’c’ + ab