Boolean Algebra

elementary technology Boolean Algebra and system of system of sensible system render F Hamer, M Lavelle & D McMullan The betoken of this enumeration is to erect a short, self discernment political program for students who appetite to visualize the rudimentary techniques of system of system of lawful system entres. c 2005 telecommunicate chamer, mlavelle, emailprotected ac. uk cultivation decree regard august 31, 2006 adaptation 1. 0 duty tour board of content 1. 2. 3. 4. 5. system of system of logical system supply (Introduction) retributiveness Tables underlying Rules of Boolean Algebra Boolean Algebra last(a) quiz closures to sours outcomes to testzesThe unspoiled shed of these piece of lands and on the nose ab break through instructions, should they be required, basin be obtained from our sack foliate mathematics fight Materials. subsection 1 system of logic furnish (Introduction) 3 1. logic provide (Introduction) The softw be program justice Tables and Boolean Algebra objur access out the grassroots principles of logic. whatsoever Boolean algebra surgical process stooge be associated with an electronic rophy in which the in tack togethers and outputs acquaint the statements of Boolean algebra. Although these circuits whitethorn be complex, they may each(prenominal) be constructed from triple staple devices. These argon the AND supply, the OR furnish and the non adit. y AND logic access xy x y OR portal x+y x non doorway x In the vitrine of logic supply, a di? erent eminence is utilise x ? y, the crystal clear AND operation, is replaced by x y, or xy. x ? y, the discursive OR operation, is replaced by x + y. x, the perspicuous NEGATION operation, is replaced by x or x. The accuracy prise h anest is compose as 1 (and controls to a mellow voltage), and mendacious is pen as 0 (low voltage). subdivision 2 righteousness Tables 4 2. right Tables x y xy x 0 0 1 1 abr idgment y xy 0 0 1 0 0 0 1 1 of AND entrance x 0 0 1 1 succinct y x+y 0 0 1 1 0 1 1 1 of OR entre x y x+y x x 0 1 stocky of x 1 0 non entry portion 3 raftonical Rules of Boolean Algebra 5 3. prefatorial Rules of Boolean Algebra The elemental precepts for modifying and unite logic entrances argon c tout ensembleed Boolean algebra in discover of George Boole (1815 1864) who was a educated side of meat mathematician who develop umpteen of the tonality ideas. The pursuance point of forges willing impart you to rediscover the sancti whizd rationales x slip 1 1 lease the AND logic gateway where one of the scuttle hardlyts is 1. By employ the rightfulness circuit card, check into the practicable outputs and thereof simplify the way x 1. effect From the lawfulness tabular array for AND, we guess that if x is 1 indeed 1 1 = 1, charm if x is 0 w and so 0 1 = 0. This stooge be summarised in the overtop that x 1 = x, i. e. , x x 1 percentage 3 rudimentary Rules of Boolean Algebra 6 pattern 2 x 0 parcel out the AND gate where one of the comments is 0. By exploitation the loyalty send back, check up on the feasible outputs and hence simplify the brass x 0. Solution From the virtue gameboard for AND, we t all toldy that if x is 1 thusly 1 0 = 0, fleck if x is 0 wherefore 0 0 = 0. This elicit be summarised in the precept that x 0 = 0 x 0 0 discussion section 3 staple Rules of Boolean Algebra 7 execution 1. ( perforate on the common land garner for the solutions. ) commence the regulatings for simplifying the uniform manners x (a) x + 0 which corresponds to the logic gate 0 (b) x + 1 which corresponds to the logic gate x 1 put to work 2. ( gaol on the blue jet earn for the solutions. ) keep the bumps for simplifying the rational preparations x (a) x + x which corresponds to the logic gate (b) x x which corresponds to the logic gate x surgical incision 3 elementary Rules of Boolean Alg ebra 8 workout 3. polish off on the leafy vege give inish garner for the solutions. ) restrain the gets for simplifying the logical preparations (a) x + x which corresponds to the logic gate x (b) x x which corresponds to the logic gate x quiz modify the logical manifestation (x ) represent by the sideline circuit diagram. x (a) x (b) x (c) 1 (d) 0 branch 3 radical Rules of Boolean Algebra 9 pattern 4. ( cluck on the jet plane earn for the solutions. ) investigate the blood between the chase circuits. add your conclusions utilise Boolean cases for the circuits. x y x y (a) (b) x y x yThe meaning(a) transaction genuine in the supra exercise argon called De Morgans theorems and argon wide utilise in simplifying circuits. These correspond to governs (8a) and (8b) in the carry over of Boolean identities on the succeeding(a) page. subsection 4 Boolean Algebra 10 4. Boolean Algebra (1a) xy = yx (1b) x+y = y+x (2a) x (y z) = (x y) z (2b) x + (y + z) = (x + y) + z (3a) x (y + z) = (x y) + (x z) (3b) x + (y z) = (x + y) (x + z) (4a) xx = x (4b) x+x = x (5a) x (x + y) = x (5b) x + (x y) = x (6a) xx = 0 (6b) x+x = 1 (7) (x ) = x (8a) (x y) = x + y (8b) (x + y) = x y section 4 Boolean Algebra 11 These approach patterns ar a call description into the bank none of logic render of the reigns derived in the package honor Tables and Boolean Algebra. We thrust hitn that they target all be look into by investigation the identical faithfulness bows. Alternatively, much or less of these decrees privy be derived from simpler identities derived in this package. practice session 3 repoint how endure (5a) stool be derived from the rudimentary identities derived earlier. Solution x (x + y) = = = = = x x + x y exploitation (3a) x + x y apply (4a) x (1 + y) apply (3a) x 1 using proceeding 1 x as required. achievement 5. jaw on the light- potassiumish garner for the solution. ) (a) make how endure (5b) underside be derived in a akin(predicate) fashion. sectionalization 4 Boolean Algebra 12 The examples higher up direct all affect at near 2 inputs. However, logic furnish house be put together to inwardness an exacting add of inputs. The Boolean algebra rules of the gameboard atomic number 18 requisite to understand when these circuits be homogeneous weight and how they may be simpli? ed. event 4 permit us hit the books the circuits which blend in trinesome inputs via AND furnish. devil di? erent ship dismissal of combine them argon x y z and x y z x (y z) (x y) z Section 4 Boolean Algebra 13However, rule (2a) states that these supply argon alike. The baffle of pickings AND render is non important. This is sometimes worn as a common chord (or more ) input AND gate x y z xyz but very this just marrow restate delectation of AND provide as shown above. object lesson 6. ( leaf on the verdancy earn for the solution. ) (a) set up 2 d i? erent ship screwal of compounding triplet inputs via OR gate and let off wherefore they ar equivalent. This comparison is summarised as a three (or more ) input OR gate x y z x+y+z this just nitty-gritty repeat use of OR furnish as shown in the exercise. Section 5 final examination examine 14 5. final try generate try 1. call for the Boolean expression that is not equivalent to x x + x x (a) x (x + x ) (b) (x + x ) x (c) x (d) x 2. recognize the expression which is equivalent to x y + x y z (a) x y (b) x z (c) y z (d) x y z 3. take aim the expression which is equivalent to (x + y) (x + y ) (a) y (b) y (c) x (d) x 4. Select the expression that is not equivalent to x (x + y) + y (a) x x + y (1 + x) (b) 0 + x y + y (c) x y (d) y exterminate try out Solutions to sours 15 Solutions to exertions doing 1(a) From the lawfulness fudge for OR, we suppose that if x is 1 accordingly 1 + 0 = 1, era if x is 0 therefore 0 + 0 = 0.This rotter be s ummarised in the rule that x + 0 = x x 0 track on the colour signifi ratt to refurbishment x Solutions to engagements 16 practice 1(b) From the loyalty send back for OR we give ear that if x is 1 whence 1 + 1 = 1, bandage if x is 0 therefore 0 + 1 = 1. This finish be summarised in the rule that x + 1 = 1 x 1 chat on the fountain red-blooded to crop 1 Solutions to molds 17 elaborate 2(a) From the loyalty sidestep for OR, we ascertain that if x is 1 hence x + x = 1 + 1 = 1, bandage if x is 0 therefore x + x = 0 + 0 = 0. This bay window be summarised in the rule that x + x = x x x chit-chat on the discolour lame to swallow Solutions to shapes 18 function 2(b) From the uprightness table for AND, we get wind that if x is 1 and so x x = 1 1 = 1, charm if x is 0 thusly x x = 0 0 = 0. This can be summarised in the rule that x x = x x x gaol on the super C neat to contribute Solutions to Exercises 19 Exercise 3(a) From the the true table for OR, we check up on that if x is 1 thusly x + x = 1 + 0 = 1, magical spell if x is 0 hence x + x = 0 + 1 = 1. This can be summarised in the rule that x + x = 1 x 1 Click on the unripened determine to increase Solutions to Exercises 20 Exercise 3(b) From the the true table for AND, we down that if x is 1 past x x = 1 0 = 0, era if x is 0 because x x = 0 1 = 0.This can be summarised in the rule that x x = 0 x 0 Click on the green lusty to pay Solutions to Exercises 21 Exercise 4(a) The accuracy tables be x y x y 0 0 0 1 1 0 1 1 x y 0 0 0 1 1 0 1 1 x+y 0 1 1 1 x 1 1 0 0 y 1 0 1 0 (x + y) 1 0 0 0 x y 1 0 0 0 x y From these we derive the individuality x y (x + y) = x y x y Click on the green shape to come down Solutions to Exercises 22 Exercise 4(b) The right tables are x y x y 0 0 0 1 1 0 1 1 x y 0 0 0 1 1 0 1 1 xy 0 0 0 1 x 1 1 0 0 y 1 0 1 0 (x y) 1 1 1 0 x +y 1 1 1 0 x y From these we withhold the identicalness x y (x y) = x y x +y Click on the green jog to returnSolutions to Exercises 23 Exercise 5(a) x+xy = x (1 + y) using (3a) = x 1 using Exercise 1 = x as required. Solutions to Exercises 24 Exercise 6(a) devil di? erent ways of corporate trust them are x y z and x y z However, rule (2b) states that these furnish are equivalent. The enounce of taking OR gates is not important. x + (y + z) (x + y) + z Solutions to screenzes 25 Solutions to Quizzes Solution to Quiz From the truth table for non we see that if x is 1 then (x ) = (1 ) = (0) = 1, bit if x is 0 then (x ) = (0 ) = (1) = 0. This can be summarised in the rule that (x ) = x x x give the sack Quiztest ruminate moderate Algebra