DOCSLIB.ORG

Boolean Algebra, Logic Gates Today's Lecture the Big Picture What We

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Boolean Algebra, Logic Gates Today's Lecture the Big Picture What We Today’s Lecture • No Recitation Friday • Projects (groups of 2 or 3) Outline Basics of Logic Design: • Building the building blocks… Boolean Algebra, Logic Gates • Logic Design Truth tables, Boolean functions, Gates and Circuits Computer Science 104 Reading Appendix B © Alvin R. Lebeck CPS 104 2 The Big Picture What We’ve Done, Where We’re Going • The Five Classic Components of a Computer Application Top Down Operating Software Processor/CPU System Input Compiler Firmware Interface Between Control HW and SW Instruction Set Memory CPU Memory I/O system Architecture , Digital Design Memory, I/O Datapath Output Circuit Design Hardware Bottom UP to CPU © Alvin R. Lebeck CPS 104 3 © Alvin R. Lebeck CPS 104 4 Digital Design Boolean Algebra • Logic Design, Switching Circuits, Digital Logic • Boolean functions have arguments that take two Recall: Everything is built from transistors values ({T,F} or {1,0}) and they return a single or a set of ({T,F} or {1,0}) value(s). • A transistor is a switch • Boolean functions can always be represented by a • It is either on or off table called a “ Truth Table ” • On or off can represent True or False • Example: F: {0,1} 3 -> {0,1} 2 Given a bunch of bits (0 or 1)… a b c f 1f2 • Is this instruction a lw or a beq? 0 0 0 0 1 • What register do I read? 0 0 1 1 1 0 1 0 1 0 • How do I add two numbers? 0 1 1 0 0 • Need a method to reason about complex expressions 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 © Alvin R. Lebeck CPS 104 5 © Alvin R. Lebeck CPS 104 6 Boolean Functions Boolean Functions and Expressions • Example Boolean Functions: NOT, AND, OR, XOR, . • Boolean algebra notation : Use * for AND, + for OR, ~ for NOT. a NOT(a) a b AND(a,b) a b OR(a,b) NOT is also written as A’ and A 0 1 0 0 0 0 0 0 • Using the above notation we can write Boolean 1 0 0 1 0 0 1 1 expressions for functions 1 0 0 1 0 1 1 1 1 1 1 1 F(A, B, C) = (A * B) + (~A * C) a b XOR(a,b) a b XNOR(a,b) a b NOR(a,b) 0 0 0 0 0 1 0 0 1 • We can evaluate the Boolean expression with all 0 1 1 0 1 0 0 1 0 possible argument values to construct a truth table. 1 0 1 1 0 0 1 0 0 1 1 0 1 1 1 1 1 0 • What is truth table for F? © Alvin R. Lebeck CPS 104 7 © Alvin R. Lebeck CPS 104 8 Boolean Function Simplification Boolean Function Simplification • Boolean expressions can be simplified by using the following rules (bitwise logical): A*A = A a b c f 1f2 A* 0 = 0 0 0 0 0 1 f1 = ~a *~b *c + ~a *b*c + a *~b *c + a *b*c A*1 = A 0 0 1 1 1 A*~A = 0 0 1 0 0 0 0 1 1 1 0 f2 = ~a*~b*~c + ~a*~b*c + a*b*~c + a*b*c A+A = A 1 0 0 0 0 A+0 = A 1 0 1 1 0 A+1 = 1 1 1 0 0 1 A+~A = 1 1 1 1 1 1 Simplify these functions... A*B = B*A A*(B+C) = (B+C)*A = A*B + A*C © Alvin R. Lebeck CPS 104 10 © Alvin R. Lebeck CPS 104 11 Boolean Functions and Expressions Boolean Functions and Expressions • The Fundamental Theorem of Boolean Algebra : • Example-2: Every Boolean function can be written in disjunctive normal form as an OR of ANDs ( Sum-of products ) of a b c f 1f2 it’s arguments or their complements. 0 0 0 0 1 f1 = ~a *~b *c + ~a *b*~c + a *~b *~c + a *b*c 0 0 1 1 1 “Proof:” Write the truth table, construct sum-of- 0 1 0 1 0 product from the table. 0 1 1 0 0 f2 = ~a*~b*~c + ~a*~b*c + a*b*~c + a*b*c 1 0 0 1 0 1 1 0 0 1 a b XNOR(a,b) 1 1 1 1 1 0 0 1 0 1 0 XNOR = (~a * ~b) + (a * b) 1 0 0 1 1 1 © Alvin R. Lebeck CPS 104 13 © Alvin R. Lebeck CPS 104 14 Applying the Theory Boolean Gates • Lots of good theory • Gates are electronic devices that implement simple • Can reason about complex boolean expressions Boolean functions • Now we have to make it real… Examples a a AND(a,b) OR(a,b) a NOT(a) b b a XOR(a,b) a NAND(a,b) b b a NOR(a,b) a XNOR(a,b) b b © Alvin R. Lebeck CPS 104 15 © Alvin R. Lebeck CPS 104 16 Reality Check Boolean Functions, Gates and Circuits • Basic 1 or 2 Input Boolean Gate 1- 4 Transistors • Circuits are made from a network of gates. (function compositions). Pentium III a b XOR(a,b) a 0 0 0 • Processor Core 9.5 Million Transistors XOR(a,b) F = ~a*b + ~b*a b 0 1 1 • Total: 28 Million Transistors 1 0 1 Pentium 4 1 1 0 • Total: 42 Million Transistors a Core2 Duo (two processors) • Total: 290 Million Transistors F b Core2 Duo Extreme (4 processors, 8MB cache) • Total: 590 Million Transistors © Alvin R. Lebeck CPS 104 17 © Alvin R. Lebeck CPS 104 18 Digital Design Examples Design Example Input: 2 bits representing an unsigned number (n) • Consider machine with 4 registers 2 Output: n as 4-bit unsigned binary number • Given 2-bit input (register specifier, I 1, I 0) • Want one of 4 output bits (O 3-O0) to be 1 E.g., allows a single register to be accessed Input: 2 bits representing an unsigned number (n) • What is the circuit for this? Output: 3-n as unsigned binary number © Alvin R. Lebeck CPS 104 19 © Alvin R. Lebeck CPS 104 20 More Design Examples Parity Example • X is a 3-bit quantity • The parity code of a binary word counts the number 1. Write a logic function that is true if and only if X contains at of ones in a word. If there are an even number of least two 1s. ones the parity code is 0, if there are an odd number 2. Implement the logic function from problem 1. using only AND, of ones the parity code is 1. For example, the parity OR and NOT gates. (Note there are no constraints on the of 0101 is 0, and the parity of 1101 is 1. number of gate inputs.) By implement, I mean draw the circuit diagram. • Construct the truth table for a function that computes the parity of a four-bit word . Implement 3. Write a logic function that is true if and only if X, when this function using AND, OR and NOT gates. (Note interpreted as an unsigned binary number, is greater than the there are no constraints on the number of gate number 4. inputs.) 4. Implement the logic function from problem 3. using only AND, OR and NOT gates. (Note there are no constraints on the number of gate inputs.) © Alvin R. Lebeck CPS 104 21 © Alvin R. Lebeck CPS 104 22 Circuit Example: Decoder Circuit Example: 2x1 MUX Q3 Multiplexor (MUX) selects from one of many inputs I1 I0 Q0 Q1 Q2 Q3 a 1 y Q2 0 0 1 0 0 0 MUX(A, B, S) = (A * S) + (B * ~S) b 0 0 1 0 1 0 0 Q s 1 1 0 0 0 1 0 1 1 0 0 0 1 B Q0 Gate 1 Y = (A * S) + (B * ~S) Gate 3 A Gate 2 S I1 I0 © Alvin R. Lebeck CPS 104 23 © Alvin R. Lebeck CPS 104 24 Example 4x1 MUX Arithmetic and Logical Operations in ISA • What operations are there? a 1 • How do we implement them? 3 b 0 a Consider a 1-bit Adder 1 y 2 0 b y c 1 c 1 d 0 d 0 s0 s1 2 S © Alvin R. Lebeck CPS 104 25 © Alvin R. Lebeck CPS 104 26 Summary • Boolean Algebra & functions • Logic gates (AND, OR, NOT, etc) • Multiplexors Reading • Appendix B © Alvin R. Lebeck CPS 104 27.
Load more

Recommended publications
  • Laced Boolean Functions and Subset Sum Problems in Finite Fields
    Laced Boolean functions and subset sum problems in finite fields David Canright1, Sugata Gangopadhyay2 Subhamoy Maitra3, Pantelimon Stanic˘ a˘1 1 Department of Applied Mathematics, Naval Postgraduate School Monterey, CA 93943{5216, USA; fdcanright,[email protected] 2 Department of Mathematics, Indian Institute of Technology Roorkee 247667 INDIA; [email protected] 3 Applied Statistics Unit, Indian Statistical Institute 203 B. T. Road, Calcutta 700 108, INDIA; [email protected] March 13, 2011 Abstract In this paper, we investigate some algebraic and combinatorial properties of a special Boolean function on n variables, defined us- ing weighted sums in the residue ring modulo the least prime p ≥ n. We also give further evidence to a question raised by Shparlinski re- garding this function, by computing accurately the Boolean sensitivity, thus settling the question for prime number values p = n. Finally, we propose a generalization of these functions, which we call laced func- tions, and compute the weight of one such, for every value of n. Mathematics Subject Classification: 06E30,11B65,11D45,11D72 Key Words: Boolean functions; Hamming weight; Subset sum problems; residues modulo primes. 1 1 Introduction Being interested in read-once branching programs, Savicky and Zak [7] were led to the definition and investigation, from a complexity point of view, of a special Boolean function based on weighted sums in the residue ring modulo a prime p. Later on, a modification of the same function was used by Sauerhoff [6] to show that quantum read-once branching programs are exponentially more powerful than classical read-once branching programs. Shparlinski [8] used exponential sums methods to find bounds on the Fourier coefficients, and he posed several open questions, which are the motivation of this work.
    [Show full text]
  • Analysis of Boolean Functions and Its Applications to Topics Such As Property Testing, Voting, Pseudorandomness, Gaussian Geometry and the Hardness of Approximation
    Analysis of Boolean Functions Notes from a series of lectures by Ryan O’Donnell Guest lecture by Per Austrin Barbados Workshop on Computational Complexity February 26th – March 4th, 2012 Organized by Denis Th´erien Scribe notes by Li-Yang Tan arXiv:1205.0314v1 [cs.CC] 2 May 2012 Contents 1 Linearity testing and Arrow’s theorem 3 1.1 TheFourierexpansion ............................. 3 1.2 Blum-Luby-Rubinfeld. .. .. .. .. .. .. .. .. 7 1.3 Votingandinfluence .............................. 9 1.4 Noise stability and Arrow’s theorem . ..... 12 2 Noise stability and small set expansion 15 2.1 Sheppard’s formula and Stabρ(MAJ)...................... 15 2.2 Thenoisyhypercubegraph. 16 2.3 Bonami’slemma................................. 18 3 KKL and quasirandomness 20 3.1 Smallsetexpansion ............................... 20 3.2 Kahn-Kalai-Linial ............................... 21 3.3 Dictator versus Quasirandom tests . ..... 22 4 CSPs and hardness of approximation 26 4.1 Constraint satisfaction problems . ...... 26 4.2 Berry-Ess´een................................... 27 5 Majority Is Stablest 30 5.1 Borell’s isoperimetric inequality . ....... 30 5.2 ProofoutlineofMIST ............................. 32 5.3 Theinvarianceprinciple . 33 6 Testing dictators and UGC-hardness 37 1 Linearity testing and Arrow’s theorem Monday, 27th February 2012 Rn Open Problem [Guy86, HK92]: Let a with a 2 = 1. Prove Prx 1,1 n [ a, x • ∈ k k ∈{− } |h i| ≤ 1] 1 . ≥ 2 Open Problem (S. Srinivasan): Suppose g : 1, 1 n 2 , 1 where g(x) 2 , 1 if • {− } →± 3 ∈ 3 n x n and g(x) 1, 2 if n x n . Prove deg( f)=Ω(n). i=1 i ≥ 2 ∈ − − 3 i=1 i ≤− 2 P P In this workshop we will study the analysis of boolean functions and its applications to topics such as property testing, voting, pseudorandomness, Gaussian geometry and the hardness of approximation.
    [Show full text]
  • Probabilistic Boolean Logic, Arithmetic and Architectures
    PROBABILISTIC BOOLEAN LOGIC, ARITHMETIC AND ARCHITECTURES A Thesis Presented to The Academic Faculty by Lakshmi Narasimhan Barath Chakrapani In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the School of Computer Science, College of Computing Georgia Institute of Technology December 2008 PROBABILISTIC BOOLEAN LOGIC, ARITHMETIC AND ARCHITECTURES Approved by: Professor Krishna V. Palem, Advisor Professor Trevor Mudge School of Computer Science, College Department of Electrical Engineering of Computing and Computer Science Georgia Institute of Technology University of Michigan, Ann Arbor Professor Sung Kyu Lim Professor Sudhakar Yalamanchili School of Electrical and Computer School of Electrical and Computer Engineering Engineering Georgia Institute of Technology Georgia Institute of Technology Professor Gabriel H. Loh Date Approved: 24 March 2008 College of Computing Georgia Institute of Technology To my parents The source of my existence, inspiration and strength. iii ACKNOWLEDGEMENTS आचायातर् ्पादमादे पादं िशंयः ःवमेधया। पादं सॄचारयः पादं कालबमेणच॥ “One fourth (of knowledge) from the teacher, one fourth from self study, one fourth from fellow students and one fourth in due time” 1 Many people have played a profound role in the successful completion of this disser- tation and I first apologize to those whose help I might have failed to acknowledge. I express my sincere gratitude for everything you have done for me. I express my gratitude to Professor Krisha V. Palem, for his energy, support and guidance throughout the course of my graduate studies. Several key results per- taining to the semantic model and the properties of probabilistic Boolean logic were due to his brilliant insights.
    [Show full text]
  • Predicate Logic
    Predicate Logic Laura Kovács Recap: Boolean Algebra and Propositional Logic 0, 1 True, False (other notation: t, f) boolean variables a ∈{0,1} atomic formulas (atoms) p∈{True,False} boolean operators ¬, ∧, ∨, fl, ñ logical connectives ¬, ∧, ∨, fl, ñ boolean functions: propositional formulas: • 0 and 1 are boolean functions; • True and False are propositional formulas; • boolean variables are boolean functions; • atomic formulas are propositional formulas; • if a is a boolean function, • if a is a propositional formula, then ¬a is a boolean function; then ¬a is a propositional formula; • if a and b are boolean functions, • if a and b are propositional formulas, then a∧b, a∨b, aflb, añb are boolean functions. then a∧b, a∨b, aflb, añb are propositional formulas. truth value of a boolean function truth value of a propositional formula (truth tables) (truth tables) Recap: Boolean Algebra and Propositional Logic 0, 1 True, False (other notation: t, f) boolean variables a ∈{0,1} atomic formulas (atoms) p∈{True,False} boolean operators ¬, ∧, ∨, fl, ñ logical connectives ¬, ∧, ∨, fl, ñ boolean functions: propositional formulas (propositions, Aussagen ): • 0 and 1 are boolean functions; • True and False are propositional formulas; • boolean variables are boolean functions; • atomic formulas are propositional formulas; • if a is a boolean function, • if a is a propositional formula, then ¬a is a boolean function; then ¬a is a propositional formula; • if a and b are boolean functions, • if a and b are propositional formulas, then a∧b, a∨b, aflb, añb are
    [Show full text]
  • Satisfiability 6 the Decision Problem 7
    Satisfiability Difficult Problems Dealing with SAT Implementation Klaus Sutner Carnegie Mellon University 2020/02/04 Finding Hard Problems 2 Entscheidungsproblem to the Rescue 3 The Entscheidungsproblem is solved when one knows a pro- cedure by which one can decide in a finite number of operations So where would be look for hard problems, something that is eminently whether a given logical expression is generally valid or is satis- decidable but appears to be outside of P? And, we’d like the problem to fiable. The solution of the Entscheidungsproblem is of funda- be practical, not some monster from CRT. mental importance for the theory of all fields, the theorems of which are at all capable of logical development from finitely many The Circuit Value Problem is a good indicator for the right direction: axioms. evaluating Boolean expressions is polynomial time, but relatively difficult D. Hilbert within P. So can we push CVP a little bit to force it outside of P, just a little bit? In a sense, [the Entscheidungsproblem] is the most general Say, up into EXP1? problem of mathematics. J. Herbrand Exploiting Difficulty 4 Scaling Back 5 Herbrand is right, of course. As a research project this sounds like a Taking a clue from CVP, how about asking questions about Boolean fiasco. formulae, rather than first-order? But: we can turn adversity into an asset, and use (some version of) the Probably the most natural question that comes to mind here is Entscheidungsproblem as the epitome of a hard problem. Is ϕ(x1, . , xn) a tautology? The original Entscheidungsproblem would presumable have included arbitrary first-order questions about number theory.
    [Show full text]
  • Problems and Comments on Boolean Algebras Rosen, Fifth Edition: Chapter 10; Sixth Edition: Chapter 11 Boolean Functions
    Problems and Comments on Boolean Algebras Rosen, Fifth Edition: Chapter 10; Sixth Edition: Chapter 11 Boolean Functions Section 10. 1, Problems: 1, 2, 3, 4, 10, 11, 29, 36, 37 (fifth edition); Section 11.1, Problems: 1, 2, 5, 6, 12, 13, 31, 40, 41 (sixth edition) The notation ""forOR is bad and misleading. Just think that in the context of boolean functions, the author uses instead of ∨.The integers modulo 2, that is ℤ2 0,1, have an addition where 1 1 0 while 1 ∨ 1 1. AsetA is partially ordered by a binary relation ≤, if this relation is reflexive, that is a ≤ a holds for every element a ∈ S,it is transitive, that is if a ≤ b and b ≤ c hold for elements a,b,c ∈ S, then one also has that a ≤ c, and ≤ is anti-symmetric, that is a ≤ b and b ≤ a can hold for elements a,b ∈ S only if a b. The subsets of any set S are partially ordered by set inclusion. that is the power set PS,⊆ is a partially ordered set. A partial ordering on S is a total ordering if for any two elements a,b of S one has that a ≤ b or b ≤ a. The natural numbers ℕ,≤ with their ordinary ordering are totally ordered. A bounded lattice L is a partially ordered set where every finite subset has a least upper bound and a greatest lower bound.The least upper bound of the empty subset is defined as 0, it is the smallest element of L.
    [Show full text]
  • Logic and Bit Operations
    Logic and bit operations • Computers represent information by bits. • A bit has two possible values, namely zero and one. This meaning of the word comes from binary digit, since zeroes and ones are the digits used in binary representations of numbers. The well-known statistician John Tuley introduced this terminology in 1946. (There were several other suggested words for a binary digit, including binit and bigit, that never were widely accepted.) • A bit can be used to represent a truth value, since there are two truth values, true and false. • A variable is called a Boolean variable if its values are either true or false. • Computer bit operations correspond to the logical connectives. We will also use the notation OR, AND and XOR for _ , ^ and exclusive _. • A bit string is a sequence of zero or more bits. The length of the string is the number of bits in the string. 1 • We can extend bit operations to bit strings. We define bitwise OR, bitwise AND and bitwise XOR of two strings of the same length to be the strings that have as their bits the OR, AND and XOR of the corresponding bits in the two strings. • Example: Find the bitwise OR, bitwise AND and bitwise XOR of the bit strings 01101 10110 11000 11101 Solution: The bitwise OR is 11101 11111 The bitwise AND is 01000 10100 and the bitwise XOR is 10101 01011 2 Boolean algebra • The circuits in computers and other electronic devices have inputs, each of which is either a 0 or a 1, and produce outputs that are also 0s and 1s.
    [Show full text]
  • Propositional Logic
    Reasoning About Programs Panagiotis Manolios Northeastern University March 22, 2012 Version: 58 Copyright c 2012 by Panagiotis Manolios All rights reserved. We hereby grant permission for this publication to be used for personal or classroom use. No part of this publication may be stored in a retrieval system or transmitted in any form or by any means other personal or classroom use without the prior written permission of the author. Please contact the author for details. 2 Propositional Logic The study of logic was initiated by the ancient Greeks, who were concerned with analyzing the laws of reasoning. They wanted to fully understand what conclusions could be derived from a given set of premises. Logic was considered to be a part of philosophy for thousands of years. In fact, until the late 1800’s, no significant progress was made in the field since the time of the ancient Greeks. But then, the field of modern mathematical logic was born and a stream of powerful, important, and surprising results were obtained. For example, to answer foundational questions about mathematics, logicians had to essentially create what later became the foundations of computer science. In this class, we’ll explore some of the many connections between logic and computer science. We’ll start with propositional logic, a simple, but surprisingly powerful fragment of logic. Expressions in propositional logic can only have one of two values. We’ll use T and F to denote the two values, but other choices are possible, e.g., 1 and 0 are sometimes used. The expressions of propositional logic include: 1.
    [Show full text]
  • CS40-S13: Functional Completeness 1 Introduction
    CS40-S13: Functional Completeness Victor Amelkin [email protected] April 12, 2013 In class, we have briefly discussed what functional completeness means and how to prove that a certain system (a set) of logical operators is functionally complete. In this note, I will review what functional completeness is, how to prove that a system of logical operators is complete as well as how to prove that a system is incomplete. 1 Introduction When we talk about logical operators, we mean operators like ^, _, :, !, ⊕. Logical operators work on propositions, i.e. if we write a ^ b, then both a and b are propositions. The mentioned propositions may be either simple or compound, as in a ^ :(x ! y) { here, a is simple, while b is compound. | {z } b For a logical operator it does not matter whether it operates on simple or compound propositions; the only thing that matters is that all the arguments are propositions, i.e., either True of False. Since logical operators apply to propositions and \return" either True or False, they can be seen as boolean functions. For example, conjunction ^(x1; x2) = x1 ^ x2 is a binary boolean function, i.e., it accepts two arguments (hence, \binary") that can be either True of False and returns either True of False (hence, \boolean"). As another example, negation can be seen as a unary (accepts only one argument) boolean function :(x) = :x. We, however, rarely use the functional notation ^(x1; x2) (like in f(x)); instead, we usually use the operator notation, e.g., x1 _ x2. (Similarly, when you do algebra, you probably write a + b instead of +(a; b), unless you do algebra in Lisp.) Further, instead of talking about logical operators, we will talk about boolean functions, but, having read the material above, you should understand that these two are the same.
    [Show full text]
  • Binary Recursion
    Binary recursion 1 Unate functions ! Theorem If a cover C(f) is unate in xj, then f is unate in xj. ! Theorem If f is unate in xj, then every prime implicant of f is unate in xj. 2 1 Why are unate functions so special? ! Special Boolean algorithms for unate functions are not only fast, but also give the guaranteed minimum result. ! The algorithms presented perform the Boolean operations and simplify the result in the same time. ! In practice, many functions are partially unate. The divide and conquer strategy aimed at unateness produces very shallow recursions; algorithms are fast in practice. 3 Unate recursive paradigm works for: ! Complementation ! Irredundant cover ! Reduce ! Essential primes = + ! Example f x 1 x 2 x 2 x 3 is unate 4 2 Important facts about unate functions Proposition 1. A logic function f is monotone increasing (decreasing) in x j if and only if no prime implicant of f has a 0 (1) in the j-th position. Proof: (if) If no prime implicant has a 0 in the jth position, any prime cover C(f) is unate in x j . So f is unate in x j. (only if) Let c i is a prime with 0 at jth position -> the cube c~i obtained by replacing 0 in jth position with a 1 contains at least one vector in f . For the input v corresponding to that vertex, f(v)=0. If we change x j from 1 to 0 in v, the value of f changes from 0 to 1, so f is not monotone increasing.
    [Show full text]
  • Boolean Algebra
    Boolean Algebra Definition: A Boolean Algebra is a math construct (B,+, . , ‘, 0,1) where B is a non-empty set, + and . are binary operations in B, ‘ is a unary operation in B, 0 and 1 are special elements of B, such that: a) + and . are communative: for all x and y in B, x+y=y+x, and x.y=y.x b) + and . are associative: for all x, y and z in B, x+(y+z)=(x+y)+z, and x.(y.z)=(x.y).z c) + and . are distributive over one another: x.(y+z)=xy+xz, and x+(y.z)=(x+y).(x+z) d) Identity laws: 1.x=x.1=x and 0+x=x+0=x for all x in B e) Complementation laws: x+x’=1 and x.x’=0 for all x in B Examples: (B=set of all propositions, ∨, ∧, ¬, T, F) (B=2A, U, ∩, c, Φ,A) Theorem 1: Let (B,+, . , ‘, 0,1) be a Boolean Algebra. Then the following hold: a) x+x=x and x.x=x for all x in B b) x+1=1 and 0.x=0 for all x in B c) x+(xy)=x and x.(x+y)=x for all x and y in B Proof: a) x = x+0 Identity laws = x+xx’ Complementation laws = (x+x).(x+x’) because + is distributive over . = (x+x).1 Complementation laws = x+x Identity laws x = x.1 Identity laws = x.(x+x’) Complementation laws = x.x +x.x’ because + is distributive over . = x.x+0 Identity laws = x.x b) x+1 =x+(x+x’) Complementation laws = (x+x)+x’ + is associative = x+x’ using (a) = 1 Complementation laws 0.x =(x’.x).x Complementation laws = x’.(x.x) .
    [Show full text]
  • Boolean Logic
    Boolean Logic Lecture Notes Gert Smolka Saarland University November 22, 2016 Abstract. We study satisfiability and equivalence of boolean expressions. We show that prime trees (reduced and ordered decision trees) are a unique normal form for boolean expressions. We verify a SAT solver and a prime tree normaliser. We show that assignment equivalence agrees with the equivalence obtained from the axioms of boolean algebra. The material is presented using type-theoretic language. There is an accompa- nying Coq development complementing the mathematical presentation given in the notes. The notes also serve as introduction to formal syntax. Expressions are modelled with an inductive type and recursive functions and inductive predicates on expres- sions are studied. This includes notions like substitution and equational deduction. Advice to the student. To master the material, you want to understand both the notes and the accompanying Coq development. It is important that you can switch between the informal mathematical presentation of the notes and the formal de- velopment in Coq. You may think of the notes as the design of the theory and the Coq development as the implementation of the theory. The goal is to understand both the design and the implementation of the theory. The notes occasionally omit details that are spelled out in the Coq development (e.g., the precise definition of substitution). Make sure you can fill in the missing details. Contents 1 Boolean Operations . 3 2 SAT Solver . 4 3 Equivalence of Boolean Expressions . 7 4 Shallow and Deep Embedding . 9 5 Decision Trees . 12 6 Prime Tree Normal Form . 12 7 Significant Variables .
    [Show full text]